{"id":2290,"date":"2015-04-22T21:09:08","date_gmt":"2015-04-22T21:09:08","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=2290"},"modified":"2020-12-28T19:20:41","modified_gmt":"2020-12-28T19:20:41","slug":"equilibrium-calculations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/chapter\/equilibrium-calculations\/","title":{"raw":"Equilibrium Calculations","rendered":"Equilibrium Calculations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems<\/li>\r\n \t<li>Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp222894896\">Having covered the essential concepts of chemical equilibria in the preceding sections of this chapter, this final section will demonstrate the more practical aspect of using these concepts and appropriate mathematical strategies to perform various equilibrium calculations. These types of computations are essential to many areas of science and technology\u2014for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.<\/p>\r\n<p id=\"fs-idp292035664\">Many of the useful equilibrium calculations that will be demonstrated here require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometry of the reaction, as illustrated by decomposition of ammonia:<\/p>\r\n<p style=\"text-align: center;\">[latex]2{\\text{NH}}_{3}\\left(g\\right){\\rightleftharpoons}{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/p>\r\n<p id=\"fs-idp42253680\">As shown earlier in this chapter, this equilibrium may be established within a sealed container that initially contains either NH<sub>3<\/sub>\u00a0only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount\u00a0<em>x<\/em>:<\/p>\r\n\r\n<center>\u0394[latex][\\text{N}_2] = +x[\/latex]<\/center>\r\n<p id=\"fs-idp249176544\">the corresponding changes in the other species concentrations are<\/p>\r\n\r\n<center>\u0394[latex][\\text{H}_2] = [\/latex]\u0394[latex][\\text{N}_2]\\left(\\dfrac{3\\text{molH}_2}{1\\text{molN}_2}\\right) = +3x[\/latex]<\/center>&nbsp;\r\n\r\n<center>\u0394[latex][\\text{NH}_3] = [\/latex]\u0394[latex][\\text{N}_2]\\left(\\dfrac{2\\text{molNH}_3}{1\\text{molN}_2}\\right) = -2x[\/latex]<\/center>\r\n<p id=\"fs-idp146645120\">where the negative sign indicates a decrease in concentration.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Determining Relative Changes in Concentration<\/h3>\r\nDerive the missing terms representing concentration changes for each of the following reactions.\r\n<ol>\r\n \t<li>[latex]\\begin{array}{ccccc}{\\text{C}}_{2}{\\text{H}}_{2}\\text{(}g\\text{)}&amp;+\\hfill &amp; 2{\\text{Br}}_{2}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons&amp; {\\text{C}}_{2}{\\text{H}}_{2}{\\text{Br}}_{4}\\text{(}g\\text{)}\\hfill \\\\ x\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccccc}{\\text{I}}_{2}\\text{(}aq\\text{)}&amp;+\\hfill &amp; {\\text{I}}^{-}\\text{(}aq\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; {\\text{I}}_{3}{}^{-}\\text{(}aq\\text{)}\\hfill \\\\ &amp; &amp; &amp; &amp; x \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{cccccc}{\\text{C}}_{3}{\\text{H}}_{8}\\text{(}g\\text{)}&amp;+\\hfill &amp; 5{\\text{O}}_{2}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; 3{\\text{CO}}_{2}\\text{(}g\\text{)}+\\hfill &amp; 4{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\hfill \\\\ x \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"8310\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"8310\"]\r\n<ol>\r\n \t<li>[latex]\\begin{array}{ccccc}{\\text{C}}_{2}{\\text{H}}_{2}\\text{(}g\\text{)}&amp;+\\hfill &amp; 2{\\text{Br}}_{2}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; {\\text{C}}_{2}{\\text{H}}_{2}{\\text{Br}}_{4}\\text{(}g\\text{)}\\hfill \\\\ x\\hfill&amp; &amp; 2x\\hfill &amp; &amp; -x\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccccc}{\\text{I}}_{2}\\text{(}aq\\text{)}&amp;+\\hfill &amp; {\\text{I}}^{-}\\text{(}aq\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; {\\text{I}}_{3}{}^{-}\\text{(}aq\\text{)}\\hfill \\\\ -x\\hfill&amp; &amp; -x\\hfill &amp; &amp; x\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{cccccc}{\\text{C}}_{3}{\\text{H}}_{8}\\text{(}g\\text{)}&amp;+\\hfill &amp; 5{\\text{O}}_{2}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; 3{\\text{CO}}_{2}\\text{(}g\\text{)}+\\hfill &amp; 4{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\hfill \\\\ x\\hfill &amp; &amp; 5x\\hfill &amp; &amp; -3x\\hfill &amp; -4x\\hfill \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nComplete the changes in concentrations for each of the following reactions:\r\n<ol>\r\n \t<li>[latex]\\begin{array}{ccccc}2{\\text{SO}}_{2}\\text{(}g\\text{)}&amp;+\\hfill &amp; {\\text{O}}_{2}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; 2{\\text{SO}}_{3}\\text{(}g\\text{)}\\hfill \\\\ &amp; &amp; x &amp; \\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccc}{\\text{C}}_{4}{\\text{H}}_{8}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; 2{\\text{C}}_{2}{\\text{H}}_{4}\\text{(}g\\text{)}\\hfill \\\\ &amp; &amp; -2x\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccccccc}4{\\text{NH}}_{3}\\text{(}g\\text{)}&amp;+\\hfill &amp; 7{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; 4{\\text{NO}}_{2}\\text{(}g\\text{)}&amp;+\\hfill &amp; 6{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"707791\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"707791\"]\r\n<ol>\r\n \t<li>[latex]\\begin{array}{ccccc}2{\\text{SO}}_{2}\\text{(}g\\text{)}&amp;+\\hfill &amp; {\\text{O}}_{2}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; 2{\\text{SO}}_{3}\\text{(}g\\text{)}\\hfill \\\\ 2x &amp; &amp; x &amp;&amp; -2x &amp; \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccc}{\\text{C}}_{4}{\\text{H}}_{8}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; 2{\\text{C}}_{2}{\\text{H}}_{4}\\text{(}g\\text{)}\\hfill \\\\ x &amp; &amp; -2x\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>There are two possible solutions for this question:\r\n[latex]\\begin{array}{ccccccc}4{\\text{NH}}_{3}\\text{(}g\\text{)}&amp;+\\hfill &amp; 7{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; 4{\\text{NO}}_{2}\\text{(}g\\text{)}&amp;+\\hfill &amp; 6{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\\\ 4x&amp; &amp;7x&amp; &amp;-4x&amp; &amp;-6x \\end{array}[\/latex]\r\nor\r\n[latex]\\begin{array}{ccccccc}4{\\text{NH}}_{3}\\text{(}g\\text{)}&amp;+\\hfill &amp; 7{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\hfill &amp; \\rightleftharpoons\\hfill &amp; 4{\\text{NO}}_{2}\\text{(}g\\text{)}&amp;+\\hfill &amp; 6{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\\\ -4x&amp; &amp;-7x&amp; &amp;4x&amp; &amp;6x \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>A reaction is represented by this equation: [latex]\\text{A}\\left(aq\\right)+2\\text{B}\\left(aq\\right)\\rightleftharpoons2\\text{C}\\left(aq\\right){K}_{c}=1\\times {10}^{3}[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Write the mathematical expression for the equilibrium constant.<\/li>\r\n \t<li>Using concentrations \u22641 <em>M<\/em>, make up two sets of concentrations that describe a mixture of A, B, and C at equilibrium.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>A reaction is represented by this equation: [latex]2\\text{W}\\left(aq\\right)\\rightleftharpoons\\text{X}\\left(aq\\right)+2\\text{Y}\\left(aq\\right){K}_{c}=5\\times {10}^{-4}[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Write the mathematical expression for the equilibrium constant.<\/li>\r\n \t<li>Using concentrations of \u22641 <em>M<\/em>, make up two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"69852\"]Show Solution to Question 1[\/reveal-answer]\r\n[hidden-answer a=\"69852\"]\r\n\r\n[latex]{K}_{c}=\\dfrac{{\\left[\\text{C}\\right]}^{2}}{\\left[\\text{A}\\right]{\\left[\\text{B}\\right]}^{2}}\\text{.}[\/latex] There are many different sets of equilibrium concentrations; two are [A] = 0.1 <em>M<\/em>, [B] = 0.1 <em>M<\/em>, [C] = 1 <em>M<\/em>; and [A] = 0.01, [B] = 0.250, [C] = 0.791.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Calculations of an Equilibrium Constant<\/h2>\r\nThe equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the K expression, as was illustrated by Example 2. A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the reactant and product concentrations\u00a0<em>initially<\/em>\u00a0present, for how they\u00a0<em>change<\/em>\u00a0as the reaction proceeds, and for what they are when the system reaches\u00a0<em>equilibrium<\/em>. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0Calculation of an Equilibrium Constant<\/h3>\r\nIodine molecules react reversibly with iodide ions to produce triiodide ions.\r\n<p style=\"text-align: center;\">[latex]{\\text{I}}_{2}\\left(aq\\right)+{\\text{I}}^{-}\\left(aq\\right)\\rightleftharpoons{\\text{I}}_{3}{}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nIf a solution with the concentrations of I<sub>2<\/sub> and I<sup>\u2212<\/sup> both equal to 1.000 \u00d7 10<sup>\u22123<\/sup><em>M<\/em> before reaction gives an equilibrium concentration of I<sub>2<\/sub> of 6.61 \u00d7 10<sup>\u22124<\/sup><em>M<\/em>, what is the equilibrium constant for the reaction?\r\n[reveal-answer q=\"536062\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"536062\"]\r\n\r\nTo calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:\r\n<p style=\"text-align: center;\">[latex]{K}_{c}=\\dfrac{\\left[{\\text{I}}_{3}{}^{-}\\right]}{\\left[{I}_{2}\\right]\\left[{\\text{I}}^{-}\\right]}[\/latex]<\/p>\r\nProvided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an ICE table.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/openstax.org\/resources\/efdf4733dd94e860584f66ae6c1d1c2f4b2418a8\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, \u201cI subscript 2 plus sign I superscript negative sign equilibrium arrow I subscript 3 superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column has the following: 1.000 times 10 to the negative third power, negative x, [ I subscript 2 ] subscript i minus x. The second column has the following: 1.000 times 10 to the negative third power, negative x, [ I superscript negative sign ] subscript i minus x. The third column has the following: 0, positive x, [ I superscript negative sign ] subscript i plus x.\" width=\"599\" height=\"198\" \/>\r\n\r\nAt equilibrium the concentration of I<sub>2<\/sub> is 6.61 \u00d7 10<sup>\u22124<\/sup><em>M<\/em> so that\r\n<p style=\"text-align: center;\">[latex]1.000\\times {10}^{-3}-x=6.61\\times {10}^{-4}[\/latex]\r\n[latex]x=1.000\\times {10}^{-3}-6.61\\times {10}^{-4}[\/latex]\r\n[latex]=3.39\\times {10}^{-4}M[\/latex]<\/p>\r\n<p id=\"fs-idp187828096\">The ICE table may now be updated with numerical values for all its concentrations:<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/openstax.org\/resources\/d73aca40ea1b3b0cc341a9ecc2941b22f0754aa9\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, \u201cI subscript 2 plus sign I superscript negative sign equilibrium arrow I subscript 3 superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column has the following: 1.000 times 10 to the negative third power, negative 3.39 times 10 to the negative fourth power, 6.61 times 10 to the negative fourth power. The second column has the following: 1.000 times 10 to the negative third power, negative 3.39 times 10 to the negative fourth power, 6.61 times 10 to the negative fourth power. The third column has the following: 0, positive 3.39 times 10 to the negative fourth power, 3.39 times 10 to the negative fourth power.\" width=\"600\" height=\"162\" \/>\r\n\r\nFinally, substitute the equilibrium concentrations into the\u00a0<em data-effect=\"italics\">K<\/em>\u00a0expression and solve:\r\n<p style=\"text-align: center;\">[latex]{K}_{c}={Q}_{c}=\\dfrac{\\left[{\\text{I}}_{3}{}^{-}\\right]}{\\left[{I}_{2}\\right]\\left[{\\text{I}}^{-}\\right]}[\/latex]\r\n[latex]=\\dfrac{3.39\\times {10}^{-4}M}{\\left(6.61\\times {10}^{-4}M\\right)\\left(6.61\\times {10}^{-4}M\\right)}=776[\/latex]<\/p>\r\n\r\n<h4>Check Your Learning<\/h4>\r\nEthanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.\r\n<p style=\"text-align: center;\">[latex]{\\text{C}}_{2}{\\text{H}}_{5}\\text{OH}+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\rightleftharpoons{\\text{CH}}_{3}{\\text{CO}}_{2}{\\text{C}}_{2}{\\text{H}}_{5}+{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\r\nWhen 1 mol each of C<sub>2<\/sub>H<sub>5<\/sub>OH and CH<sub>3<\/sub>CO<sub>2<\/sub>H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when [latex]\\frac{1}{3}[\/latex] mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (<em>Note:<\/em> Water is not a solvent in this reaction.)\r\n[reveal-answer q=\"619052\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"619052\"]<em>K<\/em><sub>c<\/sub> = 4[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>What is the value of the equilibrium constant at 500 \u00b0C for the formation of NH<sub>3<\/sub> according to the following equation? [latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]\r\nAn equilibrium mixture of NH<sub>3<\/sub>(<em>g<\/em>), H<sub>2<\/sub>(<em>g<\/em>), and N<sub>2<\/sub>(<em>g<\/em>) at 500 \u00b0C was found to contain 1.35 <em>M<\/em> H<sub>2<\/sub>, 1.15 <em>M<\/em> N<sub>2<\/sub>, and 4.12 \u00d7 10<sup>-1<\/sup><em>M<\/em> NH<sub>3<\/sub>.<\/li>\r\n \t<li>Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.\r\n[latex]{\\text{CH}}_{4}\\left(g\\right)+{\\text{H}}_{2}\\text{O}\\left(g\\right)\\rightleftharpoons3{\\text{H}}_{2}\\left(g\\right)+\\text{CO}\\left(g\\right)[\/latex]\r\nWhat is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH<sub>4<\/sub>, 0.126 <em>M<\/em>; H<sub>2<\/sub>O, 0.242 <em>M<\/em>; CO, 0.126 <em>M<\/em>; H<sub>2<\/sub> 1.15 <em>M<\/em>, at a temperature of 760 \u00b0C?<\/li>\r\n \t<li>A 0.72-mol sample of PCl<sub>5<\/sub> is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl<sub>3<\/sub>(<em>g<\/em>) and 0.40 mol of Cl<sub>2<\/sub>(<em>g<\/em>). Calculate the value of the equilibrium constant for the decomposition of PCl<sub>5<\/sub> to PCl<sub>3<\/sub> and Cl<sub>2<\/sub> at this temperature.<\/li>\r\n \t<li>At 1 atm and 25 \u00b0C, NO<sub>2<\/sub> with an initial concentration of 1.00 <em>M<\/em> is 3.3 \u00d7 10<sup>-3<\/sup>% decomposed into NO and O<sub>2<\/sub>. Calculate the value of the equilibrium constant for the reaction\r\n[latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NO}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>Calculate the value of the equilibrium constant <em>K<sub>P<\/sub><\/em> for the reaction [latex]2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NOCl}\\left(g\\right)[\/latex] from these equilibrium pressures: NO, 0.050 atm; Cl<sub>2<\/sub>, 0.30 atm; NOCl, 1.2 atm.<\/li>\r\n \t<li>When heated, iodine vapor dissociates according to this equation:\u00a0[latex]{\\text{I}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{I}\\left(g\\right)[\/latex]\r\nAt 1274 K, a sample exhibits a partial pressure of I<sub>2<\/sub> of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, <em>K<sub>P<\/sub><\/em>, for the decomposition at 1274 K.<\/li>\r\n \t<li>A sample of ammonium chloride was heated in a closed container:\u00a0[latex]{\\text{NH}}_{4}\\text{Cl}\\left(s\\right)\\rightleftharpoons{\\text{NH}}_{3}\\left(g\\right)+\\text{HCl}\\left(g\\right)[\/latex]\r\nAt equilibrium, the pressure of NH<sub>3<\/sub>(<em>g<\/em>) was found to be 1.75 atm. What is the value of the equilibrium constant <em>K<sub>P<\/sub><\/em> for the decomposition at this temperature?<\/li>\r\n \t<li>At a temperature of 60 \u00b0C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant <em>K<sub>P<\/sub><\/em> for the transformation at 60 \u00b0C?\r\n[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"141837\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"141837\"]\r\n\r\n1.\u00a0The reaction may be written as\r\n<p style=\"text-align: center;\">[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/p>\r\nThe equilibrium constant for the reaction is\r\n<p style=\"text-align: center;\">[latex]{K}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}=\\frac{{\\left(4.12\\times {10}^{-1}\\right)}^{2}}{\\left(1.15\\right){\\left(1.35\\right)}^{3}}=\\frac{{\\left(0.170\\right)}^{2}}{\\left(1.15\\right){\\left(2.46\\right)}^{3}}0.0600=6.00\\times10^{-2}[\/latex]<\/p>\r\n3.\u00a0The decomposition of PCl<sub>5<\/sub> to PCl<sub>3<\/sub> and Cl<sub>2<\/sub> is given as\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{PCl}}_{5}\\left(g\\right)\\rightleftharpoons{\\text{PCl}}_{3}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\hfill \\\\{K}_{c}=\\frac{\\left[{\\text{PCl}}_{3}\\right]\\left[{\\text{Cl}}_{2}\\right]}{\\left[{\\text{PCl}}_{5}\\right]}\\hfill \\end{array}[\/latex]<\/p>\r\nLet <em>x<\/em> = change in [PCl<sub>5<\/sub>].\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[PCl<sub>5<\/sub>]<\/th>\r\n<th>[PCl<sub>3<\/sub>]<\/th>\r\n<th>[Cl<sub>2<\/sub>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (M)<\/th>\r\n<td>0.72<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (M)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium concentration (M)<\/th>\r\n<td>0.72\u00a0\u2212\u00a0<em>x<\/em> = 0.32<\/td>\r\n<td>0 +\u00a0<em>x<\/em> = 0.40<\/td>\r\n<td>0 +\u00a0<em>x<\/em> = 0.40<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center;\">[latex]{K}_{c}=\\frac{\\left(0.40\\right)\\left(0.40\\right)}{\\left(0.32\\right)}=0.50[\/latex]<\/p>\r\n5. \u00a0The equilibrium equation is\r\n<p style=\"text-align: center;\">[latex]{K}_{P}=\\frac{{\\left[\\text{NOCl}\\right]}^{2}}{{\\left[\\text{NO}\\right]}^{2}\\left[{\\text{Cl}}_{2}\\right]}=\\frac{{\\left(1.2\\right)}^{2}}{{\\left(0.050\\right)}^{2}\\left(0.30\\right)}=\\frac{1.44}{\\left(2.5\\times {10}^{-3}\\right)\\left(0.30\\right)}=1.9\\times {10}^{3}[\/latex]<\/p>\r\n7.\u00a0Because the decomposition must generate the same pressure of HCl as NH<sub>3<\/sub>, 1.75 atm of HCl must be present.\r\n<p style=\"text-align: center;\">[latex]{K}_{p}={P}_{{\\text{NH}}_{3}}{P}_{\\text{HCl}}=\\left(1.75\\text{atm}\\right)\\left(1.75\\text{atm}\\right)=3.06[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Calculation of a Missing Equilibrium Concentration<\/h3>\r\nWhen the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated. A computation of this sort is illustrated in the next example exercise.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0Calculation of a Missing Equilibrium Concentration<\/h3>\r\nNitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 \u00b0C, the value of the equilibrium constant for the reaction, [latex]{\\text{N}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NO}\\left(g\\right)[\/latex], is 4.1 \u00d7 10<sup>\u22124<\/sup>. Find the concentration of NO(<em>g<\/em>) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N<sub>2<\/sub>] = 0.036 mol\/L and [O<sub>2<\/sub>] 0.0089 mol\/L.\r\n\r\n[reveal-answer q=\"844327\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"844327\"]\r\n\r\nWe are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{ }{K}_{c}&amp;=&amp;\\dfrac{{\\left[\\text{NO}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]}\\\\{\\left[\\text{NO}\\right]}^{2}&amp;=&amp;{K}_{c}\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]\\\\\\left[\\text{NO}\\right]&amp;=&amp;\\sqrt{{K}_{c}\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]}\\\\ &amp;=&amp;\\sqrt{\\left(4.1\\times {10}^{-4}\\right)\\left(0.036\\right)\\left(0.0089\\right)}\\\\ &amp;=&amp;\\sqrt{1.31\\times {10}^{-7}}\\\\ &amp;=&amp;3.6\\times {10}^{-4}\\end{array}[\/latex]<\/p>\r\nThus [NO] is 3.6 \u00d7 10<sup>\u22124<\/sup> mol\/L at equilibrium under these conditions.\r\n\r\nWe can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ }{K}_{c}&amp;=&amp;\\dfrac{{\\left[\\text{NO}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]}\\\\&amp;=&amp;\\dfrac{{\\left(3.6\\times {10}^{-4}\\right)}^{2}}{\\left(0.036\\right)\\left(0.0089\\right)}\\\\&amp;=&amp;4.0\\times {10}^{-4}\\end{array}[\/latex]<\/p>\r\nThe answer checks; our calculated value gives the equilibrium constant within the error associated with the significant figures in the problem.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nThe equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 \u00d7 10<sup>\u22122<\/sup>. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 <em>M<\/em> and 2.09 <em>M<\/em>, respectively.\r\n\r\n[reveal-answer q=\"705150\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"705150\"]\r\n\r\n1.53 mol\/L[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>Analysis of the gases in a sealed reaction vessel containing NH<sub>3<\/sub>, N<sub>2<\/sub>, and H<sub>2<\/sub> at equilibrium at 400 \u00b0C established the concentration of N<sub>2<\/sub> to be 1.2 <em>M<\/em> and the concentration of H<sub>2<\/sub> to be 0.24 <em>M<\/em>.\r\n[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right){K}_{c}=0.50\\text{ at }400^{\\circ}\\text{C}[\/latex]\r\nCalculate the equilibrium molar concentration of NH<sub>3<\/sub>.<\/li>\r\n \t<li>Carbon reacts with water vapor at elevated temperatures.\r\n[latex]\\text{C}\\left(s\\right)+{\\text{H}}_{2}\\text{O}\\left(g\\right)\\rightleftharpoons\\text{CO}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right){K}_{c}=0.2\\text{ at }1000^{\\circ}\\text{C}[\/latex]\r\nWhat is the concentration of CO in an equilibrium mixture with [H<sub>2<\/sub>O] = 0.500 <em>M<\/em> at 1000 \u00b0C?<\/li>\r\n \t<li>Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.\r\n[latex]\\text{CoO}\\left(s\\right)+\\text{CO}\\left(g\\right)\\rightleftharpoons\\text{Co}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right){K}_{c}=4.90\\times {10}^{2}\\text{ at }550^{\\circ}\\text{C}[\/latex]\r\nWhat concentration of CO remains in an equilibrium mixture with [CO<sub>2<\/sub>] = 0.100 <em>M<\/em>?