{"id":2257,"date":"2015-04-22T21:07:52","date_gmt":"2015-04-22T21:07:52","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=2257"},"modified":"2020-12-28T17:59:23","modified_gmt":"2020-12-28T17:59:23","slug":"equilibrium-constants","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/chapter\/equilibrium-constants\/","title":{"raw":"Equilibrium Constants","rendered":"Equilibrium Constants"},"content":{"raw":"<div>\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions<\/li>\r\n \t<li>Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures<\/li>\r\n \t<li>Relate the magnitude of an equilibrium constant to properties of the chemical system<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe status of a reversible reaction is conveniently assessed by evaluating its\u00a0<strong>reaction quotient (<em>Q<\/em>)<\/strong>. For a reversible reaction described by\r\n<p style=\"text-align: center;\">[latex]m\\text{A}+n\\text{B}+\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]<\/p>\r\nthe reaction quotient is derived directly from the stoichiometry of the balanced equation as\r\n<p style=\"text-align: center;\">[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[\\text{C}\\right]}^{x}{\\left[\\text{D}\\right]}^{y}}{{\\left[\\text{A}\\right]}^{m}{\\left[\\text{B}\\right]}^{n}}[\/latex]<\/p>\r\n<p id=\"fs-idm7959200\">where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:<\/p>\r\n\r\n<center>[latex]Q_p=\\dfrac{\\text{P}^x_C\\text{P}^y_D}{\\text{P}^m_A\\text{P}^n_B}[\/latex]<\/center>\r\n<p id=\"fs-idm352046000\">Note that the reaction quotient equations above are a simplification of more rigorous expressions that use\u00a0<em>relative<\/em>\u00a0values for concentrations and pressures rather than\u00a0<em>absolute<\/em>\u00a0values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing\u00a0<em>Q<\/em>. In most cases, this will introduce only modest errors in calculations involving reaction quotients.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Writing Reaction Quotient Expressions<\/h3>\r\nWrite the expression for the reaction quotient for each of the following reactions:\r\n<ol>\r\n \t<li>[latex]3{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{O}}_{3}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]4{\\text{NH}}_{3}\\left(g\\right)+7{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons4{\\text{NO}}_{2}\\left(g\\right)+6{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"549251\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"549251\"]\r\n<ol>\r\n \t<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{O}_{3}\\right]}^{2}}{{\\left[{O}_{2}\\right]}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{\\text{NO}}_{2}\\right]}^{4}{\\left[{\\text{H}}_{2}\\text{O}\\right]}^{6}}{{\\left[{\\text{NH}}_{3}\\right]}^{4}{\\left[{\\text{O}}_{2}\\right]}^{7}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nWrite the expression for the reaction quotient for each of the following reactions:\r\n<ol>\r\n \t<li>[latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{C}}_{4}{\\text{H}}_{8}\\left(g\\right)\\rightleftharpoons2{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]2{\\text{C}}_{4}{\\text{H}}_{10}\\left(g\\right)+13{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons8{\\text{CO}}_{2}\\left(g\\right)+10{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"418960\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"418960\"]\r\n<ol>\r\n \t<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{\\text{SO}}_{3}\\right]}^{2}}{{\\left[{\\text{SO}}_{2}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{\\text{C}}_{2}{\\text{H}}_{4}\\right]}^{2}}{\\left[{\\text{C}}_{4}{\\text{H}}_{8}\\right]}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{\\text{CO}}_{2}\\right]}^{8}{\\left[{\\text{H}}_{2}\\text{O}\\right]}^{10}}{{\\left[{\\text{C}}_{4}{\\text{H}}_{10}\\right]}^{2}{\\left[{\\text{O}}_{2}\\right]}^{13}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n[caption id=\"\" align=\"alignnone\" width=\"1578\"]<img src=\"https:\/\/openstax.org\/resources\/0d8e21124b0a1678b7934ee5640581db37567ac5\" alt=\"Four graphs are shown and labeled, \u201ca,\u201d \u201cb,\u201d \u201cc,\u201d and \u201cd.\u201d All four graphs have a vertical dotted line running through the middle labeled, \u201cEquilibrium is reached.\u201d The y-axis on graph a is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph a. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts high on the y-axis, ends midway down the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it starts and ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph b is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph b. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts low on the y-axis, ends midway up the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph c is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph c. This curve begins at the bottom of the y-axis and rises steeply up near the top of the y-axis, then levels off into a horizontal line. The top point of this line is labeled, \u201ck.\u201d The y-axis on graph d is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph d. This curve begins near the edge of the top of the y-axis and falls steeply toward the x-axis, then levels off into a horizontal line. The bottom point of this line is labeled, \u201ck.\u201d\" width=\"1578\" height=\"1353\" \/> FIgure 1. Changes in concentrations and Qc for a chemical equilibrium achieved beginning with (a) a mixture of reactants only and (b) products only.[\/caption]\r\nThe numerical value of\u00a0<em>Q<\/em>\u00a0varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction\u2019s status. To illustrate this point, consider the oxidation of sulfur dioxide:<center>[latex]2\\text{SO}_2(g) + \\text{O}_2(g) \\rightleftharpoons 2\\text{SO}_3(g) [\/latex]<\/center>\r\n<p id=\"fs-idp138846544\">Two different experimental scenarios are depicted in Figure 1, one in which this reaction is initiated with a mixture of reactants only, SO<sub>2<\/sub>\u00a0and O<sub>2<\/sub>, and another that begins with only product, SO<sub>3<\/sub>. For the reaction that begins with a mixture of reactants only,\u00a0<em>Q<\/em>\u00a0is initially equal to zero:<\/p>\r\n\r\n<center>[latex]Q_c = \\dfrac{[\\text{SO}_3]^2}{[\\text{SO}_2]^2[\\text{O}_2]} = \\dfrac{0^2}{[\\text{SO}_2]^2[\\text{O}_2]} = 0 [\/latex]<\/center>\r\n<p id=\"fs-idm375739520\">As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of\u00a0<em>Q<sub>c<\/sub><\/em>), product concentration increases (as does the numerator of\u00a0<em>Q<sub>c<\/sub><\/em>), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of\u00a0<em>Q<sub>c<\/sub><\/em>.<\/p>\r\n<p id=\"fs-idm378445456\">If the reaction begins with only product present, the value of\u00a0<em>Q<sub>c<\/sub><\/em>\u00a0is initially undefined (immeasurably large, or infinite):<\/p>\r\n\r\n<center>[latex]Q_c = \\dfrac{[\\text{SO}_3]^2}{[\\text{SO}_2]^2[\\text{O}_2]} = \\dfrac{[\\text{SO}_3]^2}{0} \\rightarrow \\infty [\/latex]<\/center>\r\n<p id=\"fs-idm374784256\">In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of\u00a0<em>Q<sub>c<\/sub><\/em>\u00a0decrease with time, the reactant concentrations and the denominator of\u00a0<em>Q<sub>c<\/sub><\/em>\u00a0increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.<\/p>\r\nThe constant value of\u00a0<em>Q<\/em>\u00a0exhibited by a system at equilibrium is called the\u00a0<strong>equilibrium constant<\/strong>,\u00a0<strong><em>K<\/em><\/strong>:\r\n\r\n<center>[latex]K=Q \\qquad\\text{at equilibrium}[\/latex]<\/center>\r\n<p id=\"fs-idm328721280\">Comparison of the data plots in Figure 1 shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the\u00a0<strong>law of mass action<\/strong>: At a given temperature, the reaction quotient for a system at equilibrium is constant.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0Evaluating a Reaction Quotient<\/h3>\r\nGaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:\r\n<p style=\"text-align: center;\">[latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)[\/latex]<\/p>\r\nWhen 0.10 mol NO<sub>2<\/sub> is added to a 1.0-L flask at 25 \u00b0C, the concentration changes so that at equilibrium, [NO<sub>2<\/sub>] = 0.016 <em>M<\/em> and [N<sub>2<\/sub>O<sub>4<\/sub>] = 0.042 <em>M<\/em>.\r\n<ol>\r\n \t<li>What is the value of the reaction quotient before any reaction occurs?<\/li>\r\n \t<li>What is the value of the equilibrium constant for the reaction?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"219715\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"219715\"]\r\n<ol>\r\n \t<li>Before any product is formed, [latex]\\left[{\\text{NO}}_{2}\\right]=\\dfrac{0.10\\text{mol}}{1.0\\text{L}}=0.10M[\/latex], and [N<sub>2<\/sub>O<sub>4<\/sub>] = 0 <em>M<\/em>. Thus,\r\n[latex]{Q}_{c}=\\dfrac{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}=\\dfrac{0}{{0.10}^{2}}=0[\/latex]<\/li>\r\n \t<li>At equilibrium, the value of the equilibrium constant is equal to the value of the reaction quotient. At equilibrium, [latex]{K}_{c}={Q}_{c}=\\dfrac{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}=\\dfrac{0.042}{{0.016}^{2}}=1.6\\times {10}^{2}\\text{.}[\/latex] The equilibrium constant is 1.6 \u00d7 10<sup>2<\/sup>.