{"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":"2016-08-09T04:04:09","modified_gmt":"2016-08-09T04:04:09","slug":"equilibrium-constants-missing-formulas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/chapter\/equilibrium-constants-missing-formulas\/","title":{"raw":"Equilibrium Constants","rendered":"Equilibrium Constants"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>LEARNING OBJECTIVES<\/h3>\r\nBy the end of this module, you will be able to:\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\n<p id=\"fs-idp120703584\">Now that we have a symbol [latex]\\rightleftharpoons[\/latex] to designate reversible reactions, we will need a way to express mathematically how the amounts of reactants and products affect the equilibrium of the system. A general equation for a reversible reaction may be written as follows:<\/p>\r\n\r\n<div id=\"fs-idp5986304\" data-type=\"equation\">[latex]m\\text{A}+n\\text{B}+\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]<\/div>\r\n<p id=\"fs-idp11973152\">We can write the <span data-type=\"term\">reaction quotient (<em data-effect=\"italics\">Q<\/em>)<\/span> for this equation. When evaluated using concentrations, it is called <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>. We use brackets to indicate molar concentrations of reactants and products.<\/p>\r\n\r\n<div id=\"fs-idp44297776\" data-type=\"equation\">[latex]{Q}_{c}=\\frac{{\\left[\\text{C}\\right]}^{x}{\\left[\\text{D}\\right]}^{y}}{{\\left[\\text{A}\\right]}^{m}{\\left[\\text{B}\\right]}^{n}}[\/latex]<\/div>\r\n<p id=\"fs-idm7959200\">The reaction quotient is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation. For example, the reaction quotient for the reversible reaction [latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)[\/latex] is given by this expression:<\/p>\r\n\r\n<div id=\"fs-idp23922448\" data-type=\"equation\">[latex]{Q}_{c}=\\frac{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}[\/latex]<\/div>\r\n<div id=\"fs-idp168372720\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 1<\/h3>\r\n<h4 id=\"fs-idp68042176\"><span data-type=\"title\">Writing Reaction Quotient Expressions<\/span><\/h4>\r\nWrite the expression for the reaction quotient for each of the following reactions:\r\n<p id=\"fs-idp246365472\">(a) [latex]3{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{O}}_{3}\\left(g\\right)[\/latex]<\/p>\r\n<p id=\"fs-idp142495760\">(b) [latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/p>\r\n<p id=\"fs-idp82472256\">(c) [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]<\/p>\r\n\r\n<h4 id=\"fs-idp63878256\"><span data-type=\"title\">Solution<\/span><\/h4>\r\n(a) [latex]{Q}_{c}=\\frac{{\\left[{O}_{3}\\right]}^{2}}{{\\left[{O}_{2}\\right]}^{3}}[\/latex]\r\n<p id=\"fs-idp177788304\">(b) [latex]{Q}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}[\/latex]<\/p>\r\n<p id=\"fs-idp171443200\">(c) [latex]{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]<\/p>\r\n\r\n<h4 id=\"fs-idp17156768\"><span data-type=\"title\">Check Your Learning<\/span><\/h4>\r\nWrite the expression for the reaction quotient for each of the following reactions:\r\n<p id=\"fs-idp35394896\">(a) [latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right)[\/latex]<\/p>\r\n<p id=\"fs-idp63748560\">(b) [latex]{\\text{C}}_{4}{\\text{H}}_{8}\\left(g\\right)\\rightleftharpoons2{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)[\/latex]<\/p>\r\n<p id=\"fs-idp105907744\">(c) [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]<\/p>\r\n\r\n<div id=\"fs-idp167282768\" data-type=\"note\">\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp7633728\" style=\"text-align: right;\">(a) [latex]{Q}_{c}=\\frac{{\\left[{\\text{SO}}_{3}\\right]}^{2}}{{\\left[{\\text{SO}}_{2}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]}[\/latex]<\/p>\r\n<p style=\"text-align: right;\">(b) [latex]{Q}_{c}=\\frac{{\\left[{\\text{C}}_{2}{\\text{H}}_{4}\\right]}^{2}}{\\left[{\\text{C}}_{4}{\\text{H}}_{8}\\right]}[\/latex]<\/p>\r\n<p style=\"text-align: right;\">(c) [latex]{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]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idp138846544\">The numeric value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> for a given reaction varies; it depends on the concentrations of products and reactants present at the time when <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> is determined. When pure reactants are mixed, <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> is initially zero because there are no products present at that point. As the reaction proceeds, the value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> increases as the concentrations of the products increase and the concentrations of the reactants simultaneously decrease Figure 1. When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change.<\/p>\r\n\r\n[caption id=\"attachment_4880\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-4880\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/08\/23214441\/CNX_Chem_13_02_quotient-1024x850.jpg\" alt=\"Three graphs are shown and labeled, \u201ca,\u201d \u201cb,\u201d and \u201cc.\u201d All three 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\" width=\"1024\" height=\"850\" \/> Figure 1. (a) The change in the concentrations of reactants and products is depicted as the 2SO2(g)+O2(g)\u21cc2SO3(g) reaction approaches equilibrium. (b) The change in concentrations of reactants and products is depicted as the reaction 2SO3(g)\u21cc2SO2(g)+O2(g) approaches equilibrium. (c) The graph shows the change in the value of the reaction quotient as the reaction approaches equilibrium.[\/caption]\r\n<p id=\"fs-idm122432\">When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the <span data-type=\"term\">equilibrium constant (<em data-effect=\"italics\">K<\/em>)<\/span> of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as <em data-effect=\"italics\">K<sub>c<\/sub><\/em>.<\/p>\r\n<p id=\"fs-idp279896080\">That a reaction quotient always assumes the same value at equilibrium can be expressed as:<\/p>\r\n\r\n<div id=\"fs-idp266400656\" data-type=\"equation\">[latex]{Q}_{c}\\text{at equilibrium}={K}_{c}=\\frac{{\\left[\\text{C}\\right]}^{x}{\\left[\\text{D}\\right]}^{y}\\ldots }{{\\left[\\text{A}\\right]}^{m}{\\left[\\text{B}\\right]}^{n}\\ldots}[\/latex]<\/div>\r\n<p id=\"fs-idp73447152\">This equation is a mathematical statement of the <span data-type=\"term\">law of mass action<\/span>: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.<\/p>\r\n\r\n<div id=\"fs-idp54423808\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 2<\/h3>\r\n<h4 id=\"fs-idp23994768\"><span data-type=\"title\">Evaluating a Reaction Quotient<\/span><\/h4>\r\nGaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:\r\n<div id=\"fs-idp64726272\" data-type=\"equation\">[latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)[\/latex]<\/div>\r\n<p id=\"fs-idp17381824\">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 data-effect=\"italics\">M<\/em> and [N<sub>2<\/sub>O<sub>4<\/sub>] = 0.042 <em data-effect=\"italics\">M<\/em>.<\/p>\r\n<p id=\"fs-idp89605104\">(a) What is the value of the reaction quotient before any reaction occurs?<\/p>\r\n<p id=\"fs-idp84731776\">(b) What is the value of the equilibrium constant for the reaction?<\/p>\r\n\r\n<h4 id=\"fs-idp179491728\"><span data-type=\"title\">Solution<\/span><\/h4>\r\n(a) Before any product is formed, [latex]\\left[{\\text{NO}}_{2}\\right]=\\frac{0.10\\text{mol}}{1.0\\text{L}}=0.10M[\/latex], and [N<sub>2<\/sub>O<sub>4<\/sub>] = 0 <em data-effect=\"italics\">M<\/em>. Thus,\r\n<div id=\"fs-idp173627248\" data-type=\"equation\">[latex]{Q}_{c}=\\frac{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}=\\frac{0}{{0.10}^{2}}=0[\/latex]<\/div>\r\n<p id=\"fs-idp110962640\">(b) At equilibrium, the value of the equilibrium constant is equal to the value of the reaction quotient. At equilibrium, [latex]{K}_{c}={Q}_{c}=\\frac{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}=\\frac{0.042}{{0.016}^{2}}=1.6\\times {10}^{2}\\text{.}[\/latex] The equilibrium constant is 1.6 \u00d7 10<sup>2<\/sup>.<\/p>\r\n<p id=\"fs-idp120754816\">Note that dimensional analysis would suggest the unit for this <em data-effect=\"italics\">K<sub>c<\/sub><\/em> value should be <em data-effect=\"italics\">M<\/em><sup>-1<\/sup>. However, it is common practice to omit units for <em data-effect=\"italics\">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 data-effect=\"italics\">K<sub>c<\/sub><\/em> values are truly unitless.<\/p>\r\n\r\n<h4 id=\"fs-idp284528080\"><span data-type=\"title\">Check Your Learning<\/span><\/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 data-effect=\"italics\">M<\/em>, [O<sub>2<\/sub>] = 0.35 <em data-effect=\"italics\">M<\/em>, and [SO<sub>3<\/sub>] = 1.1 <em data-effect=\"italics\">M<\/em>. What is the value of the equilibrium constant, <em data-effect=\"italics\">K<sub>c<\/sub><\/em>?