<\/li>\r\n \t<li>A student solved the following problem and found [N<sub>2<\/sub>O<sub>4<\/sub>] = 0.16 <em>M<\/em> at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of N<sub>2<\/sub>O<sub>4<\/sub> in a mixture formed from a sample of NO<sub>2<\/sub> with a concentration of 0.10 <em>M<\/em>?\r\n[latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right){K}_{c}=160[\/latex]<\/li>\r\n \t<li>A student solved the following problem and found the equilibrium concentrations to be [SO<sub>2<\/sub>] = 0.590 <em>M<\/em>, [O<sub>2<\/sub>] = 0.0450 <em>M<\/em>, and [SO<sub>3<\/sub>] = 0.260 <em>M<\/em>. How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 \u00b0C:\r\n[latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right){K}_{c}=4.32[\/latex]\r\nWhat are the equilibrium concentrations of all species in a mixture that was prepared with [SO<sub>3<\/sub>] = 0.500 <em>M<\/em>, [SO<sub>2<\/sub>] = 0 <em>M<\/em>, and [O<sub>2<\/sub>] = 0.350 <em>M<\/em><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"51339\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"51339\"]\r\n\r\n1. \u00a0Write the equilibrium constant expression and solve for [NH<sub>3<\/sub>].\r\n<p style=\"text-align: center;\">[latex]{K}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[1.2\\right]{\\left[0.24\\right]}^{3}}=0.50[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[NH<sub>3<\/sub>]<sup>2<\/sup> = 1.2 \u00d7 (0.24)<sup>3<\/sup> \u00d7 0.50 = 0.0083<\/p>\r\n<p style=\"text-align: center;\">[NH<sub>3<\/sub>] = 9.1 \u00d7 10<sup>-2<\/sup><em>M<\/em><\/p>\r\n3.\u00a0Write the equilibrium constant expression and solve for [CO].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{K}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]}{\\left[\\text{CO}\\right]}=\\frac{0.100}{\\left[\\text{CO}\\right]}=4.90\\times {10}^{2}\\hfill \\\\ \\left[\\text{CO}\\right]=\\frac{0.100}{4.90\\times {10}^{2}}=2.0\\times {0}^{-4}M\\hfill \\end{array}[\/latex]<\/p>\r\n5. Calculate <em>Q<\/em> based on the calculated concentrations and see if it is equal to <em>K<sub>c<\/sub><\/em>. Because <em>Q<\/em> does equal 4.32, the system must be at equilibrium.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Calculation of Changes in Concentration<\/h3>\r\n<p id=\"fs-idp100361248\">Perhaps the most challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:<\/p>\r\n\r\n<ol id=\"fs-idm126911232\" type=\"1\">\r\n \t<li>Identify the direction in which the reaction will proceed to reach equilibrium.<\/li>\r\n \t<li>Develop an ICE table.<\/li>\r\n \t<li>Calculate the concentration changes and, subsequently, the equilibrium concentrations.<\/li>\r\n \t<li>Confirm the calculated equilibrium concentrations.<\/li>\r\n<\/ol>\r\n<p id=\"fs-idm215588032\">The last two example exercises of this chapter demonstrate the application of this strategy.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 4:\u00a0Calculation of Concentration Changes as a Reaction Goes to Equilibrium<\/h3>\r\nUnder certain conditions, the equilibrium constant for the decomposition of PCl<sub>5<\/sub>(<em>g<\/em>) into PCl<sub>3<\/sub>(<em>g<\/em>) and Cl<sub>2<\/sub>(<em>g<\/em>) is 0.0211. What are the equilibrium concentrations of PCl<sub>5<\/sub>, PCl<sub>3<\/sub>, and Cl<sub>2<\/sub> if the initial concentration of PCl<sub>5<\/sub> was 1.00 <em>M<\/em>?\r\n\r\n[reveal-answer q=\"113380\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"113380\"]\r\n\r\nUse the stepwise process described above.\r\n<h4>Step 1: Determine the direction the reaction proceeds.<\/h4>\r\nThe balanced equation for the decomposition of PCl<sub>5<\/sub> is\r\n<p style=\"text-align: center;\">[latex]{\\text{PCl}}_{5}\\left(g\\right)\\rightleftharpoons{\\text{PCl}}_{3}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)[\/latex]<\/p>\r\nBecause we have no products initially, <em>Q<sub>c<\/sub><\/em> = 0 and the reaction will proceed to the right.\r\n<h4>Step 2: Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.<\/h4>\r\n<em data-effect=\"italics\">Develop an ICE table.<\/em>\r\n\r\n<img class=\"\" src=\"https:\/\/openstax.org\/resources\/6f80346599e8e7ccb11c6959f89464108531b29d\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, \u201cP C l subscript 5 equilibrium arrow P C l subscript 3 plus C l subscript 2.\u201d Under the second column is a subgroup of three rows and three columns. The first column has the following: 1.00, negative x, 1.00 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.\" width=\"600\" height=\"162\" \/>\r\n<h4>Step 3: Solve for the change and the equilibrium concentrations.<\/h4>\r\n<p id=\"fs-idp178328384\"><em data-effect=\"italics\">Solve for the change and the equilibrium concentrations.<\/em><\/p>\r\nSubstituting the equilibrium concentrations into the equilibrium constant equation gives\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{K}_{c}&amp;=&amp;\\dfrac{\\left[{\\text{PCl}}_{3}\\right]\\left[{\\text{Cl}}_{2}\\right]}{\\left[{\\text{PCl}}_{5}\\right]}=0.0211\\\\{}&amp;=&amp;\\dfrac{\\left(x\\right)\\left(x\\right)}{\\left(1.00-x\\right)}\\\\0.0211&amp;=&amp;\\dfrac{\\left(x\\right)\\left(x\\right)}{\\left(1.00-x\\right)}\\\\0.0211\\left(1.00-x\\right)&amp;=&amp;{x}^{2}\\\\{x}^{2}+0.0211 x-0.0211&amp;=&amp;0\\end{array}[\/latex]<\/p>\r\n<a class=\"target-chapter\" href=\".\/chapter\/essential-mathematics\/\" target=\"_blank\" rel=\"noopener\">Essential Mathematics<\/a> shows us an equation of the form <em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em> = 0 can be rearranged to solve for <em>x<\/em>:\r\n<p style=\"text-align: center;\">[latex]x=\\dfrac{-b\\pm\\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/p>\r\nIn this case, [latex]a = 1, b = 0.0211[\/latex], and [latex]c = \u22120.0211[\/latex]. Substituting the appropriate values for <em>a<\/em>, <em>b<\/em>, and <em>c<\/em> yields:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{x}&amp;=&amp;\\dfrac{-0.0211\\pm\\sqrt{\\left(0.0211\\right)^{2}-4\\left(1\\right)\\left(-0.0211\\right)}}{2\\left(1\\right)}\\\\{}&amp;=&amp;\\dfrac{-0.0211\\pm\\sqrt{\\left(4.45\\times{10}^{-4}\\right)+\\left(8.44\\times{10}^{-2}\\right)}}{2}\\\\{}&amp;=&amp;\\dfrac{-0.0211\\pm0.291}{2}\\end{array}[\/latex]<\/p>\r\nThe two roots of the quadratic are, therefore,\r\n<p style=\"text-align: center;\">[latex]x=\\dfrac{-0.0211+0.291}{2}=0.135[\/latex] or [latex]x=\\dfrac{-0.0211-0.291}{2}=-0.156[\/latex].<\/p>\r\n<p id=\"fs-idp226605088\">For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so\u00a0<em data-effect=\"italics\">x<\/em>\u00a0= 0.135\u00a0<em data-effect=\"italics\">M<\/em>.<\/p>\r\n<p id=\"fs-idp66450592\">The equilibrium concentrations are<\/p>\r\n\r\n<ul>\r\n \t<li>[latex]\\left[{\\text{PCl}}_{5}\\right]=1.00-0.135=0.87M[\/latex]<\/li>\r\n \t<li>[latex]\\left[{\\text{PCl}}_{3}\\right]=x=0.135M[\/latex]<\/li>\r\n \t<li>[latex]\\left[{\\text{Cl}}_{2}\\right]=x=0.135M[\/latex]<\/li>\r\n<\/ul>\r\n<h4>Step 4: Check the arithmetic.<\/h4>\r\n<p id=\"fs-idm43994720\"><em data-effect=\"italics\">Confirm the calculated equilibrium concentrations.<\/em><\/p>\r\n<p id=\"fs-idp274032256\">Substitution into the expression for\u00a0<em data-effect=\"italics\">K<sub>c<\/sub><\/em>\u00a0(to check the calculation) gives<\/p>\r\n<p style=\"text-align: center;\">[latex]{K}_{c}=\\dfrac{\\left[{\\text{PCl}}_{3}\\right]\\left[{\\text{Cl}}_{2}\\right]}{\\left[{\\text{PCl}}_{5}\\right]}=\\dfrac{\\left(0.135\\right)\\left(0.135\\right)}{0.87}=0.021[\/latex]<\/p>\r\nThe equilibrium constant calculated from the equilibrium concentrations is equal to the value of <em>K<sub>c<\/sub><\/em> given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nAcetic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>H, reacts with ethanol, C<sub>2<\/sub>H<sub>5<\/sub>OH, to form water and ethyl acetate, CH<sub>3<\/sub>CO<sub>2<\/sub>C<sub>2<\/sub>H<sub>5<\/sub>.\r\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}+{\\text{C}}_{2}{\\text{H}}_{5}\\text{OH}\\rightleftharpoons{\\text{CH}}_{3}{\\text{CO}}_{2}{\\text{C}}_{2}{\\text{H}}_{5}+{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\r\nThe equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations when a mixture that is 0.15 <em>M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>H, 0.15 <em>M<\/em> in C<sub>2<\/sub>H<sub>5<\/sub>OH, 0.40 <em>M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>C<sub>2<\/sub>H<sub>5<\/sub>, and 0.40 <em>M<\/em> in H<sub>2<\/sub>O are mixed in enough dioxane to make 1.0 L of solution?\r\n\r\n[reveal-answer q=\"703405\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"703405\"][CH<sub>3<\/sub>CO<sub>2<\/sub>H] = 0.18\u00a0<em>M<\/em>, [C<sub>2<\/sub>H<sub>5<\/sub>OH] = 0.37\u00a0<em>M<\/em>, [CH<sub>3<\/sub>CO<sub>2<\/sub>C<sub>2<\/sub>H<sub>5<\/sub>] = 0.37 <em>M<\/em>, [H<sub>2<\/sub>O] = 0.37 <em>M<\/em>[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nA 1.00-L flask is filled with 1.00 moles of H<sub>2<\/sub> and 2.00 moles of I<sub>2<\/sub>. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H<sub>2<\/sub>, I<sub>2<\/sub>, and HI in moles\/L?\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{I}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{HI}\\left(g\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"604530\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"604530\"][H<sub>2<\/sub>] = 0.06 <em>M<\/em>, [I<sub>2<\/sub>] = 1.06 <em>M<\/em>, [HI] = 1.88 <em>M<\/em>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\"><header>\r\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE 5:\u00a0<\/span>Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption<\/h3>\r\n<\/header>What are the concentrations at equilibrium of a 0.15\u00a0<em>M<\/em>\u00a0solution of HCN?\r\n\r\n[latex]\\text{HCN}(aq) \\rightleftharpoons \\text{H}^+ (aq) + \\text{CN}^- (aq)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_c = 4.9 \\text{x} 10^{-10}[\/latex]\r\n\r\n[reveal-answer q=\"703415\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"703415\"]\r\nUsing \u201c<em>x<\/em>\u201d to represent the concentration of each product at equilibrium gives this ICE table.<span id=\"fs-idm90952112\" class=\"scaled-down\"><img id=\"11\" class=\"alignnone\" src=\"https:\/\/openstax.org\/resources\/52ab1350d0394ec1eeb1d18fdc50a25d91e97593\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium ( M ). The second column has the header, \u201cH C N ( a q ) equilibrium arrow H superscript plus sign ( a q ) plus C N subscript negative sign ( a q ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.15, negative x, 0.15 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.\" width=\"601\" height=\"162\" \/><\/span>\r\n<p id=\"fs-idp22895280\">Substitute the equilibrium concentration terms into the\u00a0<em>K<sub>c<\/sub><\/em>\u00a0expression<\/p>\r\n<p style=\"text-align: center;\">[latex]K_c = \\dfrac{(x)(x)}{0.15-x}[\/latex]<\/p>\r\n<p id=\"fs-idp77336512\">rearrange to the quadratic form and solve for\u00a0<em>x<\/em><\/p>\r\n<p style=\"text-align: center;\">[latex]x^2 +4.9 \\times 10^{-10} - 7.35 \\times 10^{-11} = 0[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x = 8.56 \\text{x} 10^{-6}M \\text{(3 sig. figs)} = 8.6 \\text{x} 10^{-6}M \\text{(2 sig. figs)}[\/latex]<\/p>\r\n<p id=\"fs-idm36129424\">Thus [H<sup>+<\/sup>] = [CN<sup>\u2013<\/sup>] =\u00a0<em>x<\/em>\u00a0= 8.6\u00a0<span class=\"os-math-in-para\"><span id=\"MathJax-Element-83-Frame\" class=\"MathJax\" style=\"overflow: initial; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; letter-spacing: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px;\" role=\"presentation\"><span id=\"MathJax-Span-2767\" class=\"math\"><span id=\"MathJax-Span-2768\" class=\"mrow\"><span id=\"MathJax-Span-2769\" class=\"semantics\"><span id=\"MathJax-Span-2770\" class=\"mrow\"><span id=\"MathJax-Span-2771\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><span class=\"MJX_Assistive_MathML\" role=\"presentation\">\u00d7<\/span><\/span><\/span>\u00a010<sup>\u20136<\/sup>\u00a0<em>M<\/em>\u00a0and [HCN] = 0.15 \u2013\u00a0<em>x<\/em>\u00a0= 0.15\u00a0<em>M<\/em>.<\/p>\r\n<p id=\"fs-idm8990656\">Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small\u00a0<em>K<\/em>), and so the initial concentration experiences a negligible change:<\/p>\r\n<p style=\"text-align: center;\">if [latex]x \\ll 0.15 M, \\text{then} (0.15-x)[\/latex] \u2248 [latex]0.15[\/latex]<\/p>\r\n<p class=\"body\" style=\"text-align: left;\">This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]K_c = \\dfrac{(x)(x)}{0.15-x} [\/latex] \u2248 [latex]\\dfrac{x^2}{0.15}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]4.9 \\times 10^{-10} = \\dfrac{x^2}{0.15}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x^2 = (0.15)(4.9\\times 10^{-10}) = 7.4\\times 10^{-11}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x = \\sqrt{7.4\\times 10^{-11}} = 8.6\\times 10^{-6}M[\/latex]<\/p>\r\nThe value of\u00a0<em>x<\/em>\u00a0calculated is, indeed, much less than the initial concentration\r\n<p style=\"text-align: center;\">[latex]8.6\\times{10}^{-6}\\ll{0.15}[\/latex]<\/p>\r\nand so the approximation was justified. If this simplified approach were to yield a value for\u00a0<em>x<\/em>\u00a0that did\u00a0<em>not<\/em>\u00a0justify the approximation, the calculation would need to be repeated without making the approximation.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nWhat are the equilibrium concentrations in a 0.25\u00a0<em>M<\/em>\u00a0NH<sub>3<\/sub>\u00a0solution?\r\n\r\n[latex] \\text{NH}_3(aq) + \\text{H}_2\\text{O}(l) \\rightleftharpoons \\text{NH}_4+(aq) +\\text{OH}-(aq)\\qquad{K}_{c} = 1.8 \\times 10^{-5}[\/latex]\r\n\r\n<section><\/section><section>[reveal-answer q=\"703416\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"703416\"]\r\n<div class=\"os-note-body\"><center>[latex][\\text{OH}-] = [\\text{NH}_4+] = 0.0021 M; [\\text{NH}_3] = 0.25 M[\/latex]<\/center>[\/hidden-answer]<\/div>\r\n<\/section><\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{llll}2{\\text{SO}}_{3}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; 2{\\text{SO}}_{2}\\left(g\\right)+\\hfill &amp; {\\text{O}}_{2}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill &amp; &amp; \\text{ }\\hfill &amp; +x\\hfill \\\\ \\text{ }\\hfill &amp; &amp; \\text{ }\\hfill &amp; 0.125M\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{lllll}4{\\text{NH}}_{3}\\left(g\\right)\\hfill &amp; +3{\\text{O}}_{2}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; 2{\\text{N}}_{2}\\left(g\\right)+\\hfill &amp; 6{\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill &amp; 3x\\hfill &amp; &amp; \\text{ }\\hfill &amp; \\text{ }\\hfill \\\\ \\text{ }\\hfill &amp; 0.24M\\hfill &amp; &amp; \\text{ }\\hfill &amp; \\text{ }\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>Change in pressure:\r\n[latex]\\begin{array}{llll}2{\\text{CH}}_{4}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; {\\text{C}}_{2}{\\text{H}}_{2}\\left(g\\right)+\\hfill &amp; 3{\\text{H}}_{2}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill &amp; &amp; x\\hfill &amp; \\text{ }\\hfill \\\\ \\text{ }\\hfill &amp; &amp; 25\\text{torr}\\hfill &amp; \\text{ }\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>Change in pressure:\r\n[latex]\\begin{array}{lllll}{\\text{CH}}_{4}\\left(g\\right)+\\hfill &amp; {\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; \\text{CO}\\left(g\\right)+\\hfill &amp; 3{\\text{H}}_{2}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill &amp; x\\hfill &amp; &amp; \\text{ }\\hfill &amp; \\text{ }\\hfill \\\\ \\text{ }\\hfill &amp; 5\\text{atm}\\hfill &amp; &amp; \\text{ }\\hfill &amp; \\text{ }\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{llll}{\\text{NH}}_{4}\\text{Cl}\\left(s\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; {\\text{NH}}_{3}\\left(g\\right)+\\hfill &amp; \\text{HCl}\\left(g\\right)\\hfill \\\\ &amp; &amp; x\\hfill &amp; \\text{ }\\hfill \\\\ &amp; &amp; \\hfill 1.03\\times {10}^{-4}M\\hfill &amp; \\text{ }\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>change in pressure:\r\n[latex]\\begin{array}{cccc}\\text{Ni}\\left(s\\right)+\\hfill &amp; 4\\text{CO}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; \\text{Ni}{\\left(\\text{CO}\\right)}_{4}\\left(g\\right)\\hfill \\\\ &amp; 4x\\hfill &amp; &amp; \\text{ }\\hfill \\\\ &amp; \\hfill 0.40\\text{atm}\\hfill &amp; &amp; \\text{ }\\hfill \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{cccc}2{\\text{H}}_{2}\\left(g\\right)+\\hfill &amp; {\\text{O}}_{2}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; 2{\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill &amp; \\text{ }\\hfill &amp; &amp; +2x\\hfill \\\\ \\text{ }\\hfill &amp; \\text{ }\\hfill &amp; &amp; 1.50M\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccccc}{\\text{CS}}_{2}\\left(g\\right)+\\hfill &amp; 4{\\text{H}}_{2}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; {\\text{CH}}_{4}\\left(g\\right)+\\hfill &amp; 2{\\text{H}}_{2}\\text{S}\\left(g\\right)\\hfill \\\\ x\\hfill &amp; \\text{ }\\hfill &amp; &amp; \\text{ }\\hfill &amp; \\text{ }\\hfill \\\\ 0.020M\\hfill &amp; \\text{ }\\hfill &amp; &amp; \\text{ }\\hfill &amp; \\text{ }\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>Change in pressure:\r\n[latex]\\begin{array}{cccc}{\\text{H}}_{2}\\left(g\\right)+\\hfill &amp; {\\text{Cl}}_{2}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; 2\\text{HCl}\\left(g\\right)\\hfill \\\\ x\\hfill &amp; \\text{ }\\hfill &amp; &amp; \\text{ }\\hfill \\\\ 1.50\\text{atm}\\hfill &amp; \\text{ }\\hfill &amp; &amp; \\text{ }\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>Change in pressure:\r\n[latex]\\begin{array}{ccccc}2{\\text{NH}}_{3}\\left(g\\right)\\hfill &amp; +2{\\text{O}}_{2}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; {\\text{N}}_{2}\\text{O}\\left(g\\right)+\\hfill &amp; 3{\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill &amp; \\text{ }\\hfill &amp; &amp; \\text{ }\\hfill &amp; x\\hfill \\\\ \\text{ }\\hfill &amp; \\text{ }\\hfill &amp; &amp; \\text{ }\\hfill &amp; 60.6\\text{torr}\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{cccc}{\\text{NH}}_{4}\\text{HS}\\left(s\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; {\\text{NH}}_{3}\\left(g\\right)+\\hfill &amp; {\\text{H}}_{2}\\text{S}\\left(g\\right)\\hfill \\\\ &amp; &amp; x\\hfill &amp; \\text{ }\\hfill \\\\ &amp; &amp; 9.8\\times {10}^{-6}M\\hfill &amp; \\text{ }\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>Change in pressure:\r\n[latex]\\begin{array}{cccc}\\text{Fe}\\left(s\\right)+\\hfill &amp; 5\\text{CO}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; \\text{Fe}{\\left(\\text{CO}\\right)}_{4}\\left(g\\right)\\hfill \\\\ &amp; \\text{ }\\hfill &amp; &amp; x\\hfill \\\\ &amp; \\text{ }\\hfill &amp; &amp; 0.012\\text{atm}\\hfill \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Why are there no changes specified for Ni in question\u00a011 , part (f)? What property of Ni does change?<\/li>\r\n \t<li>Why are there no changes specified for NH<sub>4<\/sub>HS in question 12, part (e)? What property of NH<sub>4<\/sub>HS does change?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"381796\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"381796\"]\r\n\r\n1.\u00a0The changes in concentrations (or pressure, if requested)\u00a0are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{llll}2{\\text{SO}}_{3}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; 2{\\text{SO}}_{2}\\left(g\\right)+\\hfill &amp; {\\text{O}}_{2}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill -2x &amp; &amp;2x\\hfill &amp; +x\\hfill \\\\ -0.250M\\hfill &amp; &amp; 0.250M\\hfill &amp; 0.125M\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{lllll}4{\\text{NH}}_{3}\\left(g\\right)\\hfill &amp; +3{\\text{O}}_{2}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; 2{\\text{N}}_{2}\\left(g\\right)+\\hfill &amp; 6{\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill \\\\ 4x\\hfill &amp; 3x\\hfill &amp; &amp; -2x\\hfill &amp; -6x\\hfill \\\\ 0.32M\\hfill &amp; 0.24M\\hfill &amp; &amp;-0.16M\\hfill &amp; -0.48M\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>Change in pressure:\r\n[latex]\\begin{array}{llll}2{\\text{CH}}_{4}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; {\\text{C}}_{2}{\\text{H}}_{2}\\left(g\\right)+\\hfill &amp; 3{\\text{H}}_{2}\\left(g\\right)\\hfill \\\\ -2x\\hfill &amp; &amp; x\\hfill &amp; 3x\\hfill \\\\ -50\\text{ torr}\\hfill &amp; &amp; 25\\text{ torr}\\hfill &amp; 75\\text{ torr}\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>Change in pressure:\r\n[latex]\\begin{array}{lllll}{\\text{CH}}_{4}\\left(g\\right)+\\hfill &amp; {\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; \\text{CO}\\left(g\\right)+\\hfill &amp; 3{\\text{H}}_{2}\\left(g\\right)\\hfill \\\\ x\\hfill &amp; x\\hfill &amp; &amp; -x\\hfill &amp; -3x\\hfill \\\\ 5\\text{ atm}\\hfill &amp; 5\\text{atm}\\hfill &amp; &amp; -5\\text{ atm}\\hfill &amp; -15\\text{ atm}\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{llll}{\\text{NH}}_{4}\\text{Cl}\\left(s\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; {\\text{NH}}_{3}\\left(g\\right)+\\hfill &amp; \\text{HCl}\\left(g\\right)\\hfill \\\\ &amp; &amp; x\\hfill &amp; x\\hfill \\\\ &amp; &amp; \\hfill 1.03\\times {10}^{-4}M\\hfill &amp; 1.03\\times10^{-4}M\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>change in pressure:\r\n[latex]\\begin{array}{cccc}\\text{Ni}\\left(s\\right)+\\hfill &amp; 4\\text{CO}\\left(g\\right)\\hfill &amp; \\rightleftharpoons\\hfill &amp; \\text{Ni}{\\left(\\text{CO}\\right)}_{4}\\left(g\\right)\\hfill \\\\ &amp; 4x\\hfill &amp; &amp; x\\hfill \\\\ &amp; \\hfill 0.40\\text{atm}\\hfill &amp; &amp; 0.1\\text{ atm}\\hfill \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n3.