<\/li>\r\n<\/ol>\r\nNote that dimensional analysis would suggest the unit for this <em>K<sub>c<\/sub><\/em> value should be <em>M<\/em><sup>\u22121<\/sup>. However, it is common practice to omit units for <em>K<sub>c<\/sub><\/em> values computed as described here, since it is the magnitude of an equilibrium constant that relays useful information. As will be discussed later in this module, the rigorous approach to computing equilibrium constants uses dimensionless quantities derived from concentrations instead of actual concentrations, and so <em>K<sub>c<\/sub><\/em> values are truly unitless.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nFor the reaction [latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right)[\/latex], the concentrations at equilibrium are [SO<sub>2<\/sub>] = 0.90 <em>M<\/em>, [O<sub>2<\/sub>] = 0.35 <em>M<\/em>, and [SO<sub>3<\/sub>] = 1.1 <em>M<\/em>. What is the value of the equilibrium constant, <em>K<sub>c<\/sub><\/em>?\r\n\r\n[reveal-answer q=\"417701\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"417701\"]<em>K<sub>c<\/sub> =<\/em> 4.3[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-idm80100288\">By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large\u00a0<em>K<\/em>\u00a0will reach equilibrium when most of the reactant has been converted to product, whereas a small\u00a0<em>K<\/em>\u00a0indicates the reaction achieves equilibrium after very little reactant has been converted. It\u2019s important to keep in mind that the magnitude of\u00a0<em>K<\/em>\u00a0does\u00a0<em>not<\/em>\u00a0indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.<\/p>\r\n<p id=\"fs-idm377606544\">The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and\/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing\u00a0<em>Q<\/em>\u00a0to\u00a0<em>K<\/em>\u00a0for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.<\/p>\r\n<p id=\"fs-idm323833328\">To further illustrate this important point, consider the reversible reaction shown below:<\/p>\r\n<p style=\"text-align: center;\">[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)\\qquad{K}_{c}=0.640\\qquad\\text{T}=800^{\\circ}\\text{C}[\/latex]<\/p>\r\nThe bar charts in Figure 2 represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction\u2019s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"1300\"]<img src=\"https:\/\/openstax.org\/resources\/b5b006532a3ad063e3740dc609ab8ae5870ed3b6\" alt=\"Two sets of bar graphs are shown. The left is labeled, \u201cBefore reaction,\u201d and the right is labeled, \u201cAt equilibrium.\u201d Both graphs have y-axes labeled, \u201cConcentration ( M ),\u201d and three bars on the x-axes labeled, \u201cMixture 1,\u201d \u201cMixture 2,\u201d and \u201cMixture 3.\u201d The y-axis has a scale beginning at 0.00 and ending at 0.10, with measurement increments of 0.02. The bars on the graphs are color coded and a key is provided with a legend. Red is labeled, \u201cC O;\u201d blue is labeled, \u201cH subscript 2 O;\u201d green is labeled, \u201cC O subscript 2,\u201d and yellow is labeled, \u201cH subscript 2.\u201d The graph on the left shows the red bar for mixture one just above 0.02, labeled \u201c0.0243,\u201d and the blue bar near 0.05, labeled \u201c0.0243.\u201d For mixture two, the green bar is near 0.05, labeled \u201c0.0468,\u201d and the yellow bar is near 0.09, labeled \u201c0.0468.\u201d For mixture 3, the red bar is near 0.01, labeled \u201c0.0330,\u201d the blue bar is slightly above that, labeled \u201c0.190,\u201d with green and yellow topping it off at 0.02. Green is labeled \u201c0.00175\u201d and yellow is labeled \u201c0.00160.\u201d On the right graph, the bar for mixture one shows the red bar slightly above 0.01, labeled \u201c0.0135,\u201d the blue bar stacked on it rising slightly above 0.02, labeled \u201c0.0135,\u201d the green rising near 0.04, labeled \u201c0.0108,\u201d and the yellow bar reaching near 0.05, labeled \u201c0.0108.\u201d A label above this bar reads, \u201cQ equals 0.640.\u201d The bar for mixture two shows the red bar slightly above 0.02, labeled \u201c0.0260,\u201d the blue bar stacked on it rising near 0.05, labeled \u201c0.0260,\u201d the green rising near 0.07, labeled \u201c0.0208,\u201d and the yellow bar reaching near 0.10, labeled \u201c0.0208.\u201d A label above this bar reads \u201cQ equals 0.640.\u201d The bar for mixture three shows the red bar near 0.01, labeled \u201c0.0231,\u201d the blue bar stacked on it rising slightly above 0.01, labeled \u201c0.00909,\u201d the green rising near 0.02, labeled \u201c0.0115,\u201d and the yellow bar reaching 0.02, labeled \u201c0.0117.\u201d A label above this bar reads \u201cQ equals 0.640\u201d.\" width=\"1300\" height=\"617\" \/> Figure 2. Figure 13.6 Compositions of three mixtures before (Q<sub>c<\/sub> \u2260 K<sub>c<\/sub>) and after (Q<sub>c<\/sub> = K<sub>c<\/sub>) equilibrium is established for the reaction CO(g) + H<sub>2<\/sub>O(g) \u21cc CO<sub>2<\/sub>(g) + H<sub>2<\/sub>(g).[\/caption]\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0Predicting the Direction of Reaction<\/h3>\r\nGiven below are the starting concentrations of reactants and products for three experiments involving this reaction:\r\n<p style=\"text-align: center;\">[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[latex]{K}_{c}=0.64[\/latex]<\/p>\r\nDetermine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.\r\n<table id=\"fs-idp70024256\" class=\"medium unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row, and it labels each column, \u201cReactants \/ Products,\u201d \u201cExperiment 1,\u201d \u201cExperiment 2,\u201d and \u201cExperiment 3.\u201d Under the \u201cReactants \/ Products\u201d column are: [ C O ] subscript i; [ H subscript 2 O ] subscript i; [ C O subscript 2 ] subscript i; [ H subscript 2 ] subscript i. Under the \u201cExperiment 1\u201d column are the numbers: 0.0203 M; 0.0203 M; 0.0040 M; and 0.0040 M. Under the \u201cExperiment 2\u201d column are the numbers: 0.011 M; 0.0011 M; 0.037 M; and 0.046 M. Under the \u201cExperiment 3\u201d column are the numbers: 0.0094 M; 0.0025 M; 0.0015 M; 0.0076 M.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Reactants\/Products<\/th>\r\n<th>Experiment 1<\/th>\r\n<th>Experiment 2<\/th>\r\n<th>Experiment 3<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[CO]<sub>i<\/sub><\/td>\r\n<td>0.0203 <em>M<\/em><\/td>\r\n<td>0.011 <em>M<\/em><\/td>\r\n<td>0.0094 <em>M<\/em><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[H<sub>2<\/sub>O]<sub>i<\/sub><\/td>\r\n<td>0.0203 <em>M<\/em><\/td>\r\n<td>0.0011 <em>M<\/em><\/td>\r\n<td>0.0025 <em>M<\/em><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[CO<sub>2<\/sub>]<sub>i<\/sub><\/td>\r\n<td>0.0040 <em>M<\/em><\/td>\r\n<td>0.037 <em>M<\/em><\/td>\r\n<td>0.0015 <em>M<\/em><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[H<sub>2<\/sub>]<sub>i<\/sub><\/td>\r\n<td>0.0040 <em>M<\/em><\/td>\r\n<td>0.046 <em>M<\/em><\/td>\r\n<td>0.0076 <em>M<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"289231\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"289231\"]\r\n<h4>Experiment 1<\/h4>\r\n[latex]\\displaystyle{Q}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]\\left[{\\text{H}}_{2}\\right]}{\\left[\\text{CO}\\right]\\left[{\\text{H}}_{2}\\text{O}\\right]}=\\frac{\\left(0.0040\\right)\\left(0.0040\\right)}{\\left(0.0203\\right)\\left(0.0203\\right)}=0.039\\text{.}[\/latex]\r\n\r\n<em>Q<sub>c<\/sub><\/em> &lt; <em>K<sub>c<\/sub><\/em> (0.039 &lt; 0.64)\r\n\r\nThe reaction will shift to the right in the forward direction.\r\n<h4>Experiment 2<\/h4>\r\n[latex]\\displaystyle{Q}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]\\left[{\\text{H}}_{2}\\right]}{\\left[\\text{CO}\\right]\\left[{\\text{H}}_{2}\\text{O}\\right]}=\\frac{\\left(0.037\\right)\\left(0.046\\right)}{\\left(0.011\\right)\\left(0.0011\\right)}=1.4\\times {10}^{2}[\/latex]\r\n\r\n<em>Q<sub>c<\/sub><\/em> &gt; <em>K<sub>c<\/sub><\/em> (140 &gt; 0.64)\r\n\r\nThe reaction will shift to the left in the reverse direction.\r\n<h4>Experiment 3<\/h4>\r\n[latex]\\displaystyle{Q}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]\\left[{\\text{H}}_{2}\\right]}{\\left[\\text{CO}\\right]\\left[{\\text{H}}_{2}\\text{O}\\right]}=\\frac{\\left(0.0015\\right)\\left(0.0076\\right)}{\\left(0.0094\\right)\\left(0.0025\\right)}=0.48[\/latex]\r\n\r\n<em>Q<sub>c<\/sub><\/em> &lt; <em>K<sub>c<\/sub><\/em> (0.48 &lt; 0.64)\r\n\r\nThe reaction will shift to the right.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nCalculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.\r\n<ol>\r\n \t<li>A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl:\r\n[latex]2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NOCl}\\left(g\\right)\\qquad{K}_{c}=4.6\\times {10}^{4}[\/latex]<\/li>\r\n \t<li>A 5.0-L flask containing 17 g of NH<sub>3<\/sub>, 14 g of N<sub>2<\/sub>, and 12 g of H<sub>2<\/sub>:\r\n[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)\\qquad{K}_{c}=0.060[\/latex]<\/li>\r\n \t<li>A 2.00-L flask containing 230 g of SO<sub>3<\/sub>(g):\r\n[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\qquad{K}_{c}=0.230[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"45817\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"45817\"]\r\n<ol>\r\n \t<li><em>Q<sub>c<\/sub><\/em> = 6.45 \u00d7 10<sup>3<\/sup>, shifts right\/forward<\/li>\r\n \t<li><em>Q<sub>c<\/sub><\/em> = 0.23, shifts left\/reverse<\/li>\r\n \t<li><em>Q<sub>c<\/sub><\/em> = 0, shifts right\/forward<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Homogeneous Equilibria<\/h2>\r\nA homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). By this definition, homogenous equilibria take place in <em>solutions<\/em>. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:\r\n<ul>\r\n \t<li>[latex]{\\text{C}}_{2}{\\text{H}}_{2}\\left(aq\\right)+2{\\text{Br}}_{2}\\left(aq\\right)\\rightleftharpoons{\\text{C}}_{2}{\\text{H}}_{2}{\\text{Br}}_{4}\\left(aq\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{C}}_{2}{\\text{H}}_{2}{\\text{Br}}_{4}\\right]}{\\left[{\\text{C}}_{2}{\\text{H}}_{2}\\right]{\\left[{\\text{Br}}_{2}\\right]}^{2}}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{I}}_{2}\\left(aq\\right)+{\\text{I}}^{-}\\left(aq\\right)\\rightleftharpoons{\\text{I}}_{3}{}^{-}\\left(aq\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{I}}_{3}{}^{-}\\right]}{\\left[{\\text{I}}_{2}\\right]\\left[{\\text{I}}^{-}\\right]}[\/latex]<\/li>\r\n \t<li>[latex]\\text{HF}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{F}}^{\\text{-}}\\left(aq\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{F}}^{-}\\right]}{\\left[\\text{HF}\\right]}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{NH}}_{4}{}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}[\/latex]<\/li>\r\n<\/ul>\r\n<p id=\"fs-idp11719264\">These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is\u00a0<em>not<\/em>\u00a0included in the reaction quotient. The reason for this omission is related to the more rigorous form of the\u00a0<em>Q<\/em>\u00a0(or\u00a0<em>K<\/em>) expression mentioned previously in this chapter, in which\u00a0<em>relative concentrations for liquids and solids are equal to 1 and needn\u2019t be included<\/em>. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.<\/p>\r\n<p id=\"fs-idp11805840\">The equilibria below all involve gas-phase solutions:<\/p>\r\n\r\n<ul>\r\n \t<li>[latex]{\\text{C}}_{2}{\\text{H}}_{6}\\left(g\\right)\\rightleftharpoons{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{C}}_{2}{\\text{H}}_{4}\\right]\\left[{\\text{H}}_{2}\\right]}{\\left[{\\text{C}}_{2}{\\text{H}}_{6}\\right]}[\/latex]<\/li>\r\n \t<li>[latex]3{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{O}}_{3}\\left(g\\right)\\qquad{K}_{c}=\\frac{{\\left[{\\text{O}}_{3}\\right]}^{2}}{{\\left[{\\text{O}}_{2}\\right]}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)\\qquad{K}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons3{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(g\\right)\\qquad{K}_{c}=\\frac{{\\left[{\\text{CO}}_{2}\\right]}^{3}{\\left[{\\text{H}}_{2}\\text{O}\\right]}^{4}}{\\left[{\\text{C}}_{3}{\\text{H}}_{8}\\right]{\\left[{\\text{O}}_{2}\\right]}^{5}}[\/latex]<\/li>\r\n<\/ul>\r\nNote that the concentration of H<sub>2<\/sub>O(<em>g<\/em>) has been included in the last example because water is not the solvent in this gas-phase reaction and its concentration (and activity) changes.\r\n\r\nFor gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (<em>K<sub>c<\/sub><\/em>) or partial pressures (<em>K<sub>p<\/sub><\/em>) of the reactants and products. A relation between these two\u00a0<em>K<\/em>\u00a0values may be simply derived from the ideal gas equation and the definition of molarity:\r\n<p style=\"text-align: center;\">[latex]\\begin{array} { }PV&amp;=&amp;nRT \\\\ P&amp;=&amp;\\left(\\dfrac{n}{V}\\right)RT\\\\ &amp;=&amp;MRT\\end{array}[\/latex]<\/p>\r\nwhere P is partial pressure, V is volume, n is molar amount, R is the gas constant, T is temperature, and M is molar concentration.\r\n\r\nFor the gas-phase reaction [latex]m\\text{A}+n\\text{B}\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ }{K}_{P}&amp;=&amp;\\dfrac{{\\left({P}_{C}\\right)}^{x}{\\left({P}_{D}\\right)}^{y}}{{\\left({P}_{A}\\right)}^{m}{\\left({P}_{B}\\right)}^{n}}\\\\ &amp;=&amp;\\dfrac{{\\left(\\left[\\text{C}\\right]\\times RT\\right)}^{x}{\\left(\\left[\\text{D}\\right]\\times RT\\right)}^{y}}{{\\left(\\left[\\text{A}\\right]\\times RT\\right)}^{m}{\\left(\\left[\\text{B}\\right]\\times RT\\right)}^{n}}\\\\ &amp;=&amp;\\dfrac{{\\left[\\text{C}\\right]}^{x}{\\left[\\text{D}\\right]}^{y}}{{\\left[\\text{A}\\right]}^{m}{\\left[\\text{B}\\right]}^{n}}\\times \\dfrac{{\\left(RT\\right)}^{x+y}}{{\\left(RT\\right)}^{m+n}}\\\\ &amp;=&amp;{K}_{c}{\\left(RT\\right)}^{\\left(x+y\\right)-\\left(m+n\\right)}\\\\ &amp;=&amp;{K}_{c}{\\left(RT\\right)}^{\\Delta n}\\end{array}[\/latex]<\/p>\r\nAnd so, the relationship between <em>K<sub>c<\/sub><\/em> and <em>K<sub>P<\/sub><\/em> is\r\n<p style=\"text-align: center;\">[latex]{K}_{P}={K}_{c}{\\left(RT\\right)}^{\\Delta n}[\/latex]<\/p>\r\nwhere \u0394<em>n<\/em> is the difference between the sum of the molar amounts of product and reactant fases, in this case:\r\n<p style=\"text-align: center;\">[latex]\\Delta n=\\left(x+y\\right)-\\left(m+n\\right)[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 4:\u00a0Calculation of <em>K<sub>P<\/sub><\/em><\/h3>\r\nWrite the equations for the conversion of <em>K<sub>c<\/sub><\/em> to <em>K<sub>P<\/sub><\/em> for each of the following reactions:\r\n<ol>\r\n \t<li>[latex]{\\text{C}}_{2}{\\text{H}}_{6}\\left(g\\right)\\rightleftharpoons{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[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]<\/li>\r\n \t<li>[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/li>\r\n \t<li><em>K<sub>c<\/sub><\/em> is equal to 0.28 for the following reaction at 900 \u00b0C:\u00a0[latex]{\\text{CS}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CH}}_{4}\\left(g\\right)+2{\\text{H}}_{2}\\text{S}\\left(g\\right)[\/latex]\r\nWhat is <em>K<sub>P<\/sub><\/em> at this temperature?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"63243\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"63243\"]\r\n<ol>\r\n \t<li>\u0394<em>n<\/em> = (2) \u2212 (1) = 1 <em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>\u0394<em>n<\/em><\/sup> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>1<\/sup> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<\/li>\r\n \t<li>\u0394<em>n<\/em> = (2) \u2212 (2) = 0 <em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>\u0394<em>n<\/em><\/sup> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>0<\/sup> = <em>K<sub>c<\/sub><\/em><\/li>\r\n \t<li>\u0394<em>n<\/em> = (2) \u2212 (1 + 3) = -2 <em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>\u0394<em>n<\/em><\/sup> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>\u22122<\/sup> = [latex]\\frac{{K}_{c}}{{\\left(RT\\right)}^{2}}[\/latex]<\/li>\r\n \t<li><em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em> (RT) <sup>\u0394<em>n<\/em><\/sup> = (0.28)[(0.0821)(1173)]<sup>\u22122<\/sup> = 3.0 \u00d7 10<sup>\u22125<\/sup><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nWrite the equations for the conversion of <em>K<sub>c<\/sub><\/em> to <em>K<sub>P<\/sub><\/em> for each of the following reactions, which occur in the gas phase:\r\n<ol>\r\n \t<li>[latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)\\rightleftharpoons2{\\text{NO}}_{2}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons3{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>At 227 \u00b0C, the following reaction has <em>K<sub>c<\/sub><\/em> = 0.0952:\u00a0[latex]{\\text{CH}}_{3}\\text{OH}\\left(g\\right)\\rightleftharpoons\\text{CO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)[\/latex]\r\nWhat would be the value of <em>K<sub>P<\/sub><\/em> at this temperature?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"849860\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"849860\"]\r\n<ol>\r\n \t<li><em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em>(<em>RT<\/em>)<sup>\u22121<\/sup><\/li>\r\n \t<li><em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em>(<em>RT<\/em>)<\/li>\r\n \t<li><em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em>(<em>RT<\/em>)<\/li>\r\n \t<li>160 or 1.6 \u00d7 10<sup>2<\/sup><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Heterogeneous Equilibria<\/h2>\r\nA\u00a0<span id=\"term568\">heterogeneous equilibrium<\/span>\u00a0involves reactants and products in two or more different phases, as illustrated by the following examples:\r\n<ul>\r\n \t<li>[latex]{\\text{PbCl}}_{2}\\left(s\\right)\\rightleftharpoons{\\text{Pb}}^{2+}\\left(aq\\right)+2{\\text{Cl}}^{-}\\left(aq\\right)\\qquad{K}_{c}=\\left[{\\text{Pb}}^{2+}\\right]{\\left[{\\text{Cl}}^{-}\\right]}^{2}[\/latex]<\/li>\r\n \t<li>[latex]\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CaCO}}_{3}\\left(s\\right)\\qquad{K}_{c}=\\frac{1}{\\left[{\\text{CO}}_{2}\\right]}[\/latex]<\/li>\r\n \t<li>[latex]\\text{C}\\left(s\\right)+2\\text{S}\\left(g\\right)\\rightleftharpoons{\\text{CS}}_{2}\\left(g\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{CS}}_{2}\\right]}{{\\left[\\text{S}\\right]}^{2}}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{Br}}_{2}\\left(l\\right)\\rightleftharpoons{\\text{Br}}_{2}\\left(g\\right)\\qquad{K}_{c}=\\left[{\\text{Br}}_{2}\\right][\/latex]<\/li>\r\n<\/ul>\r\nTwo of the above examples include terms for gaseous species only in their equilibrium constants, and so\u00a0<em>K<sub>p<\/sub><\/em>\u00a0expressions may also be written:\r\n<ul>\r\n \t<li>[latex]\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CaCO}}_{3}\\left(s\\right)\\qquad{K}_{P}=\\frac{1}{{P}_{{\\text{CO}}_{2}}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{C}\\left(s\\right)+2\\text{S}\\left(g\\right)\\rightleftharpoons{\\text{CS}}_{2}\\left(g\\right)\\qquad{K}_{P}=\\frac{{P}_{C{S}_{2}}}{{\\left({P}_{S}\\right)}^{2}}[\/latex]<\/li>\r\n<\/ul>\r\n<h3>Coupled Equilibria<\/h3>\r\n<p id=\"fs-idm374944688\">The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more\u00a0<em>coupled<\/em>\u00a0equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.