\r\n<div id=\"fs-idp10573648\" data-type=\"note\">\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong><em data-effect=\"italics\">K<sub>c<\/sub> =<\/em> 4.3<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm80100288\">The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A large value for <em data-effect=\"italics\">K<sub>c<\/sub><\/em> indicates that equilibrium is attained only after the reactants have been largely converted into products. A small value of <em data-effect=\"italics\">K<sub>c<\/sub><\/em>\u2014much less than 1\u2014indicates that equilibrium is attained when only a small proportion of the reactants have been converted into products.<\/p>\r\n<p id=\"fs-idp246311264\">Once a value of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> is known for a reaction, it can be used to predict directional shifts when compared to the value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>. A system that is not at equilibrium will proceed in the direction that establishes equilibrium. The data in Figure 13.7\u00a0illustrate this. When heated to a consistent temperature, 800 \u00b0C, different starting mixtures of CO, H<sub>2<\/sub>O, CO<sub>2<\/sub>, and H<sub>2<\/sub> react to reach compositions adhering to the same equilibrium (the value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> changes until it equals the value of <em data-effect=\"italics\">K<sub>c<\/sub><\/em>). This value is 0.640, the equilibrium constant for the reaction under these conditions.<\/p>\r\n\r\n<div id=\"fs-idp63923872\" data-type=\"equation\">[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){K}_{c}=0.640\\text{T}=800\\text{\\textdegree C}[\/latex]<\/div>\r\n<p id=\"fs-idp94527104\">It is important to recognize that an equilibrium can be established starting either from reactants or from products, or from a mixture of both. For example, equilibrium was established from Mixture 2 in Figure 13.7\u00a0when the products of the reaction were heated in a closed container. In fact, one technique used to determine whether a reaction is truly at equilibrium is to approach equilibrium starting with reactants in one experiment and starting with products in another. If the same value of the reaction quotient is observed when the concentrations stop changing in both experiments, then we may be certain that the system has reached equilibrium.<\/p>\r\n\r\n\r\n[caption id=\"attachment_4881\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-4881\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/08\/23214443\/CNX_Chem_13_02_mixtures-1024x440.jpg\" 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 and the blue bar near 0.05. For mixture two, the green bar is near 0.05, and the yellow bar is near 0.09. For mixture 3, the red bar is near 0.01, the blue bar is slightly above that with green and yellow topping it off at 0.02. On the right graph, the bar for mixture one shows the red bar slightly above 0.01, the blue bar stacked on it rising slightly above 0.02, the green rising near 0.04, and the yellow bar reaching near 0.05. A label above this bar reads, \u201cQ equals 0.640.\u201d The bar for mixture two shows the red bar slightly above 0.02, the blue bar stacked on it rising near 0.05, the green rising near 0.07, and the yellow bar reaching near 0.10. A label above this bar reads \u201cQ equals 0.640.\u201d The bar for mixture three shows the red bar near 0.01, the blue bar stacked on it rising slightly above 0.01, the green rising near 0.02, and the yellow bar reaching 0.02. A label above this bar reads \u201cQ equals 0.640\u201d.\" width=\"1024\" height=\"440\" \/> Figure 2. Concentrations of three mixtures are shown before and after reaching equilibrium at 800 \u00b0C for the so-called water gas shift reaction: CO(g)+H2O(g)\u21ccCO2(g)+H2(g).[\/caption]\r\n\r\n<div id=\"fs-idp174376240\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 3<\/h3>\r\n<h4 id=\"fs-idp60883568\"><span data-type=\"title\">Predicting the Direction of Reaction<\/span><\/h4>\r\nGiven below are the starting concentrations of reactants and products for three experiments involving this reaction:\r\n<div id=\"fs-idm31469104\" data-type=\"equation\">[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]<\/div>\r\n<div id=\"fs-idp89310352\" data-type=\"equation\">[latex]{K}_{c}=0.64[\/latex]<\/div>\r\n<p id=\"fs-idp162489744\">Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.<\/p>\r\n\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.\" data-label=\"\">\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 data-effect=\"italics\">M<\/em><\/td>\r\n<td>0.011 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td>0.0094 <em data-effect=\"italics\">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 data-effect=\"italics\">M<\/em><\/td>\r\n<td>0.0011 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td>0.0025 <em data-effect=\"italics\">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 data-effect=\"italics\">M<\/em><\/td>\r\n<td>0.037 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td>0.0015 <em data-effect=\"italics\">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 data-effect=\"italics\">M<\/em><\/td>\r\n<td>0.046 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td>0.0076 <em data-effect=\"italics\">M<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4 id=\"fs-idp3711408\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nExperiment 1:\r\n<div id=\"fs-idp24553712\" data-type=\"equation\">[latex]{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]<\/div>\r\n<p id=\"fs-idm1136592\"><em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &lt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (0.039 &lt; 0.64)<\/p>\r\n<p id=\"fs-idp267578784\">The reaction will shift to the right.<\/p>\r\n<p id=\"fs-idp56047184\">Experiment 2:<\/p>\r\n\r\n<div id=\"fs-idp15286912\" data-type=\"equation\">[latex]{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]<\/div>\r\n<p id=\"fs-idp3859856\"><em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &gt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (140 &gt; 0.64)<\/p>\r\n<p id=\"fs-idp68828672\">The reaction will shift to the left.<\/p>\r\n<p id=\"fs-idp101164320\">Experiment 3:<\/p>\r\n\r\n<div id=\"fs-idp74212144\" data-type=\"equation\">[latex]{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]<\/div>\r\n<p id=\"fs-idp70179520\"><em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &lt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (0.48 &lt; 0.64)<\/p>\r\n<p id=\"fs-idp155462992\">The reaction will shift to the right.<\/p>\r\n\r\n<h4 id=\"fs-idm76258576\"><span data-type=\"title\">Check Your Learning<\/span><\/h4>\r\nCalculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.\r\n<p id=\"fs-idp59277392\">(a) A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl:<\/p>\r\n\r\n<div id=\"fs-idp56406272\" data-type=\"equation\">[latex]2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NOCl}\\left(g\\right){K}_{c}=4.6\\times {10}^{4}\\text{.}[\/latex]<\/div>\r\n<p id=\"fs-idp278668928\">(b) 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>:<\/p>\r\n\r\n<div id=\"fs-idp278669312\" data-type=\"equation\">[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right){K}_{c}=0.060[\/latex]<\/div>\r\n<p id=\"fs-idp110625216\">(c) A 2.00-L flask containing 230 g of SO<sub>3<\/sub>(g):<\/p>\r\n\r\n<div id=\"fs-idp294747856\" data-type=\"equation\">[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right){K}_{c}=0.230[\/latex]<\/div>\r\n<div id=\"fs-idp32992592\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong>(a) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 6.45 \u00d7 103, shifts right. (b) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 0.12, shifts left. (c) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 0, shifts right<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idp168987520\">In Example 2, it was mentioned that the common practice is to omit units when evaluating reaction quotients and equilibrium constants. It should be pointed out that using concentrations in these computations is a convenient but simplified approach that sometimes leads to results that seemingly conflict with the law of mass action. For example, equilibria involving aqueous ions often exhibit equilibrium constants that vary quite significantly (are <em data-effect=\"italics\">not<\/em> constant) at high solution concentrations. This may be avoided by computing <em data-effect=\"italics\">K<sub>c<\/sub><\/em> values using the <em data-effect=\"italics\">activities<\/em> of the reactants and products in the equilibrium system instead of their concentrations. The <span data-type=\"term\">activity<\/span> of a substance is a measure of its effective concentration under specified conditions. While a detailed discussion of this important quantity is beyond the scope of an introductory text, it is necessary to be aware of a few important aspects:<\/p>\r\n\r\n<ul id=\"fs-idp284227328\" data-bullet-style=\"bullet\">\r\n \t<li>Activities are dimensionless (unitless) quantities and are in essence \u201cadjusted\u201d concentrations.<\/li>\r\n \t<li>For relatively dilute solutions, a substance's activity and its molar concentration are roughly equal.<\/li>\r\n \t<li>Activities for pure condensed phases (solids and liquids) are equal to 1.