\u00a0Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>Assume that the change in concentration of N<sub>2<\/sub>O<sub>4<\/sub> is small enough to be neglected in the following problem.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N<sub>2<\/sub>O<sub>4<\/sub> with chloroform as the solvent.\r\n[latex]{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)\\rightleftharpoons2{\\text{NO}}_{2}\\left(g\\right){K}_{c}=1.07\\times {10}^{-5}[\/latex] in chloroform<\/li>\r\n \t<li>Show that the change is small enough to be neglected.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assume that the change in concentration of COCl<sub>2<\/sub> is small enough to be neglected in the following problem.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl<sub>2<\/sub> with an initial concentration of 0.3166 <em>M<\/em>.\r\n[latex]{\\text{COCl}}_{2}\\left(g\\right)\\rightleftharpoons\\text{CO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right){K}_{c}=2.2\\times {10}^{-10}[\/latex]<\/li>\r\n \t<li>Show that the change is small enough to be neglected.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assume that the change in pressure of H<sub>2<\/sub>S is small enough to be neglected in the following problem.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H<sub>2<\/sub>S with an initial pressure of 0.824 atm.\r\n[latex]2{\\text{H}}_{2}\\text{S}\\left(g\\right)\\rightleftharpoons2{\\text{H}}_{2}\\left(g\\right)+{\\text{S}}_{2}\\left(g\\right){K}_{P}=2.2\\times {10}^{-6}[\/latex]<\/li>\r\n \t<li>Show that the change is small enough to be neglected.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What are all concentrations after a mixture that contains [H<sub>2<\/sub>O] = 1.00 <em>M<\/em> and [Cl<sub>2<\/sub>O] = 1.00 <em>M<\/em> comes to equilibrium at 25 \u00b0C?\r\n[latex]{\\text{H}}_{2}\\text{O}\\left(g\\right)+{\\text{Cl}}_{2}\\text{O}\\left(g\\right)\\rightleftharpoons2\\text{HOCl}\\left(g\\right){K}_{c}=0.0900[\/latex]<\/li>\r\n \t<li>What are the concentrations of PCl<sub>5<\/sub>, PCl<sub>3<\/sub>, and Cl<sub>2<\/sub> in an equilibrium mixture produced by the decomposition of a sample of pure PCl<sub>5<\/sub> with [PCl<sub>5<\/sub>] = 2.00 <em>M<\/em>?\r\n[latex]{\\text{PCl}}_{5}\\left(g\\right)\\rightleftharpoons{\\text{PCl}}_{3}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right){K}_{c}=0.0211[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"955866\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"955866\"]\r\n\r\n1. (a) Write the starting conditions, change, and equilibrium constant in tabular form.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[NO<sub>2<\/sub>]<\/th>\r\n<th>[N<sub>2<\/sub>O<sub>4<\/sub>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (M)<\/th>\r\n<td>0<\/td>\r\n<td>0.129<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (M)<\/th>\r\n<td>+2<em>x<\/em><\/td>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium concentration (M)<\/th>\r\n<td>2<em>x<\/em><\/td>\r\n<td>0.129\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince <em>K<\/em> is very small, ignore <em>x<\/em> in comparison with 0.129 <em>M<\/em>. The equilibrium expression is\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} {K}_{c}=1.07\\times {10}^{-5}=\\frac{{\\left[{\\text{NO}}_{2}\\right]}^{2}}{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}=\\frac{{\\left(2x\\right)}^{2}}{\\left(0.129-x\\right)}=\\frac{{\\left(2x\\right)}^{2}}{0.129}\\hfill \\\\ {x}^{2}=\\frac{0.129\\times 1.07\\times {10}^{-5}}{4}=3.45\\times {10}^{-7}\\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em> = 5.87 \u00d7 10<sup>\u22124<\/sup><\/p>\r\nThe concentrations are:\r\n<ul>\r\n \t<li>[NO<sub>2<\/sub>] = 2<em>x<\/em> = 5.87 \u00d7 10<sup>-4<\/sup> = 1.17 \u00d7 10<sup>-3<\/sup><em>M<\/em><\/li>\r\n \t<li>[N<sub>2<\/sub>O<sub>4<\/sub>] = 0.129 - <em>x<\/em> = 0.129 - 5.87 \u00d7 10<sup>-4<\/sup> = 0.128 <em>M<\/em><\/li>\r\n<\/ul>\r\n(b) Percent error [latex]=\\frac{5.87\\times {10}^{-4}}{0.129}\\times 100\\%=0.455\\%[\/latex]. The change in concentration of N<sub>2<\/sub>O<sub>4<\/sub> is far less than the 5% maximum allowed.\r\n\r\n3. \u00a0(a) Write the balanced equation, and then set up a table with initial pressures and the changed pressures using <em>x<\/em> as the change in pressure. The simplest way to find the coefficients for the <em>x<\/em> values is to use the coefficient in the balanced equation.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212416\/CNX_Chem_13_04_ICETable7_img1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( a t m ), Change ( a t m ), Equilibrium pressure ( a t m ). The second column has the header, \u201c2 H subscript 2 S ( g ) equilibrium arrow 2 H subscript 2 ( g ) plus S subscript 2 ( g ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.824, negative 2 x, 0.824 minus 2 x. The second column has the following: 0, negative 2 x, positive 2 x. The third column has the following: 0, negative 2 x, positive 2 x.\" \/>\r\n<p style=\"text-align: center;\">[latex]{K}_{P}=2.2\\times {10}^{-6}=\\frac{\\left({P}_{{\\text{S}}_{2}}\\right){\\left({P}_{{\\text{H}}_{2}}\\right)}^{2}}{{\\left({P}_{{\\text{H}}_{2}\\text{S}}\\right)}^{2}}=\\frac{\\left[x\\right]{\\left[2x\\right]}^{2}}{{\\left[0.824-2x\\right]}^{2}}[\/latex]<\/p>\r\nSince the value of <em>K<\/em><sub>p<\/sub> is much smaller than 0.824, the 2<em>x<\/em> term in 0.824 \u2212 2<em>x<\/em> is deemed negligible and, therefore, can be dropped.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{ }2.2\\times {10}^{-6}=\\frac{\\left[x\\right]{\\left[2x\\right]}^{2}}{{\\left[0.824\\right]}^{2}}\\hfill \\\\ 2.2\\times {10}^{-6}=\\frac{4{\\text{X}}^{3}}{\\left[0.679\\right]}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>1.494 \u00d7 10<sup>\u22126<\/sup> = 4<em>x<\/em><sup>3<\/sup><\/li>\r\n \t<li>3.73 \u00d7 10<sup>\u22127<\/sup> = <em>x<\/em><sup>3<\/sup><\/li>\r\n \t<li>7.20 \u00d7 10<sup>\u22123<\/sup> = <em>x<\/em><\/li>\r\n<\/ul>\r\nFinal equilibrium pressures:\r\n<ul>\r\n \t<li>[H<sub>2<\/sub>S] = 0.824 \u2212 2<em>x<\/em> = 0.824 \u2212 2(7.20 \u00d7 10<sup>\u22123<\/sup>) = 0.824 - 0.0144 = 0.810 atm<\/li>\r\n \t<li>[H<sub>2<\/sub>] = 2<em>x<\/em> = 2(7.2 \u00d7 10<sup>\u22123<\/sup>) = 0.014 atm<\/li>\r\n \t<li>[S<sub>2<\/sub>] = [<em>x<\/em>] = 0.0072 atm<\/li>\r\n<\/ul>\r\n(b) The 2<em>x<\/em> is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2<em>x<\/em> is [latex]\\frac{0.014}{0.824}\\times 100\\%=1.7\\%[\/latex], which is less than allowed by the \u201c5% test.\u201d The error is, indeed, negligible.\r\n\r\n5.\u00a0As all species are gases and are in <em>M<\/em> concentration units, a simple <em>K<sub>c<\/sub><\/em> equilibrium can be solved using the balanced equation:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212419\/CNX_Chem_13_04_ICETable9_img1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, \u201cP C l subscript 5 ( g ) equilibrium arrow P C l subscript 3 ( g ) plus C l subscript 2 ( g ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 2.00, negative x, 2.00 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.\" width=\"883\" height=\"239\" \/>\r\n<p style=\"text-align: center;\">[latex]{K}_{c}=0.0211=\\frac{\\left[{\\text{PCl}}_{3}\\right]\\left[{\\text{Cl}}_{2}\\right]}{\\left[{\\text{PCl}}_{5}\\right]}=\\frac{\\left[x\\right]\\left[x\\right]}{\\left[2.00-x\\right]}[\/latex]<\/p>\r\nAs the value of <em>K<sub>c<\/sub><\/em> is substantial when compared with 2.00 <em>M<\/em>, the <strong><em>x<\/em><\/strong> terms in 2.00 \u2212 <strong><em>x<\/em><\/strong> cannot be disregarded. Thus, <strong><em>x<\/em><\/strong> must be solved by using the quadratic formula.\r\n<p style=\"text-align: center;\">[latex]0.0211=\\frac{{x}^{2}}{\\left[2.00-x\\right]}=0.0422-0.0211x={x}^{2}[\/latex]<\/p>\r\nBegin by arranging the terms in the form of the quadratic equation:\r\n<ul>\r\n \t<li><em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em> = 0<\/li>\r\n \t<li><em>x<\/em><sup>2<\/sup> + 0.0211<em>x<\/em> \u2212 0.0422 = 0<\/li>\r\n<\/ul>\r\nNext, solve for <strong><em>x<\/em><\/strong> using the quadratic formula.\r\n<p style=\"text-align: center;\">[latex]x=\\frac{\\text{-}b\\pm\\sqrt{{b}^{2}-4ac}}{2a}=\\frac{-0.0211\\pm\\sqrt{{\\left(0.0211\\right)}^{2}-4\\left(1\\right)\\left(-0.0422\\right)}}{2\\left(1\\right)}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]=\\frac{-0.0211\\pm\\sqrt{0.0004452+0.1688}}{2}=\\frac{-0.0211\\pm 0.4114}{2}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">= 0.195 <em>M<\/em> or \u22120.216 <em>M<\/em><\/p>\r\nThe process of dissociation renders only positive quantities. Thus, <em>x<\/em> must be a positive value when factored into the solution so as to guarantee a realistic result. The final equilibrium concentrations are: [PCl<sub>3<\/sub>] = 2.00 \u2212 <em>x<\/em> = 2.00 \u2212 0.195 = 1.80 <em>M<\/em>; [PC<sub>3<\/sub>] = [Cl<sub>2<\/sub>] = <em>x<\/em> = 0.195 <em>M<\/em>; [PCl<sub>3<\/sub>] = [Cl<sub>2<\/sub>] = <em>x<\/em> = 0.195 <em>M<\/em>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nCalculating values for equilibrium constants and\/or equilibrium concentrations is of practical benefit to many applications. A mathematical strategy that uses initial concentrations, changes in concentrations, and equilibrium concentrations (and goes by the acronym ICE) is useful for several types of equilibrium calculations.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H<sub>2<\/sub> and 1.25 mol of I<sub>2<\/sub> in a 5.00-L flask at 448 \u00b0C.\r\n[latex]{\\text{H}}_{2}+{\\text{I}}_{2}\\rightleftharpoons2\\text{HI}{K}_{c}=50.2\\text{ at }448^{\\circ}\\text{C}[\/latex]<\/li>\r\n \t<li>What is the pressure of BrCl in an equilibrium mixture of Cl<sub>2<\/sub>, Br<sub>2<\/sub>, and BrCl if the pressure of Cl<sub>2<\/sub> in the mixture is 0.115 atm and the pressure of Br<sub>2<\/sub> in the mixture is 0.450 atm?\r\n[latex]{\\text{Cl}}_{2}\\left(g\\right)+{\\text{Br}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{BrCl}\\left(g\\right){K}_{P}=4.7\\times {10}^{-2}[\/latex]<\/li>\r\n \t<li>What is the pressure of CO<sub>2<\/sub> in a mixture at equilibrium that contains 0.50 atm H<sub>2<\/sub>, 2.0 atm of H<sub>2<\/sub>O, and 1.0 atm of CO at 990 \u00b0C?\r\n[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right)+\\text{CO}\\left(g\\right){K}_{P}=1.6\\text{ at }990^{\\circ}\\text{C}[\/latex]<\/li>\r\n \t<li>Sodium sulfate 10-hydrate, Na<sub>2<\/sub>SO<sub>4<\/sub> [latex]\\cdot [\/latex] 10H<sub>2<\/sub>O, dehydrates according to the equation\r\n[latex]{\\text{Na}}_{2}{\\text{SO}}_{4}\\cdot 10{\\text{H}}_{2}\\text{O}\\left(s\\right)\\rightleftharpoons{\\text{Na}}_{2}{\\text{SO}}_{4}\\left(s\\right)+10{\\text{H}}_{2}\\text{O}\\left(g\\right){K}_{P}=4.08\\times {10}^{-25}\\text{ at }25^{\\circ}\\text{C}[\/latex]\r\nWhat is the pressure of water vapor at equilibrium with a mixture of Na<sub>2<\/sub>SO<sub>4<\/sub> [latex]\\cdot [\/latex] 10H<sub>2<\/sub>O and NaSO<sub>4<\/sub>?<\/li>\r\n \t<li>Calcium chloride 6-hydrate, CaCl<sub>2<\/sub> [latex]\\cdot [\/latex] 6H<sub>2<\/sub>O, dehydrates according to the equation\r\n[latex]{\\text{CaCl}}_{2}\\cdot 6{\\text{H}}_{2}\\text{O}\\left(s\\right)\\rightleftharpoons{\\text{CaCl}}_{2}\\left(s\\right)+6{\\text{H}}_{2}\\text{O}\\left(g\\right){K}_{P}=5.09\\times {10}^{-44}\\text{ at }25^{\\circ}\\text{C}[\/latex]\r\nWhat is the pressure of water vapor at equilibrium with a mixture of CaCl<sub>2<\/sub> [latex]\\cdot [\/latex] 6H<sub>2<\/sub>O and CaCl<sub>2<\/sub>?<\/li>\r\n \t<li>Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl<sub>2<\/sub> produced when a sample of NOCl with a pressure of 0.500 atm comes to equilibrium according to this reaction:\r\n[latex]2\\text{NOCl}\\left(g\\right)\\rightleftharpoons 2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right){K}_{P}=4.0\\times {10}^{-4}[\/latex]<\/li>\r\n \t<li>Calculate the equilibrium concentrations of NO, O<sub>2<\/sub>, and NO<sub>2<\/sub> in a mixture at 250 \u00b0C that results from the reaction of 0.20 <em>M<\/em> NO and 0.10 <em>M<\/em> O<sub>2<\/sub>. (Hint: <em>K<\/em> is large; assume the reaction goes to completion then comes back to equilibrium.)\r\n[latex]2\\text{NO}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NO}}_{2}\\left(g\\right){K}_{c}=2.3\\times {10}^{5}\\text{ at }250^{\\circ}\\text{C}[\/latex]<\/li>\r\n \t<li>Calculate the equilibrium concentrations that result when 0.25 <em>M<\/em> O<sub>2<\/sub> and 1.0 <em>M<\/em> HCl react and come to equilibrium. (Hint: <em>K<sub>c<\/sub><\/em> is large; assume the reaction goes to completion, then comes back to equilibrium.)\r\n[latex]4\\text{HCl}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{Cl}}_{2}\\left(g\\right)+2{\\text{H}}_{2}\\text{O}\\left(g\\right){K}_{c}=3.1\\times {10}^{13}[\/latex]<\/li>\r\n \t<li>One of the important reactions in the formation of smog is represented by the equation\r\n[latex]{\\text{O}}_{3}\\left(g\\right)+\\text{NO}\\left(g\\right)\\rightleftharpoons{\\text{NO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right){K}_{P}=6.0\\times {10}^{34}[\/latex]\r\nWhat is the pressure of O<sub>3<\/sub> remaining after a mixture of O<sub>3<\/sub> with a pressure of 1.2 \u00d7 10<sup>-8<\/sup> atm and NO with a pressure of 1.2 \u00d7 10<sup>-8<\/sup> atm comes to equilibrium? (Hint: <em>K<sub>P<\/sub><\/em> is large; assume the reaction goes to completion then comes back to equilibrium.)<\/li>\r\n \t<li>Calculate the pressures of NO, Cl<sub>2<\/sub>, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 4.0 atm NO and 2.0 atm Cl<sub>2<\/sub>. (Hint: <em>K<sub>P<\/sub><\/em> is small; assume the reverse reaction goes to completion then comes back to equilibrium.)<\/li>\r\n \t<li>Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H<sub>2<\/sub> and 63.5 g of iodine at 448 \u00b0C.\r\n[latex]{\\text{H}}_{2}+{\\text{I}}_{2}\\rightleftharpoons2\\text{HI}{K}_{c}=50.2\\text{ at }448^{\\circ}\\text{C}[\/latex]<\/li>\r\n \t<li>Butane exists as two isomers, <em>n<\/em>-butane and isobutane.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212432\/CNX_Chem_13_04_butane_img1.jpg\" alt=\"Three Lewis structures are shown. The first is labeled, \u201cn dash Butane,\u201d and has a C H subscript 3 single bonded to a C H subscript 2 group. This C H subscript 2 group is single bonded to another C H subscript 2 group which is single bonded to a C H subscript 3 group. The second is labeled, \u201ciso dash Butane,\u201d and is composed of a C H group single bonded to three C H subscript 3 groups. The third structure shows a chain of atoms: \u201cC H subscript 3, C H subscript 2, C H subscript 2, C H subscript 3,\u201d a double-headed arrow, then a carbon atom single bonded to three C H subscript 3 groups as well as a hydrogen atom.\" \/>\r\n<em>K<sub>P<\/sub><\/em> = 2.5 at 25 \u00b0CWhat is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?<\/li>\r\n \t<li>What is the minimum mass of CaCO<sub>3<\/sub> required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (<em>K<sub>c<\/sub><\/em>) is 0.050 for the decomposition reaction of CaCO<sub>3<\/sub> at that temperature?\r\n[latex]{\\text{CaCO}}_{3}\\left(s\\right)\\rightleftharpoons\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>The equilibrium constant (<em>K<sub>c<\/sub><\/em>) for this reaction is 1.60 at 990 \u00b0C:\u00a0[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right)+\\text{CO}\\left(g\\right)[\/latex]\r\nCalculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H<sub>2<\/sub>, 2.00 mol of CO<sub>2<\/sub>, 0.750 mol of H<sub>2<\/sub>O, and 1.00 mol of CO to a 5.00-L container at 990 \u00b0C.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"303841\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"303841\"]\r\n\r\n2.\u00a0Write the equilibrium constant expression and solve for <em>P<\/em><sub>BrCl<\/sub>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{K}_{P}=\\frac{{\\left({P}_{\\text{BrCl}}\\right)}^{2}}{{P}_{{\\text{Cl}}_{2}}{P}_{{\\text{Br}}_{2}}}=\\frac{{\\left({P}_{\\text{BrCl}}\\right)}^{2}}{\\left(0.115\\right)\\left(0.450\\right)}=4.7\\times {10}^{-2}\\\\ {\\left({P}_{\\text{BrCl}}\\right)}^{2}=0.115\\times 0.450\\times 4.7\\times {10}^{-2}=2.43\\times {10}^{-3}\\text{atm}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>P<\/em><sub>BrCl<\/sub> = 4.9 \u00d7 10<sup>-2<\/sup> atm<\/p>\r\n4.\u00a0Because two of the substances involved in the equilibrium are solids, their activities are 1, and their pressures are constant and do not appear in the equilibrium expression.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{K}_{p}=4.08\\times {10}^{-25}={\\left({P}_{{\\text{H}}_{2}\\text{O}}\\right)}^{10}\\\\ {P}_{{\\text{H}}_{2}\\text{O}}=\\sqrt[10]{4.08\\times {10}^{-25}}=3.64\\times {10}^{-3}\\text{atm}\\end{array}[\/latex]<\/p>\r\n7.\u00a0As the equilibrium constant for the reaction is large at \u2248 10<sup>5<\/sup>, first assume that the reaction is complete\u2014that is, that one or both reactants are completely consumed.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212422\/CNX_Chem_13_04_ICETable11_img1.jpg\" alt=\"This table has two main columns and four rows. The first cell in the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, \u201c2 N O ( g ) plus O subscript 2 ( g ) right-facing arrow 2 N O subscript 2 ( g ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.20, 2 x, 0. The second column has the following: 0.10, x, 0. The third column has the following: 0, 2 x, 0.20.\" width=\"878\" height=\"236\" \/>\r\n\r\nAs only NO<sub>2<\/sub> exists at this stage, the reaction will establish equilibrium:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212423\/CNX_Chem_13_04_ICETable12_img1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium ( M ). The second column has the header, \u201c2 N O subscript 2 ( g ) equilibrium arrow 2 N O ( g ) plus O subscript 2 ( g ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.20, negative 2 x, 0.20 minus 2 x. The second column has the following: 0, 2 x, 2 x. The third column has the following: 0, x, x.\" width=\"879\" height=\"238\" \/>\r\n\r\nAs this equilibrium is the reverse of that originally given, the new <em>K<sub>c<\/sub><\/em> will be the inverse of the given <em>K<sub>c<\/sub><\/em>\u2014that is, [latex]\\frac{1}{{K}_{c}}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}{K}_{\\text{new}}\\hfill &amp; =\\frac{1}{{K}_{c}}=\\frac{{\\left[\\text{NO}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}\\hfill \\\\ \\hfill &amp; =\\frac{1}{2.3\\times {10}^{5}}=4.35\\times {10}^{-6}=\\frac{{\\left[2x\\right]}^{2}\\left[x\\right]}{{\\left[0.20-2x\\right]}^{2}}\\hfill \\end{array}[\/latex]<\/p>\r\nBecause the new equilibrium constant is small compared with the 0.2 <em>M<\/em> term, the 2<em>x<\/em> can be dropped and the equilibrium solved.\r\n<p style=\"text-align: center;\">[latex]4.35\\times {10}^{-6}=\\frac{\\left(4{x}^{2}\\right)x}{\\left[0.04\\right]}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">4<em>x<\/em><sup>3<\/sup> = 1.74 \u00d7 10<sup>\u22127<\/sup><\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em><sup>3<\/sup> = 4.35 \u00d7 10<sup>\u22128<\/sup><\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em> = 3.52 \u00d7 10<sup>\u22123<\/sup><\/p>\r\nFinal equilibrium concentrations:\r\n<ul>\r\n \t<li>[NO<sub>2<\/sub>] = 0.20 \u2212 2<em>x<\/em> = 0.20 \u2212 2(0.00352) = 0.19 <em>M<\/em><\/li>\r\n \t<li>[NO] = 2<em>x<\/em> = 2(0.00352) = 0.0070 <em>M<\/em><\/li>\r\n \t<li>[O<sub>2<\/sub>] = <em>x<\/em> = 0.0035 <em>M<\/em><\/li>\r\n<\/ul>\r\n2<em>x<\/em> is small compared to 0.20 <em>M<\/em> because\r\n<p style=\"text-align: center;\">[latex]\\left(\\frac{2x}{0.20}\\right)\\times 100\\%=\\left(\\frac{0.00352}{0.20}\\right)\\times 100\\%=1.7\\%[\/latex] which is less than the maximum 5% allowed.<\/p>\r\n9.\u00a0Assume that the reaction goes to completion, where one or both reactants are completely used up:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212428\/CNX_Chem_13_04_ICETable15_img1.jpg\" alt=\"This table has two main columns and four rows. The first cell in the first column does not have a heading and then has the following: Initial pressure ( a t m ), Change ( a t m ), Equilibrium pressure ( a t m ). The second column has the header, \u201cO subscript 3 ( g ) plus N O ( g ) right-facing arrow N O subscript 2 ( g ) plus O subscript 2 ( g ).\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 1.2 times ten to the negative 8, negative x, 0. The second column has the following: 1.2 times ten to the negative 8, negative x, 0. The third column has the following: 0, positive x, 1.2 times ten to the negative 8. The fourth column as the following: 0, positive x, 1.2 times ten to the negative 8.\" width=\"878\" height=\"177\" \/>\r\n\r\nNow assume the equilibrium will be established by the reaction moving toward the reactants:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212429\/CNX_Chem_13_04_ICETable16_img1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( a t m ), Change ( a t m ), Equilibrium pressure ( a t m ). The second column has the header, \u201cO subscript 3 ( g ) plus N O ( g ) equilibrium arrow N O subscript 2 ( g ) plus O subscript 2 ( g ).\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0, positive x, x. The second column has the following: 0, positive x, x. The third column has the following: 1.2 times 10 to the negative eighth power, negative x, 1.2 times 10 to the negative eighth power minus x.\" width=\"881\" height=\"179\" \/>\r\n<p style=\"text-align: center;\">[latex]{K}_{P}=\\frac{{P}_{{\\text{NO}}_{2}}{P}_{{\\text{O}}_{2}}}{{P}_{{\\text{O}}_{3}}{P}_{\\text{NO}}}=\\frac{\\left(1.2\\times {10}^{-8}-x\\right)\\left(1.2\\times {10}^{-8}-x\\right)}{\\left(x\\right)\\left(x\\right)}[\/latex]<\/p>\r\nAs <em>K<sub>P<\/sub><\/em> is much larger than 1.2 \u00d7 10<sup>\u22128<\/sup>, the X terms in the expression 1.2 \u00d7 10<sup>\u22128<\/sup>\u00a0\u2212 <em>x<\/em> are relatively negligible and, therefore, can be dropped.