<\/p>\r\n<p id=\"fs-idm328298096\">1. Changing the direction of a chemical equation essentially swaps the identities of \u201creactants\u201d and \u201cproducts,\u201d and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.<\/p>\r\n\r\n<center>[latex] \\text{A} \\rightleftharpoons \\text{B} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_c = \\dfrac{[B]}{[A]} [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{B} \\rightleftharpoons \\text{A} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_c' = \\dfrac{[A]}{[B]} [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{K}_{c'} = \\dfrac{1}{\\text{K}_c} [\/latex]<\/center>\r\n<p id=\"fs-idm336251504\">2. Changing the stoichiometric coefficients in an equation by some factor\u00a0<em>x<\/em>\u00a0results in an exponential change in the equilibrium constant by that same factor:<\/p>\r\n\r\n<center>[latex] \\text{A} \\rightleftharpoons \\text{B} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_{c} = \\dfrac{[B]}{[A]} [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{xA} \\rightleftharpoons \\text{xB} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_c' = \\dfrac{[B]^x}{[A]^x} [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{K}_{c'} = \\text{K}_c^x [\/latex]<\/center>\r\n<p id=\"fs-idm351795744\">3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction\u2019s K values:<\/p>\r\n\r\n<center>[latex] \\text{A} \\rightleftharpoons \\text{B} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_{c1} = \\dfrac{[B]}{[A]} [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{B} \\rightleftharpoons \\text{C} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_{c2} = \\dfrac{[C]}{[B]} [\/latex]<\/center>\r\n<p id=\"fs-idm374676944\">The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:<\/p>\r\n\r\n<center>[latex] \\text{A} + \\text{B} \\rightleftharpoons \\text{B} + \\text{C} [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{A} + \\cancel{\\text{B}} \\rightleftharpoons \\cancel{\\text{B}} + \\text{C} [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{A} \\rightleftharpoons \\text{C} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_c' = \\dfrac{[C]}{[A]} [\/latex]<\/center>\r\n<p id=\"fs-idm380030416\">Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:<\/p>\r\n\r\n<center>[latex] \\text{K}_{c1}\\text{K}_{c2} = \\dfrac{[B]}{[A]} \\text{x} \\dfrac{[C]}{[B]} = \\dfrac{[\\cancel{\\text{B}}][C]}{[A][\\cancel{\\text{B}}]} = \\dfrac{[C]}{[A]} = \\text{K}_c' [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{K}_c' = \\text{K}_{c1}\\text{K}_{c2} [\/latex]<\/center>\r\n<p id=\"fs-idm329039744\">Example 5 demonstrates the use of this strategy in describing coupled equilibrium processes.<\/p>\r\n\r\n<div id=\"fs-idm384162720\" class=\"ui-has-child-title\"><header>\r\n<div class=\"textbox examples\"><header>\r\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE\u00a0<\/span><span class=\"os-number\">5:\u00a0<\/span>Equilibrium Constants for Coupled Reactions<\/h3>\r\n<\/header><section>\r\n<div class=\"body\">\r\n\r\nA mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 \u00b0C:\r\n\r\n<center>[latex] 2\\text{NH}_3(g) + 3\\text{I}_2(g) \\rightleftharpoons \\text{N}_2(g) + 6\\text{HI}(g) [\/latex]<\/center>\r\n<p id=\"fs-idm493175904\">Use the information below to calculate Kc for this reaction.<\/p>\r\n\r\n<center>[latex] \\text{N}_2(g) + 3\\text{H}_2(g) \\rightleftharpoons 2\\text{NH}_3(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_{c1} = 0.50 \\text{at } 400^{\\circ} \\text{C} [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{H}_2(g) + \\text{I}_2(g) \\rightleftharpoons 2\\text{HI}(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_{c2} = 50 \\text{at } 400^{\\circ} \\text{C} [\/latex]<\/center>\r\n<h4>Solution<\/h4>\r\n[reveal-answer q=\"549255\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"549255\"]\r\nThe equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.\r\n<p id=\"fs-idm385085648\">Reverse the first coupled reaction equation:<\/p>\r\n\r\n<center>[latex] 2\\text{NH}_3(g) \\rightleftharpoons \\text{N}_2(g) + 3\\text{H}_2(g) [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_{c1'} = \\frac{1}{\\text{K}_{c1}} = \\frac{1}{0.50} = 2.0[\/latex]<\/center>\r\n<p id=\"fs-idm339773312\">Multiply the second coupled reaction by 3:<\/p>\r\n\r\n<center>[latex] 3\\text{H}_2(g) + 3\\text{I}_2(g) \\rightleftharpoons 6\\text{HI}(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_{c2'} = \\text{K}_{c2}^3 = 50^3 = 1.2 \\text{x} 10^5 [\/latex]<\/center>\r\n<p id=\"fs-idm372296992\">Finally, add the two revised equations:<\/p>\r\n\r\n<center>[latex] 2\\text{NH}_3(g) + \\cancel{3\\text{H}_2(g)} + 3\\text{I}_2(g) \\rightleftharpoons \\text{N}_2(g) + \\cancel{3\\text{H}_2(g)} + 6\\text{HI}(g) [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] 2\\text{NH}_3(g) + 3\\text{I}_2(g) \\rightleftharpoons \\text{N}_2(g) + 6\\text{HI}(g) [\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{K}_c = \\text{K}_{c1'}\\text{K}_{c2'} = (2.0)(1.2 \\text{x} 10^5) = 2.5\\text{x} 10^5 [\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/div>\r\n<h4>Check Your Learning<\/h4>\r\nUse the provided information to calculate Kc for the following reaction at 550 \u00b0C:\r\n\r\n<center>[latex] \\text{H}_2(g) + \\text{CO}_2(g) \\rightleftharpoons \\text{CO}(g) + \\text{H}_2\\text{O}(g) [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_c = ?[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{CoO}(s) + \\text{CO}(g) \\rightleftharpoons \\text{Co}(s) + \\text{CO}_2(g) [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_{c1} = 490[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex] \\text{CoO}(s) + \\text{H}_2(g) \\rightleftharpoons \\text{Co}(s) + \\text{H}_2\\text{O}(g) [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex] \\text{K}_{c1} = 67[\/latex]<\/center>\r\n<h4>Answer:<\/h4>\r\n[reveal-answer q=\"549256\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"549256\"]\r\n\r\nK<sub>c<\/sub>\u00a0= 0.14\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>\r\n<\/header><\/div>\r\n<\/div>\r\nhttps:\/\/www.youtube.com\/watch?v=xfGlEXWDRZE\r\n\r\nYou can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/TheEquilibriumConstant_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"The Equilibrium Constant\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\n<p id=\"fs-idp165520928\">The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient,\u00a0<em>Q<\/em>. For a reaction at equilibrium, the composition is constant, and\u00a0<em>Q<\/em>\u00a0is called the equilibrium constant,\u00a0<em>K<\/em>.<\/p>\r\n<p id=\"fs-idp116522496\">A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases.<\/p>\r\n\r\n<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li>[latex]Q=\\dfrac{{\\left[\\text{C}\\right]}^{\\text{x}}{\\left[\\text{D}\\right]}^{\\text{y}}}{{\\left[\\text{A}\\right]}^{\\text{m}}{\\left[\\text{B}\\right]}^{\\text{n}}}\\text{ where }m\\text{A}+n\\text{B}\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]<\/li>\r\n \t<li>[latex]{Q}_{P}=\\dfrac{{\\left({P}_{C}\\right)}^{x}{\\left({P}_{D}\\right)}^{y}}{{\\left({P}_{A}\\right)}^{m}{\\left({P}_{B}\\right)}^{n}}\\text{ where }m\\text{A}+n\\text{B}\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]<\/li>\r\n \t<li>[latex]P=MRT[\/latex]<\/li>\r\n \t<li>[latex]{K}_{P}={K}_{c}{\\left(RT\\right)}^{\\Delta n}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>equilibrium constant (<em>K<\/em>): <\/strong>value of the reaction quotient for a system at equilibrium\r\n\r\n<strong>heterogeneous equilibria: <\/strong>equilibria between reactants and products in different phases\r\n\r\n<strong>homogeneous equilibria: <\/strong>equilibria within a single phase\r\n\r\n<strong><em>K<sub>c<\/sub><\/em>:<em>\u00a0<\/em><\/strong>equilibrium constant for reactions based on concentrations of reactants and products\r\n\r\n<strong><em>K<sub>P<\/sub><\/em>:<\/strong> equilibrium constant for gas-phase reactions based on partial pressures of reactants and products\r\n\r\n<strong>law of mass action: <\/strong>when a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant\r\n\r\n<strong>reaction quotient (<em>Q<\/em>): <\/strong>ratio of the product of molar concentrations (or pressures) of the products to that of the reactants, each concentration (or pressure) being raised to the power equal to the coefficient in the equation","rendered":"<div>\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions<\/li>\n<li>Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures<\/li>\n<li>Relate the magnitude of an equilibrium constant to properties of the chemical system<\/li>\n<\/ul>\n<\/div>\n<p>The status of a reversible reaction is conveniently assessed by evaluating its\u00a0<strong>reaction quotient (<em>Q<\/em>)<\/strong>. For a reversible reaction described by<\/p>\n<p style=\"text-align: center;\">[latex]m\\text{A}+n\\text{B}+\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]<\/p>\n<p>the reaction quotient is derived directly from the stoichiometry of the balanced equation as<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[\\text{C}\\right]}^{x}{\\left[\\text{D}\\right]}^{y}}{{\\left[\\text{A}\\right]}^{m}{\\left[\\text{B}\\right]}^{n}}[\/latex]<\/p>\n<p id=\"fs-idm7959200\">where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:<\/p>\n<div style=\"text-align: center;\">[latex]Q_p=\\dfrac{\\text{P}^x_C\\text{P}^y_D}{\\text{P}^m_A\\text{P}^n_B}[\/latex]<\/div>\n<p id=\"fs-idm352046000\">Note that the reaction quotient equations above are a simplification of more rigorous expressions that use\u00a0<em>relative<\/em>\u00a0values for concentrations and pressures rather than\u00a0<em>absolute<\/em>\u00a0values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing\u00a0<em>Q<\/em>. In most cases, this will introduce only modest errors in calculations involving reaction quotients.