<\/li>\r\n<\/ul>\r\n<p id=\"fs-idp43748544\">As a consequence of this last consideration, <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> and <em data-effect=\"italics\">K<sub>c<\/sub><\/em> expressions do not contain terms for solids or liquids (being numerically equal to 1, these terms have no effect on the expression's value)<em data-effect=\"italics\">.<\/em> Several examples of equilibria yielding such expressions will be encountered below.<\/p>\r\n\r\n<section id=\"fs-idp86160096\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Homogeneous Equilibria<\/h2>\r\n<p id=\"fs-idm54871216\">A <span data-type=\"term\">homogeneous equilibrium<\/span> is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). In this chapter, we will concentrate on the two most common types of homogeneous equilibria: those occurring in liquid-phase solutions and those involving exclusively gaseous species. Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both. Several examples are provided below.<\/p>\r\n\r\n<div id=\"fs-idp38922960\" data-type=\"equation\">[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){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]<\/div>\r\n<div id=\"fs-idp25144000\" data-type=\"equation\">[latex]{\\text{I}}_{2}\\left(aq\\right)+{\\text{I}}^{-}\\left(aq\\right)\\rightleftharpoons{\\text{I}}_{3}{}^{-}\\left(aq\\right){K}_{c}=\\frac{\\left[{\\text{I}}_{3}{}^{-}\\right]}{\\left[{\\text{I}}_{2}\\right]\\left[{\\text{I}}^{-}\\right]}[\/latex]<\/div>\r\n<div id=\"fs-idm76151744\" data-type=\"equation\">[latex]{\\text{Hg}}_{2}{}^{2+}\\left(aq\\right)+{\\text{NO}}_{3}{}^{-}\\left(aq\\right)+3{\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)\\rightleftharpoons2{\\text{Hg}}^{2+}\\left(aq\\right)+{\\text{HNO}}_{2}\\left(aq\\right)+4{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\r\n<div id=\"fs-idp122601856\" data-type=\"equation\">[latex]{K}_{c}=\\frac{{\\left[{\\text{Hg}}^{2+}\\right]}^{2}\\left[{\\text{HNO}}_{2}\\right]}{\\left[{\\text{Hg}}_{2}{}^{2+}\\right]\\left[{\\text{NO}}_{3}{}^{-}\\right]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{3}}[\/latex]<\/div>\r\n<div id=\"fs-idp101805696\" data-type=\"equation\">[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){K}_{c}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{F}}^{-}\\right]}{\\left[\\text{HF}\\right]}[\/latex]<\/div>\r\n<div id=\"fs-idp15332976\" data-type=\"equation\">[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){K}_{c}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}[\/latex]<\/div>\r\n<p id=\"fs-idp11719264\">In each of these examples, the equilibrium system is an aqueous solution, as denoted by the <em data-effect=\"italics\">aq<\/em> annotations on the solute formulas. Since H<sub>2<\/sub>O(<em data-effect=\"italics\">l<\/em>) is the solvent for these solutions, its concentration does not appear as a term in the <em data-effect=\"italics\">K<sub>c<\/sub><\/em> expression, as discussed above, even though it may also appear as a reactant or product in the chemical equation.<\/p>\r\n<p id=\"fs-idp11805840\">Reactions in which all reactants and products are gases represent a second class of homogeneous equilibria. We use molar concentrations in the following examples, but we will see shortly that partial pressures of the gases may be used as well.<\/p>\r\n\r\n<div id=\"fs-idp17340352\" data-type=\"equation\">[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){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]<\/div>\r\n<div id=\"fs-idp100807584\" data-type=\"equation\">[latex]3{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{O}}_{3}\\left(g\\right){K}_{c}=\\frac{{\\left[{\\text{O}}_{3}\\right]}^{2}}{{\\left[{\\text{O}}_{2}\\right]}^{3}}[\/latex]<\/div>\r\n<div id=\"fs-idp68583712\" data-type=\"equation\">[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right){K}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}[\/latex]<\/div>\r\n<div id=\"fs-idp86498496\" data-type=\"equation\">[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){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]<\/div>\r\n<p id=\"fs-idp58545648\">Note that the concentration of H<sub>2<\/sub>O(<em data-effect=\"italics\">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>\r\n<p id=\"fs-idp165025968\">Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the reaction quotient because the partial pressure of a gas is directly proportional to its concentration at constant temperature. This relationship can be derived from the ideal gas equation, where <em data-effect=\"italics\">M<\/em> is the molar concentration of gas, [latex]\\frac{n}{V}\\text{.}[\/latex]<\/p>\r\n\r\n<div id=\"fs-idp111709808\" data-type=\"equation\">[latex]PV=nRT[\/latex]<\/div>\r\n<div id=\"fs-idm75268224\" data-type=\"equation\">[latex]P=\\left(\\frac{n}{V}\\right)RT[\/latex]<\/div>\r\n<div id=\"fs-idp44758960\" data-type=\"equation\">[latex]=MRT[\/latex]<\/div>\r\n<p id=\"fs-idp123497808\">Thus, at constant temperature, the pressure of a gas is directly proportional to its concentration.<\/p>\r\n<p id=\"fs-idp43798512\">Using the partial pressures of the gases, we can write the reaction quotient for the system [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] by following the same guidelines for deriving concentration-based expressions:<\/p>\r\n\r\n<div id=\"fs-idp279715456\" data-type=\"equation\">[latex]{Q}_{P}=\\frac{{P}_{{\\text{C}}_{2}{\\text{H}}_{4}}{P}_{{\\text{H}}_{2}}}{{P}_{{\\text{C}}_{2}{\\text{H}}_{6}}}[\/latex]<\/div>\r\n<p id=\"fs-idp180873632\">In this equation we use <em data-effect=\"italics\">Q<sub>P<\/sub><\/em> to indicate a reaction quotient written with partial pressures: [latex]{P}_{{\\text{C}}_{2}{\\text{H}}_{6}}[\/latex] is the partial pressure of C<sub>2<\/sub>H<sub>6<\/sub>; [latex]{P}_{{\\text{H}}_{2}}[\/latex], the partial pressure of H<sub>2<\/sub>; and [latex]{P}_{{\\text{C}}_{2}{\\text{H}}_{6}}[\/latex], the partial pressure of C<sub>2<\/sub>H<sub>4<\/sub>. At equilibrium:<\/p>\r\n\r\n<div id=\"fs-idp116676256\" data-type=\"equation\">[latex]{K}_{P}={Q}_{P}=\\frac{{P}_{{\\text{C}}_{2}{\\text{H}}_{4}}{P}_{{\\text{H}}_{2}}}{{P}_{{\\text{C}}_{2}{\\text{H}}_{6}}}[\/latex]<\/div>\r\n<p id=\"fs-idp155025056\">The subscript <em data-effect=\"italics\">P<\/em> in the symbol <span data-type=\"term\"><em data-effect=\"italics\">K<sub>P<\/sub><\/em><\/span> designates an equilibrium constant derived using partial pressures instead of concentrations. The equilibrium constant, <em data-effect=\"italics\">K<sub>P<\/sub><\/em>, is still a constant, but its numeric value may differ from the equilibrium constant found for the same reaction by using concentrations.<\/p>\r\n<p id=\"fs-idp122706016\">Conversion between a value for <span data-type=\"term\"><em data-effect=\"italics\">K<sub>c<\/sub><\/em><\/span>, an equilibrium constant expressed in terms of concentrations, and a value for <em data-effect=\"italics\">K<sub>P<\/sub><\/em>, an equilibrium constant expressed in terms of pressures, is straightforward (a <em data-effect=\"italics\">K<\/em> or <em data-effect=\"italics\">Q<\/em> without a subscript could be either concentration or pressure).<\/p>\r\n<p id=\"fs-idp160302688\">The equation relating <em data-effect=\"italics\">K<sub>c<\/sub><\/em> and <em data-effect=\"italics\">K<sub>P<\/sub><\/em> is derived as follows. For the gas-phase reaction [latex]m\\text{A}+n\\text{B}\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]:<\/p>\r\n\r\n<div id=\"fs-idp171727376\" data-type=\"equation\">[latex]{K}_{P}=\\frac{{\\left({P}_{C}\\right)}^{x}{\\left({P}_{D}\\right)}^{y}}{{\\left({P}_{A}\\right)}^{m}{\\left({P}_{B}\\right)}^{n}}[\/latex]<\/div>\r\n<div id=\"fs-idp89891888\" data-type=\"equation\">[latex]=\\frac{{\\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}}[\/latex]<\/div>\r\n<div id=\"fs-idp72794512\" data-type=\"equation\">[latex]=\\frac{{\\left[\\text{C}\\right]}^{x}{\\left[\\text{D}\\right]}^{y}}{{\\left[\\text{A}\\right]}^{m}{\\left[\\text{B}\\right]}^{n}}\\times \\frac{{\\left(RT\\right)}^{x+y}}{{\\left(RT\\right)}^{m+n}}[\/latex]<\/div>\r\n<div id=\"fs-idp96527968\" data-type=\"equation\">[latex]={K}_{c}{\\left(RT\\right)}^{\\left(x+y\\right)-\\left(m+n\\right)}[\/latex]<\/div>\r\n<div id=\"fs-idp66307536\" data-type=\"equation\">[latex]={K}_{c}{\\left(RT\\right)}^{\\Delta n}[\/latex]<\/div>\r\n<p id=\"fs-idp93962000\">The relationship between <em data-effect=\"italics\">K<sub>c<\/sub><\/em> and <em data-effect=\"italics\">K<sub>P<\/sub><\/em> is<\/p>\r\n\r\n<div id=\"fs-idp142684400\" data-type=\"equation\">[latex]{K}_{P}={K}_{c}{\\left(RT\\right)}^{\\Delta n}[\/latex]<\/div>\r\n<p id=\"fs-idp38982608\">In this equation, \u0394<em data-effect=\"italics\">n<\/em> is the difference between the sum of the coefficients of the <em data-effect=\"italics\">gaseous<\/em> products and the sum of the coefficients of the <em data-effect=\"italics\">gaseous<\/em> reactants in the reaction (the change in moles of gas between the reactants and the products). For the gas-phase reaction [latex]m\\text{A}+n\\text{B}\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-idp126597184\" data-type=\"equation\">[latex]\\Delta n=\\left(x+y\\right)-\\left(m+n\\right)[\/latex]<\/div>\r\n<div id=\"fs-idp74619616\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 4<\/h3>\r\n<h4 id=\"fs-idp74619872\"><span data-type=\"title\">Calculation of <em data-effect=\"italics\">K<sub>P<\/sub><\/em><\/span><\/h4>\r\nWrite the equations for the conversion of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> to <em data-effect=\"italics\">K<sub>P<\/sub><\/em> for each of the following reactions:\r\n<p id=\"fs-idp61114160\">(a) [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]<\/p>\r\n<p id=\"fs-idp127890016\">(b) [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]<\/p>\r\n<p id=\"fs-idp98159632\">(c) [latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/p>\r\n<p id=\"fs-idp175509632\">(d) <em data-effect=\"italics\">K<sub>c<\/sub><\/em> is equal to 0.28 for the following reaction at 900 \u00b0C:<\/p>\r\n\r\n<div id=\"fs-idp277886816\" data-type=\"equation\">[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]<\/div>\r\n<p id=\"fs-idp237503488\">What is <em data-effect=\"italics\">K<sub>P<\/sub><\/em> at this temperature?