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{ }6.0\\times {10}^{34}=\\frac{\\left(1.2\\times {10}^{-8}\\right)\\left(1.2\\times {10}^{-8}\\right)}{\\left(x\\right)\\left(x\\right)}\\hfill \\\\ {x}^{2}=\\frac{\\left(1.44\\times {10}^{-16}\\right)}{6.0\\times {10}^{34}}=2.4\\times {10}^{-51}\\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em> = 4.9 \u00d7 10<sup>\u221226<\/sup> atm<\/p>\r\n<p style=\"text-align: center;\">[latex]{P}_{{\\text{O}}_{3}}=x=4.9\\times {10}^{-26}\\text{atm}[\/latex]<\/p>\r\nClearly <em>x<\/em> is much smaller than 1.2 \u00d7 10<sup>\u22124<\/sup> so the assumption is valid.\r\n\r\n11.\u00a0The number of moles of I<sub>2<\/sub> is\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{ }\\text{mol}=\\frac{63.5\\text{g}}{253.809\\text{g}{\\text{mol}}^{-1}}=0.250\\text{mol}{\\text{I}}_{2}\\hfill \\\\ {K}_{c}=\\frac{{\\left[\\text{HI}\\right]}^{2}}{\\left[{\\text{H}}_{2}\\right]\\left[{\\text{I}}_{2}\\right]}\\hfill \\end{array}[\/latex]<\/p>\r\nThe unit for each concentration term is moles per liter. If the volume were known for this exercise, the number of moles in each term should be divided by this volume. However, there are two terms in the numerator and two terms in the denominator, so these volumes cancel one another. Consequently, for any expression with the same number of numerator terms as denominator terms, the number for moles can be used in place of moles per liter. In this exercise, the volume is not needed even though it is given.\r\n<p style=\"text-align: center;\">(mol HI)<sup>2<\/sup> = <em>K<\/em> \u00d7 mol H<sub>2<\/sub> \u00d7 mol I<sub>2<\/sub><\/p>\r\n<p style=\"text-align: center;\">= 50.2 \u00d7 1.25 mol \u00d7 0.250 mol<\/p>\r\n<p style=\"text-align: center;\">= 15.7 mol<sup>2<\/sup><\/p>\r\n<p style=\"text-align: center;\">mol HI = [latex]\\sqrt{15.7{\\text{mol}}^{2}}=3.96\\text{mol}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">Mass(HI) = 3.96 mol \u00d7 127.9124 g\/mol = 507 g<\/p>\r\n13. At equilibrium the concentration of CO<sub>2<\/sub> = <em>K<sub>c<\/sub><\/em> = 0.50 <em>M<\/em>. The number of moles CO<sub>2<\/sub> in the system is then mol CO<sub>2<\/sub> = 6.5 L \u00d7 0.50 mol\/L = 3.3 mol. The minimum moles of CaCO<sub>2<\/sub> required is then just more than:\r\n<p style=\"text-align: center;\">[latex]3.3\\text{mol}{\\text{CO}}_{2}\\times \\frac{1\\text{mol}{\\text{CaCO}}_{3}}{1\\text{mol}{\\text{CO}}_{2}}=3.3\\text{mol}{\\text{CaCO}}_{3}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]3.3\\text{mol}{\\text{CO}}_{2}\\times \\frac{1\\text{mol}{\\text{CaCO}}_{3}}{100.1\\text{g}{\\text{CaCO}}_{3}}=330\\text{g}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>At 25 \u00b0C and at 1 atm, the partial pressures in an equilibrium mixture of N<sub>2<\/sub>O<sub>4<\/sub> and NO<sub>2<\/sub> are [latex]{\\text{P}}_{{\\text{N}}_{2}{\\text{O}}_{4}}=0.70\\text{atm}[\/latex] and [latex]{\\text{P}}_{{\\text{NO}}_{2}}=0.30\\text{atm.}[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Predict how the pressures of NO<sub>2<\/sub> and N<sub>2<\/sub>O<sub>4<\/sub> will change if the total pressure increases to 9.0 atm. Will they increase, decrease, or remain the same?<\/li>\r\n \t<li>Calculate the partial pressures of NO<sub>2<\/sub> and N<sub>2<\/sub>O<sub>4<\/sub> when they are at equilibrium at 9.0 atm and 25 \u00b0C.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>In a 3.0-L vessel, the following equilibrium partial pressures are measured: N<sub>2<\/sub>, 190 torr; H<sub>2<\/sub>, 317 torr; NH<sub>3<\/sub>, 1.00 \u00d7 10<sup>3<\/sup> torr:\u00a0[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>How will the partial pressures of H<sub>2<\/sub>, N<sub>2<\/sub>, and NH<sub>3<\/sub> change if H<sub>2<\/sub> is removed from the system? Will they increase, decrease, or remain the same?<\/li>\r\n \t<li>Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The equilibrium constant (<em>K<sub>c<\/sub><\/em>) for this reaction is 5.0 at a given temperature:\u00a0[latex]\\text{CO}\\left(g\\right)+{\\text{H}}_{2}\\text{O}\\left(g\\right)\\rightleftharpoons{\\text{CO}}_{2}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right)[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H<sub>2<\/sub> in a liter. How many moles of CO<sub>2<\/sub> were there in the equilibrium mixture?<\/li>\r\n \t<li>Maintaining the same temperature, additional H<sub>2<\/sub> was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H<sub>2<\/sub> in a liter. How many moles of CO<sub>2<\/sub> were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Antimony pentachloride decomposes according to this equation:\u00a0[latex]{\\text{SbCl}}_{5}\\left(g\\right)\\rightleftharpoons{\\text{SbCl}}_{3}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)[\/latex]\r\nAn equilibrium mixture in a 5.00-L flask at 448 \u00b0C contains 3.85 g of SbCl<sub>5<\/sub>, 9.14 g of SbCl<sub>3<\/sub>, and 2.84 g of Cl<sub>2<\/sub>. How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?<\/li>\r\n \t<li>Consider the reaction between H<sub>2<\/sub> and O<sub>2<\/sub> at 1000 K\r\n[latex]2H_{2}(g)+O_{2}(g)\\rightleftharpoons{2H_{2}O(g)}[\/latex] [latex]K_{P}=\\frac{(P_{H_{2}O})^{2}}{(P_{O_{2}})(P_{H_{2}})^{3}}=1.33\\times{10^{20}}[\/latex]\r\nIf 0.500 atm of H2 and 0.500 atm of O2 are allowed to come to equilibrium at this temperature, what are the partial pressures of the components?<\/li>\r\n \t<li>An equilibrium is established according to the following equation\r\n[latex]{\\text{Hg}}_{2}{}^{2+}\\left(aq\\right)+{\\text{NO}}_{3}{}^{-}\\left(aq\\right)+3{\\text{H}}^{+}\\left(aq\\right)\\rightleftharpoons2{\\text{Hg}}^{2+}\\left(aq\\right)+{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right){K}_{c}=4.6[\/latex]\r\nWhat will happen in a solution that is 0.20 <em>M<\/em> each in [latex]{\\text{Hg}}_{2}{}^{2+}[\/latex], [latex]{\\text{NO}}_{3}{}^{-}[\/latex], H<sup>+<\/sup>, Hg<sup>2+<\/sup>, and HNO<sub>2<\/sub>?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\text{Hg}}_{2}{}^{2+}[\/latex] will be oxidized and [latex]{\\text{NO}}_{3}{}^{-}[\/latex] reduced.<\/li>\r\n \t<li>[latex]{\\text{Hg}}_{2}{}^{2+}[\/latex] will be reduced and [latex]{\\text{NO}}_{3}{}^{-}[\/latex] oxidized.<\/li>\r\n \t<li>Hg<sup>2+<\/sup> will be oxidized and HNO<sub>2<\/sub> reduced.<\/li>\r\n \t<li>Hg<sup>2+<\/sup> will be reduced and HNO<sub>2<\/sub> oxidized.<\/li>\r\n \t<li>There will be no change, because all reactants and products have an activity of 1.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Consider the equilibrium: [latex]4{\\text{NO}}_{2}\\left(g\\right)+6{\\text{H}}_{2}\\text{O}\\left(g\\right)\\rightleftharpoons4{\\text{NH}}_{3}\\left(g\\right)+7{\\text{O}}_{2}\\left(g\\right)[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>What is the expression for the equilibrium constant (<em>K<sub>c<\/sub><\/em>) of the reaction?<\/li>\r\n \t<li>How must the concentration of NH<sub>3<\/sub> change to reach equilibrium if the reaction quotient is less than the equilibrium constant?<\/li>\r\n \t<li>If the reaction were at equilibrium, how would a decrease in pressure (from an increase in the volume of the reaction vessel) affect the pressure of NO<sub>2<\/sub>?<\/li>\r\n \t<li>If the change in the pressure of NO<sub>2<\/sub> is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O<sub>2<\/sub> change?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO<sub>2<\/sub>), is partially regulated by the concentration of H<sub>3<\/sub>O<sup>+<\/sup> and dissolved CO<sub>2<\/sub> in the blood. Although the equilibrium is complicated, it can be summarized as\r\n[latex]{\\text{HbO}}_{2}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CO}}_{2}-\\text{Hb}-{\\text{H}}^{+}+{\\text{O}}_{2}\\left(g\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Write the equilibrium constant expression for this reaction.<\/li>\r\n \t<li>Explain why the production of lactic acid and CO<sub>2<\/sub> in a muscle during exertion stimulates release of O<sub>2<\/sub> from the oxyhemoglobin in the blood passing through the muscle.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.\r\n[latex]{\\text{C}}_{12}{\\text{H}}_{22}{\\text{O}}_{11}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightarrow{\\text{C}}_{6}{\\text{H}}_{12}{\\text{O}}_{6}\\left(aq\\right)+{\\text{C}}_{6}{\\text{H}}_{12}{\\text{O}}_{6}\\left(aq\\right)[\/latex]\r\nRate = <em>k<\/em>[C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>]\r\nIn neutral solution, <em>k<\/em> = 2.1 \u00d7 10<sup>-11<\/sup>\/s at 27 \u00b0C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation\u2014the products of the reaction, glucose and fructose, have the same molecular formulas, C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 \u00d7 10<sup>5<\/sup> at 27 \u00b0C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 <em>M<\/em> aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1).<\/li>\r\n \t<li>The density of trifluoroacetic acid vapor was determined at 118.1 \u00b0C and 468.5 torr, and found to be 2.784 g\/L. Calculate <em>K<sub>c<\/sub><\/em> for the association of the acid.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212440\/CNX_Chem_13_04_trifluoro_img1.jpg\" alt=\"Two Lewis structures are shown in a reaction. The first structure, which is condensed, reads, \u201c2 C F subscript 3 C O subscript 2 H ( g ),\u201d and is followed by a double-headed arrow. The second structure shows a partially condensed hexagonal ring shape. From the left side, in a clockwise manner, it reads \u201cC F subscript 3 C, single bond, O, single bond, H, dotted line bond, O, double bond, C F subscript 3 C ( g ), single bond, O, single bond, H, dotted line bond, O, double bond back to the starting compound.\u201d\" \/><\/li>\r\n \t<li>Liquid N<sub>2<\/sub>O<sub>3<\/sub> is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO<sub>2<\/sub>. At 25 \u00b0C, a value of <em>K<sub>P<\/sub><\/em> = 1.91 has been established for this decomposition. If 0.236 moles of N<sub>2<\/sub>O<sub>3<\/sub> are placed in a 1.52-L vessel at 25 \u00b0C, calculate the equilibrium partial pressures of N<sub>2<\/sub>O<sub>3<\/sub>(<em>g<\/em>), NO<sub>2<\/sub>(<em>g<\/em>), and NO(<em>g<\/em>).<\/li>\r\n \t<li>A 1.00-L vessel at 400 \u00b0C contains the following equilibrium concentrations: N<sub>2<\/sub>, 1.00 <em>M<\/em>; H<sub>2<\/sub>, 0.50 <em>M<\/em>; and NH<sub>3<\/sub>, 0.25 <em>M<\/em>. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 <em>M<\/em>?<\/li>\r\n \t<li>A 0.010 <em>M<\/em> solution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 \u00b0C. A 0.010 <em>M<\/em> solution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Which acid has the larger equilibrium constant for ionization\r\nHA [latex]\\left[\\text{HA}\\left(aq\\right)\\rightleftharpoons{\\text{A}}^{-}\\left(aq\\right)+{\\text{H}}^{+}\\left(aq\\right)\\right][\/latex] or HB [latex]\\left[\\text{HB}\\left(aq\\right)\\rightleftharpoons{\\text{H}}^{+}\\left(aq\\right)+{\\text{B}}^{-}\\left(aq\\right)\\right][\/latex] ?<\/li>\r\n \t<li>What are the equilibrium constants for the ionization of these acids?\r\n(Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the conjugate base (A<sup>-<\/sup> or B<sup>-<\/sup>), and the hydrogen ion (H<sup>+<\/sup>). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"580949\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"580949\"]\r\n\r\n1.\u00a0(a) The reaction is [latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)\\text{.}[\/latex] At equilibrium, [latex]{K}_{P}=\\frac{\\left({P}_{{\\text{N}}_{2}{\\text{O}}_{4}}\\right)}{{\\left({P}_{{\\text{NO}}_{2}}\\right)}^{2}}\\text{.}[\/latex] The value of <em>K<sub>P<\/sub><\/em> must remain the same when the pressure increases to 9.0 atm. Both gases must increase in pressure.\r\n\r\n(b) [latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)\\text{.}[\/latex] At equilibrium, [latex]{K}_{P}=\\frac{{P}_{{\\text{N}}_{2}{\\text{O}}_{4}}}{{\\left({P}_{{\\text{NO}}_{2}}\\right)}^{2}}=\\frac{0.70}{{\\left(0.30\\right)}^{2}}=\\frac{0.70}{0.09}=7.78.[\/latex] For a total pressure of 9.0 atm, the pressure of N<sub>2<\/sub>O<sub>4<\/sub> is 9.0 - <em>x<\/em>; that of NO is <em>x<\/em>.\r\n\r\n[latex]{K}_{P}=\\frac{9.0-x}{{x}^{2}}=7.78[\/latex]\r\n\r\n7.78<em>x<\/em><sup>2<\/sup> + <em>x<\/em> \u2212 9.0 = 0\r\n\r\nUse the quadratic expression, where\r\n\r\n[latex]x=\\frac{\\text{-}b\\pm \\sqrt{{b}^{2}-4ac}}{2a}=\\frac{-1\\pm \\sqrt{1+4\\left(7.78\\right)\\left(-9.0\\right)}}{2\\left(7.78\\right)}.[\/latex]\r\n\r\n[latex]=\\frac{-1\\pm 16.765}{15.56}=1.013[\/latex]\r\n\r\nFor the plus sign (the negative sign gives a negative pressure which is impossible)\r\n\r\n[latex]x=\\frac{15.765}{15.56}=1.013[\/latex]\r\n\r\nThe other answer is extraneous. The pressures are [latex]{P}_{{\\text{N}}_{2}{\\text{O}}_{4}}[\/latex] = 8.0 atm and [latex]{P}_{{\\text{NO}}_{2}}[\/latex] = 1.0 atm\r\n\r\n3.\u00a0(a) For the above reaction, [latex]{K}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]\\left[{\\text{H}}_{2}\\right]}{\\left[\\text{CO}\\right]\\left[{\\text{H}}_{2}\\text{O}\\right]}=5.0.[\/latex] The concentrations at equilibrium are 0.20 <em>M<\/em> CO, 0.30 <em>M<\/em> H<sub>2<\/sub>O, and 0.90 <em>M<\/em> H<sub>2<\/sub>. Substitution gives [latex]K=5.0=\\frac{\\left[{\\text{CO}}_{2}\\right]\\left[0.90\\right]}{\\left[0.20\\right]\\left[0.30\\right]};[\/latex] [latex]\\left[{\\text{CO}}_{2}\\right]=\\frac{5.0\\left(0.20\\right)\\left(0.30\\right)}{0.90}=0.33M;[\/latex] Amount of CO<sub>2<\/sub> = 0.33 mol \u00d7 1 = 0.33 mol.\r\n\r\n(b) At the particular temperature of reaction, <em>K<sub>c<\/sub><\/em> remains constant at 5.0. The new concentrations are 0.40 <em>M<\/em> CO, 0.30 <em>M<\/em> H<sub>2<\/sub>O, and 1.2 <em>M<\/em> H<sub>2<\/sub>.\r\n\r\n[latex]\\frac{\\left[{\\text{CO}}_{2}\\right]\\left(1.2\\right)}{\\left(0.40\\right)\\left(0.30\\right)}=5.0[\/latex]\r\n\r\n[latex]{\\left[\\text{CO}\\right]}^{2}=0.50M[\/latex]\r\n\r\nAmount of CO<sub>2<\/sub> = 0.50 mol \u00d7 1 = 0.50 mol. Added H<sub>2<\/sub> forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H<sub>2<\/sub> is added.\r\n\r\n5.\u00a0P<sub>H<sub>2<\/sub><\/sub> = 8.64 \u00d7 10<sup>\u221211<\/sup>atm\r\n\r\nP<sub>O<sub>2<\/sub><\/sub> = 0.250atm\r\n\r\nP<sub>H<sub>2<\/sub>O<\/sub> = 0.500 atm\r\n\r\n7.\u00a0(a) [latex]{K}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{4}{\\left[{\\text{O}}_{2}\\right]}^{7}}{{\\left[{\\text{NO}}_{2}\\right]}^{4}{\\left[{\\text{H}}_{2}\\text{O}\\right]}^{6}}\\text{.}[\/latex] (b) Because [NH<sub>3<\/sub>] is in the numerator of <em>K<sub>c<\/sub><\/em>, [NH<sub>3<\/sub>] must increase for <em>Q<sub>c<\/sub><\/em> to reach <em>K<sub>c<\/sub><\/em>. (c) That decrease in pressure would decrease [NO<sub>2<\/sub>] because the NO<sub>2<\/sub> must convert to NH<sub>3<\/sub> to reduce the effects of the expansion and the consequent relative decrease in products. (d) The relative pressures are controlled by the stoichiometry of the reaction.\r\n\r\n[latex]{P}_{{\\text{O}}_{2}}=\\frac{7}{4}{P}_{{\\text{NO}}_{2}}=\\frac{7}{4}\\left(28\\text{torr}\\right)=49\\text{torr}[\/latex]\r\n\r\n9.\u00a0[fructose] = 0.15 <em>M<\/em>\r\n\r\n11. Write the balanced equilibrium expression. With all of the species as gases, it is a straightforward <em>K<sub>P<\/sub><\/em> problem to solve. However, <em>all<\/em> species must be converted to pressures, from other units related to concentration. For N<sub>2<\/sub>O<sub>3<\/sub>, with 0.236 mol in 1.52 L at 25 \u00baC: <em>PV<\/em> = <em>nRT<\/em>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}P&amp; =&amp; \\frac{n}{V}RT\\\\ &amp; =&amp; \\frac{0.236\\cancel{\\text{mol}}}{1.52\\cancel{\\text{L}}}\\times \\frac{\\left(0.08206\\cancel{\\text{L}}\\text{atm}\\right)\\left(298.15\\cancel{\\text{K}}\\right)}{\\cancel{\\text{mol}}\\cancel{\\text{K}}}\\\\ &amp; =&amp; 3.80\\text{atm}\\end{array}[\/latex]<\/p>\r\nWrite the balanced equation and the equilibrium changes:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212443\/CNX_Chem_13_04_ICETable26_img1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( a t m ), Change ( a t m ), Equilibrium pressure ( a t m ). The second column has the header, \u201cN subscript 2 O subscript 3 ( g ) equilibrium arrow N O ( g ) plus N O subscript 2 ( g ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 3.80, negative x, 3.80 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.\" width=\"879\" height=\"238\" \/>\r\n<p style=\"text-align: center;\">[latex]{K}_{P}=\\frac{\\left({P}_{\\text{NO}}\\right)\\left({P}_{{\\text{NO}}_{2}}\\right)}{\\left({P}_{{\\text{N}}_{2}{\\text{O}}_{3}}\\right)}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]1.91\\text{atm}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(3.80-x\\right)}[\/latex]<\/p>\r\nBecause 3.80 is substantial when compared with the equilibrium constant, the value of X must be considered. A quadratic equation must be solved:\r\n<p style=\"text-align: center;\">[latex]1.91=\\frac{{x}^{2}}{\\left(3.80-x\\right)}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">7.258 \u2212 1.91<em>x<\/em> = <em>x<\/em><sup>2<\/sup><\/p>\r\n<p style=\"text-align: center;\">0 = <em>x<\/em><sup>2<\/sup> +1.91<em>x<\/em>\u00a0\u2013 7.258<\/p>\r\n<p style=\"text-align: center;\">0 = <em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em><\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}x&amp; =&amp; \\frac{\\text{-}b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\\\ &amp; =&amp; \\frac{-1.91\\pm \\sqrt{{\\left(1.91\\right)}^{2}-4\\left(1\\right)\\left(-7.258\\right)}}{2\\left(1\\right)}\\\\ &amp; =&amp; \\frac{-1.91\\pm \\sqrt{3.648+29.032}}{2}\\\\ &amp; =&amp; \\frac{-1.91\\pm \\sqrt{32.680}}{2}\\\\ &amp; =&amp; \\frac{-1.91\\pm 5.717}{2}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">= 1.90 atm or -3.81 atm<\/p>\r\nAs negative pressure is not possible in this case, we use the positive value only. The final pressures are: [latex]{P}_{{\\text{N}}_{2}{\\text{O}}_{3}}[\/latex] = 3.80 \u2212 <em>x<\/em> = 3.80 \u2212 1.90 = 1.90 atm and [latex]{P}_{\\text{NO}}={P}_{{\\text{NO}}_{2}}={P}_{{\\text{NO}}_{2}}x=1.90\\text{atm}[\/latex]\r\n\r\n13. \u00a0(a) Recall that osmotic pressure is a <em>colligative<\/em> property, depending on the total number of particles present in the system. The osmotic pressure equation is \u03c0 = <em>iMRT<\/em>. Given that the two acids face identical reaction conditions, the osmotic pressure (\u03c0) will increase only with a corresponding (&gt;1) in the degree of ionization (<em>i<\/em>). Because both acids produce two ions per molecule dissociated, the number of particles in solution depends on the size of their respective equilibrium constants (<em>K<sub>c<\/sub><\/em>(HA) and <em>K<sub>c<\/sub><\/em>(HB)):\r\n<p style=\"text-align: center;\">[latex]\\text{HA}\\rightleftharpoons{\\text{H}}^{+}+{\\text{A}}^{-}\\text{HB}\\rightleftharpoons{\\text{H}}^{+}+{\\text{B}}^{-}[\/latex]<\/p>\r\nBecause HB has a greater osmotic pressure (0.345 atm) than HA (0.293 atm), HB must produce more particles. Therefore, HB ionizes to a greater degree and has the larger <em>K<sub>c<\/sub><\/em>.\r\n\r\n(b) Determine the value of <em>i<\/em> for each acid. For HA,\r\n<p style=\"text-align: center;\">[latex]\\pi =iMRT=0.293\\text{atm}=i\\left(0.010M\\right)\\left(\\frac{0.08206\\text{L atm}}{\\text{mol K}}\\right)\\left(298.15\\text{K}\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>i<\/em><sub>HY<\/sub> = 1.2<\/p>\r\n<p style=\"text-align: center;\">For HB,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\pi =iMRT=0.345\\text{atm}=i\\left(0.010M\\right)\\left(\\frac{0.08206\\text{L atm}}{\\text{mol K}}\\right)\\left(298.15\\text{K}\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>i<\/em><sub>HZ<\/sub> = 1.4<\/p>\r\nThe <em>i<\/em> values mean that for every mole of originally undissociated species, <em>i<\/em> moles particles are produced in solution. The dissociations can be expressed in tabular form. For HA,\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212444\/CNX_Chem_13_04_ICETable27_img1.jpg\" alt=\"This table has two main columns and four fours. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium pressure ( M ). The second column has the header, \u201cH A equilibrium H superscript positive sign plus A superscript negative sign.\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.100, negative x, 0.100 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.\" width=\"879\" height=\"238\" \/>\r\n<p style=\"text-align: center;\">0.012 <em>M<\/em> = <em>xM<\/em> + <em>xM<\/em> + (0.010 \u2212 <em>x<\/em>) <em>M<\/em><\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em> = 0.002 <em>M<\/em><\/p>\r\nFor HB, the only difference from the table for HA is the value of <em>i<\/em><sub>HB<\/sub>. <em>x<\/em> = 0.004 <em>M<\/em>.\r\n\r\nNow calculate the values of <em>K<sub>c<\/sub><\/em>(HA) and <em>K<sub>c<\/sub><\/em>(HB).\r\n\r\nFor HA,\r\n<p style=\"text-align: center;\">[latex]{K}_{c}\\left(\\text{HA}\\right)=\\frac{\\left[x\\right]\\left[x\\right]}{\\left[0.010-x\\right]}=\\frac{\\left(0.002\\right)\\left(0.002\\right)}{0.008}=5\\times {10}^{-4}[\/latex]<\/p>\r\nFor HB,\r\n<p style=\"text-align: center;\">[latex]{K}_{c}\\left(\\text{HB}\\right)=\\frac{\\left[x\\right]\\left[x\\right]}{\\left[0.010-x\\right]}=\\frac{\\left(0.004\\right)\\left(0.004\\right)}{0.006}=3\\times {10}^{-3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems<\/li>\n<li>Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp222894896\">Having covered the essential concepts of chemical equilibria in the preceding sections of this chapter, this final section will demonstrate the more practical aspect of using these concepts and appropriate mathematical strategies to perform various equilibrium calculations. These types of computations are essential to many areas of science and technology\u2014for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.