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0Writing Reaction Quotient Expressions<\/h3>\n<p>Write the expression for the reaction quotient for each of the following reactions:<\/p>\n<ol>\n<li>[latex]3{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{O}}_{3}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]4{\\text{NH}}_{3}\\left(g\\right)+7{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons4{\\text{NO}}_{2}\\left(g\\right)+6{\\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=\"q549251\">Show Solution<\/span><\/p>\n<div id=\"q549251\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{O}_{3}\\right]}^{2}}{{\\left[{O}_{2}\\right]}^{3}}[\/latex]<\/li>\n<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}[\/latex]<\/li>\n<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{\\text{NO}}_{2}\\right]}^{4}{\\left[{\\text{H}}_{2}\\text{O}\\right]}^{6}}{{\\left[{\\text{NH}}_{3}\\right]}^{4}{\\left[{\\text{O}}_{2}\\right]}^{7}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Write the expression for the reaction quotient for each of the following reactions:<\/p>\n<ol>\n<li>[latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{C}}_{4}{\\text{H}}_{8}\\left(g\\right)\\rightleftharpoons2{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]2{\\text{C}}_{4}{\\text{H}}_{10}\\left(g\\right)+13{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons8{\\text{CO}}_{2}\\left(g\\right)+10{\\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=\"q418960\">Show Solution<\/span><\/p>\n<div id=\"q418960\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{\\text{SO}}_{3}\\right]}^{2}}{{\\left[{\\text{SO}}_{2}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]}[\/latex]<\/li>\n<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{\\text{C}}_{2}{\\text{H}}_{4}\\right]}^{2}}{\\left[{\\text{C}}_{4}{\\text{H}}_{8}\\right]}[\/latex]<\/li>\n<li>[latex]\\displaystyle{Q}_{c}=\\frac{{\\left[{\\text{CO}}_{2}\\right]}^{8}{\\left[{\\text{H}}_{2}\\text{O}\\right]}^{10}}{{\\left[{\\text{C}}_{4}{\\text{H}}_{10}\\right]}^{2}{\\left[{\\text{O}}_{2}\\right]}^{13}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div style=\"width: 1588px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/openstax.org\/resources\/0d8e21124b0a1678b7934ee5640581db37567ac5\" alt=\"Four graphs are shown and labeled, \u201ca,\u201d \u201cb,\u201d \u201cc,\u201d and \u201cd.\u201d All four graphs have a vertical dotted line running through the middle labeled, \u201cEquilibrium is reached.\u201d The y-axis on graph a is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph a. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts high on the y-axis, ends midway down the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it starts and ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph b is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph b. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts low on the y-axis, ends midway up the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph c is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph c. This curve begins at the bottom of the y-axis and rises steeply up near the top of the y-axis, then levels off into a horizontal line. The top point of this line is labeled, \u201ck.\u201d The y-axis on graph d is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph d. This curve begins near the edge of the top of the y-axis and falls steeply toward the x-axis, then levels off into a horizontal line. The bottom point of this line is labeled, \u201ck.\u201d\" width=\"1578\" height=\"1353\" \/><\/p>\n<p class=\"wp-caption-text\">FIgure 1. Changes in concentrations and Qc for a chemical equilibrium achieved beginning with (a) a mixture of reactants only and (b) products only.<\/p>\n<\/div>\n<p>The numerical value of\u00a0<em>Q<\/em>\u00a0varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction\u2019s status. To illustrate this point, consider the oxidation of sulfur dioxide:<\/p>\n<div style=\"text-align: center;\">[latex]2\\text{SO}_2(g) + \\text{O}_2(g) \\rightleftharpoons 2\\text{SO}_3(g)[\/latex]<\/div>\n<p id=\"fs-idp138846544\">Two different experimental scenarios are depicted in Figure 1, one in which this reaction is initiated with a mixture of reactants only, SO<sub>2<\/sub>\u00a0and O<sub>2<\/sub>, and another that begins with only product, SO<sub>3<\/sub>. For the reaction that begins with a mixture of reactants only,\u00a0<em>Q<\/em>\u00a0is initially equal to zero:<\/p>\n<div style=\"text-align: center;\">[latex]Q_c = \\dfrac{[\\text{SO}_3]^2}{[\\text{SO}_2]^2[\\text{O}_2]} = \\dfrac{0^2}{[\\text{SO}_2]^2[\\text{O}_2]} = 0[\/latex]<\/div>\n<p id=\"fs-idm375739520\">As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of\u00a0<em>Q<sub>c<\/sub><\/em>), product concentration increases (as does the numerator of\u00a0<em>Q<sub>c<\/sub><\/em>), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of\u00a0<em>Q<sub>c<\/sub><\/em>.<\/p>\n<p id=\"fs-idm378445456\">If the reaction begins with only product present, the value of\u00a0<em>Q<sub>c<\/sub><\/em>\u00a0is initially undefined (immeasurably large, or infinite):<\/p>\n<div style=\"text-align: center;\">[latex]Q_c = \\dfrac{[\\text{SO}_3]^2}{[\\text{SO}_2]^2[\\text{O}_2]} = \\dfrac{[\\text{SO}_3]^2}{0} \\rightarrow \\infty[\/latex]<\/div>\n<p id=\"fs-idm374784256\">In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of\u00a0<em>Q<sub>c<\/sub><\/em>\u00a0decrease with time, the reactant concentrations and the denominator of\u00a0<em>Q<sub>c<\/sub><\/em>\u00a0increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.<\/p>\n<p>The constant value of\u00a0<em>Q<\/em>\u00a0exhibited by a system at equilibrium is called the\u00a0<strong>equilibrium constant<\/strong>,\u00a0<strong><em>K<\/em><\/strong>:<\/p>\n<div style=\"text-align: center;\">[latex]K=Q \\qquad\\text{at equilibrium}[\/latex]<\/div>\n<p id=\"fs-idm328721280\">Comparison of the data plots in Figure 1 shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the\u00a0<strong>law of mass action<\/strong>: At a given temperature, the reaction quotient for a system at equilibrium is constant.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0Evaluating a Reaction Quotient<\/h3>\n<p>Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:<\/p>\n<p style=\"text-align: center;\">[latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)[\/latex]<\/p>\n<p>When 0.10 mol NO<sub>2<\/sub> is added to a 1.0-L flask at 25 \u00b0C, the concentration changes so that at equilibrium, [NO<sub>2<\/sub>] = 0.016 <em>M<\/em> and [N<sub>2<\/sub>O<sub>4<\/sub>] = 0.042 <em>M<\/em>.<\/p>\n<ol>\n<li>What is the value of the reaction quotient before any reaction occurs?<\/li>\n<li>What is the value of the equilibrium constant for the reaction?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q219715\">Show Solution<\/span><\/p>\n<div id=\"q219715\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Before any product is formed, [latex]\\left[{\\text{NO}}_{2}\\right]=\\dfrac{0.10\\text{mol}}{1.0\\text{L}}=0.10M[\/latex], and [N<sub>2<\/sub>O<sub>4<\/sub>] = 0 <em>M<\/em>. Thus,<br \/>\n[latex]{Q}_{c}=\\dfrac{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}=\\dfrac{0}{{0.10}^{2}}=0[\/latex]<\/li>\n<li>At equilibrium, the value of the equilibrium constant is equal to the value of the reaction quotient. At equilibrium, [latex]{K}_{c}={Q}_{c}=\\dfrac{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}=\\dfrac{0.042}{{0.016}^{2}}=1.6\\times {10}^{2}\\text{.}[\/latex] The equilibrium constant is 1.6 \u00d7 10<sup>2<\/sup>.<\/li>\n<\/ol>\n<p>Note that dimensional analysis would suggest the unit for this <em>K<sub>c<\/sub><\/em> value should be <em>M<\/em><sup>\u22121<\/sup>. However, it is common practice to omit units for <em>K<sub>c<\/sub><\/em> values computed as described here, since it is the magnitude of an equilibrium constant that relays useful information. As will be discussed later in this module, the rigorous approach to computing equilibrium constants uses dimensionless quantities derived from concentrations instead of actual concentrations, and so <em>K<sub>c<\/sub><\/em> values are truly unitless.