<\/p>\r\n\r\n<h4 id=\"fs-idp22632784\"><span data-type=\"title\">Solution<\/span><\/h4>\r\n(a) \u0394<em data-effect=\"italics\">n<\/em> = (2) - (1) = 1 <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>\u0394<em data-effect=\"italics\">n<\/em><\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>1<\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)\r\n<p id=\"fs-idp15495888\">(b) \u0394<em data-effect=\"italics\">n<\/em> = (2) - (2) = 0 <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>\u0394<em data-effect=\"italics\">n<\/em><\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>0<\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em><\/p>\r\n<p id=\"fs-idp64674800\">(c) \u0394<em data-effect=\"italics\">n<\/em> = (2) - (1 + 3) = -2 <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>\u0394<em data-effect=\"italics\">n<\/em><\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>-2<\/sup> = [latex]\\frac{{K}_{c}}{{\\left(RT\\right)}^{2}}[\/latex]<\/p>\r\n<p id=\"fs-idp100458064\">(d) <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (RT) <sup>\u0394<em data-effect=\"italics\">n<\/em><\/sup> = (0.28)[(0.0821)(1173)]<sup>-2<\/sup> = 3.0 \u00d7 10<sup>-5<\/sup><\/p>\r\n\r\n<h4 id=\"fs-idp112448864\"><span data-type=\"title\">Check Your Learning<\/span><\/h4>\r\nWrite the equations for the conversion of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> to <em data-effect=\"italics\">K<sub>P<\/sub><\/em> for each of the following reactions, which occur in the gas phase:\r\n<p id=\"fs-idp181427520\">(a) [latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right)[\/latex]<\/p>\r\n<p id=\"fs-idp162294032\">(b) [latex]{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)\\rightleftharpoons2{\\text{NO}}_{2}\\left(g\\right)[\/latex]<\/p>\r\n<p id=\"fs-idp103215280\">(c) [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]<\/p>\r\n<p id=\"fs-idp64364800\">(d) At 227 \u00b0C, the following reaction has <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = 0.0952:<\/p>\r\n\r\n<div id=\"fs-idp23547856\" data-type=\"equation\">[latex]{\\text{CH}}_{3}\\text{OH}\\left(g\\right)\\rightleftharpoons\\text{CO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/div>\r\n<p id=\"fs-idp131624768\">What would be the value of <em data-effect=\"italics\">K<sub>P<\/sub><\/em> at this temperature?<\/p>\r\n\r\n<div id=\"fs-idp33474544\" data-type=\"note\">\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong>(a) <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>)<sup>-1<\/sup>; (b) <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>); (c) <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>); (d) 160 or 1.6 \u00d7 10<sup>2<\/sup><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-idp112997072\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Heterogeneous Equilibria<\/h2>\r\n<p id=\"fs-idp27217312\">A <span data-type=\"term\">heterogeneous equilibrium<\/span> is a system in which reactants and products are found in two or more phases. The phases may be any combination of solid, liquid, or gas phases, and solutions. When dealing with these equilibria, remember that solids and pure liquids do not appear in equilibrium constant expressions (the activities of pure solids, pure liquids, and solvents are 1).<\/p>\r\n<p id=\"fs-idp134404832\">Some heterogeneous equilibria involve chemical changes; for example:<\/p>\r\n\r\n<div id=\"fs-idp129666976\" data-type=\"equation\">[latex]{\\text{PbCl}}_{2}\\left(s\\right)\\rightleftharpoons{\\text{Pb}}^{2+}\\left(aq\\right)+2{\\text{Cl}}^{-}\\left(aq\\right){K}_{c}=\\left[{\\text{Pb}}^{2+}\\right]{\\left[{\\text{Cl}}^{-}\\right]}^{2}[\/latex]<\/div>\r\n<div id=\"fs-idp89064704\" data-type=\"equation\">[latex]\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CaCO}}_{3}\\left(s\\right){K}_{c}=\\frac{1}{\\left[{\\text{CO}}_{2}\\right]}[\/latex]<\/div>\r\n<div id=\"fs-idp172202816\" data-type=\"equation\">[latex]\\text{C}\\left(s\\right)+2\\text{S}\\left(g\\right)\\rightleftharpoons{\\text{CS}}_{2}\\left(g\\right){K}_{c}=\\frac{\\left[{\\text{CS}}_{2}\\right]}{{\\left[\\text{S}\\right]}^{2}}[\/latex]<\/div>\r\n<p id=\"fs-idp84722304\">Other heterogeneous equilibria involve phase changes, for example, the evaporation of liquid bromine, as shown in the following equation:<\/p>\r\n\r\n<div id=\"fs-idp64021520\" data-type=\"equation\">[latex]{\\text{Br}}_{2}\\left(l\\right)\\rightleftharpoons{\\text{Br}}_{2}\\left(g\\right){K}_{c}=\\left[{\\text{Br}}_{2}\\right][\/latex]<\/div>\r\n<p id=\"fs-idp8357744\">We can write equations for reaction quotients of heterogeneous equilibria that involve gases, using partial pressures instead of concentrations. Two examples are:<\/p>\r\n\r\n<div id=\"fs-idp11158544\" data-type=\"equation\">[latex]\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CaCO}}_{3}\\left(s\\right){K}_{P}=\\frac{1}{{P}_{{\\text{CO}}_{2}}}[\/latex]<\/div>\r\n<div id=\"fs-idp269174800\" data-type=\"equation\">[latex]\\text{C}\\left(s\\right)+2\\text{S}\\left(g\\right)\\rightleftharpoons{\\text{CS}}_{2}\\left(g\\right){K}_{P}=\\frac{{P}_{C{S}_{2}}}{{\\left({P}_{S}\\right)}^{2}}[\/latex]<\/div>\r\n<\/section><section id=\"fs-idp284371824\" class=\"summary\" data-depth=\"1\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h2>Key Concepts and Summary<\/h2>\r\n<p id=\"fs-idp165520928\">For any reaction that is at equilibrium, the reaction quotient <em data-effect=\"italics\">Q<\/em> is equal to the equilibrium constant <em data-effect=\"italics\">K<\/em> for the reaction. If a reactant or product is a pure solid, a pure liquid, or the solvent in a dilute solution, the concentration of this component does not appear in the expression for the equilibrium constant. At equilibrium, the values of the concentrations of the reactants and products are constant. Their particular values may vary depending on conditions, but the value of the reaction quotient will always equal <em data-effect=\"italics\">K<\/em> (<em data-effect=\"italics\">K<sub>c<\/sub><\/em> when using concentrations or <em data-effect=\"italics\">K<sub>P<\/sub><\/em> when using partial pressures).<\/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. We can decide whether a reaction is at equilibrium by comparing the reaction quotient with the equilibrium constant for the reaction.<\/p>\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idp116522752\" class=\"key-equations\" data-depth=\"1\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Equations<\/h3>\r\n<ul id=\"fs-idp16702192\" data-bullet-style=\"bullet\">\r\n \t<li>[latex]Q=\\frac{{\\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}=\\frac{{\\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><em data-effect=\"italics\">P<\/em> = <em data-effect=\"italics\">MRT<\/em><\/li>\r\n \t<li><em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>)<sup>\u0394<em data-effect=\"italics\">n<\/em><\/sup><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section><section id=\"fs-idp16341104\" class=\"exercises\" data-depth=\"1\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Chemistry End of Chapter Exercises<\/h3>\r\n<div id=\"fs-idp109971312\" data-type=\"exercise\">\r\n<div id=\"fs-idp60916480\" data-type=\"problem\">\r\n<ol>\r\n \t<li>Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.<\/li>\r\n \t<li>Explain why an equilibrium between Br<sub>2<\/sub>(<em data-effect=\"italics\">l<\/em>) and Br<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) would not be established if the container were not a closed vessel shown in Figure 3.\r\n\r\n[caption id=\"attachment_2253\" align=\"alignnone\" width=\"300\"]<img class=\"wp-image-2253\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212357\/CNX_Chem_13_01_bromine1.jpg\" alt=\"A glass container is shown that is filled with an orange-brown gas and a small amount of dark orange liquid.\" width=\"300\" height=\"354\" \/> Figure 3. A sample of liquid bromine at equilibrium with bromine vapor in a closed container. (credit: http:\/\/images-of-elements.com\/bromine.php)[\/caption]<\/li>\r\n \t<li>If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO<sub>2<\/sub> or with pure N<sub>2<\/sub>O<sub>4<\/sub>?\r\n[latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg<sub>2<\/sub>Cl<sub>2<\/sub>, AgCl, PbCl<sub>2<\/sub>, and CuCl.