<\/p>\n<p id=\"fs-idp292035664\">Many of the useful equilibrium calculations that will be demonstrated here require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometry of the reaction, as illustrated by decomposition of ammonia:<\/p>\n<p style=\"text-align: center;\">[latex]2{\\text{NH}}_{3}\\left(g\\right){\\rightleftharpoons}{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/p>\n<p id=\"fs-idp42253680\">As shown earlier in this chapter, this equilibrium may be established within a sealed container that initially contains either NH<sub>3<\/sub>\u00a0only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount\u00a0<em>x<\/em>:<\/p>\n<div style=\"text-align: center;\">\u0394[latex][\\text{N}_2] = +x[\/latex]<\/div>\n<p id=\"fs-idp249176544\">the corresponding changes in the other species concentrations are<\/p>\n<div style=\"text-align: center;\">\u0394[latex][\\text{H}_2] =[\/latex]\u0394[latex][\\text{N}_2]\\left(\\dfrac{3\\text{molH}_2}{1\\text{molN}_2}\\right) = +3x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">\u0394[latex][\\text{NH}_3] =[\/latex]\u0394[latex][\\text{N}_2]\\left(\\dfrac{2\\text{molNH}_3}{1\\text{molN}_2}\\right) = -2x[\/latex]<\/div>\n<p id=\"fs-idp146645120\">where the negative sign indicates a decrease in concentration.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0Determining Relative Changes in Concentration<\/h3>\n<p>Derive the missing terms representing concentration changes for each of the following reactions.<\/p>\n<ol>\n<li>[latex]\\begin{array}{ccccc}{\\text{C}}_{2}{\\text{H}}_{2}\\text{(}g\\text{)}&+\\hfill & 2{\\text{Br}}_{2}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons& {\\text{C}}_{2}{\\text{H}}_{2}{\\text{Br}}_{4}\\text{(}g\\text{)}\\hfill \\\\ x\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccccc}{\\text{I}}_{2}\\text{(}aq\\text{)}&+\\hfill & {\\text{I}}^{-}\\text{(}aq\\text{)}\\hfill & \\rightleftharpoons\\hfill & {\\text{I}}_{3}{}^{-}\\text{(}aq\\text{)}\\hfill \\\\ & & & & x \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{cccccc}{\\text{C}}_{3}{\\text{H}}_{8}\\text{(}g\\text{)}&+\\hfill & 5{\\text{O}}_{2}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons\\hfill & 3{\\text{CO}}_{2}\\text{(}g\\text{)}+\\hfill & 4{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\hfill \\\\ x \\end{array}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q8310\">Show Solution<\/span><\/p>\n<div id=\"q8310\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\begin{array}{ccccc}{\\text{C}}_{2}{\\text{H}}_{2}\\text{(}g\\text{)}&+\\hfill & 2{\\text{Br}}_{2}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons\\hfill & {\\text{C}}_{2}{\\text{H}}_{2}{\\text{Br}}_{4}\\text{(}g\\text{)}\\hfill \\\\ x\\hfill& & 2x\\hfill & & -x\\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccccc}{\\text{I}}_{2}\\text{(}aq\\text{)}&+\\hfill & {\\text{I}}^{-}\\text{(}aq\\text{)}\\hfill & \\rightleftharpoons\\hfill & {\\text{I}}_{3}{}^{-}\\text{(}aq\\text{)}\\hfill \\\\ -x\\hfill& & -x\\hfill & & x\\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{cccccc}{\\text{C}}_{3}{\\text{H}}_{8}\\text{(}g\\text{)}&+\\hfill & 5{\\text{O}}_{2}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons\\hfill & 3{\\text{CO}}_{2}\\text{(}g\\text{)}+\\hfill & 4{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\hfill \\\\ x\\hfill & & 5x\\hfill & & -3x\\hfill & -4x\\hfill \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Complete the changes in concentrations for each of the following reactions:<\/p>\n<ol>\n<li>[latex]\\begin{array}{ccccc}2{\\text{SO}}_{2}\\text{(}g\\text{)}&+\\hfill & {\\text{O}}_{2}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons\\hfill & 2{\\text{SO}}_{3}\\text{(}g\\text{)}\\hfill \\\\ & & x & \\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccc}{\\text{C}}_{4}{\\text{H}}_{8}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons\\hfill & 2{\\text{C}}_{2}{\\text{H}}_{4}\\text{(}g\\text{)}\\hfill \\\\ & & -2x\\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccccccc}4{\\text{NH}}_{3}\\text{(}g\\text{)}&+\\hfill & 7{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons\\hfill & 4{\\text{NO}}_{2}\\text{(}g\\text{)}&+\\hfill & 6{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\end{array}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q707791\">Show Solution<\/span><\/p>\n<div id=\"q707791\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\begin{array}{ccccc}2{\\text{SO}}_{2}\\text{(}g\\text{)}&+\\hfill & {\\text{O}}_{2}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons\\hfill & 2{\\text{SO}}_{3}\\text{(}g\\text{)}\\hfill \\\\ 2x & & x && -2x & \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccc}{\\text{C}}_{4}{\\text{H}}_{8}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons\\hfill & 2{\\text{C}}_{2}{\\text{H}}_{4}\\text{(}g\\text{)}\\hfill \\\\ x & & -2x\\hfill \\end{array}[\/latex]<\/li>\n<li>There are two possible solutions for this question:<br \/>\n[latex]\\begin{array}{ccccccc}4{\\text{NH}}_{3}\\text{(}g\\text{)}&+\\hfill & 7{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons\\hfill & 4{\\text{NO}}_{2}\\text{(}g\\text{)}&+\\hfill & 6{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\\\ 4x& &7x& &-4x& &-6x \\end{array}[\/latex]<br \/>\nor<br \/>\n[latex]\\begin{array}{ccccccc}4{\\text{NH}}_{3}\\text{(}g\\text{)}&+\\hfill & 7{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\hfill & \\rightleftharpoons\\hfill & 4{\\text{NO}}_{2}\\text{(}g\\text{)}&+\\hfill & 6{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}\\\\ -4x& &-7x& &4x& &6x \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>A reaction is represented by this equation: [latex]\\text{A}\\left(aq\\right)+2\\text{B}\\left(aq\\right)\\rightleftharpoons2\\text{C}\\left(aq\\right){K}_{c}=1\\times {10}^{3}[\/latex]\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Write the mathematical expression for the equilibrium constant.<\/li>\n<li>Using concentrations \u22641 <em>M<\/em>, make up two sets of concentrations that describe a mixture of A, B, and C at equilibrium.<\/li>\n<\/ol>\n<\/li>\n<li>A reaction is represented by this equation: [latex]2\\text{W}\\left(aq\\right)\\rightleftharpoons\\text{X}\\left(aq\\right)+2\\text{Y}\\left(aq\\right){K}_{c}=5\\times {10}^{-4}[\/latex]\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Write the mathematical expression for the equilibrium constant.<\/li>\n<li>Using concentrations of \u22641 <em>M<\/em>, make up two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q69852\">Show Solution to Question 1<\/span><\/p>\n<div id=\"q69852\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{K}_{c}=\\dfrac{{\\left[\\text{C}\\right]}^{2}}{\\left[\\text{A}\\right]{\\left[\\text{B}\\right]}^{2}}\\text{.}[\/latex] There are many different sets of equilibrium concentrations; two are [A] = 0.1 <em>M<\/em>, [B] = 0.1 <em>M<\/em>, [C] = 1 <em>M<\/em>; and [A] = 0.01, [B] = 0.250, [C] = 0.791.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Calculations of an Equilibrium Constant<\/h2>\n<p>The equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the K expression, as was illustrated by Example 2. A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the reactant and product concentrations\u00a0<em>initially<\/em>\u00a0present, for how they\u00a0<em>change<\/em>\u00a0as the reaction proceeds, and for what they are when the system reaches\u00a0<em>equilibrium<\/em>. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0Calculation of an Equilibrium Constant<\/h3>\n<p>Iodine molecules react reversibly with iodide ions to produce triiodide ions.<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{I}}_{2}\\left(aq\\right)+{\\text{I}}^{-}\\left(aq\\right)\\rightleftharpoons{\\text{I}}_{3}{}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>If a solution with the concentrations of I<sub>2<\/sub> and I<sup>\u2212<\/sup> both equal to 1.000 \u00d7 10<sup>\u22123<\/sup><em>M<\/em> before reaction gives an equilibrium concentration of I<sub>2<\/sub> of 6.61 \u00d7 10<sup>\u22124<\/sup><em>M<\/em>, what is the equilibrium constant for the reaction?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q536062\">Show Solution<\/span><\/p>\n<div id=\"q536062\" class=\"hidden-answer\" style=\"display: none\">\n<p>To calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{c}=\\dfrac{\\left[{\\text{I}}_{3}{}^{-}\\right]}{\\left[{I}_{2}\\right]\\left[{\\text{I}}^{-}\\right]}[\/latex]<\/p>\n<p>Provided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an ICE table.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/openstax.org\/resources\/efdf4733dd94e860584f66ae6c1d1c2f4b2418a8\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, \u201cI subscript 2 plus sign I superscript negative sign equilibrium arrow I subscript 3 superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column has the following: 1.000 times 10 to the negative third power, negative x, &#091; I subscript 2 &#093; subscript i minus x. The second column has the following: 1.000 times 10 to the negative third power, negative x, &#091; I superscript negative sign &#093; subscript i minus x. The third column has the following: 0, positive x, &#091; I superscript negative sign &#093; subscript i plus x.\" width=\"599\" height=\"198\" \/><\/p>\n<p>At equilibrium the concentration of I<sub>2<\/sub> is 6.61 \u00d7 10<sup>\u22124<\/sup><em>M<\/em> so that<\/p>\n<p style=\"text-align: center;\">[latex]1.000\\times {10}^{-3}-x=6.61\\times {10}^{-4}[\/latex]<br \/>\n[latex]x=1.000\\times {10}^{-3}-6.61\\times {10}^{-4}[\/latex]<br \/>\n[latex]=3.39\\times {10}^{-4}M[\/latex]<\/p>\n<p id=\"fs-idp187828096\">The ICE table may now be updated with numerical values for all its concentrations:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/openstax.org\/resources\/d73aca40ea1b3b0cc341a9ecc2941b22f0754aa9\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, \u201cI subscript 2 plus sign I superscript negative sign equilibrium arrow I subscript 3 superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column has the following: 1.000 times 10 to the negative third power, negative 3.39 times 10 to the negative fourth power, 6.61 times 10 to the negative fourth power. The second column has the following: 1.000 times 10 to the negative third power, negative 3.39 times 10 to the negative fourth power, 6.61 times 10 to the negative fourth power. The third column has the following: 0, positive 3.39 times 10 to the negative fourth power, 3.39 times 10 to the negative fourth power.\" width=\"600\" height=\"162\" \/><\/p>\n<p>Finally, substitute the equilibrium concentrations into the\u00a0<em data-effect=\"italics\">K<\/em>\u00a0expression and solve:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{c}={Q}_{c}=\\dfrac{\\left[{\\text{I}}_{3}{}^{-}\\right]}{\\left[{I}_{2}\\right]\\left[{\\text{I}}^{-}\\right]}[\/latex]<br \/>\n[latex]=\\dfrac{3.39\\times {10}^{-4}M}{\\left(6.61\\times {10}^{-4}M\\right)\\left(6.61\\times {10}^{-4}M\\right)}=776[\/latex]<\/p>\n<h4>Check Your Learning<\/h4>\n<p>Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{C}}_{2}{\\text{H}}_{5}\\text{OH}+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\rightleftharpoons{\\text{CH}}_{3}{\\text{CO}}_{2}{\\text{C}}_{2}{\\text{H}}_{5}+{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\n<p>When 1 mol each of C<sub>2<\/sub>H<sub>5<\/sub>OH and CH<sub>3<\/sub>CO<sub>2<\/sub>H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when [latex]\\frac{1}{3}[\/latex] mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (<em>Note:<\/em> Water is not a solvent in this reaction.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q619052\">Show Solution<\/span><\/p>\n<div id=\"q619052\" class=\"hidden-answer\" style=\"display: none\"><\/div>\n<\/div>\n<p><em>K<\/em><sub>c<\/sub> = 4<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>What is the value of the equilibrium constant at 500 \u00b0C for the formation of NH<sub>3<\/sub> according to the following equation? [latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<br \/>\nAn equilibrium mixture of NH<sub>3<\/sub>(<em>g<\/em>), H<sub>2<\/sub>(<em>g<\/em>), and N<sub>2<\/sub>(<em>g<\/em>) at 500 \u00b0C was found to contain 1.35 <em>M<\/em> H<sub>2<\/sub>, 1.15 <em>M<\/em> N<sub>2<\/sub>, and 4.12 \u00d7 10<sup>-1<\/sup><em>M<\/em> NH<sub>3<\/sub>.<\/li>\n<li>Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.<br \/>\n[latex]{\\text{CH}}_{4}\\left(g\\right)+{\\text{H}}_{2}\\text{O}\\left(g\\right)\\rightleftharpoons3{\\text{H}}_{2}\\left(g\\right)+\\text{CO}\\left(g\\right)[\/latex]<br \/>\nWhat is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH<sub>4<\/sub>, 0.126 <em>M<\/em>; H<sub>2<\/sub>O, 0.242 <em>M<\/em>; CO, 0.126 <em>M<\/em>; H<sub>2<\/sub> 1.15 <em>M<\/em>, at a temperature of 760 \u00b0C?<\/li>\n<li>A 0.72-mol sample of PCl<sub>5<\/sub> is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl<sub>3<\/sub>(<em>g<\/em>) and 0.40 mol of Cl<sub>2<\/sub>(<em>g<\/em>). Calculate the value of the equilibrium constant for the decomposition of PCl<sub>5<\/sub> to PCl<sub>3<\/sub> and Cl<sub>2<\/sub> at this temperature.<\/li>\n<li>At 1 atm and 25 \u00b0C, NO<sub>2<\/sub> with an initial concentration of 1.00 <em>M<\/em> is 3.3 \u00d7 10<sup>-3<\/sup>% decomposed into NO and O<sub>2<\/sub>. Calculate the value of the equilibrium constant for the reaction<br \/>\n[latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NO}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)[\/latex]<\/li>\n<li>Calculate the value of the equilibrium constant <em>K<sub>P<\/sub><\/em> for the reaction [latex]2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NOCl}\\left(g\\right)[\/latex] from these equilibrium pressures: NO, 0.050 atm; Cl<sub>2<\/sub>, 0.30 atm; NOCl, 1.2 atm.<\/li>\n<li>When heated, iodine vapor dissociates according to this equation:\u00a0[latex]{\\text{I}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{I}\\left(g\\right)[\/latex]<br \/>\nAt 1274 K, a sample exhibits a partial pressure of I<sub>2<\/sub> of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, <em>K<sub>P<\/sub><\/em>, for the decomposition at 1274 K.<\/li>\n<li>A sample of ammonium chloride was heated in a closed container:\u00a0[latex]{\\text{NH}}_{4}\\text{Cl}\\left(s\\right)\\rightleftharpoons{\\text{NH}}_{3}\\left(g\\right)+\\text{HCl}\\left(g\\right)[\/latex]<br \/>\nAt equilibrium, the pressure of NH<sub>3<\/sub>(<em>g<\/em>) was found to be 1.75 atm. What is the value of the equilibrium constant <em>K<sub>P<\/sub><\/em> for the decomposition at this temperature?<\/li>\n<li>At a temperature of 60 \u00b0C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant <em>K<sub>P<\/sub><\/em> for the transformation at 60 \u00b0C?<br \/>\n[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q141837\">Show Selected Solutions<\/span><\/p>\n<div id=\"q141837\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\u00a0The reaction may be written as<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/p>\n<p>The equilibrium constant for the reaction is<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}=\\frac{{\\left(4.12\\times {10}^{-1}\\right)}^{2}}{\\left(1.15\\right){\\left(1.35\\right)}^{3}}=\\frac{{\\left(0.170\\right)}^{2}}{\\left(1.15\\right){\\left(2.46\\right)}^{3}}0.0600=6.00\\times10^{-2}[\/latex]<\/p>\n<p>3.\u00a0The decomposition of PCl<sub>5<\/sub> to PCl<sub>3<\/sub> and Cl<sub>2<\/sub> is given as<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{PCl}}_{5}\\left(g\\right)\\rightleftharpoons{\\text{PCl}}_{3}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\hfill \\\\{K}_{c}=\\frac{\\left[{\\text{PCl}}_{3}\\right]\\left[{\\text{Cl}}_{2}\\right]}{\\left[{\\text{PCl}}_{5}\\right]}\\hfill \\end{array}[\/latex]<\/p>\n<p>Let <em>x<\/em> = change in [PCl<sub>5<\/sub>].<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[PCl<sub>5<\/sub>]<\/th>\n<th>[PCl<sub>3<\/sub>]<\/th>\n<th>[Cl<sub>2<\/sub>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (M)<\/th>\n<td>0.72<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (M)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium concentration (M)<\/th>\n<td>0.72\u00a0\u2212\u00a0<em>x<\/em> = 0.32<\/td>\n<td>0 +\u00a0<em>x<\/em> = 0.40<\/td>\n<td>0 +\u00a0<em>x<\/em> = 0.40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">[latex]{K}_{c}=\\frac{\\left(0.40\\right)\\left(0.40\\right)}{\\left(0.32\\right)}=0.50[\/latex]<\/p>\n<p>5. \u00a0The equilibrium equation is<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{P}=\\frac{{\\left[\\text{NOCl}\\right]}^{2}}{{\\left[\\text{NO}\\right]}^{2}\\left[{\\text{Cl}}_{2}\\right]}=\\frac{{\\left(1.2\\right)}^{2}}{{\\left(0.050\\right)}^{2}\\left(0.30\\right)}=\\frac{1.44}{\\left(2.5\\times {10}^{-3}\\right)\\left(0.30\\right)}=1.9\\times {10}^{3}[\/latex]<\/p>\n<p>7.\u00a0Because the decomposition must generate the same pressure of HCl as NH<sub>3<\/sub>, 1.75 atm of HCl must be present.<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{p}={P}_{{\\text{NH}}_{3}}{P}_{\\text{HCl}}=\\left(1.75\\text{atm}\\right)\\left(1.75\\text{atm}\\right)=3.06[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Calculation of a Missing Equilibrium Concentration<\/h3>\n<p>When the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated. A computation of this sort is illustrated in the next example exercise.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3:\u00a0Calculation of a Missing Equilibrium Concentration<\/h3>\n<p>Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 \u00b0C, the value of the equilibrium constant for the reaction, [latex]{\\text{N}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NO}\\left(g\\right)[\/latex], is 4.1 \u00d7 10<sup>\u22124<\/sup>. Find the concentration of NO(<em>g<\/em>) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N<sub>2<\/sub>] = 0.036 mol\/L and [O<sub>2<\/sub>] 0.0089 mol\/L.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q844327\">Show Solution<\/span><\/p>\n<div id=\"q844327\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{ }{K}_{c}&=&\\dfrac{{\\left[\\text{NO}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]}\\\\{\\left[\\text{NO}\\right]}^{2}&=&{K}_{c}\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]\\\\\\left[\\text{NO}\\right]&=&\\sqrt{{K}_{c}\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]}\\\\ &=&\\sqrt{\\left(4.1\\times {10}^{-4}\\right)\\left(0.036\\right)\\left(0.0089\\right)}\\\\ &=&\\sqrt{1.31\\times {10}^{-7}}\\\\ &=&3.6\\times {10}^{-4}\\end{array}[\/latex]<\/p>\n<p>Thus [NO] is 3.6 \u00d7 10<sup>\u22124<\/sup> mol\/L at equilibrium under these conditions.<\/p>\n<p>We can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ }{K}_{c}&=&\\dfrac{{\\left[\\text{NO}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]}\\\\&=&\\dfrac{{\\left(3.6\\times {10}^{-4}\\right)}^{2}}{\\left(0.036\\right)\\left(0.0089\\right)}\\\\&=&4.0\\times {10}^{-4}\\end{array}[\/latex]<\/p>\n<p>The answer checks; our calculated value gives the equilibrium constant within the error associated with the significant figures in the problem.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 \u00d7 10<sup>\u22122<\/sup>. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 <em>M<\/em> and 2.09 <em>M<\/em>, respectively.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q705150\">Show Solution<\/span><\/p>\n<div id=\"q705150\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.53 mol\/L<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>Analysis of the gases in a sealed reaction vessel containing NH<sub>3<\/sub>, N<sub>2<\/sub>, and H<sub>2<\/sub> at equilibrium at 400 \u00b0C established the concentration of N<sub>2<\/sub> to be 1.2 <em>M<\/em> and the concentration of H<sub>2<\/sub> to be 0.24 <em>M<\/em>.<br \/>\n[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right){K}_{c}=0.50\\text{ at }400^{\\circ}\\text{C}[\/latex]<br \/>\nCalculate the equilibrium molar concentration of NH<sub>3<\/sub>.<\/li>\n<li>Carbon reacts with water vapor at elevated temperatures.<br \/>\n[latex]\\text{C}\\left(s\\right)+{\\text{H}}_{2}\\text{O}\\left(g\\right)\\rightleftharpoons\\text{CO}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right){K}_{c}=0.2\\text{ at }1000^{\\circ}\\text{C}[\/latex]<br \/>\nWhat is the concentration of CO in an equilibrium mixture with [H<sub>2<\/sub>O] = 0.500 <em>M<\/em> at 1000 \u00b0C?<\/li>\n<li>Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.<br \/>\n[latex]\\text{CoO}\\left(s\\right)+\\text{CO}\\left(g\\right)\\rightleftharpoons\\text{Co}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right){K}_{c}=4.90\\times {10}^{2}\\text{ at }550^{\\circ}\\text{C}[\/latex]<br \/>\nWhat concentration of CO remains in an equilibrium mixture with [CO<sub>2<\/sub>] = 0.100 <em>M<\/em>?<\/li>\n<li>A student solved the following problem and found [N<sub>2<\/sub>O<sub>4<\/sub>] = 0.16 <em>M<\/em> at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of N<sub>2<\/sub>O<sub>4<\/sub> in a mixture formed from a sample of NO<sub>2<\/sub> with a concentration of 0.10 <em>M<\/em>?