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>For the reaction [latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right)[\/latex], the concentrations at equilibrium are [SO<sub>2<\/sub>] = 0.90 <em>M<\/em>, [O<sub>2<\/sub>] = 0.35 <em>M<\/em>, and [SO<sub>3<\/sub>] = 1.1 <em>M<\/em>. What is the value of the equilibrium constant, <em>K<sub>c<\/sub><\/em>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q417701\">Show Solution<\/span><\/p>\n<div id=\"q417701\" class=\"hidden-answer\" style=\"display: none\"><em>K<sub>c<\/sub> =<\/em> 4.3<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-idm80100288\">By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large\u00a0<em>K<\/em>\u00a0will reach equilibrium when most of the reactant has been converted to product, whereas a small\u00a0<em>K<\/em>\u00a0indicates the reaction achieves equilibrium after very little reactant has been converted. It\u2019s important to keep in mind that the magnitude of\u00a0<em>K<\/em>\u00a0does\u00a0<em>not<\/em>\u00a0indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.<\/p>\n<p id=\"fs-idm377606544\">The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and\/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing\u00a0<em>Q<\/em>\u00a0to\u00a0<em>K<\/em>\u00a0for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.<\/p>\n<p id=\"fs-idm323833328\">To further illustrate this important point, consider the reversible reaction shown below:<\/p>\n<p style=\"text-align: center;\">[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)\\qquad{K}_{c}=0.640\\qquad\\text{T}=800^{\\circ}\\text{C}[\/latex]<\/p>\n<p>The bar charts in Figure 2 represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction\u2019s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.<\/p>\n<div style=\"width: 1310px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/openstax.org\/resources\/b5b006532a3ad063e3740dc609ab8ae5870ed3b6\" alt=\"Two sets of bar graphs are shown. The left is labeled, \u201cBefore reaction,\u201d and the right is labeled, \u201cAt equilibrium.\u201d Both graphs have y-axes labeled, \u201cConcentration ( M ),\u201d and three bars on the x-axes labeled, \u201cMixture 1,\u201d \u201cMixture 2,\u201d and \u201cMixture 3.\u201d The y-axis has a scale beginning at 0.00 and ending at 0.10, with measurement increments of 0.02. The bars on the graphs are color coded and a key is provided with a legend. Red is labeled, \u201cC O;\u201d blue is labeled, \u201cH subscript 2 O;\u201d green is labeled, \u201cC O subscript 2,\u201d and yellow is labeled, \u201cH subscript 2.\u201d The graph on the left shows the red bar for mixture one just above 0.02, labeled \u201c0.0243,\u201d and the blue bar near 0.05, labeled \u201c0.0243.\u201d For mixture two, the green bar is near 0.05, labeled \u201c0.0468,\u201d and the yellow bar is near 0.09, labeled \u201c0.0468.\u201d For mixture 3, the red bar is near 0.01, labeled \u201c0.0330,\u201d the blue bar is slightly above that, labeled \u201c0.190,\u201d with green and yellow topping it off at 0.02. Green is labeled \u201c0.00175\u201d and yellow is labeled \u201c0.00160.\u201d On the right graph, the bar for mixture one shows the red bar slightly above 0.01, labeled \u201c0.0135,\u201d the blue bar stacked on it rising slightly above 0.02, labeled \u201c0.0135,\u201d the green rising near 0.04, labeled \u201c0.0108,\u201d and the yellow bar reaching near 0.05, labeled \u201c0.0108.\u201d A label above this bar reads, \u201cQ equals 0.640.\u201d The bar for mixture two shows the red bar slightly above 0.02, labeled \u201c0.0260,\u201d the blue bar stacked on it rising near 0.05, labeled \u201c0.0260,\u201d the green rising near 0.07, labeled \u201c0.0208,\u201d and the yellow bar reaching near 0.10, labeled \u201c0.0208.\u201d A label above this bar reads \u201cQ equals 0.640.\u201d The bar for mixture three shows the red bar near 0.01, labeled \u201c0.0231,\u201d the blue bar stacked on it rising slightly above 0.01, labeled \u201c0.00909,\u201d the green rising near 0.02, labeled \u201c0.0115,\u201d and the yellow bar reaching 0.02, labeled \u201c0.0117.\u201d A label above this bar reads \u201cQ equals 0.640\u201d.\" width=\"1300\" height=\"617\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Figure 13.6 Compositions of three mixtures before (Q<sub>c<\/sub> \u2260 K<sub>c<\/sub>) and after (Q<sub>c<\/sub> = K<sub>c<\/sub>) equilibrium is established for the reaction CO(g) + H<sub>2<\/sub>O(g) \u21cc CO<sub>2<\/sub>(g) + H<sub>2<\/sub>(g).<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3:\u00a0Predicting the Direction of Reaction<\/h3>\n<p>Given below are the starting concentrations of reactants and products for three experiments involving this reaction:<\/p>\n<p style=\"text-align: center;\">[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]<br \/>\n[latex]{K}_{c}=0.64[\/latex]<\/p>\n<p>Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.<\/p>\n<table id=\"fs-idp70024256\" class=\"medium unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row, and it labels each column, \u201cReactants \/ Products,\u201d \u201cExperiment 1,\u201d \u201cExperiment 2,\u201d and \u201cExperiment 3.\u201d Under the \u201cReactants \/ Products\u201d column are: [ C O ] subscript i; [ H subscript 2 O ] subscript i; [ C O subscript 2 ] subscript i; [ H subscript 2 ] subscript i. Under the \u201cExperiment 1\u201d column are the numbers: 0.0203 M; 0.0203 M; 0.0040 M; and 0.0040 M. Under the \u201cExperiment 2\u201d column are the numbers: 0.011 M; 0.0011 M; 0.037 M; and 0.046 M. Under the \u201cExperiment 3\u201d column are the numbers: 0.0094 M; 0.0025 M; 0.0015 M; 0.0076 M.\">\n<thead>\n<tr valign=\"top\">\n<th>Reactants\/Products<\/th>\n<th>Experiment 1<\/th>\n<th>Experiment 2<\/th>\n<th>Experiment 3<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[CO]<sub>i<\/sub><\/td>\n<td>0.0203 <em>M<\/em><\/td>\n<td>0.011 <em>M<\/em><\/td>\n<td>0.0094 <em>M<\/em><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[H<sub>2<\/sub>O]<sub>i<\/sub><\/td>\n<td>0.0203 <em>M<\/em><\/td>\n<td>0.0011 <em>M<\/em><\/td>\n<td>0.0025 <em>M<\/em><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[CO<sub>2<\/sub>]<sub>i<\/sub><\/td>\n<td>0.0040 <em>M<\/em><\/td>\n<td>0.037 <em>M<\/em><\/td>\n<td>0.0015 <em>M<\/em><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[H<sub>2<\/sub>]<sub>i<\/sub><\/td>\n<td>0.0040 <em>M<\/em><\/td>\n<td>0.046 <em>M<\/em><\/td>\n<td>0.0076 <em>M<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q289231\">Show Solution<\/span><\/p>\n<div id=\"q289231\" class=\"hidden-answer\" style=\"display: none\">\n<h4>Experiment 1<\/h4>\n<p>[latex]\\displaystyle{Q}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]\\left[{\\text{H}}_{2}\\right]}{\\left[\\text{CO}\\right]\\left[{\\text{H}}_{2}\\text{O}\\right]}=\\frac{\\left(0.0040\\right)\\left(0.0040\\right)}{\\left(0.0203\\right)\\left(0.0203\\right)}=0.039\\text{.}[\/latex]<\/p>\n<p><em>Q<sub>c<\/sub><\/em> &lt; <em>K<sub>c<\/sub><\/em> (0.039 &lt; 0.64)<\/p>\n<p>The reaction will shift to the right in the forward direction.<\/p>\n<h4>Experiment 2<\/h4>\n<p>[latex]\\displaystyle{Q}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]\\left[{\\text{H}}_{2}\\right]}{\\left[\\text{CO}\\right]\\left[{\\text{H}}_{2}\\text{O}\\right]}=\\frac{\\left(0.037\\right)\\left(0.046\\right)}{\\left(0.011\\right)\\left(0.0011\\right)}=1.4\\times {10}^{2}[\/latex]<\/p>\n<p><em>Q<sub>c<\/sub><\/em> &gt; <em>K<sub>c<\/sub><\/em> (140 &gt; 0.64)<\/p>\n<p>The reaction will shift to the left in the reverse direction.<\/p>\n<h4>Experiment 3<\/h4>\n<p>[latex]\\displaystyle{Q}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]\\left[{\\text{H}}_{2}\\right]}{\\left[\\text{CO}\\right]\\left[{\\text{H}}_{2}\\text{O}\\right]}=\\frac{\\left(0.0015\\right)\\left(0.0076\\right)}{\\left(0.0094\\right)\\left(0.0025\\right)}=0.48[\/latex]<\/p>\n<p><em>Q<sub>c<\/sub><\/em> &lt; <em>K<sub>c<\/sub><\/em> (0.48 &lt; 0.64)<\/p>\n<p>The reaction will shift to the right.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.<\/p>\n<ol>\n<li>A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl:<br \/>\n[latex]2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NOCl}\\left(g\\right)\\qquad{K}_{c}=4.6\\times {10}^{4}[\/latex]<\/li>\n<li>A 5.0-L flask containing 17 g of NH<sub>3<\/sub>, 14 g of N<sub>2<\/sub>, and 12 g of H<sub>2<\/sub>:<br \/>\n[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)\\qquad{K}_{c}=0.060[\/latex]<\/li>\n<li>A 2.00-L flask containing 230 g of SO<sub>3<\/sub>(g):<br \/>\n[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\qquad{K}_{c}=0.230[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q45817\">Show Solution<\/span><\/p>\n<div id=\"q45817\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li><em>Q<sub>c<\/sub><\/em> = 6.45 \u00d7 10<sup>3<\/sup>, shifts right\/forward<\/li>\n<li><em>Q<sub>c<\/sub><\/em> = 0.23, shifts left\/reverse<\/li>\n<li><em>Q<sub>c<\/sub><\/em> = 0, shifts right\/forward<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Homogeneous Equilibria<\/h2>\n<p>A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). By this definition, homogenous equilibria take place in <em>solutions<\/em>. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:<\/p>\n<ul>\n<li>[latex]{\\text{C}}_{2}{\\text{H}}_{2}\\left(aq\\right)+2{\\text{Br}}_{2}\\left(aq\\right)\\rightleftharpoons{\\text{C}}_{2}{\\text{H}}_{2}{\\text{Br}}_{4}\\left(aq\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{C}}_{2}{\\text{H}}_{2}{\\text{Br}}_{4}\\right]}{\\left[{\\text{C}}_{2}{\\text{H}}_{2}\\right]{\\left[{\\text{Br}}_{2}\\right]}^{2}}[\/latex]<\/li>\n<li>[latex]{\\text{I}}_{2}\\left(aq\\right)+{\\text{I}}^{-}\\left(aq\\right)\\rightleftharpoons{\\text{I}}_{3}{}^{-}\\left(aq\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{I}}_{3}{}^{-}\\right]}{\\left[{\\text{I}}_{2}\\right]\\left[{\\text{I}}^{-}\\right]}[\/latex]<\/li>\n<li>[latex]\\text{HF}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{F}}^{\\text{-}}\\left(aq\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{F}}^{-}\\right]}{\\left[\\text{HF}\\right]}[\/latex]<\/li>\n<li>[latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{NH}}_{4}{}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-idp11719264\">These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is\u00a0<em>not<\/em>\u00a0included in the reaction quotient. The reason for this omission is related to the more rigorous form of the\u00a0<em>Q<\/em>\u00a0(or\u00a0<em>K<\/em>) expression mentioned previously in this chapter, in which\u00a0<em>relative concentrations for liquids and solids are equal to 1 and needn\u2019t be included<\/em>. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.<\/p>\n<p id=\"fs-idp11805840\">The equilibria below all involve gas-phase solutions:<\/p>\n<ul>\n<li>[latex]{\\text{C}}_{2}{\\text{H}}_{6}\\left(g\\right)\\rightleftharpoons{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{C}}_{2}{\\text{H}}_{4}\\right]\\left[{\\text{H}}_{2}\\right]}{\\left[{\\text{C}}_{2}{\\text{H}}_{6}\\right]}[\/latex]<\/li>\n<li>[latex]3{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{O}}_{3}\\left(g\\right)\\qquad{K}_{c}=\\frac{{\\left[{\\text{O}}_{3}\\right]}^{2}}{{\\left[{\\text{O}}_{2}\\right]}^{3}}[\/latex]<\/li>\n<li>[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)\\qquad{K}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}[\/latex]<\/li>\n<li>[latex]{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons3{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(g\\right)\\qquad{K}_{c}=\\frac{{\\left[{\\text{CO}}_{2}\\right]}^{3}{\\left[{\\text{H}}_{2}\\text{O}\\right]}^{4}}{\\left[{\\text{C}}_{3}{\\text{H}}_{8}\\right]{\\left[{\\text{O}}_{2}\\right]}^{5}}[\/latex]<\/li>\n<\/ul>\n<p>Note that the concentration of H<sub>2<\/sub>O(<em>g<\/em>) has been included in the last example because water is not the solvent in this gas-phase reaction and its concentration (and activity) changes.<\/p>\n<p>For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (<em>K<sub>c<\/sub><\/em>) or partial pressures (<em>K<sub>p<\/sub><\/em>) of the reactants and products. A relation between these two\u00a0<em>K<\/em>\u00a0values may be simply derived from the ideal gas equation and the definition of molarity:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array} { }PV&=&nRT \\\\ P&=&\\left(\\dfrac{n}{V}\\right)RT\\\\ &=&MRT\\end{array}[\/latex]<\/p>\n<p>where P is partial pressure, V is volume, n is molar amount, R is the gas constant, T is temperature, and M is molar concentration.<\/p>\n<p>For the gas-phase reaction [latex]m\\text{A}+n\\text{B}\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ }{K}_{P}&=&\\dfrac{{\\left({P}_{C}\\right)}^{x}{\\left({P}_{D}\\right)}^{y}}{{\\left({P}_{A}\\right)}^{m}{\\left({P}_{B}\\right)}^{n}}\\\\ &=&\\dfrac{{\\left(\\left[\\text{C}\\right]\\times RT\\right)}^{x}{\\left(\\left[\\text{D}\\right]\\times RT\\right)}^{y}}{{\\left(\\left[\\text{A}\\right]\\times RT\\right)}^{m}{\\left(\\left[\\text{B}\\right]\\times RT\\right)}^{n}}\\\\ &=&\\dfrac{{\\left[\\text{C}\\right]}^{x}{\\left[\\text{D}\\right]}^{y}}{{\\left[\\text{A}\\right]}^{m}{\\left[\\text{B}\\right]}^{n}}\\times \\dfrac{{\\left(RT\\right)}^{x+y}}{{\\left(RT\\right)}^{m+n}}\\\\ &=&{K}_{c}{\\left(RT\\right)}^{\\left(x+y\\right)-\\left(m+n\\right)}\\\\ &=&{K}_{c}{\\left(RT\\right)}^{\\Delta n}\\end{array}[\/latex]<\/p>\n<p>And so, the relationship between <em>K<sub>c<\/sub><\/em> and <em>K<sub>P<\/sub><\/em> is<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{P}={K}_{c}{\\left(RT\\right)}^{\\Delta n}[\/latex]<\/p>\n<p>where \u0394<em>n<\/em> is the difference between the sum of the molar amounts of product and reactant fases, in this case:<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta n=\\left(x+y\\right)-\\left(m+n\\right)[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4:\u00a0Calculation of <em>K<sub>P<\/sub><\/em><\/h3>\n<p>Write the equations for the conversion of <em>K<sub>c<\/sub><\/em> to <em>K<sub>P<\/sub><\/em> for each of the following reactions:<\/p>\n<ol>\n<li>[latex]{\\text{C}}_{2}{\\text{H}}_{6}\\left(g\\right)\\rightleftharpoons{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/li>\n<li>[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]<\/li>\n<li>[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/li>\n<li><em>K<sub>c<\/sub><\/em> is equal to 0.28 for the following reaction at 900 \u00b0C:\u00a0[latex]{\\text{CS}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CH}}_{4}\\left(g\\right)+2{\\text{H}}_{2}\\text{S}\\left(g\\right)[\/latex]<br \/>\nWhat is <em>K<sub>P<\/sub><\/em> at this temperature?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q63243\">Show Solution<\/span><\/p>\n<div id=\"q63243\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>\u0394<em>n<\/em> = (2) \u2212 (1) = 1 <em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>\u0394<em>n<\/em><\/sup> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>1<\/sup> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<\/li>\n<li>\u0394<em>n<\/em> = (2) \u2212 (2) = 0 <em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>\u0394<em>n<\/em><\/sup> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>0<\/sup> = <em>K<sub>c<\/sub><\/em><\/li>\n<li>\u0394<em>n<\/em> = (2) \u2212 (1 + 3) = -2 <em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>\u0394<em>n<\/em><\/sup> = <em>K<sub>c<\/sub><\/em> (<em>RT<\/em>)<sup>\u22122<\/sup> = [latex]\\frac{{K}_{c}}{{\\left(RT\\right)}^{2}}[\/latex]<\/li>\n<li><em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em> (RT) <sup>\u0394<em>n<\/em><\/sup> = (0.28)[(0.0821)(1173)]<sup>\u22122<\/sup> = 3.0 \u00d7 10<sup>\u22125<\/sup><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Write the equations for the conversion of <em>K<sub>c<\/sub><\/em> to <em>K<sub>P<\/sub><\/em> for each of the following reactions, which occur in the gas phase:<\/p>\n<ol>\n<li>[latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)\\rightleftharpoons2{\\text{NO}}_{2}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons3{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\n<li>At 227 \u00b0C, the following reaction has <em>K<sub>c<\/sub><\/em> = 0.0952:\u00a0[latex]{\\text{CH}}_{3}\\text{OH}\\left(g\\right)\\rightleftharpoons\\text{CO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)[\/latex]<br \/>\nWhat would be the value of <em>K<sub>P<\/sub><\/em> at this temperature?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q849860\">Show Solution<\/span><\/p>\n<div id=\"q849860\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li><em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em>(<em>RT<\/em>)<sup>\u22121<\/sup><\/li>\n<li><em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em>(<em>RT<\/em>)<\/li>\n<li><em>K<sub>P<\/sub><\/em> = <em>K<sub>c<\/sub><\/em>(<em>RT<\/em>)<\/li>\n<li>160 or 1.6 \u00d7 10<sup>2<\/sup><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Heterogeneous Equilibria<\/h2>\n<p>A\u00a0<span id=\"term568\">heterogeneous equilibrium<\/span>\u00a0involves reactants and products in two or more different phases, as illustrated by the following examples:<\/p>\n<ul>\n<li>[latex]{\\text{PbCl}}_{2}\\left(s\\right)\\rightleftharpoons{\\text{Pb}}^{2+}\\left(aq\\right)+2{\\text{Cl}}^{-}\\left(aq\\right)\\qquad{K}_{c}=\\left[{\\text{Pb}}^{2+}\\right]{\\left[{\\text{Cl}}^{-}\\right]}^{2}[\/latex]<\/li>\n<li>[latex]\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CaCO}}_{3}\\left(s\\right)\\qquad{K}_{c}=\\frac{1}{\\left[{\\text{CO}}_{2}\\right]}[\/latex]<\/li>\n<li>[latex]\\text{C}\\left(s\\right)+2\\text{S}\\left(g\\right)\\rightleftharpoons{\\text{CS}}_{2}\\left(g\\right)\\qquad{K}_{c}=\\frac{\\left[{\\text{CS}}_{2}\\right]}{{\\left[\\text{S}\\right]}^{2}}[\/latex]<\/li>\n<li>[latex]{\\text{Br}}_{2}\\left(l\\right)\\rightleftharpoons{\\text{Br}}_{2}\\left(g\\right)\\qquad{K}_{c}=\\left[{\\text{Br}}_{2}\\right][\/latex]<\/li>\n<\/ul>\n<p>Two of the above examples include terms for gaseous species only in their equilibrium constants, and so\u00a0<em>K<sub>p<\/sub><\/em>\u00a0expressions may also be written:<\/p>\n<ul>\n<li>[latex]\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CaCO}}_{3}\\left(s\\right)\\qquad{K}_{P}=\\frac{1}{{P}_{{\\text{CO}}_{2}}}[\/latex]<\/li>\n<li>[latex]\\text{C}\\left(s\\right)+2\\text{S}\\left(g\\right)\\rightleftharpoons{\\text{CS}}_{2}\\left(g\\right)\\qquad{K}_{P}=\\frac{{P}_{C{S}_{2}}}{{\\left({P}_{S}\\right)}^{2}}[\/latex]<\/li>\n<\/ul>\n<h3>Coupled Equilibria<\/h3>\n<p id=\"fs-idm374944688\">The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more\u00a0<em>coupled<\/em>\u00a0equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.