\r\n<ol>\r\n \t<li>Write the expression for the equilibrium constant for the reaction represented by the equation [latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons{\\text{Ag}}^{+}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)\\text{.}[\/latex] Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481? Explain your answer.<\/li>\r\n \t<li>Write the expression for the equilibrium constant for the reaction represented by the equation [latex]{\\text{Pb}}^{2+}\\left(aq\\right)+2{\\text{Cl}}^{-}\\left(aq\\right)\\rightleftharpoons{\\text{PbCl}}_{2}\\left(s\\right)\\text{.}[\/latex] Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481? Explain your answer.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates\u2014except those of the ammonium ion and the alkali metals\u2014are insoluble.\r\n<ol>\r\n \t<li>Write the expression for the equilibrium constant for the reaction represented by the equation [latex]{\\text{CaCO}}_{3}\\left(s\\right)\\rightleftharpoons{\\text{Ca}}^{2+}\\left(aq\\right)+{\\text{CO}}_{3}{}^{-}\\left(aq\\right)\\text{.}[\/latex] Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481? Explain your answer.<\/li>\r\n \t<li>Write the expression for the equilibrium constant for the reaction represented by the equation [latex]3{\\text{Ba}}^{2+}\\left(aq\\right)+2{\\text{PO}}_{4}{}^{3-}\\left(aq\\right)\\rightleftharpoons{\\text{Ba}}_{3}{\\left({\\text{PO}}_{4}\\right)}_{2}\\left(s\\right)\\text{.}[\/latex] Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481? Explain your answer.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: [latex]3{\\text{C}}_{2}{\\text{H}}_{2}\\left(g\\right)\\rightarrow{\\text{C}}_{6}{\\text{H}}_{6}\\left(g\\right)\\text{.}[\/latex] Which value of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> would make this reaction most useful commercially? <em data-effect=\"italics\">K<sub>c<\/sub><\/em> \u2248 0.01, <em data-effect=\"italics\">K<sub>c<\/sub><\/em> \u2248 1, or <em data-effect=\"italics\">K<sub>c<\/sub><\/em> \u2248 10. Explain your answer.<\/li>\r\n \t<li>Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation [latex]\\text{KI}\\left(aq\\right)+{\\text{I}}_{2}\\left(aq\\right)\\rightleftharpoons{\\text{KI}}_{3}\\left(aq\\right)[\/latex] give the same expression for the reaction quotient. KI<sub>3<\/sub> is composed of the ions K<sup>+<\/sup> and [latex]{\\text{I}}_{3}{}^{\\text{-}}\\text{.}[\/latex]\r\nThe reaction quotient for the complete form of the equation is [latex]{Q}_{c}=\\frac{\\left[{\\text{KI}}_{3}\\right]}{\\left[\\text{KI}\\right]\\left[{\\text{I}}_{2}\\right]}\\text{.}[\/latex]<\/li>\r\n \t<li>For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481 for a titration reaction?<\/li>\r\n \t<li>For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481 for a useful precipitation reaction?<\/li>\r\n \t<li>Write the mathematical expression for the reaction quotient, <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>, for each of the following reactions:\r\n<ol>\r\n \t<li>[latex]{\\text{CH}}_{4}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CH}}_{3}\\text{Cl}\\left(g\\right)+\\text{HCl}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{N}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NO}\\left(g\\right)[\/latex]<\/li>\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{BaSO}}_{3}\\left(s\\right)\\rightleftharpoons\\text{BaO}\\left(s\\right)+{\\text{SO}}_{2}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{P}}_{4}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{P}}_{4}{\\text{O}}_{10}\\left(s\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{Br}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{Br}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{CH}}_{4}\\left(g\\right)+2{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CO}}_{2}\\left(g\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{CuSO}}_{4}\\cdot 5{\\text{H}}_{2}\\text{O}\\left(s\\right)\\rightleftharpoons{\\text{CuSO}}_{4}\\left(s\\right)+5{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Write the mathematical expression for the reaction quotient, <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>, for each of the following reactions:\r\n<ol>\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)+5{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons4\\text{NO}\\left(g\\right)+6{\\text{H}}_{2}\\text{O}\\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{CO}}_{2}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons\\text{CO}\\left(g\\right)+{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{NH}}_{4}\\text{Cl}\\left(s\\right)\\rightleftharpoons{\\text{NH}}_{3}\\left(g\\right)+\\text{HCl}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]2\\text{Pb}{\\left({\\text{NO}}_{3}\\right)}_{2}\\left(s\\right)\\rightleftharpoons2\\text{PbO}\\left(s\\right)+4{\\text{NO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]2{\\text{H}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{S}}_{8}\\left(g\\right)\\rightleftharpoons8\\text{S}\\left(g\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.\r\n<ol>\r\n \t<li>[latex]2{\\text{NH}}_{3}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right){K}_{c}=17\\text{;}[\/latex] [NH<sub>3<\/sub>] = 0.20 <em data-effect=\"italics\">M<\/em>, [N<sub>2<\/sub>] = 1.00 <em data-effect=\"italics\">M<\/em>, [H<sub>2<\/sub>] = 1.00 <em data-effect=\"italics\">M\r\n<\/em><\/li>\r\n \t<li>[latex]2{\\text{NH}}_{3}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right){K}_{P}=6.8\\times {10}^{4}\\text{;}[\/latex] initial pressures: NH<sub>3<\/sub> = 3.0 atm, N<sub>2<\/sub> = 2.0 atm, H<sub>2<\/sub> = 1.0 atm<\/li>\r\n \t<li>[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\text{(}g\\text{)}{K}_{c}=0.230\\text{;}[\/latex] [SO<sub>3<\/sub>] = 0.00 <em data-effect=\"italics\">M<\/em>, [SO<sub>2<\/sub>] = 1.00 <em data-effect=\"italics\">M<\/em>, [O<sub>2<\/sub>] = 1.00 <em data-effect=\"italics\">M\r\n<\/em><\/li>\r\n \t<li>[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right){K}_{P}=16.5\\text{;}[\/latex] initial pressures: SO<sub>3<\/sub> = 1.00 atm, SO<sub>2<\/sub> = 1.00 atm, O<sub>2<\/sub> = 1.00 atm<\/li>\r\n \t<li>[latex]2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NOCl}\\left(g\\right){K}_{c}=4.6\\times {10}^{4}\\text{;}[\/latex] [NO] = 1.00 <em data-effect=\"italics\">M<\/em>, [Cl<sub>2<\/sub>] = 1.00 <em data-effect=\"italics\">M<\/em>, [NOCl] = 0 <em data-effect=\"italics\">M\r\n<\/em><\/li>\r\n \t<li>[latex]{\\text{N}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NO}\\left(g\\right){K}_{P}=0.050\\text{;}[\/latex] initial pressures: NO = 10.0 atm, N<sub>2<\/sub> = O<sub>2<\/sub> = 5 atm<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.\r\n<ol>\r\n \t<li>[latex]2{\\text{NH}}_{3}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right){K}_{c}=17\\text{;}[\/latex] [NH<sub>3<\/sub>] = 0.50 <em data-effect=\"italics\">M<\/em>, [N<sub>2<\/sub>] = 0.15 <em data-effect=\"italics\">M<\/em>, [H<sub>2<\/sub>] = 0.12 <em data-effect=\"italics\">M\r\n<\/em><\/li>\r\n \t<li>[latex]2{\\text{NH}}_{3}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right){K}_{P}=6.8\\times {10}^{4}\\text{;}[\/latex] initial pressures: NH<sub>3<\/sub> = 2.00 atm, N<sub>2<\/sub> = 10.00 atm, H<sub>2<\/sub> = 10.00 atm<\/li>\r\n \t<li>[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right){K}_{c}=0.230\\text{;}[\/latex] [SO<sub>3<\/sub>] = 2.00 <em data-effect=\"italics\">M<\/em>, [SO<sub>2<\/sub>] = 2.00 <em data-effect=\"italics\">M<\/em>, [O<sub>2<\/sub>] = 2.00 <em data-effect=\"italics\">M\r\n<\/em><\/li>\r\n \t<li>[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right){K}_{P}=6.5\\text{atm;}[\/latex] initial pressures: SO<sub>2<\/sub> = 1.00 atm, O<sub>2<\/sub> = 1.130 atm, SO<sub>3<\/sub> = 0 atm<\/li>\r\n \t<li>[latex]2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NOCl}\\left(g\\right){K}_{P}=2.5\\times {10}^{3}\\text{;}[\/latex] initial pressures: NO = 1.00 atm, Cl<sub>2<\/sub> = 1.00 atm, NOCl = 0 atm<\/li>\r\n \t<li>[latex]{\\text{N}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NO}\\left(g\\right){K}_{c}=0.050\\text{;}[\/latex] [N<sub>2<\/sub>] = 0.100 <em data-effect=\"italics\">M<\/em>, [O<sub>2<\/sub>] = 0.200 <em data-effect=\"italics\">M<\/em>, [NO] = 1.00 <em data-effect=\"italics\">M<\/em><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The following reaction has <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = 4.50 \u00d7 10<sup>-5<\/sup> at 720 K.\r\n[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]\r\nIf a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? <em data-effect=\"italics\">P<\/em>(NH<sub>3<\/sub>) = 93 atm, <em data-effect=\"italics\">P<\/em>(N<sub>2<\/sub>) = 48 atm, and <em data-effect=\"italics\">P<\/em>(H<sub>2<\/sub>) = 52<\/li>\r\n \t<li>Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?<\/li>\r\n \t<li>Which of the systems described in question 10 give homogeneous equilibria? Which give heterogeneous equilibria?<\/li>\r\n \t<li>Which of the systems described in question 11\u00a0give homogeneous equilibria? Which give heterogeneous equilibria?<\/li>\r\n \t<li>For which of the reactions in question 10\u00a0does <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (calculated using concentrations) equal <em data-effect=\"italics\">K<sub>P<\/sub><\/em> (calculated using pressures)?