<br \/>\n[latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right){K}_{c}=160[\/latex]<\/li>\n<li>A student solved the following problem and found the equilibrium concentrations to be [SO<sub>2<\/sub>] = 0.590 <em>M<\/em>, [O<sub>2<\/sub>] = 0.0450 <em>M<\/em>, and [SO<sub>3<\/sub>] = 0.260 <em>M<\/em>. How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 \u00b0C:<br \/>\n[latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right){K}_{c}=4.32[\/latex]<br \/>\nWhat are the equilibrium concentrations of all species in a mixture that was prepared with [SO<sub>3<\/sub>] = 0.500 <em>M<\/em>, [SO<sub>2<\/sub>] = 0 <em>M<\/em>, and [O<sub>2<\/sub>] = 0.350 <em>M<\/em><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q51339\">Show Selected Solutions<\/span><\/p>\n<div id=\"q51339\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. \u00a0Write the equilibrium constant expression and solve for [NH<sub>3<\/sub>].<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[1.2\\right]{\\left[0.24\\right]}^{3}}=0.50[\/latex]<\/p>\n<p style=\"text-align: center;\">[NH<sub>3<\/sub>]<sup>2<\/sup> = 1.2 \u00d7 (0.24)<sup>3<\/sup> \u00d7 0.50 = 0.0083<\/p>\n<p style=\"text-align: center;\">[NH<sub>3<\/sub>] = 9.1 \u00d7 10<sup>-2<\/sup><em>M<\/em><\/p>\n<p>3.\u00a0Write the equilibrium constant expression and solve for [CO].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{K}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]}{\\left[\\text{CO}\\right]}=\\frac{0.100}{\\left[\\text{CO}\\right]}=4.90\\times {10}^{2}\\hfill \\\\ \\left[\\text{CO}\\right]=\\frac{0.100}{4.90\\times {10}^{2}}=2.0\\times {0}^{-4}M\\hfill \\end{array}[\/latex]<\/p>\n<p>5. Calculate <em>Q<\/em> based on the calculated concentrations and see if it is equal to <em>K<sub>c<\/sub><\/em>. Because <em>Q<\/em> does equal 4.32, the system must be at equilibrium.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Calculation of Changes in Concentration<\/h3>\n<p id=\"fs-idp100361248\">Perhaps the most challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:<\/p>\n<ol id=\"fs-idm126911232\" type=\"1\">\n<li>Identify the direction in which the reaction will proceed to reach equilibrium.<\/li>\n<li>Develop an ICE table.<\/li>\n<li>Calculate the concentration changes and, subsequently, the equilibrium concentrations.<\/li>\n<li>Confirm the calculated equilibrium concentrations.<\/li>\n<\/ol>\n<p id=\"fs-idm215588032\">The last two example exercises of this chapter demonstrate the application of this strategy.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4:\u00a0Calculation of Concentration Changes as a Reaction Goes to Equilibrium<\/h3>\n<p>Under certain conditions, the equilibrium constant for the decomposition of PCl<sub>5<\/sub>(<em>g<\/em>) into PCl<sub>3<\/sub>(<em>g<\/em>) and Cl<sub>2<\/sub>(<em>g<\/em>) is 0.0211. What are the equilibrium concentrations of PCl<sub>5<\/sub>, PCl<sub>3<\/sub>, and Cl<sub>2<\/sub> if the initial concentration of PCl<sub>5<\/sub> was 1.00 <em>M<\/em>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q113380\">Show Solution<\/span><\/p>\n<div id=\"q113380\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the stepwise process described above.<\/p>\n<h4>Step 1: Determine the direction the reaction proceeds.<\/h4>\n<p>The balanced equation for the decomposition of PCl<sub>5<\/sub> is<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{PCl}}_{5}\\left(g\\right)\\rightleftharpoons{\\text{PCl}}_{3}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)[\/latex]<\/p>\n<p>Because we have no products initially, <em>Q<sub>c<\/sub><\/em> = 0 and the reaction will proceed to the right.<\/p>\n<h4>Step 2: Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.<\/h4>\n<p><em data-effect=\"italics\">Develop an ICE table.<\/em><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/openstax.org\/resources\/6f80346599e8e7ccb11c6959f89464108531b29d\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, \u201cP C l subscript 5 equilibrium arrow P C l subscript 3 plus C l subscript 2.\u201d Under the second column is a subgroup of three rows and three columns. The first column has the following: 1.00, negative x, 1.00 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.\" width=\"600\" height=\"162\" \/><\/p>\n<h4>Step 3: Solve for the change and the equilibrium concentrations.<\/h4>\n<p id=\"fs-idp178328384\"><em data-effect=\"italics\">Solve for the change and the equilibrium concentrations.<\/em><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium constant equation gives<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{K}_{c}&=&\\dfrac{\\left[{\\text{PCl}}_{3}\\right]\\left[{\\text{Cl}}_{2}\\right]}{\\left[{\\text{PCl}}_{5}\\right]}=0.0211\\\\{}&=&\\dfrac{\\left(x\\right)\\left(x\\right)}{\\left(1.00-x\\right)}\\\\0.0211&=&\\dfrac{\\left(x\\right)\\left(x\\right)}{\\left(1.00-x\\right)}\\\\0.0211\\left(1.00-x\\right)&=&{x}^{2}\\\\{x}^{2}+0.0211 x-0.0211&=&0\\end{array}[\/latex]<\/p>\n<p><a class=\"target-chapter\" href=\".\/chapter\/essential-mathematics\/\" target=\"_blank\" rel=\"noopener\">Essential Mathematics<\/a> shows us an equation of the form <em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em> = 0 can be rearranged to solve for <em>x<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]x=\\dfrac{-b\\pm\\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/p>\n<p>In this case, [latex]a = 1, b = 0.0211[\/latex], and [latex]c = \u22120.0211[\/latex]. Substituting the appropriate values for <em>a<\/em>, <em>b<\/em>, and <em>c<\/em> yields:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{x}&=&\\dfrac{-0.0211\\pm\\sqrt{\\left(0.0211\\right)^{2}-4\\left(1\\right)\\left(-0.0211\\right)}}{2\\left(1\\right)}\\\\{}&=&\\dfrac{-0.0211\\pm\\sqrt{\\left(4.45\\times{10}^{-4}\\right)+\\left(8.44\\times{10}^{-2}\\right)}}{2}\\\\{}&=&\\dfrac{-0.0211\\pm0.291}{2}\\end{array}[\/latex]<\/p>\n<p>The two roots of the quadratic are, therefore,<\/p>\n<p style=\"text-align: center;\">[latex]x=\\dfrac{-0.0211+0.291}{2}=0.135[\/latex] or [latex]x=\\dfrac{-0.0211-0.291}{2}=-0.156[\/latex].<\/p>\n<p id=\"fs-idp226605088\">For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so\u00a0<em data-effect=\"italics\">x<\/em>\u00a0= 0.135\u00a0<em data-effect=\"italics\">M<\/em>.<\/p>\n<p id=\"fs-idp66450592\">The equilibrium concentrations are<\/p>\n<ul>\n<li>[latex]\\left[{\\text{PCl}}_{5}\\right]=1.00-0.135=0.87M[\/latex]<\/li>\n<li>[latex]\\left[{\\text{PCl}}_{3}\\right]=x=0.135M[\/latex]<\/li>\n<li>[latex]\\left[{\\text{Cl}}_{2}\\right]=x=0.135M[\/latex]<\/li>\n<\/ul>\n<h4>Step 4: Check the arithmetic.<\/h4>\n<p id=\"fs-idm43994720\"><em data-effect=\"italics\">Confirm the calculated equilibrium concentrations.<\/em><\/p>\n<p id=\"fs-idp274032256\">Substitution into the expression for\u00a0<em data-effect=\"italics\">K<sub>c<\/sub><\/em>\u00a0(to check the calculation) gives<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{c}=\\dfrac{\\left[{\\text{PCl}}_{3}\\right]\\left[{\\text{Cl}}_{2}\\right]}{\\left[{\\text{PCl}}_{5}\\right]}=\\dfrac{\\left(0.135\\right)\\left(0.135\\right)}{0.87}=0.021[\/latex]<\/p>\n<p>The equilibrium constant calculated from the equilibrium concentrations is equal to the value of <em>K<sub>c<\/sub><\/em> given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Acetic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>H, reacts with ethanol, C<sub>2<\/sub>H<sub>5<\/sub>OH, to form water and ethyl acetate, CH<sub>3<\/sub>CO<sub>2<\/sub>C<sub>2<\/sub>H<sub>5<\/sub>.<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}+{\\text{C}}_{2}{\\text{H}}_{5}\\text{OH}\\rightleftharpoons{\\text{CH}}_{3}{\\text{CO}}_{2}{\\text{C}}_{2}{\\text{H}}_{5}+{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\n<p>The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations when a mixture that is 0.15 <em>M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>H, 0.15 <em>M<\/em> in C<sub>2<\/sub>H<sub>5<\/sub>OH, 0.40 <em>M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>C<sub>2<\/sub>H<sub>5<\/sub>, and 0.40 <em>M<\/em> in H<sub>2<\/sub>O are mixed in enough dioxane to make 1.0 L of solution?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q703405\">Show Solution<\/span><\/p>\n<div id=\"q703405\" class=\"hidden-answer\" style=\"display: none\">[CH<sub>3<\/sub>CO<sub>2<\/sub>H] = 0.18\u00a0<em>M<\/em>, [C<sub>2<\/sub>H<sub>5<\/sub>OH] = 0.37\u00a0<em>M<\/em>, [CH<sub>3<\/sub>CO<sub>2<\/sub>C<sub>2<\/sub>H<sub>5<\/sub>] = 0.37 <em>M<\/em>, [H<sub>2<\/sub>O] = 0.37 <em>M<\/em><\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>A 1.00-L flask is filled with 1.00 moles of H<sub>2<\/sub> and 2.00 moles of I<sub>2<\/sub>. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H<sub>2<\/sub>, I<sub>2<\/sub>, and HI in moles\/L?<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{I}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{HI}\\left(g\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q604530\">Show Solution<\/span><\/p>\n<div id=\"q604530\" class=\"hidden-answer\" style=\"display: none\">[H<sub>2<\/sub>] = 0.06 <em>M<\/em>, [I<sub>2<\/sub>] = 1.06 <em>M<\/em>, [HI] = 1.88 <em>M<\/em><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<header>\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE 5:\u00a0<\/span>Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption<\/h3>\n<\/header>\n<p>What are the concentrations at equilibrium of a 0.15\u00a0<em>M<\/em>\u00a0solution of HCN?<\/p>\n<p>[latex]\\text{HCN}(aq) \\rightleftharpoons \\text{H}^+ (aq) + \\text{CN}^- (aq)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_c = 4.9 \\text{x} 10^{-10}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q703415\">Show Solution<\/span><\/p>\n<div id=\"q703415\" class=\"hidden-answer\" style=\"display: none\">\nUsing \u201c<em>x<\/em>\u201d to represent the concentration of each product at equilibrium gives this ICE table.<span id=\"fs-idm90952112\" class=\"scaled-down\"><img loading=\"lazy\" decoding=\"async\" id=\"11\" class=\"alignnone\" src=\"https:\/\/openstax.org\/resources\/52ab1350d0394ec1eeb1d18fdc50a25d91e97593\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium ( M ). The second column has the header, \u201cH C N ( a q ) equilibrium arrow H superscript plus sign ( a q ) plus C N subscript negative sign ( a q ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.15, negative x, 0.15 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.\" width=\"601\" height=\"162\" \/><\/span><\/p>\n<p id=\"fs-idp22895280\">Substitute the equilibrium concentration terms into the\u00a0<em>K<sub>c<\/sub><\/em>\u00a0expression<\/p>\n<p style=\"text-align: center;\">[latex]K_c = \\dfrac{(x)(x)}{0.15-x}[\/latex]<\/p>\n<p id=\"fs-idp77336512\">rearrange to the quadratic form and solve for\u00a0<em>x<\/em><\/p>\n<p style=\"text-align: center;\">[latex]x^2 +4.9 \\times 10^{-10} - 7.35 \\times 10^{-11} = 0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x = 8.56 \\text{x} 10^{-6}M \\text{(3 sig. figs)} = 8.6 \\text{x} 10^{-6}M \\text{(2 sig. figs)}[\/latex]<\/p>\n<p id=\"fs-idm36129424\">Thus [H<sup>+<\/sup>] = [CN<sup>\u2013<\/sup>] =\u00a0<em>x<\/em>\u00a0= 8.6\u00a0<span class=\"os-math-in-para\"><span id=\"MathJax-Element-83-Frame\" class=\"MathJax\" style=\"overflow: initial; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; letter-spacing: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px;\" role=\"presentation\"><span id=\"MathJax-Span-2767\" class=\"math\"><span id=\"MathJax-Span-2768\" class=\"mrow\"><span id=\"MathJax-Span-2769\" class=\"semantics\"><span id=\"MathJax-Span-2770\" class=\"mrow\"><span id=\"MathJax-Span-2771\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><span class=\"MJX_Assistive_MathML\" role=\"presentation\">\u00d7<\/span><\/span><\/span>\u00a010<sup>\u20136<\/sup>\u00a0<em>M<\/em>\u00a0and [HCN] = 0.15 \u2013\u00a0<em>x<\/em>\u00a0= 0.15\u00a0<em>M<\/em>.<\/p>\n<p id=\"fs-idm8990656\">Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small\u00a0<em>K<\/em>), and so the initial concentration experiences a negligible change:<\/p>\n<p style=\"text-align: center;\">if [latex]x \\ll 0.15 M, \\text{then} (0.15-x)[\/latex] \u2248 [latex]0.15[\/latex]<\/p>\n<p class=\"body\" style=\"text-align: left;\">This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:<\/p>\n<p style=\"text-align: center;\">[latex]K_c = \\dfrac{(x)(x)}{0.15-x}[\/latex] \u2248 [latex]\\dfrac{x^2}{0.15}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]4.9 \\times 10^{-10} = \\dfrac{x^2}{0.15}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x^2 = (0.15)(4.9\\times 10^{-10}) = 7.4\\times 10^{-11}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x = \\sqrt{7.4\\times 10^{-11}} = 8.6\\times 10^{-6}M[\/latex]<\/p>\n<p>The value of\u00a0<em>x<\/em>\u00a0calculated is, indeed, much less than the initial concentration<\/p>\n<p style=\"text-align: center;\">[latex]8.6\\times{10}^{-6}\\ll{0.15}[\/latex]<\/p>\n<p>and so the approximation was justified. If this simplified approach were to yield a value for\u00a0<em>x<\/em>\u00a0that did\u00a0<em>not<\/em>\u00a0justify the approximation, the calculation would need to be repeated without making the approximation.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>What are the equilibrium concentrations in a 0.25\u00a0<em>M<\/em>\u00a0NH<sub>3<\/sub>\u00a0solution?<\/p>\n<p>[latex]\\text{NH}_3(aq) + \\text{H}_2\\text{O}(l) \\rightleftharpoons \\text{NH}_4+(aq) +\\text{OH}-(aq)\\qquad{K}_{c} = 1.8 \\times 10^{-5}[\/latex]<\/p>\n<section><\/section>\n<section>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q703416\">Show Solution<\/span><\/p>\n<div id=\"q703416\" class=\"hidden-answer\" style=\"display: none\">\n<div class=\"os-note-body\">\n<div style=\"text-align: center;\">[latex][\\text{OH}-] = [\\text{NH}_4+] = 0.0021 M; [\\text{NH}_3] = 0.25 M[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{llll}2{\\text{SO}}_{3}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & 2{\\text{SO}}_{2}\\left(g\\right)+\\hfill & {\\text{O}}_{2}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill & & \\text{ }\\hfill & +x\\hfill \\\\ \\text{ }\\hfill & & \\text{ }\\hfill & 0.125M\\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{lllll}4{\\text{NH}}_{3}\\left(g\\right)\\hfill & +3{\\text{O}}_{2}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & 2{\\text{N}}_{2}\\left(g\\right)+\\hfill & 6{\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill & 3x\\hfill & & \\text{ }\\hfill & \\text{ }\\hfill \\\\ \\text{ }\\hfill & 0.24M\\hfill & & \\text{ }\\hfill & \\text{ }\\hfill \\end{array}[\/latex]<\/li>\n<li>Change in pressure:<br \/>\n[latex]\\begin{array}{llll}2{\\text{CH}}_{4}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & {\\text{C}}_{2}{\\text{H}}_{2}\\left(g\\right)+\\hfill & 3{\\text{H}}_{2}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill & & x\\hfill & \\text{ }\\hfill \\\\ \\text{ }\\hfill & & 25\\text{torr}\\hfill & \\text{ }\\hfill \\end{array}[\/latex]<\/li>\n<li>Change in pressure:<br \/>\n[latex]\\begin{array}{lllll}{\\text{CH}}_{4}\\left(g\\right)+\\hfill & {\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & \\text{CO}\\left(g\\right)+\\hfill & 3{\\text{H}}_{2}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill & x\\hfill & & \\text{ }\\hfill & \\text{ }\\hfill \\\\ \\text{ }\\hfill & 5\\text{atm}\\hfill & & \\text{ }\\hfill & \\text{ }\\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{llll}{\\text{NH}}_{4}\\text{Cl}\\left(s\\right)\\hfill & \\rightleftharpoons\\hfill & {\\text{NH}}_{3}\\left(g\\right)+\\hfill & \\text{HCl}\\left(g\\right)\\hfill \\\\ & & x\\hfill & \\text{ }\\hfill \\\\ & & \\hfill 1.03\\times {10}^{-4}M\\hfill & \\text{ }\\hfill \\end{array}[\/latex]<\/li>\n<li>change in pressure:<br \/>\n[latex]\\begin{array}{cccc}\\text{Ni}\\left(s\\right)+\\hfill & 4\\text{CO}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & \\text{Ni}{\\left(\\text{CO}\\right)}_{4}\\left(g\\right)\\hfill \\\\ & 4x\\hfill & & \\text{ }\\hfill \\\\ & \\hfill 0.40\\text{atm}\\hfill & & \\text{ }\\hfill \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{cccc}2{\\text{H}}_{2}\\left(g\\right)+\\hfill & {\\text{O}}_{2}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & 2{\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill & \\text{ }\\hfill & & +2x\\hfill \\\\ \\text{ }\\hfill & \\text{ }\\hfill & & 1.50M\\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccccc}{\\text{CS}}_{2}\\left(g\\right)+\\hfill & 4{\\text{H}}_{2}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & {\\text{CH}}_{4}\\left(g\\right)+\\hfill & 2{\\text{H}}_{2}\\text{S}\\left(g\\right)\\hfill \\\\ x\\hfill & \\text{ }\\hfill & & \\text{ }\\hfill & \\text{ }\\hfill \\\\ 0.020M\\hfill & \\text{ }\\hfill & & \\text{ }\\hfill & \\text{ }\\hfill \\end{array}[\/latex]<\/li>\n<li>Change in pressure:<br \/>\n[latex]\\begin{array}{cccc}{\\text{H}}_{2}\\left(g\\right)+\\hfill & {\\text{Cl}}_{2}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & 2\\text{HCl}\\left(g\\right)\\hfill \\\\ x\\hfill & \\text{ }\\hfill & & \\text{ }\\hfill \\\\ 1.50\\text{atm}\\hfill & \\text{ }\\hfill & & \\text{ }\\hfill \\end{array}[\/latex]<\/li>\n<li>Change in pressure:<br \/>\n[latex]\\begin{array}{ccccc}2{\\text{NH}}_{3}\\left(g\\right)\\hfill & +2{\\text{O}}_{2}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & {\\text{N}}_{2}\\text{O}\\left(g\\right)+\\hfill & 3{\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill & \\text{ }\\hfill & & \\text{ }\\hfill & x\\hfill \\\\ \\text{ }\\hfill & \\text{ }\\hfill & & \\text{ }\\hfill & 60.6\\text{torr}\\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{cccc}{\\text{NH}}_{4}\\text{HS}\\left(s\\right)\\hfill & \\rightleftharpoons\\hfill & {\\text{NH}}_{3}\\left(g\\right)+\\hfill & {\\text{H}}_{2}\\text{S}\\left(g\\right)\\hfill \\\\ & & x\\hfill & \\text{ }\\hfill \\\\ & & 9.8\\times {10}^{-6}M\\hfill & \\text{ }\\hfill \\end{array}[\/latex]<\/li>\n<li>Change in pressure:<br \/>\n[latex]\\begin{array}{cccc}\\text{Fe}\\left(s\\right)+\\hfill & 5\\text{CO}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & \\text{Fe}{\\left(\\text{CO}\\right)}_{4}\\left(g\\right)\\hfill \\\\ & \\text{ }\\hfill & & x\\hfill \\\\ & \\text{ }\\hfill & & 0.012\\text{atm}\\hfill \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Why are there no changes specified for Ni in question\u00a011 , part (f)? What property of Ni does change?<\/li>\n<li>Why are there no changes specified for NH<sub>4<\/sub>HS in question 12, part (e)? What property of NH<sub>4<\/sub>HS does change?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q381796\">Show Selected Solutions<\/span><\/p>\n<div id=\"q381796\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\u00a0The changes in concentrations (or pressure, if requested)\u00a0are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{llll}2{\\text{SO}}_{3}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & 2{\\text{SO}}_{2}\\left(g\\right)+\\hfill & {\\text{O}}_{2}\\left(g\\right)\\hfill \\\\ \\text{ }\\hfill -2x & &2x\\hfill & +x\\hfill \\\\ -0.250M\\hfill & & 0.250M\\hfill & 0.125M\\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{lllll}4{\\text{NH}}_{3}\\left(g\\right)\\hfill & +3{\\text{O}}_{2}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & 2{\\text{N}}_{2}\\left(g\\right)+\\hfill & 6{\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill \\\\ 4x\\hfill & 3x\\hfill & & -2x\\hfill & -6x\\hfill \\\\ 0.32M\\hfill & 0.24M\\hfill & &-0.16M\\hfill & -0.48M\\hfill \\end{array}[\/latex]<\/li>\n<li>Change in pressure:<br \/>\n[latex]\\begin{array}{llll}2{\\text{CH}}_{4}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & {\\text{C}}_{2}{\\text{H}}_{2}\\left(g\\right)+\\hfill & 3{\\text{H}}_{2}\\left(g\\right)\\hfill \\\\ -2x\\hfill & & x\\hfill & 3x\\hfill \\\\ -50\\text{ torr}\\hfill & & 25\\text{ torr}\\hfill & 75\\text{ torr}\\hfill \\end{array}[\/latex]<\/li>\n<li>Change in pressure:<br \/>\n[latex]\\begin{array}{lllll}{\\text{CH}}_{4}\\left(g\\right)+\\hfill & {\\text{H}}_{2}\\text{O}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & \\text{CO}\\left(g\\right)+\\hfill & 3{\\text{H}}_{2}\\left(g\\right)\\hfill \\\\ x\\hfill & x\\hfill & & -x\\hfill & -3x\\hfill \\\\ 5\\text{ atm}\\hfill & 5\\text{atm}\\hfill & & -5\\text{ atm}\\hfill & -15\\text{ atm}\\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{llll}{\\text{NH}}_{4}\\text{Cl}\\left(s\\right)\\hfill & \\rightleftharpoons\\hfill & {\\text{NH}}_{3}\\left(g\\right)+\\hfill & \\text{HCl}\\left(g\\right)\\hfill \\\\ & & x\\hfill & x\\hfill \\\\ & & \\hfill 1.03\\times {10}^{-4}M\\hfill & 1.03\\times10^{-4}M\\hfill \\end{array}[\/latex]<\/li>\n<li>change in pressure:<br \/>\n[latex]\\begin{array}{cccc}\\text{Ni}\\left(s\\right)+\\hfill & 4\\text{CO}\\left(g\\right)\\hfill & \\rightleftharpoons\\hfill & \\text{Ni}{\\left(\\text{CO}\\right)}_{4}\\left(g\\right)\\hfill \\\\ & 4x\\hfill & & x\\hfill \\\\ & \\hfill 0.40\\text{atm}\\hfill & & 0.1\\text{ atm}\\hfill \\end{array}[\/latex]<\/li>\n<\/ol>\n<p>3.\u00a0Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>Assume that the change in concentration of N<sub>2<\/sub>O<sub>4<\/sub> is small enough to be neglected in the following problem.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N<sub>2<\/sub>O<sub>4<\/sub> with chloroform as the solvent.<br \/>\n[latex]{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)\\rightleftharpoons2{\\text{NO}}_{2}\\left(g\\right){K}_{c}=1.07\\times {10}^{-5}[\/latex] in chloroform<\/li>\n<li>Show that the change is small enough to be neglected.<\/li>\n<\/ol>\n<\/li>\n<li>Assume that the change in concentration of COCl<sub>2<\/sub> is small enough to be neglected in the following problem.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl<sub>2<\/sub> with an initial concentration of 0.3166 <em>M<\/em>.<br \/>\n[latex]{\\text{COCl}}_{2}\\left(g\\right)\\rightleftharpoons\\text{CO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right){K}_{c}=2.2\\times {10}^{-10}[\/latex]<\/li>\n<li>Show that the change is small enough to be neglected.<\/li>\n<\/ol>\n<\/li>\n<li>Assume that the change in pressure of H<sub>2<\/sub>S is small enough to be neglected in the following problem.