<\/p>\n<p id=\"fs-idm328298096\">1. Changing the direction of a chemical equation essentially swaps the identities of \u201creactants\u201d and \u201cproducts,\u201d and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\text{A} \\rightleftharpoons \\text{B}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_c = \\dfrac{[B]}{[A]}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{B} \\rightleftharpoons \\text{A}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_c' = \\dfrac{[A]}{[B]}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{K}_{c'} = \\dfrac{1}{\\text{K}_c}[\/latex]<\/div>\n<p id=\"fs-idm336251504\">2. Changing the stoichiometric coefficients in an equation by some factor\u00a0<em>x<\/em>\u00a0results in an exponential change in the equilibrium constant by that same factor:<\/p>\n<div style=\"text-align: center;\">[latex]\\text{A} \\rightleftharpoons \\text{B}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_{c} = \\dfrac{[B]}{[A]}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{xA} \\rightleftharpoons \\text{xB}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_c' = \\dfrac{[B]^x}{[A]^x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{K}_{c'} = \\text{K}_c^x[\/latex]<\/div>\n<p id=\"fs-idm351795744\">3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction\u2019s K values:<\/p>\n<div style=\"text-align: center;\">[latex]\\text{A} \\rightleftharpoons \\text{B}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_{c1} = \\dfrac{[B]}{[A]}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{B} \\rightleftharpoons \\text{C}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_{c2} = \\dfrac{[C]}{[B]}[\/latex]<\/div>\n<p id=\"fs-idm374676944\">The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:<\/p>\n<div style=\"text-align: center;\">[latex]\\text{A} + \\text{B} \\rightleftharpoons \\text{B} + \\text{C}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{A} + \\cancel{\\text{B}} \\rightleftharpoons \\cancel{\\text{B}} + \\text{C}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{A} \\rightleftharpoons \\text{C}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_c' = \\dfrac{[C]}{[A]}[\/latex]<\/div>\n<p id=\"fs-idm380030416\">Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:<\/p>\n<div style=\"text-align: center;\">[latex]\\text{K}_{c1}\\text{K}_{c2} = \\dfrac{[B]}{[A]} \\text{x} \\dfrac{[C]}{[B]} = \\dfrac{[\\cancel{\\text{B}}][C]}{[A][\\cancel{\\text{B}}]} = \\dfrac{[C]}{[A]} = \\text{K}_c'[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{K}_c' = \\text{K}_{c1}\\text{K}_{c2}[\/latex]<\/div>\n<p id=\"fs-idm329039744\">Example 5 demonstrates the use of this strategy in describing coupled equilibrium processes.<\/p>\n<div id=\"fs-idm384162720\" class=\"ui-has-child-title\">\n<header>\n<div class=\"textbox examples\"><\/div>\n<\/header>\n<header>\n<h3 class=\"os-title\"><span class=\"os-title-label\">EXAMPLE\u00a0<\/span><span class=\"os-number\">5:\u00a0<\/span>Equilibrium Constants for Coupled Reactions<\/h3>\n<\/header>\n<section>\n<div class=\"body\">\n<p>A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 \u00b0C:<\/p>\n<div style=\"text-align: center;\">[latex]2\\text{NH}_3(g) + 3\\text{I}_2(g) \\rightleftharpoons \\text{N}_2(g) + 6\\text{HI}(g)[\/latex]<\/div>\n<p id=\"fs-idm493175904\">Use the information below to calculate Kc for this reaction.<\/p>\n<div style=\"text-align: center;\">[latex]\\text{N}_2(g) + 3\\text{H}_2(g) \\rightleftharpoons 2\\text{NH}_3(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_{c1} = 0.50 \\text{at } 400^{\\circ} \\text{C}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{H}_2(g) + \\text{I}_2(g) \\rightleftharpoons 2\\text{HI}(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_{c2} = 50 \\text{at } 400^{\\circ} \\text{C}[\/latex]<\/div>\n<h4>Solution<\/h4>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q549255\">Show Solution<\/span><\/p>\n<div id=\"q549255\" class=\"hidden-answer\" style=\"display: none\">\nThe equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.<\/p>\n<p id=\"fs-idm385085648\">Reverse the first coupled reaction equation:<\/p>\n<div style=\"text-align: center;\">[latex]2\\text{NH}_3(g) \\rightleftharpoons \\text{N}_2(g) + 3\\text{H}_2(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_{c1'} = \\frac{1}{\\text{K}_{c1}} = \\frac{1}{0.50} = 2.0[\/latex]<\/div>\n<p id=\"fs-idm339773312\">Multiply the second coupled reaction by 3:<\/p>\n<div style=\"text-align: center;\">[latex]3\\text{H}_2(g) + 3\\text{I}_2(g) \\rightleftharpoons 6\\text{HI}(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_{c2'} = \\text{K}_{c2}^3 = 50^3 = 1.2 \\text{x} 10^5[\/latex]<\/div>\n<p id=\"fs-idm372296992\">Finally, add the two revised equations:<\/p>\n<div style=\"text-align: center;\">[latex]2\\text{NH}_3(g) + \\cancel{3\\text{H}_2(g)} + 3\\text{I}_2(g) \\rightleftharpoons \\text{N}_2(g) + \\cancel{3\\text{H}_2(g)} + 6\\text{HI}(g)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]2\\text{NH}_3(g) + 3\\text{I}_2(g) \\rightleftharpoons \\text{N}_2(g) + 6\\text{HI}(g)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{K}_c = \\text{K}_{c1'}\\text{K}_{c2'} = (2.0)(1.2 \\text{x} 10^5) = 2.5\\text{x} 10^5[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Use the provided information to calculate Kc for the following reaction at 550 \u00b0C:<\/p>\n<div style=\"text-align: center;\">[latex]\\text{H}_2(g) + \\text{CO}_2(g) \\rightleftharpoons \\text{CO}(g) + \\text{H}_2\\text{O}(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_c = ?[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{CoO}(s) + \\text{CO}(g) \\rightleftharpoons \\text{Co}(s) + \\text{CO}_2(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_{c1} = 490[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{CoO}(s) + \\text{H}_2(g) \\rightleftharpoons \\text{Co}(s) + \\text{H}_2\\text{O}(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{K}_{c1} = 67[\/latex]<\/div>\n<h4>Answer:<\/h4>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q549256\">Show Solution<\/span><\/p>\n<div id=\"q549256\" class=\"hidden-answer\" style=\"display: none\">\n<p>K<sub>c<\/sub>\u00a0= 0.14<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"The Equilibrium Constant\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xfGlEXWDRZE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/TheEquilibriumConstant_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;The Equilibrium Constant&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p id=\"fs-idp165520928\">The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient,\u00a0<em>Q<\/em>. For a reaction at equilibrium, the composition is constant, and\u00a0<em>Q<\/em>\u00a0is called the equilibrium constant,\u00a0<em>K<\/em>.<\/p>\n<p id=\"fs-idp116522496\">A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases.<\/p>\n<h4>Key Equations<\/h4>\n<ul>\n<li>[latex]Q=\\dfrac{{\\left[\\text{C}\\right]}^{\\text{x}}{\\left[\\text{D}\\right]}^{\\text{y}}}{{\\left[\\text{A}\\right]}^{\\text{m}}{\\left[\\text{B}\\right]}^{\\text{n}}}\\text{ where }m\\text{A}+n\\text{B}\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]<\/li>\n<li>[latex]{Q}_{P}=\\dfrac{{\\left({P}_{C}\\right)}^{x}{\\left({P}_{D}\\right)}^{y}}{{\\left({P}_{A}\\right)}^{m}{\\left({P}_{B}\\right)}^{n}}\\text{ where }m\\text{A}+n\\text{B}\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]<\/li>\n<li>[latex]P=MRT[\/latex]<\/li>\n<li>[latex]{K}_{P}={K}_{c}{\\left(RT\\right)}^{\\Delta n}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>equilibrium constant (<em>K<\/em>): <\/strong>value of the reaction quotient for a system at equilibrium<\/p>\n<p><strong>heterogeneous equilibria: <\/strong>equilibria between reactants and products in different phases<\/p>\n<p><strong>homogeneous equilibria: <\/strong>equilibria within a single phase<\/p>\n<p><strong><em>K<sub>c<\/sub><\/em>:<em>\u00a0<\/em><\/strong>equilibrium constant for reactions based on concentrations of reactants and products<\/p>\n<p><strong><em>K<sub>P<\/sub><\/em>:<\/strong> equilibrium constant for gas-phase reactions based on partial pressures of reactants and products<\/p>\n<p><strong>law of mass action: <\/strong>when a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant<\/p>\n<p><strong>reaction quotient (<em>Q<\/em>): <\/strong>ratio of the product of molar concentrations (or pressures) of the products to that of the reactants, each concentration (or pressure) being raised to the power equal to the coefficient in the equation<\/p>\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-2257\">\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 class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>The Equilibrium Constant. <strong>Authored by<\/strong>: Bozeman Science. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/xfGlEXWDRZE\">https:\/\/youtu.be\/xfGlEXWDRZE<\/a>. <strong>License<\/strong>: <em>Other<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/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":3,"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\"},{\"type\":\"copyrighted_video\",\"description\":\"The Equilibrium Constant\",\"author\":\"Bozeman Science\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/xfGlEXWDRZE\",\"project\":\"\",\"license\":\"other\",\"license_terms\":\"Standard YouTube 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