<\/li>\r\n \t<li>For which of the reactions in question 11\u00a0does <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (calculated using concentrations) equal <em data-effect=\"italics\">K<sub>P<\/sub><\/em> (calculated using pressures)?<\/li>\r\n \t<li>Convert the values of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> to values of <em data-effect=\"italics\">K<sub>P<\/sub><\/em> or the values of <em data-effect=\"italics\">K<sub>P<\/sub><\/em> to values of <em data-effect=\"italics\">K<sub>c<\/sub><\/em>.\r\n<ol>\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){K}_{c}=0.50\\text{at}400\\text{\\textdegree C}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{H}}_{2}+{\\text{I}}_{2}\\rightleftharpoons2\\text{HI}{K}_{c}=50.2\\text{at}448\\text{\\textdegree C}[\/latex]<\/li>\r\n \t<li>[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\\text{\\textdegree C}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right){K}_{P}=0.122\\text{at}50\\text{\\textdegree C}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Convert the values of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> to values of <em data-effect=\"italics\">K<sub>P<\/sub><\/em> or the values of <em data-effect=\"italics\">K<sub>P<\/sub><\/em> to values of <em data-effect=\"italics\">K<sub>c<\/sub><\/em>.\r\n<ol>\r\n \t<li>[latex]{\\text{Cl}}_{2}\\left(g\\right)+{\\text{Br}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{BrCl}\\left(g\\right){K}_{c}=4.7\\times {10}^{-2}\\text{at}25\\text{\\textdegree C}[\/latex]<\/li>\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){K}_{P}=48.2\\text{at}500\\text{\\textdegree C}[\/latex]<\/li>\r\n \t<li>[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\\text{\\textdegree C}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right){K}_{P}=0.196\\text{at}60\\text{\\textdegree C}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What is the value of the equilibrium constant expression for the change [latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex] at 30 \u00b0C?<\/li>\r\n \t<li>Write the expression of the reaction quotient for the ionization of HOCN in water.<\/li>\r\n \t<li>Write the reaction quotient expression for the ionization of NH<sub>3<\/sub> in water.<\/li>\r\n \t<li>What is the approximate value of the equilibrium constant <em data-effect=\"italics\">K<sub>P<\/sub><\/em> for the change [latex]{\\text{C}}_{2}{\\text{H}}_{5}{\\text{OC}}_{2}{\\text{H}}_{5}\\left(l\\right)\\rightleftharpoons{\\text{C}}_{2}{\\text{H}}_{5}{\\text{OC}}_{2}{\\text{H}}_{5}\\left(g\\right)[\/latex] at 25 \u00b0C. (Vapor pressure was described in the previous chapter on liquids and solids; refer back to this chapter to find the relevant information needed to solve this problem.)<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"bcc-box bcc-info\">\r\n<h4>Selected Answers<\/h4>\r\n2.\u00a0Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle, because the system is not closed; one of the components of the equilibrium, the Br<sub>2<\/sub> vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase.\r\n\r\n4. (a) <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = [Ag+][Cl-] &lt; 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 <em data-effect=\"italics\">M<\/em>;\r\n\r\n(b) [latex]{K}_{c}=\\frac{1}{\\left[{\\text{Pb}}^{2+}\\right]{\\left[{\\text{Cl}}^{-}\\right]}^{2}}[\/latex] &gt; 1 because PbCl<sub>2<\/sub> is insoluble and formation of the solid will reduce the concentration of ions to a low level (&lt;1 <em data-effect=\"italics\">M<\/em>).\r\n\r\n6.\u00a0Since [latex]{K}_{c}=\\frac{\\left[{\\text{C}}_{6}{\\text{H}}_{6}\\right]}{{\\left[{\\text{C}}_{2}{\\text{H}}_{2}\\right]}^{3}}[\/latex], a value of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> \u2248 10 means that C<sub>6<\/sub>H<sub>6<\/sub> predominates over C<sub>2<\/sub>H<sub>2<\/sub>. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.\r\n\r\n8.\u00a0<em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1; the product must be formed in overwhelmingly large proportions.\r\n\r\n10.\u00a0(a) [latex]{Q}_{c}=\\frac{\\left[{\\text{CH}}_{3}\\text{Cl}\\right]\\left[\\text{HCl}\\right]}{\\left[{\\text{CH}}_{4}\\right]\\left[{\\text{Cl}}_{2}\\right]}\\text{;}[\/latex] (b) [latex]{Q}_{c}=\\frac{{\\left[\\text{NO}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]}\\text{;}[\/latex] (c) [latex]{Q}_{c}=\\frac{{\\left[{\\text{SO}}_{3}\\right]}^{2}}{{\\left[{\\text{SO}}_{2}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]}\\text{;}[\/latex] (d) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = [SO<sub>2<\/sub>]; (e) [latex]{Q}_{c}=\\frac{1}{\\left[{\\text{P}}_{4}\\right]{\\left[{\\text{O}}_{2}\\right]}^{5}}\\text{;}[\/latex] (f) [latex]{Q}_{c}=\\frac{{\\left[\\text{Br}\\right]}^{2}}{\\left[{\\text{Br}}_{2}\\right]}\\text{;}[\/latex] (g) [latex]{Q}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]}{\\left[{\\text{CH}}_{4}\\right]{\\left[{\\text{O}}_{2}\\right]}^{2}}\\text{;}[\/latex] (h) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = [H<sub>2<\/sub>O]<sup>5<\/sup>\r\n\r\n12.\u00a0(a) [latex]{Q}_{c}=\\frac{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}{{\\left[{\\text{NH}}_{3}\\right]}^{2}}=\\frac{\\left(1.00\\right){\\left(1.00\\right)}^{3}}{{\\left(0.20\\right)}^{2}}=25[\/latex]\u00a0<em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &lt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em>, proceeds left;\r\n(b) [latex]{Q}_{P}=\\frac{{P}_{{\\text{N}}_{2}}{\\left({P}_{{\\text{H}}_{2}}\\right)}^{3}}{{\\left({P}_{{\\text{NH}}_{3}}\\right)}^{2}}=\\frac{\\left(2.0\\right){\\left(1.0\\right)}^{3}}{{\\left(3.0\\right)}^{2}}=0.22[\/latex]\u00a0<em data-effect=\"italics\">Q<sub>P<\/sub><\/em> &lt; <em data-effect=\"italics\">K<sub>P<\/sub><\/em>, proceeds right;\r\n(c) [latex]{Q}_{c}=\\frac{{\\left[{\\text{SO}}_{2}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]}{{\\left[{\\text{SO}}_{3}\\right]}^{2}}=\\frac{{\\left(1.00\\right)}^{2}\\left(1.00\\right)}{\\left(0\\right)}=\\text{undefined}[\/latex]\u00a0<em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &gt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em>, proceeds left;\r\n(d) [latex]{Q}_{P}=\\frac{{\\left({P}_{{\\text{SO}}_{2}}\\right)}^{2}{P}_{{\\text{O}}_{2}}}{{\\left({P}_{{\\text{SO}}_{3}}\\right)}^{2}}=\\frac{{\\left(1.00\\right)}^{2}\\left(1.00\\right)}{{\\left(1.00\\right)}^{2}}=1.00[\/latex]\u00a0<em data-effect=\"italics\">Q<sub>P<\/sub><\/em> &gt; <em data-effect=\"italics\">K<sub>P<\/sub><\/em>, proceeds right;\r\n(e) [latex]{Q}_{P}=\\frac{{\\left({P}_{\\text{NOCl}}\\right)}^{2}}{{\\left({P}_{\\text{NO}}\\right)}^{2}{P}_{{\\text{Cl}}_{2}}}=\\frac{{\\left(0\\right)}^{2}}{{\\left(1.00\\right)}^{2}\\left(1.00\\right)}=0[\/latex]\u00a0<em data-effect=\"italics\">Q<sub>P<\/sub><\/em> &lt; <em data-effect=\"italics\">K<sub>P<\/sub><\/em>, proceeds right;\r\n(f) [latex]{Q}_{c}=\\frac{{\\left[\\text{NO}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]}=\\frac{{\\left(10.0\\right)}^{2}}{\\left(5.00\\right)\\left(5.00\\right)}=4[\/latex] <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &gt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em>, proceeds left\r\n\r\n14.\u00a0The reaction quotient expression for this problem is [latex]{Q}_{p}=\\frac{{\\left({P}_{{\\text{NH}}_{3}}\\right)}^{2}}{\\left({P}_{{\\text{N}}_{2}}\\right){\\left({P}_{{\\text{H}}_{2}}\\right)}^{3}}\\text{.}[\/latex] Plugging in the given values of partial pressures gives [latex]\\frac{{\\left(93\\right)}^{2}}{\\left(\\left(48\\right)\\times {\\left(52\\right)}^{3}\\right)}[\/latex], so <em data-effect=\"italics\">Q<sub>p<\/sub><\/em> = 1.3 \u00d7 10<sup>-3<\/sup>. Since this value is larger than <em data-effect=\"italics\">K<sub>P<\/sub><\/em> (4.50 \u00d7 10<sup>-5<\/sup>), the system will shift toward the reactants to reach equilibrium.\r\n\r\n16.\u00a0(a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous\r\n\r\n18.\u00a0When the number of gaseous components are the same on both sides of the equilibrium expression, <em data-effect=\"italics\">K<sub>c<\/sub><\/em> will equal <em data-effect=\"italics\">K<sub>P<\/sub><\/em>. This situation occurs in (a) and (b).\r\n\r\n20.\u00a0[latex]{K}_{P}={K}_{c}{\\left(RT\\right)}^{\\Delta n}[\/latex], where \u0394<em data-effect=\"italics\">n<\/em> is the sum of gaseous products minus the sum of gaseous reactants. (a) \u0394<em data-effect=\"italics\">n<\/em> = (2) - (1 + 3) = -2, <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = 0.50[0.08206 \u00d7 673.15]<sup>-2<\/sup> = 1.6 \u00d7 10<sup>-4<\/sup>; (b) \u0394<em data-effect=\"italics\">n<\/em> = (2) - (1 + 1) = 0, <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>)<sup>0<\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = 50.2; (c) \u0394<em data-effect=\"italics\">n<\/em> = (10) - (0) = 10, <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = <em data-effect=\"italics\">K<sub>P<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>)<sup>-\u0394<em data-effect=\"italics\">n<\/em><\/sup>, <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = 4.08 \u00d7 10<sup>-25<\/sup>[0.08206 \u00d7 298.15]<sup>-10<\/sup> = 5.31 \u00d7 10<sup>-39<\/sup>; (d) \u0394<em data-effect=\"italics\">n<\/em> = (1) - (0) = 1, <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = 0.122(0.08206 \u00d7 323.15)<sup>-1<\/sup> = 4.60 \u00d7 10<sup>-3<\/sup>\r\n\r\n22.