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H<sub>2<\/sub>S with an initial pressure of 0.824 atm.<br \/>\n[latex]2{\\text{H}}_{2}\\text{S}\\left(g\\right)\\rightleftharpoons2{\\text{H}}_{2}\\left(g\\right)+{\\text{S}}_{2}\\left(g\\right){K}_{P}=2.2\\times {10}^{-6}[\/latex]<\/li>\n<li>Show that the change is small enough to be neglected.<\/li>\n<\/ol>\n<\/li>\n<li>What are all concentrations after a mixture that contains [H<sub>2<\/sub>O] = 1.00 <em>M<\/em> and [Cl<sub>2<\/sub>O] = 1.00 <em>M<\/em> comes to equilibrium at 25 \u00b0C?<br \/>\n[latex]{\\text{H}}_{2}\\text{O}\\left(g\\right)+{\\text{Cl}}_{2}\\text{O}\\left(g\\right)\\rightleftharpoons2\\text{HOCl}\\left(g\\right){K}_{c}=0.0900[\/latex]<\/li>\n<li>What are the concentrations of PCl<sub>5<\/sub>, PCl<sub>3<\/sub>, and Cl<sub>2<\/sub> in an equilibrium mixture produced by the decomposition of a sample of pure PCl<sub>5<\/sub> with [PCl<sub>5<\/sub>] = 2.00 <em>M<\/em>?<br \/>\n[latex]{\\text{PCl}}_{5}\\left(g\\right)\\rightleftharpoons{\\text{PCl}}_{3}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right){K}_{c}=0.0211[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q955866\">Show Selected Solutions<\/span><\/p>\n<div id=\"q955866\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. (a) Write the starting conditions, change, and equilibrium constant in tabular form.<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[NO<sub>2<\/sub>]<\/th>\n<th>[N<sub>2<\/sub>O<sub>4<\/sub>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (M)<\/th>\n<td>0<\/td>\n<td>0.129<\/td>\n<\/tr>\n<tr>\n<th>Change (M)<\/th>\n<td>+2<em>x<\/em><\/td>\n<td>\u2212<em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium concentration (M)<\/th>\n<td>2<em>x<\/em><\/td>\n<td>0.129\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since <em>K<\/em> is very small, ignore <em>x<\/em> in comparison with 0.129 <em>M<\/em>. The equilibrium expression is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} {K}_{c}=1.07\\times {10}^{-5}=\\frac{{\\left[{\\text{NO}}_{2}\\right]}^{2}}{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}=\\frac{{\\left(2x\\right)}^{2}}{\\left(0.129-x\\right)}=\\frac{{\\left(2x\\right)}^{2}}{0.129}\\hfill \\\\ {x}^{2}=\\frac{0.129\\times 1.07\\times {10}^{-5}}{4}=3.45\\times {10}^{-7}\\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>x<\/em> = 5.87 \u00d7 10<sup>\u22124<\/sup><\/p>\n<p>The concentrations are:<\/p>\n<ul>\n<li>[NO<sub>2<\/sub>] = 2<em>x<\/em> = 5.87 \u00d7 10<sup>-4<\/sup> = 1.17 \u00d7 10<sup>-3<\/sup><em>M<\/em><\/li>\n<li>[N<sub>2<\/sub>O<sub>4<\/sub>] = 0.129 &#8211; <em>x<\/em> = 0.129 &#8211; 5.87 \u00d7 10<sup>-4<\/sup> = 0.128 <em>M<\/em><\/li>\n<\/ul>\n<p>(b) Percent error [latex]=\\frac{5.87\\times {10}^{-4}}{0.129}\\times 100\\%=0.455\\%[\/latex]. The change in concentration of N<sub>2<\/sub>O<sub>4<\/sub> is far less than the 5% maximum allowed.<\/p>\n<p>3. \u00a0(a) Write the balanced equation, and then set up a table with initial pressures and the changed pressures using <em>x<\/em> as the change in pressure. The simplest way to find the coefficients for the <em>x<\/em> values is to use the coefficient in the balanced equation.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212416\/CNX_Chem_13_04_ICETable7_img1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( a t m ), Change ( a t m ), Equilibrium pressure ( a t m ). The second column has the header, \u201c2 H subscript 2 S ( g ) equilibrium arrow 2 H subscript 2 ( g ) plus S subscript 2 ( g ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.824, negative 2 x, 0.824 minus 2 x. The second column has the following: 0, negative 2 x, positive 2 x. The third column has the following: 0, negative 2 x, positive 2 x.\" \/><\/p>\n<p style=\"text-align: center;\">[latex]{K}_{P}=2.2\\times {10}^{-6}=\\frac{\\left({P}_{{\\text{S}}_{2}}\\right){\\left({P}_{{\\text{H}}_{2}}\\right)}^{2}}{{\\left({P}_{{\\text{H}}_{2}\\text{S}}\\right)}^{2}}=\\frac{\\left[x\\right]{\\left[2x\\right]}^{2}}{{\\left[0.824-2x\\right]}^{2}}[\/latex]<\/p>\n<p>Since the value of <em>K<\/em><sub>p<\/sub> is much smaller than 0.824, the 2<em>x<\/em> term in 0.824 \u2212 2<em>x<\/em> is deemed negligible and, therefore, can be dropped.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{ }2.2\\times {10}^{-6}=\\frac{\\left[x\\right]{\\left[2x\\right]}^{2}}{{\\left[0.824\\right]}^{2}}\\hfill \\\\ 2.2\\times {10}^{-6}=\\frac{4{\\text{X}}^{3}}{\\left[0.679\\right]}\\hfill \\end{array}[\/latex]<\/p>\n<ul>\n<li>1.494 \u00d7 10<sup>\u22126<\/sup> = 4<em>x<\/em><sup>3<\/sup><\/li>\n<li>3.73 \u00d7 10<sup>\u22127<\/sup> = <em>x<\/em><sup>3<\/sup><\/li>\n<li>7.20 \u00d7 10<sup>\u22123<\/sup> = <em>x<\/em><\/li>\n<\/ul>\n<p>Final equilibrium pressures:<\/p>\n<ul>\n<li>[H<sub>2<\/sub>S] = 0.824 \u2212 2<em>x<\/em> = 0.824 \u2212 2(7.20 \u00d7 10<sup>\u22123<\/sup>) = 0.824 &#8211; 0.0144 = 0.810 atm<\/li>\n<li>[H<sub>2<\/sub>] = 2<em>x<\/em> = 2(7.2 \u00d7 10<sup>\u22123<\/sup>) = 0.014 atm<\/li>\n<li>[S<sub>2<\/sub>] = [<em>x<\/em>] = 0.0072 atm<\/li>\n<\/ul>\n<p>(b) The 2<em>x<\/em> is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2<em>x<\/em> is [latex]\\frac{0.014}{0.824}\\times 100\\%=1.7\\%[\/latex], which is less than allowed by the \u201c5% test.\u201d The error is, indeed, negligible.<\/p>\n<p>5.\u00a0As all species are gases and are in <em>M<\/em> concentration units, a simple <em>K<sub>c<\/sub><\/em> equilibrium can be solved using the balanced equation:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212419\/CNX_Chem_13_04_ICETable9_img1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, \u201cP C l subscript 5 ( g ) equilibrium arrow P C l subscript 3 ( g ) plus C l subscript 2 ( g ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 2.00, negative x, 2.00 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.\" width=\"883\" height=\"239\" \/><\/p>\n<p style=\"text-align: center;\">[latex]{K}_{c}=0.0211=\\frac{\\left[{\\text{PCl}}_{3}\\right]\\left[{\\text{Cl}}_{2}\\right]}{\\left[{\\text{PCl}}_{5}\\right]}=\\frac{\\left[x\\right]\\left[x\\right]}{\\left[2.00-x\\right]}[\/latex]<\/p>\n<p>As the value of <em>K<sub>c<\/sub><\/em> is substantial when compared with 2.00 <em>M<\/em>, the <strong><em>x<\/em><\/strong> terms in 2.00 \u2212 <strong><em>x<\/em><\/strong> cannot be disregarded. Thus, <strong><em>x<\/em><\/strong> must be solved by using the quadratic formula.<\/p>\n<p style=\"text-align: center;\">[latex]0.0211=\\frac{{x}^{2}}{\\left[2.00-x\\right]}=0.0422-0.0211x={x}^{2}[\/latex]<\/p>\n<p>Begin by arranging the terms in the form of the quadratic equation:<\/p>\n<ul>\n<li><em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em> = 0<\/li>\n<li><em>x<\/em><sup>2<\/sup> + 0.0211<em>x<\/em> \u2212 0.0422 = 0<\/li>\n<\/ul>\n<p>Next, solve for <strong><em>x<\/em><\/strong> using the quadratic formula.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{\\text{-}b\\pm\\sqrt{{b}^{2}-4ac}}{2a}=\\frac{-0.0211\\pm\\sqrt{{\\left(0.0211\\right)}^{2}-4\\left(1\\right)\\left(-0.0422\\right)}}{2\\left(1\\right)}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]=\\frac{-0.0211\\pm\\sqrt{0.0004452+0.1688}}{2}=\\frac{-0.0211\\pm 0.4114}{2}[\/latex]<\/p>\n<p style=\"text-align: center;\">= 0.195 <em>M<\/em> or \u22120.216 <em>M<\/em><\/p>\n<p>The process of dissociation renders only positive quantities. Thus, <em>x<\/em> must be a positive value when factored into the solution so as to guarantee a realistic result. The final equilibrium concentrations are: [PCl<sub>3<\/sub>] = 2.00 \u2212 <em>x<\/em> = 2.00 \u2212 0.195 = 1.80 <em>M<\/em>; [PC<sub>3<\/sub>] = [Cl<sub>2<\/sub>] = <em>x<\/em> = 0.195 <em>M<\/em>; [PCl<sub>3<\/sub>] = [Cl<sub>2<\/sub>] = <em>x<\/em> = 0.195 <em>M<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>Calculating values for equilibrium constants and\/or equilibrium concentrations is of practical benefit to many applications. A mathematical strategy that uses initial concentrations, changes in concentrations, and equilibrium concentrations (and goes by the acronym ICE) is useful for several types of equilibrium calculations.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H<sub>2<\/sub> and 1.25 mol of I<sub>2<\/sub> in a 5.00-L flask at 448 \u00b0C.<br \/>\n[latex]{\\text{H}}_{2}+{\\text{I}}_{2}\\rightleftharpoons2\\text{HI}{K}_{c}=50.2\\text{ at }448^{\\circ}\\text{C}[\/latex]<\/li>\n<li>What is the pressure of BrCl in an equilibrium mixture of Cl<sub>2<\/sub>, Br<sub>2<\/sub>, and BrCl if the pressure of Cl<sub>2<\/sub> in the mixture is 0.115 atm and the pressure of Br<sub>2<\/sub> in the mixture is 0.450 atm?<br \/>\n[latex]{\\text{Cl}}_{2}\\left(g\\right)+{\\text{Br}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{BrCl}\\left(g\\right){K}_{P}=4.7\\times {10}^{-2}[\/latex]<\/li>\n<li>What is the pressure of CO<sub>2<\/sub> in a mixture at equilibrium that contains 0.50 atm H<sub>2<\/sub>, 2.0 atm of H<sub>2<\/sub>O, and 1.0 atm of CO at 990 \u00b0C?<br \/>\n[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right)+\\text{CO}\\left(g\\right){K}_{P}=1.6\\text{ at }990^{\\circ}\\text{C}[\/latex]<\/li>\n<li>Sodium sulfate 10-hydrate, Na<sub>2<\/sub>SO<sub>4<\/sub> [latex]\\cdot[\/latex] 10H<sub>2<\/sub>O, dehydrates according to the equation<br \/>\n[latex]{\\text{Na}}_{2}{\\text{SO}}_{4}\\cdot 10{\\text{H}}_{2}\\text{O}\\left(s\\right)\\rightleftharpoons{\\text{Na}}_{2}{\\text{SO}}_{4}\\left(s\\right)+10{\\text{H}}_{2}\\text{O}\\left(g\\right){K}_{P}=4.08\\times {10}^{-25}\\text{ at }25^{\\circ}\\text{C}[\/latex]<br \/>\nWhat is the pressure of water vapor at equilibrium with a mixture of Na<sub>2<\/sub>SO<sub>4<\/sub> [latex]\\cdot[\/latex] 10H<sub>2<\/sub>O and NaSO<sub>4<\/sub>?<\/li>\n<li>Calcium chloride 6-hydrate, CaCl<sub>2<\/sub> [latex]\\cdot[\/latex] 6H<sub>2<\/sub>O, dehydrates according to the equation<br \/>\n[latex]{\\text{CaCl}}_{2}\\cdot 6{\\text{H}}_{2}\\text{O}\\left(s\\right)\\rightleftharpoons{\\text{CaCl}}_{2}\\left(s\\right)+6{\\text{H}}_{2}\\text{O}\\left(g\\right){K}_{P}=5.09\\times {10}^{-44}\\text{ at }25^{\\circ}\\text{C}[\/latex]<br \/>\nWhat is the pressure of water vapor at equilibrium with a mixture of CaCl<sub>2<\/sub> [latex]\\cdot[\/latex] 6H<sub>2<\/sub>O and CaCl<sub>2<\/sub>?<\/li>\n<li>Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl<sub>2<\/sub> produced when a sample of NOCl with a pressure of 0.500 atm comes to equilibrium according to this reaction:<br \/>\n[latex]2\\text{NOCl}\\left(g\\right)\\rightleftharpoons 2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right){K}_{P}=4.0\\times {10}^{-4}[\/latex]<\/li>\n<li>Calculate the equilibrium concentrations of NO, O<sub>2<\/sub>, and NO<sub>2<\/sub> in a mixture at 250 \u00b0C that results from the reaction of 0.20 <em>M<\/em> NO and 0.10 <em>M<\/em> O<sub>2<\/sub>. (Hint: <em>K<\/em> is large; assume the reaction goes to completion then comes back to equilibrium.)<br \/>\n[latex]2\\text{NO}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NO}}_{2}\\left(g\\right){K}_{c}=2.3\\times {10}^{5}\\text{ at }250^{\\circ}\\text{C}[\/latex]<\/li>\n<li>Calculate the equilibrium concentrations that result when 0.25 <em>M<\/em> O<sub>2<\/sub> and 1.0 <em>M<\/em> HCl react and come to equilibrium. (Hint: <em>K<sub>c<\/sub><\/em> is large; assume the reaction goes to completion, then comes back to equilibrium.)<br \/>\n[latex]4\\text{HCl}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{Cl}}_{2}\\left(g\\right)+2{\\text{H}}_{2}\\text{O}\\left(g\\right){K}_{c}=3.1\\times {10}^{13}[\/latex]<\/li>\n<li>One of the important reactions in the formation of smog is represented by the equation<br \/>\n[latex]{\\text{O}}_{3}\\left(g\\right)+\\text{NO}\\left(g\\right)\\rightleftharpoons{\\text{NO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right){K}_{P}=6.0\\times {10}^{34}[\/latex]<br \/>\nWhat is the pressure of O<sub>3<\/sub> remaining after a mixture of O<sub>3<\/sub> with a pressure of 1.2 \u00d7 10<sup>-8<\/sup> atm and NO with a pressure of 1.2 \u00d7 10<sup>-8<\/sup> atm comes to equilibrium? (Hint: <em>K<sub>P<\/sub><\/em> is large; assume the reaction goes to completion then comes back to equilibrium.)<\/li>\n<li>Calculate the pressures of NO, Cl<sub>2<\/sub>, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 4.0 atm NO and 2.0 atm Cl<sub>2<\/sub>. (Hint: <em>K<sub>P<\/sub><\/em> is small; assume the reverse reaction goes to completion then comes back to equilibrium.)<\/li>\n<li>Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H<sub>2<\/sub> and 63.5 g of iodine at 448 \u00b0C.<br \/>\n[latex]{\\text{H}}_{2}+{\\text{I}}_{2}\\rightleftharpoons2\\text{HI}{K}_{c}=50.2\\text{ at }448^{\\circ}\\text{C}[\/latex]<\/li>\n<li>Butane exists as two isomers, <em>n<\/em>-butane and isobutane.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212432\/CNX_Chem_13_04_butane_img1.jpg\" alt=\"Three Lewis structures are shown. The first is labeled, \u201cn dash Butane,\u201d and has a C H subscript 3 single bonded to a C H subscript 2 group. This C H subscript 2 group is single bonded to another C H subscript 2 group which is single bonded to a C H subscript 3 group. The second is labeled, \u201ciso dash Butane,\u201d and is composed of a C H group single bonded to three C H subscript 3 groups. The third structure shows a chain of atoms: \u201cC H subscript 3, C H subscript 2, C H subscript 2, C H subscript 3,\u201d a double-headed arrow, then a carbon atom single bonded to three C H subscript 3 groups as well as a hydrogen atom.\" \/><br \/>\n<em>K<sub>P<\/sub><\/em> = 2.5 at 25 \u00b0CWhat is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?<\/li>\n<li>What is the minimum mass of CaCO<sub>3<\/sub> required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (<em>K<sub>c<\/sub><\/em>) is 0.050 for the decomposition reaction of CaCO<sub>3<\/sub> at that temperature?<br \/>\n[latex]{\\text{CaCO}}_{3}\\left(s\\right)\\rightleftharpoons\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]<\/li>\n<li>The equilibrium constant (<em>K<sub>c<\/sub><\/em>) for this reaction is 1.60 at 990 \u00b0C:\u00a0[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right)+\\text{CO}\\left(g\\right)[\/latex]<br \/>\nCalculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H<sub>2<\/sub>, 2.00 mol of CO<sub>2<\/sub>, 0.750 mol of H<sub>2<\/sub>O, and 1.00 mol of CO to a 5.00-L container at 990 \u00b0C.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q303841\">Show Selected Solutions<\/span><\/p>\n<div id=\"q303841\" class=\"hidden-answer\" style=\"display: none\">\n<p>2.\u00a0Write the equilibrium constant expression and solve for <em>P<\/em><sub>BrCl<\/sub>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{K}_{P}=\\frac{{\\left({P}_{\\text{BrCl}}\\right)}^{2}}{{P}_{{\\text{Cl}}_{2}}{P}_{{\\text{Br}}_{2}}}=\\frac{{\\left({P}_{\\text{BrCl}}\\right)}^{2}}{\\left(0.115\\right)\\left(0.450\\right)}=4.7\\times {10}^{-2}\\\\ {\\left({P}_{\\text{BrCl}}\\right)}^{2}=0.115\\times 0.450\\times 4.7\\times {10}^{-2}=2.43\\times {10}^{-3}\\text{atm}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>P<\/em><sub>BrCl<\/sub> = 4.9 \u00d7 10<sup>-2<\/sup> atm<\/p>\n<p>4.\u00a0Because two of the substances involved in the equilibrium are solids, their activities are 1, and their pressures are constant and do not appear in the equilibrium expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{K}_{p}=4.08\\times {10}^{-25}={\\left({P}_{{\\text{H}}_{2}\\text{O}}\\right)}^{10}\\\\ {P}_{{\\text{H}}_{2}\\text{O}}=\\sqrt[10]{4.08\\times {10}^{-25}}=3.64\\times {10}^{-3}\\text{atm}\\end{array}[\/latex]<\/p>\n<p>7.\u00a0As the equilibrium constant for the reaction is large at \u2248 10<sup>5<\/sup>, first assume that the reaction is complete\u2014that is, that one or both reactants are completely consumed.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212422\/CNX_Chem_13_04_ICETable11_img1.jpg\" alt=\"This table has two main columns and four rows. The first cell in the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header, \u201c2 N O ( g ) plus O subscript 2 ( g ) right-facing arrow 2 N O subscript 2 ( g ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.20, 2 x, 0. The second column has the following: 0.10, x, 0. The third column has the following: 0, 2 x, 0.20.\" width=\"878\" height=\"236\" \/><\/p>\n<p>As only NO<sub>2<\/sub> exists at this stage, the reaction will establish equilibrium:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212423\/CNX_Chem_13_04_ICETable12_img1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium ( M ). The second column has the header, \u201c2 N O subscript 2 ( g ) equilibrium arrow 2 N O ( g ) plus O subscript 2 ( g ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.20, negative 2 x, 0.20 minus 2 x. The second column has the following: 0, 2 x, 2 x. The third column has the following: 0, x, x.\" width=\"879\" height=\"238\" \/><\/p>\n<p>As this equilibrium is the reverse of that originally given, the new <em>K<sub>c<\/sub><\/em> will be the inverse of the given <em>K<sub>c<\/sub><\/em>\u2014that is, [latex]\\frac{1}{{K}_{c}}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}{K}_{\\text{new}}\\hfill & =\\frac{1}{{K}_{c}}=\\frac{{\\left[\\text{NO}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}\\hfill \\\\ \\hfill & =\\frac{1}{2.3\\times {10}^{5}}=4.35\\times {10}^{-6}=\\frac{{\\left[2x\\right]}^{2}\\left[x\\right]}{{\\left[0.20-2x\\right]}^{2}}\\hfill \\end{array}[\/latex]<\/p>\n<p>Because the new equilibrium constant is small compared with the 0.2 <em>M<\/em> term, the 2<em>x<\/em> can be dropped and the equilibrium solved.<\/p>\n<p style=\"text-align: center;\">[latex]4.35\\times {10}^{-6}=\\frac{\\left(4{x}^{2}\\right)x}{\\left[0.04\\right]}[\/latex]<\/p>\n<p style=\"text-align: center;\">4<em>x<\/em><sup>3<\/sup> = 1.74 \u00d7 10<sup>\u22127<\/sup><\/p>\n<p style=\"text-align: center;\"><em>x<\/em><sup>3<\/sup> = 4.35 \u00d7 10<sup>\u22128<\/sup><\/p>\n<p style=\"text-align: center;\"><em>x<\/em> = 3.52 \u00d7 10<sup>\u22123<\/sup><\/p>\n<p>Final equilibrium concentrations:<\/p>\n<ul>\n<li>[NO<sub>2<\/sub>] = 0.20 \u2212 2<em>x<\/em> = 0.20 \u2212 2(0.00352) = 0.19 <em>M<\/em><\/li>\n<li>[NO] = 2<em>x<\/em> = 2(0.00352) = 0.0070 <em>M<\/em><\/li>\n<li>[O<sub>2<\/sub>] = <em>x<\/em> = 0.0035 <em>M<\/em><\/li>\n<\/ul>\n<p>2<em>x<\/em> is small compared to 0.20 <em>M<\/em> because<\/p>\n<p style=\"text-align: center;\">[latex]\\left(\\frac{2x}{0.20}\\right)\\times 100\\%=\\left(\\frac{0.00352}{0.20}\\right)\\times 100\\%=1.7\\%[\/latex] which is less than the maximum 5% allowed.<\/p>\n<p>9.\u00a0Assume that the reaction goes to completion, where one or both reactants are completely used up:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212428\/CNX_Chem_13_04_ICETable15_img1.jpg\" alt=\"This table has two main columns and four rows. The first cell in the first column does not have a heading and then has the following: Initial pressure ( a t m ), Change ( a t m ), Equilibrium pressure ( a t m ). The second column has the header, \u201cO subscript 3 ( g ) plus N O ( g ) right-facing arrow N O subscript 2 ( g ) plus O subscript 2 ( g ).\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 1.2 times ten to the negative 8, negative x, 0. The second column has the following: 1.2 times ten to the negative 8, negative x, 0. The third column has the following: 0, positive x, 1.2 times ten to the negative 8. The fourth column as the following: 0, positive x, 1.2 times ten to the negative 8.\" width=\"878\" height=\"177\" \/><\/p>\n<p>Now assume the equilibrium will be established by the reaction moving toward the reactants:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212429\/CNX_Chem_13_04_ICETable16_img1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( a t m ), Change ( a t m ), Equilibrium pressure ( a t m ). The second column has the header, \u201cO subscript 3 ( g ) plus N O ( g ) equilibrium arrow N O subscript 2 ( g ) plus O subscript 2 ( g ).\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0, positive x, x. The second column has the following: 0, positive x, x. The third column has the following: 1.2 times 10 to the negative eighth power, negative x, 1.2 times 10 to the negative eighth power minus x.\" width=\"881\" height=\"179\" \/><\/p>\n<p style=\"text-align: center;\">[latex]{K}_{P}=\\frac{{P}_{{\\text{NO}}_{2}}{P}_{{\\text{O}}_{2}}}{{P}_{{\\text{O}}_{3}}{P}_{\\text{NO}}}=\\frac{\\left(1.2\\times {10}^{-8}-x\\right)\\left(1.2\\times {10}^{-8}-x\\right)}{\\left(x\\right)\\left(x\\right)}[\/latex]<\/p>\n<p>As <em>K<sub>P<\/sub><\/em> is much larger than 1.2 \u00d7 10<sup>\u22128<\/sup>, the X terms in the expression 1.2 \u00d7 10<sup>\u22128<\/sup>\u00a0\u2212 <em>x<\/em> are relatively negligible and, therefore, can be dropped.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{ }6.0\\times {10}^{34}=\\frac{\\left(1.2\\times {10}^{-8}\\right)\\left(1.2\\times {10}^{-8}\\right)}{\\left(x\\right)\\left(x\\right)}\\hfill \\\\ {x}^{2}=\\frac{\\left(1.44\\times {10}^{-16}\\right)}{6.0\\times {10}^{34}}=2.4\\times {10}^{-51}\\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>x<\/em> = 4.9 \u00d7 10<sup>\u221226<\/sup> atm<\/p>\n<p style=\"text-align: center;\">[latex]{P}_{{\\text{O}}_{3}}=x=4.9\\times {10}^{-26}\\text{atm}[\/latex]<\/p>\n<p>Clearly <em>x<\/em> is much smaller than 1.2 \u00d7 10<sup>\u22124<\/sup> so the assumption is valid.<\/p>\n<p>11.