\u00a0The equilibrium expression for this transformation is [latex]{K}_{P}={P}_{{\\text{H}}_{2}\\text{O}}\\text{.}[\/latex] The vapor pressure of H<sub>2<\/sub>O at 30 \u00baC is 31.8 torr. Converting to atmospheres gives [latex]3.18\\cancel{\\text{torr}}\\times \\frac{1\\text{atm}}{760\\cancel{\\text{torr}}}=0.042\\text{atm.}[\/latex] Therefore, [latex]{K}_{P}={\\text{P}}_{{\\text{H}}_{2}\\text{O}}=0.042\\text{.}[\/latex]\r\n\r\n24.\u00a0[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)\\text{.}[\/latex] Because the concentration of water is a constant, the term [H<sub>2<\/sub>O] is normally incorporated into the reaction quotient as well as the final equilibrium constant. [latex]{Q}_{c}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{HN}}_{3}\\right]}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Glossary<\/h3>\r\n<p id=\"fs-idp63345136\" data-type=\"definition\"><strong><span data-type=\"term\">equilibrium constant (<em data-effect=\"italics\">K<\/em>)\r\n<\/span><\/strong>value of the reaction quotient for a system at equilibrium<\/p>\r\n<p id=\"fs-idp102632368\" data-type=\"definition\"><strong><span data-type=\"term\">heterogeneous equilibria\r\n<\/span><\/strong>equilibria between reactants and products in different phases<\/p>\r\n<p id=\"fs-idp35066960\" data-type=\"definition\"><strong><span data-type=\"term\">homogeneous equilibria\r\n<\/span><\/strong>equilibria within a single phase<\/p>\r\n<p id=\"fs-idp55285520\" data-type=\"definition\"><strong><span data-type=\"term\"><em data-effect=\"italics\">K<sub>c\r\n<\/sub><\/em><\/span><\/strong>equilibrium constant for reactions based on concentrations of reactants and products<\/p>\r\n<p id=\"fs-idp61369840\" data-type=\"definition\"><strong><span data-type=\"term\"><em data-effect=\"italics\">K<sub>P\r\n<\/sub><\/em><\/span><\/strong>equilibrium constant for gas-phase reactions based on partial pressures of reactants and products<\/p>\r\n<p id=\"fs-idp89564720\" data-type=\"definition\"><strong><span data-type=\"term\">law of mass action\r\n<\/span><\/strong>when a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant<\/p>\r\n<p id=\"fs-idp74361936\" data-type=\"definition\"><strong><span data-type=\"term\">reaction quotient (<em data-effect=\"italics\">Q<\/em>)\r\n<\/span><\/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>\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>LEARNING OBJECTIVES<\/h3>\n<p>By the end of this module, you will be able to:<\/p>\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 id=\"fs-idp120703584\">Now that we have a symbol [latex]\\rightleftharpoons[\/latex] to designate reversible reactions, we will need a way to express mathematically how the amounts of reactants and products affect the equilibrium of the system. A general equation for a reversible reaction may be written as follows:<\/p>\n<div id=\"fs-idp5986304\" data-type=\"equation\">[latex]m\\text{A}+n\\text{B}+\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]<\/div>\n<p id=\"fs-idp11973152\">We can write the <span data-type=\"term\">reaction quotient (<em data-effect=\"italics\">Q<\/em>)<\/span> for this equation. When evaluated using concentrations, it is called <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>. We use brackets to indicate molar concentrations of reactants and products.<\/p>\n<div id=\"fs-idp44297776\" data-type=\"equation\">[latex]{Q}_{c}=\\frac{{\\left[\\text{C}\\right]}^{x}{\\left[\\text{D}\\right]}^{y}}{{\\left[\\text{A}\\right]}^{m}{\\left[\\text{B}\\right]}^{n}}[\/latex]<\/div>\n<p id=\"fs-idm7959200\">The reaction quotient is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation. For example, the reaction quotient for the reversible reaction [latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)[\/latex] is given by this expression:<\/p>\n<div id=\"fs-idp23922448\" data-type=\"equation\">[latex]{Q}_{c}=\\frac{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}[\/latex]<\/div>\n<div id=\"fs-idp168372720\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 1<\/h3>\n<h4 id=\"fs-idp68042176\"><span data-type=\"title\">Writing Reaction Quotient Expressions<\/span><\/h4>\n<p>Write the expression for the reaction quotient for each of the following reactions:<\/p>\n<p id=\"fs-idp246365472\">(a) [latex]3{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{O}}_{3}\\left(g\\right)[\/latex]<\/p>\n<p id=\"fs-idp142495760\">(b) [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 id=\"fs-idp82472256\">(c) [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]<\/p>\n<h4 id=\"fs-idp63878256\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>(a) [latex]{Q}_{c}=\\frac{{\\left[{O}_{3}\\right]}^{2}}{{\\left[{O}_{2}\\right]}^{3}}[\/latex]<\/p>\n<p id=\"fs-idp177788304\">(b) [latex]{Q}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}[\/latex]<\/p>\n<p id=\"fs-idp171443200\">(c) [latex]{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]<\/p>\n<h4 id=\"fs-idp17156768\"><span data-type=\"title\">Check Your Learning<\/span><\/h4>\n<p>Write the expression for the reaction quotient for each of the following reactions:<\/p>\n<p id=\"fs-idp35394896\">(a) [latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right)[\/latex]<\/p>\n<p id=\"fs-idp63748560\">(b) [latex]{\\text{C}}_{4}{\\text{H}}_{8}\\left(g\\right)\\rightleftharpoons2{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)[\/latex]<\/p>\n<p id=\"fs-idp105907744\">(c) [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]<\/p>\n<div id=\"fs-idp167282768\" data-type=\"note\">\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp7633728\" style=\"text-align: right;\">(a) [latex]{Q}_{c}=\\frac{{\\left[{\\text{SO}}_{3}\\right]}^{2}}{{\\left[{\\text{SO}}_{2}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]}[\/latex]<\/p>\n<p style=\"text-align: right;\">(b) [latex]{Q}_{c}=\\frac{{\\left[{\\text{C}}_{2}{\\text{H}}_{4}\\right]}^{2}}{\\left[{\\text{C}}_{4}{\\text{H}}_{8}\\right]}[\/latex]<\/p>\n<p style=\"text-align: right;\">(c) [latex]{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]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idp138846544\">The numeric value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> for a given reaction varies; it depends on the concentrations of products and reactants present at the time when <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> is determined. When pure reactants are mixed, <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> is initially zero because there are no products present at that point. As the reaction proceeds, the value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> increases as the concentrations of the products increase and the concentrations of the reactants simultaneously decrease Figure 1. When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change.<\/p>\n<div id=\"attachment_4880\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4880\" class=\"size-large wp-image-4880\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/08\/23214441\/CNX_Chem_13_02_quotient-1024x850.jpg\" alt=\"Three graphs are shown and labeled, \u201ca,\u201d \u201cb,\u201d and \u201cc.\u201d All three 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\" width=\"1024\" height=\"850\" \/><\/p>\n<p id=\"caption-attachment-4880\" class=\"wp-caption-text\">Figure 1. (a) The change in the concentrations of reactants and products is depicted as the 2SO2(g)+O2(g)\u21cc2SO3(g) reaction approaches equilibrium. (b) The change in concentrations of reactants and products is depicted as the reaction 2SO3(g)\u21cc2SO2(g)+O2(g) approaches equilibrium. (c) The graph shows the change in the value of the reaction quotient as the reaction approaches equilibrium.<\/p>\n<\/div>\n<p id=\"fs-idm122432\">When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the <span data-type=\"term\">equilibrium constant (<em data-effect=\"italics\">K<\/em>)<\/span> of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as <em data-effect=\"italics\">K<sub>c<\/sub><\/em>.<\/p>\n<p id=\"fs-idp279896080\">That a reaction quotient always assumes the same value at equilibrium can be expressed as:<\/p>\n<div id=\"fs-idp266400656\" data-type=\"equation\">[latex]{Q}_{c}\\text{at equilibrium}={K}_{c}=\\frac{{\\left[\\text{C}\\right]}^{x}{\\left[\\text{D}\\right]}^{y}\\ldots }{{\\left[\\text{A}\\right]}^{m}{\\left[\\text{B}\\right]}^{n}\\ldots}[\/latex]<\/div>\n<p id=\"fs-idp73447152\">This equation is a mathematical statement of the <span data-type=\"term\">law of mass action<\/span>: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.<\/p>\n<div id=\"fs-idp54423808\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 2<\/h3>\n<h4 id=\"fs-idp23994768\"><span data-type=\"title\">Evaluating a Reaction Quotient<\/span><\/h4>\n<p>Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:<\/p>\n<div id=\"fs-idp64726272\" data-type=\"equation\">[latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)[\/latex]<\/div>\n<p id=\"fs-idp17381824\">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 data-effect=\"italics\">M<\/em> and [N<sub>2<\/sub>O<sub>4<\/sub>] = 0.042 <em data-effect=\"italics\">M<\/em>.