\u00a0The number of moles of I<sub>2<\/sub> is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{ }\\text{mol}=\\frac{63.5\\text{g}}{253.809\\text{g}{\\text{mol}}^{-1}}=0.250\\text{mol}{\\text{I}}_{2}\\hfill \\\\ {K}_{c}=\\frac{{\\left[\\text{HI}\\right]}^{2}}{\\left[{\\text{H}}_{2}\\right]\\left[{\\text{I}}_{2}\\right]}\\hfill \\end{array}[\/latex]<\/p>\n<p>The unit for each concentration term is moles per liter. If the volume were known for this exercise, the number of moles in each term should be divided by this volume. However, there are two terms in the numerator and two terms in the denominator, so these volumes cancel one another. Consequently, for any expression with the same number of numerator terms as denominator terms, the number for moles can be used in place of moles per liter. In this exercise, the volume is not needed even though it is given.<\/p>\n<p style=\"text-align: center;\">(mol HI)<sup>2<\/sup> = <em>K<\/em> \u00d7 mol H<sub>2<\/sub> \u00d7 mol I<sub>2<\/sub><\/p>\n<p style=\"text-align: center;\">= 50.2 \u00d7 1.25 mol \u00d7 0.250 mol<\/p>\n<p style=\"text-align: center;\">= 15.7 mol<sup>2<\/sup><\/p>\n<p style=\"text-align: center;\">mol HI = [latex]\\sqrt{15.7{\\text{mol}}^{2}}=3.96\\text{mol}[\/latex]<\/p>\n<p style=\"text-align: center;\">Mass(HI) = 3.96 mol \u00d7 127.9124 g\/mol = 507 g<\/p>\n<p>13. At equilibrium the concentration of CO<sub>2<\/sub> = <em>K<sub>c<\/sub><\/em> = 0.50 <em>M<\/em>. The number of moles CO<sub>2<\/sub> in the system is then mol CO<sub>2<\/sub> = 6.5 L \u00d7 0.50 mol\/L = 3.3 mol. The minimum moles of CaCO<sub>2<\/sub> required is then just more than:<\/p>\n<p style=\"text-align: center;\">[latex]3.3\\text{mol}{\\text{CO}}_{2}\\times \\frac{1\\text{mol}{\\text{CaCO}}_{3}}{1\\text{mol}{\\text{CO}}_{2}}=3.3\\text{mol}{\\text{CaCO}}_{3}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]3.3\\text{mol}{\\text{CO}}_{2}\\times \\frac{1\\text{mol}{\\text{CaCO}}_{3}}{100.1\\text{g}{\\text{CaCO}}_{3}}=330\\text{g}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>At 25 \u00b0C and at 1 atm, the partial pressures in an equilibrium mixture of N<sub>2<\/sub>O<sub>4<\/sub> and NO<sub>2<\/sub> are [latex]{\\text{P}}_{{\\text{N}}_{2}{\\text{O}}_{4}}=0.70\\text{atm}[\/latex] and [latex]{\\text{P}}_{{\\text{NO}}_{2}}=0.30\\text{atm.}[\/latex]\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Predict how the pressures of NO<sub>2<\/sub> and N<sub>2<\/sub>O<sub>4<\/sub> will change if the total pressure increases to 9.0 atm. Will they increase, decrease, or remain the same?<\/li>\n<li>Calculate the partial pressures of NO<sub>2<\/sub> and N<sub>2<\/sub>O<sub>4<\/sub> when they are at equilibrium at 9.0 atm and 25 \u00b0C.<\/li>\n<\/ol>\n<\/li>\n<li>In a 3.0-L vessel, the following equilibrium partial pressures are measured: N<sub>2<\/sub>, 190 torr; H<sub>2<\/sub>, 317 torr; NH<sub>3<\/sub>, 1.00 \u00d7 10<sup>3<\/sup> torr:\u00a0[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]\n<ol style=\"list-style-type: lower-alpha;\">\n<li>How will the partial pressures of H<sub>2<\/sub>, N<sub>2<\/sub>, and NH<sub>3<\/sub> change if H<sub>2<\/sub> is removed from the system? Will they increase, decrease, or remain the same?<\/li>\n<li>Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.<\/li>\n<\/ol>\n<\/li>\n<li>The equilibrium constant (<em>K<sub>c<\/sub><\/em>) for this reaction is 5.0 at a given temperature:\u00a0[latex]\\text{CO}\\left(g\\right)+{\\text{H}}_{2}\\text{O}\\left(g\\right)\\rightleftharpoons{\\text{CO}}_{2}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right)[\/latex]\n<ol style=\"list-style-type: lower-alpha;\">\n<li>On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H<sub>2<\/sub> in a liter. How many moles of CO<sub>2<\/sub> were there in the equilibrium mixture?<\/li>\n<li>Maintaining the same temperature, additional H<sub>2<\/sub> was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H<sub>2<\/sub> in a liter. How many moles of CO<sub>2<\/sub> were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.<\/li>\n<\/ol>\n<\/li>\n<li>Antimony pentachloride decomposes according to this equation:\u00a0[latex]{\\text{SbCl}}_{5}\\left(g\\right)\\rightleftharpoons{\\text{SbCl}}_{3}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)[\/latex]<br \/>\nAn equilibrium mixture in a 5.00-L flask at 448 \u00b0C contains 3.85 g of SbCl<sub>5<\/sub>, 9.14 g of SbCl<sub>3<\/sub>, and 2.84 g of Cl<sub>2<\/sub>. How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?<\/li>\n<li>Consider the reaction between H<sub>2<\/sub> and O<sub>2<\/sub> at 1000 K<br \/>\n[latex]2H_{2}(g)+O_{2}(g)\\rightleftharpoons{2H_{2}O(g)}[\/latex] [latex]K_{P}=\\frac{(P_{H_{2}O})^{2}}{(P_{O_{2}})(P_{H_{2}})^{3}}=1.33\\times{10^{20}}[\/latex]<br \/>\nIf 0.500 atm of H2 and 0.500 atm of O2 are allowed to come to equilibrium at this temperature, what are the partial pressures of the components?<\/li>\n<li>An equilibrium is established according to the following equation<br \/>\n[latex]{\\text{Hg}}_{2}{}^{2+}\\left(aq\\right)+{\\text{NO}}_{3}{}^{-}\\left(aq\\right)+3{\\text{H}}^{+}\\left(aq\\right)\\rightleftharpoons2{\\text{Hg}}^{2+}\\left(aq\\right)+{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right){K}_{c}=4.6[\/latex]<br \/>\nWhat will happen in a solution that is 0.20 <em>M<\/em> each in [latex]{\\text{Hg}}_{2}{}^{2+}[\/latex], [latex]{\\text{NO}}_{3}{}^{-}[\/latex], H<sup>+<\/sup>, Hg<sup>2+<\/sup>, and HNO<sub>2<\/sub>?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\text{Hg}}_{2}{}^{2+}[\/latex] will be oxidized and [latex]{\\text{NO}}_{3}{}^{-}[\/latex] reduced.<\/li>\n<li>[latex]{\\text{Hg}}_{2}{}^{2+}[\/latex] will be reduced and [latex]{\\text{NO}}_{3}{}^{-}[\/latex] oxidized.<\/li>\n<li>Hg<sup>2+<\/sup> will be oxidized and HNO<sub>2<\/sub> reduced.<\/li>\n<li>Hg<sup>2+<\/sup> will be reduced and HNO<sub>2<\/sub> oxidized.<\/li>\n<li>There will be no change, because all reactants and products have an activity of 1.<\/li>\n<\/ol>\n<\/li>\n<li>Consider the equilibrium: [latex]4{\\text{NO}}_{2}\\left(g\\right)+6{\\text{H}}_{2}\\text{O}\\left(g\\right)\\rightleftharpoons4{\\text{NH}}_{3}\\left(g\\right)+7{\\text{O}}_{2}\\left(g\\right)[\/latex]\n<ol style=\"list-style-type: lower-alpha;\">\n<li>What is the expression for the equilibrium constant (<em>K<sub>c<\/sub><\/em>) of the reaction?<\/li>\n<li>How must the concentration of NH<sub>3<\/sub> change to reach equilibrium if the reaction quotient is less than the equilibrium constant?<\/li>\n<li>If the reaction were at equilibrium, how would a decrease in pressure (from an increase in the volume of the reaction vessel) affect the pressure of NO<sub>2<\/sub>?<\/li>\n<li>If the change in the pressure of NO<sub>2<\/sub> is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O<sub>2<\/sub> change?<\/li>\n<\/ol>\n<\/li>\n<li>The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO<sub>2<\/sub>), is partially regulated by the concentration of H<sub>3<\/sub>O<sup>+<\/sup> and dissolved CO<sub>2<\/sub> in the blood. Although the equilibrium is complicated, it can be summarized as<br \/>\n[latex]{\\text{HbO}}_{2}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CO}}_{2}-\\text{Hb}-{\\text{H}}^{+}+{\\text{O}}_{2}\\left(g\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Write the equilibrium constant expression for this reaction.<\/li>\n<li>Explain why the production of lactic acid and CO<sub>2<\/sub> in a muscle during exertion stimulates release of O<sub>2<\/sub> from the oxyhemoglobin in the blood passing through the muscle.<\/li>\n<\/ol>\n<\/li>\n<li>The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.<br \/>\n[latex]{\\text{C}}_{12}{\\text{H}}_{22}{\\text{O}}_{11}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightarrow{\\text{C}}_{6}{\\text{H}}_{12}{\\text{O}}_{6}\\left(aq\\right)+{\\text{C}}_{6}{\\text{H}}_{12}{\\text{O}}_{6}\\left(aq\\right)[\/latex]<br \/>\nRate = <em>k<\/em>[C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>]<br \/>\nIn neutral solution, <em>k<\/em> = 2.1 \u00d7 10<sup>-11<\/sup>\/s at 27 \u00b0C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation\u2014the products of the reaction, glucose and fructose, have the same molecular formulas, C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 \u00d7 10<sup>5<\/sup> at 27 \u00b0C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 <em>M<\/em> aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1).<\/li>\n<li>The density of trifluoroacetic acid vapor was determined at 118.1 \u00b0C and 468.5 torr, and found to be 2.784 g\/L. Calculate <em>K<sub>c<\/sub><\/em> for the association of the acid.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212440\/CNX_Chem_13_04_trifluoro_img1.jpg\" alt=\"Two Lewis structures are shown in a reaction. The first structure, which is condensed, reads, \u201c2 C F subscript 3 C O subscript 2 H ( g ),\u201d and is followed by a double-headed arrow. The second structure shows a partially condensed hexagonal ring shape. From the left side, in a clockwise manner, it reads \u201cC F subscript 3 C, single bond, O, single bond, H, dotted line bond, O, double bond, C F subscript 3 C ( g ), single bond, O, single bond, H, dotted line bond, O, double bond back to the starting compound.\u201d\" \/><\/li>\n<li>Liquid N<sub>2<\/sub>O<sub>3<\/sub> is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO<sub>2<\/sub>. At 25 \u00b0C, a value of <em>K<sub>P<\/sub><\/em> = 1.91 has been established for this decomposition. If 0.236 moles of N<sub>2<\/sub>O<sub>3<\/sub> are placed in a 1.52-L vessel at 25 \u00b0C, calculate the equilibrium partial pressures of N<sub>2<\/sub>O<sub>3<\/sub>(<em>g<\/em>), NO<sub>2<\/sub>(<em>g<\/em>), and NO(<em>g<\/em>).<\/li>\n<li>A 1.00-L vessel at 400 \u00b0C contains the following equilibrium concentrations: N<sub>2<\/sub>, 1.00 <em>M<\/em>; H<sub>2<\/sub>, 0.50 <em>M<\/em>; and NH<sub>3<\/sub>, 0.25 <em>M<\/em>. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 <em>M<\/em>?<\/li>\n<li>A 0.010 <em>M<\/em> solution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 \u00b0C. A 0.010 <em>M<\/em> solution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Which acid has the larger equilibrium constant for ionization<br \/>\nHA [latex]\\left[\\text{HA}\\left(aq\\right)\\rightleftharpoons{\\text{A}}^{-}\\left(aq\\right)+{\\text{H}}^{+}\\left(aq\\right)\\right][\/latex] or HB [latex]\\left[\\text{HB}\\left(aq\\right)\\rightleftharpoons{\\text{H}}^{+}\\left(aq\\right)+{\\text{B}}^{-}\\left(aq\\right)\\right][\/latex] ?<\/li>\n<li>What are the equilibrium constants for the ionization of these acids?<br \/>\n(Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the conjugate base (A<sup>&#8211;<\/sup> or B<sup>&#8211;<\/sup>), and the hydrogen ion (H<sup>+<\/sup>). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q580949\">Show Selected Solutions<\/span><\/p>\n<div id=\"q580949\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\u00a0(a) The reaction is [latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)\\text{.}[\/latex] At equilibrium, [latex]{K}_{P}=\\frac{\\left({P}_{{\\text{N}}_{2}{\\text{O}}_{4}}\\right)}{{\\left({P}_{{\\text{NO}}_{2}}\\right)}^{2}}\\text{.}[\/latex] The value of <em>K<sub>P<\/sub><\/em> must remain the same when the pressure increases to 9.0 atm. Both gases must increase in pressure.<\/p>\n<p>(b) [latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)\\text{.}[\/latex] At equilibrium, [latex]{K}_{P}=\\frac{{P}_{{\\text{N}}_{2}{\\text{O}}_{4}}}{{\\left({P}_{{\\text{NO}}_{2}}\\right)}^{2}}=\\frac{0.70}{{\\left(0.30\\right)}^{2}}=\\frac{0.70}{0.09}=7.78.[\/latex] For a total pressure of 9.0 atm, the pressure of N<sub>2<\/sub>O<sub>4<\/sub> is 9.0 &#8211; <em>x<\/em>; that of NO is <em>x<\/em>.<\/p>\n<p>[latex]{K}_{P}=\\frac{9.0-x}{{x}^{2}}=7.78[\/latex]<\/p>\n<p>7.78<em>x<\/em><sup>2<\/sup> + <em>x<\/em> \u2212 9.0 = 0<\/p>\n<p>Use the quadratic expression, where<\/p>\n<p>[latex]x=\\frac{\\text{-}b\\pm \\sqrt{{b}^{2}-4ac}}{2a}=\\frac{-1\\pm \\sqrt{1+4\\left(7.78\\right)\\left(-9.0\\right)}}{2\\left(7.78\\right)}.[\/latex]<\/p>\n<p>[latex]=\\frac{-1\\pm 16.765}{15.56}=1.013[\/latex]<\/p>\n<p>For the plus sign (the negative sign gives a negative pressure which is impossible)<\/p>\n<p>[latex]x=\\frac{15.765}{15.56}=1.013[\/latex]<\/p>\n<p>The other answer is extraneous. The pressures are [latex]{P}_{{\\text{N}}_{2}{\\text{O}}_{4}}[\/latex] = 8.0 atm and [latex]{P}_{{\\text{NO}}_{2}}[\/latex] = 1.0 atm<\/p>\n<p>3.\u00a0(a) For the above reaction, [latex]{K}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]\\left[{\\text{H}}_{2}\\right]}{\\left[\\text{CO}\\right]\\left[{\\text{H}}_{2}\\text{O}\\right]}=5.0.[\/latex] The concentrations at equilibrium are 0.20 <em>M<\/em> CO, 0.30 <em>M<\/em> H<sub>2<\/sub>O, and 0.90 <em>M<\/em> H<sub>2<\/sub>. Substitution gives [latex]K=5.0=\\frac{\\left[{\\text{CO}}_{2}\\right]\\left[0.90\\right]}{\\left[0.20\\right]\\left[0.30\\right]};[\/latex] [latex]\\left[{\\text{CO}}_{2}\\right]=\\frac{5.0\\left(0.20\\right)\\left(0.30\\right)}{0.90}=0.33M;[\/latex] Amount of CO<sub>2<\/sub> = 0.33 mol \u00d7 1 = 0.33 mol.<\/p>\n<p>(b) At the particular temperature of reaction, <em>K<sub>c<\/sub><\/em> remains constant at 5.0. The new concentrations are 0.40 <em>M<\/em> CO, 0.30 <em>M<\/em> H<sub>2<\/sub>O, and 1.2 <em>M<\/em> H<sub>2<\/sub>.<\/p>\n<p>[latex]\\frac{\\left[{\\text{CO}}_{2}\\right]\\left(1.2\\right)}{\\left(0.40\\right)\\left(0.30\\right)}=5.0[\/latex]<\/p>\n<p>[latex]{\\left[\\text{CO}\\right]}^{2}=0.50M[\/latex]<\/p>\n<p>Amount of CO<sub>2<\/sub> = 0.50 mol \u00d7 1 = 0.50 mol. Added H<sub>2<\/sub> forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H<sub>2<\/sub> is added.<\/p>\n<p>5.\u00a0P<sub>H<sub>2<\/sub><\/sub> = 8.64 \u00d7 10<sup>\u221211<\/sup>atm<\/p>\n<p>P<sub>O<sub>2<\/sub><\/sub> = 0.250atm<\/p>\n<p>P<sub>H<sub>2<\/sub>O<\/sub> = 0.500 atm<\/p>\n<p>7.\u00a0(a) [latex]{K}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{4}{\\left[{\\text{O}}_{2}\\right]}^{7}}{{\\left[{\\text{NO}}_{2}\\right]}^{4}{\\left[{\\text{H}}_{2}\\text{O}\\right]}^{6}}\\text{.}[\/latex] (b) Because [NH<sub>3<\/sub>] is in the numerator of <em>K<sub>c<\/sub><\/em>, [NH<sub>3<\/sub>] must increase for <em>Q<sub>c<\/sub><\/em> to reach <em>K<sub>c<\/sub><\/em>. (c) That decrease in pressure would decrease [NO<sub>2<\/sub>] because the NO<sub>2<\/sub> must convert to NH<sub>3<\/sub> to reduce the effects of the expansion and the consequent relative decrease in products. (d) The relative pressures are controlled by the stoichiometry of the reaction.<\/p>\n<p>[latex]{P}_{{\\text{O}}_{2}}=\\frac{7}{4}{P}_{{\\text{NO}}_{2}}=\\frac{7}{4}\\left(28\\text{torr}\\right)=49\\text{torr}[\/latex]<\/p>\n<p>9.\u00a0[fructose] = 0.15 <em>M<\/em><\/p>\n<p>11. Write the balanced equilibrium expression. With all of the species as gases, it is a straightforward <em>K<sub>P<\/sub><\/em> problem to solve. However, <em>all<\/em> species must be converted to pressures, from other units related to concentration. For N<sub>2<\/sub>O<sub>3<\/sub>, with 0.236 mol in 1.52 L at 25 \u00baC: <em>PV<\/em> = <em>nRT<\/em><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}P& =& \\frac{n}{V}RT\\\\ & =& \\frac{0.236\\cancel{\\text{mol}}}{1.52\\cancel{\\text{L}}}\\times \\frac{\\left(0.08206\\cancel{\\text{L}}\\text{atm}\\right)\\left(298.15\\cancel{\\text{K}}\\right)}{\\cancel{\\text{mol}}\\cancel{\\text{K}}}\\\\ & =& 3.80\\text{atm}\\end{array}[\/latex]<\/p>\n<p>Write the balanced equation and the equilibrium changes:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212443\/CNX_Chem_13_04_ICETable26_img1.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( a t m ), Change ( a t m ), Equilibrium pressure ( a t m ). The second column has the header, \u201cN subscript 2 O subscript 3 ( g ) equilibrium arrow N O ( g ) plus N O subscript 2 ( g ).\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 3.80, negative x, 3.80 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.\" width=\"879\" height=\"238\" \/><\/p>\n<p style=\"text-align: center;\">[latex]{K}_{P}=\\frac{\\left({P}_{\\text{NO}}\\right)\\left({P}_{{\\text{NO}}_{2}}\\right)}{\\left({P}_{{\\text{N}}_{2}{\\text{O}}_{3}}\\right)}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]1.91\\text{atm}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(3.80-x\\right)}[\/latex]<\/p>\n<p>Because 3.80 is substantial when compared with the equilibrium constant, the value of X must be considered. A quadratic equation must be solved:<\/p>\n<p style=\"text-align: center;\">[latex]1.91=\\frac{{x}^{2}}{\\left(3.80-x\\right)}[\/latex]<\/p>\n<p style=\"text-align: center;\">7.258 \u2212 1.91<em>x<\/em> = <em>x<\/em><sup>2<\/sup><\/p>\n<p style=\"text-align: center;\">0 = <em>x<\/em><sup>2<\/sup> +1.91<em>x<\/em>\u00a0\u2013 7.258<\/p>\n<p style=\"text-align: center;\">0 = <em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}x& =& \\frac{\\text{-}b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\\\ & =& \\frac{-1.91\\pm \\sqrt{{\\left(1.91\\right)}^{2}-4\\left(1\\right)\\left(-7.258\\right)}}{2\\left(1\\right)}\\\\ & =& \\frac{-1.91\\pm \\sqrt{3.648+29.032}}{2}\\\\ & =& \\frac{-1.91\\pm \\sqrt{32.680}}{2}\\\\ & =& \\frac{-1.91\\pm 5.717}{2}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">= 1.90 atm or -3.81 atm<\/p>\n<p>As negative pressure is not possible in this case, we use the positive value only. The final pressures are: [latex]{P}_{{\\text{N}}_{2}{\\text{O}}_{3}}[\/latex] = 3.80 \u2212 <em>x<\/em> = 3.80 \u2212 1.90 = 1.90 atm and [latex]{P}_{\\text{NO}}={P}_{{\\text{NO}}_{2}}={P}_{{\\text{NO}}_{2}}x=1.90\\text{atm}[\/latex]<\/p>\n<p>13. \u00a0(a) Recall that osmotic pressure is a <em>colligative<\/em> property, depending on the total number of particles present in the system. The osmotic pressure equation is \u03c0 = <em>iMRT<\/em>. Given that the two acids face identical reaction conditions, the osmotic pressure (\u03c0) will increase only with a corresponding (&gt;1) in the degree of ionization (<em>i<\/em>). Because both acids produce two ions per molecule dissociated, the number of particles in solution depends on the size of their respective equilibrium constants (<em>K<sub>c<\/sub><\/em>(HA) and <em>K<sub>c<\/sub><\/em>(HB)):<\/p>\n<p style=\"text-align: center;\">[latex]\\text{HA}\\rightleftharpoons{\\text{H}}^{+}+{\\text{A}}^{-}\\text{HB}\\rightleftharpoons{\\text{H}}^{+}+{\\text{B}}^{-}[\/latex]<\/p>\n<p>Because HB has a greater osmotic pressure (0.345 atm) than HA (0.293 atm), HB must produce more particles. Therefore, HB ionizes to a greater degree and has the larger <em>K<sub>c<\/sub><\/em>.<\/p>\n<p>(b) Determine the value of <em>i<\/em> for each acid. For HA,<\/p>\n<p style=\"text-align: center;\">[latex]\\pi =iMRT=0.293\\text{atm}=i\\left(0.010M\\right)\\left(\\frac{0.08206\\text{L atm}}{\\text{mol K}}\\right)\\left(298.15\\text{K}\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>i<\/em><sub>HY<\/sub> = 1.2<\/p>\n<p style=\"text-align: center;\">For HB,<\/p>\n<p style=\"text-align: center;\">[latex]\\pi =iMRT=0.345\\text{atm}=i\\left(0.010M\\right)\\left(\\frac{0.08206\\text{L atm}}{\\text{mol K}}\\right)\\left(298.15\\text{K}\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>i<\/em><sub>HZ<\/sub> = 1.4<\/p>\n<p>The <em>i<\/em> values mean that for every mole of originally undissociated species, <em>i<\/em> moles particles are produced in solution. The dissociations can be expressed in tabular form. For HA,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212444\/CNX_Chem_13_04_ICETable27_img1.jpg\" alt=\"This table has two main columns and four fours. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium pressure ( M ). The second column has the header, \u201cH A equilibrium H superscript positive sign plus A superscript negative sign.\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.100, negative x, 0.100 minus x. The second column has the following: 0, positive x, x. The third column has the following: 0, positive x, x.\" width=\"879\" height=\"238\" \/><\/p>\n<p style=\"text-align: center;\">0.012 <em>M<\/em> = <em>xM<\/em> + <em>xM<\/em> + (0.010 \u2212 <em>x<\/em>) <em>M<\/em><\/p>\n<p style=\"text-align: center;\"><em>x<\/em> = 0.002 <em>M<\/em><\/p>\n<p>For HB, the only difference from the table for HA is the value of <em>i<\/em><sub>HB<\/sub>. <em>x<\/em> = 0.004 <em>M<\/em>.<\/p>\n<p>Now calculate the values of <em>K<sub>c<\/sub><\/em>(HA) and <em>K<sub>c<\/sub><\/em>(HB).<\/p>\n<p>For HA,<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{c}\\left(\\text{HA}\\right)=\\frac{\\left[x\\right]\\left[x\\right]}{\\left[0.010-x\\right]}=\\frac{\\left(0.002\\right)\\left(0.002\\right)}{0.008}=5\\times {10}^{-4}[\/latex]<\/p>\n<p>For HB,<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{c}\\left(\\text{HB}\\right)=\\frac{\\left[x\\right]\\left[x\\right]}{\\left[0.010-x\\right]}=\\frac{\\left(0.004\\right)\\left(0.004\\right)}{0.006}=3\\times {10}^{-3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2290\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry 2e. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/\">https:\/\/openstax.org\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry 2e\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2290","chapter","type-chapter","status-publish","hentry"],"part":2989,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/2290","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":26,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/2290\/revisions"}],"predecessor-version":[{"id":7691,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/2290\/revisions\/7691"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/parts\/2989"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/2290\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/media?parent=2290"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapter-type?post=2290"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/contributor?post=2290"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/license?post=2290"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}