<\/p>\n<p id=\"fs-idp89605104\">(a) What is the value of the reaction quotient before any reaction occurs?<\/p>\n<p id=\"fs-idp84731776\">(b) What is the value of the equilibrium constant for the reaction?<\/p>\n<h4 id=\"fs-idp179491728\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>(a) Before any product is formed, [latex]\\left[{\\text{NO}}_{2}\\right]=\\frac{0.10\\text{mol}}{1.0\\text{L}}=0.10M[\/latex], and [N<sub>2<\/sub>O<sub>4<\/sub>] = 0 <em data-effect=\"italics\">M<\/em>. Thus,<\/p>\n<div id=\"fs-idp173627248\" data-type=\"equation\">[latex]{Q}_{c}=\\frac{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}=\\frac{0}{{0.10}^{2}}=0[\/latex]<\/div>\n<p id=\"fs-idp110962640\">(b) At equilibrium, the value of the equilibrium constant is equal to the value of the reaction quotient. At equilibrium, [latex]{K}_{c}={Q}_{c}=\\frac{\\left[{\\text{N}}_{2}{\\text{O}}_{4}\\right]}{{\\left[{\\text{NO}}_{2}\\right]}^{2}}=\\frac{0.042}{{0.016}^{2}}=1.6\\times {10}^{2}\\text{.}[\/latex] The equilibrium constant is 1.6 \u00d7 10<sup>2<\/sup>.<\/p>\n<p id=\"fs-idp120754816\">Note that dimensional analysis would suggest the unit for this <em data-effect=\"italics\">K<sub>c<\/sub><\/em> value should be <em data-effect=\"italics\">M<\/em><sup>-1<\/sup>. However, it is common practice to omit units for <em data-effect=\"italics\">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 data-effect=\"italics\">K<sub>c<\/sub><\/em> values are truly unitless.<\/p>\n<h4 id=\"fs-idp284528080\"><span data-type=\"title\">Check Your Learning<\/span><\/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 data-effect=\"italics\">M<\/em>, [O<sub>2<\/sub>] = 0.35 <em data-effect=\"italics\">M<\/em>, and [SO<sub>3<\/sub>] = 1.1 <em data-effect=\"italics\">M<\/em>. What is the value of the equilibrium constant, <em data-effect=\"italics\">K<sub>c<\/sub><\/em>?<\/p>\n<div id=\"fs-idp10573648\" data-type=\"note\">\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong><em data-effect=\"italics\">K<sub>c<\/sub> =<\/em> 4.3<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-idm80100288\">The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A large value for <em data-effect=\"italics\">K<sub>c<\/sub><\/em> indicates that equilibrium is attained only after the reactants have been largely converted into products. A small value of <em data-effect=\"italics\">K<sub>c<\/sub><\/em>\u2014much less than 1\u2014indicates that equilibrium is attained when only a small proportion of the reactants have been converted into products.<\/p>\n<p id=\"fs-idp246311264\">Once a value of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> is known for a reaction, it can be used to predict directional shifts when compared to the value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>. A system that is not at equilibrium will proceed in the direction that establishes equilibrium. The data in Figure 13.7\u00a0illustrate this. When heated to a consistent temperature, 800 \u00b0C, different starting mixtures of CO, H<sub>2<\/sub>O, CO<sub>2<\/sub>, and H<sub>2<\/sub> react to reach compositions adhering to the same equilibrium (the value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> changes until it equals the value of <em data-effect=\"italics\">K<sub>c<\/sub><\/em>). This value is 0.640, the equilibrium constant for the reaction under these conditions.<\/p>\n<div id=\"fs-idp63923872\" data-type=\"equation\">[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){K}_{c}=0.640\\text{T}=800\\text{\\textdegree C}[\/latex]<\/div>\n<p id=\"fs-idp94527104\">It is important to recognize that an equilibrium can be established starting either from reactants or from products, or from a mixture of both. For example, equilibrium was established from Mixture 2 in Figure 13.7\u00a0when the products of the reaction were heated in a closed container. In fact, one technique used to determine whether a reaction is truly at equilibrium is to approach equilibrium starting with reactants in one experiment and starting with products in another. If the same value of the reaction quotient is observed when the concentrations stop changing in both experiments, then we may be certain that the system has reached equilibrium.<\/p>\n<div id=\"attachment_4881\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4881\" class=\"size-large wp-image-4881\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/08\/23214443\/CNX_Chem_13_02_mixtures-1024x440.jpg\" 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 and the blue bar near 0.05. For mixture two, the green bar is near 0.05, and the yellow bar is near 0.09. For mixture 3, the red bar is near 0.01, the blue bar is slightly above that with green and yellow topping it off at 0.02. On the right graph, the bar for mixture one shows the red bar slightly above 0.01, the blue bar stacked on it rising slightly above 0.02, the green rising near 0.04, and the yellow bar reaching near 0.05. A label above this bar reads, \u201cQ equals 0.640.\u201d The bar for mixture two shows the red bar slightly above 0.02, the blue bar stacked on it rising near 0.05, the green rising near 0.07, and the yellow bar reaching near 0.10. A label above this bar reads \u201cQ equals 0.640.\u201d The bar for mixture three shows the red bar near 0.01, the blue bar stacked on it rising slightly above 0.01, the green rising near 0.02, and the yellow bar reaching 0.02. A label above this bar reads \u201cQ equals 0.640\u201d.\" width=\"1024\" height=\"440\" \/><\/p>\n<p id=\"caption-attachment-4881\" class=\"wp-caption-text\">Figure 2. Concentrations of three mixtures are shown before and after reaching equilibrium at 800 \u00b0C for the so-called water gas shift reaction: CO(g)+H2O(g)\u21ccCO2(g)+H2(g).<\/p>\n<\/div>\n<div id=\"fs-idp174376240\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 3<\/h3>\n<h4 id=\"fs-idp60883568\"><span data-type=\"title\">Predicting the Direction of Reaction<\/span><\/h4>\n<p>Given below are the starting concentrations of reactants and products for three experiments involving this reaction:<\/p>\n<div id=\"fs-idm31469104\" data-type=\"equation\">[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]<\/div>\n<div id=\"fs-idp89310352\" data-type=\"equation\">[latex]{K}_{c}=0.64[\/latex]<\/div>\n<p id=\"fs-idp162489744\">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.\" data-label=\"\">\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 data-effect=\"italics\">M<\/em><\/td>\n<td>0.011 <em data-effect=\"italics\">M<\/em><\/td>\n<td>0.0094 <em data-effect=\"italics\">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 data-effect=\"italics\">M<\/em><\/td>\n<td>0.0011 <em data-effect=\"italics\">M<\/em><\/td>\n<td>0.0025 <em data-effect=\"italics\">M<\/em><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[CO<sub>2<\/sub>]<sub>i<\/sub><\/td>\n<td>0.0040 <em data-effect=\"italics\">M<\/em><\/td>\n<td>0.037 <em data-effect=\"italics\">M<\/em><\/td>\n<td>0.0015 <em data-effect=\"italics\">M<\/em><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[H<sub>2<\/sub>]<sub>i<\/sub><\/td>\n<td>0.0040 <em data-effect=\"italics\">M<\/em><\/td>\n<td>0.046 <em data-effect=\"italics\">M<\/em><\/td>\n<td>0.0076 <em data-effect=\"italics\">M<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4 id=\"fs-idp3711408\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>Experiment 1:<\/p>\n<div id=\"fs-idp24553712\" data-type=\"equation\">[latex]{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]<\/div>\n<p id=\"fs-idm1136592\"><em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &lt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (0.039 &lt; 0.64)<\/p>\n<p id=\"fs-idp267578784\">The reaction will shift to the right.<\/p>\n<p id=\"fs-idp56047184\">Experiment 2:<\/p>\n<div id=\"fs-idp15286912\" data-type=\"equation\">[latex]{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]<\/div>\n<p id=\"fs-idp3859856\"><em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &gt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (140 &gt; 0.64)<\/p>\n<p id=\"fs-idp68828672\">The reaction will shift to the left.<\/p>\n<p id=\"fs-idp101164320\">Experiment 3:<\/p>\n<div id=\"fs-idp74212144\" data-type=\"equation\">[latex]{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]<\/div>\n<p id=\"fs-idp70179520\"><em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &lt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (0.48 &lt; 0.64)<\/p>\n<p id=\"fs-idp155462992\">The reaction will shift to the right.<\/p>\n<h4 id=\"fs-idm76258576\"><span data-type=\"title\">Check Your Learning<\/span><\/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<p id=\"fs-idp59277392\">(a) A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl:<\/p>\n<div id=\"fs-idp56406272\" data-type=\"equation\">[latex]2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NOCl}\\left(g\\right){K}_{c}=4.6\\times {10}^{4}\\text{.}[\/latex]<\/div>\n<p id=\"fs-idp278668928\">(b) 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>:<\/p>\n<div id=\"fs-idp278669312\" data-type=\"equation\">[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right){K}_{c}=0.060[\/latex]<\/div>\n<p id=\"fs-idp110625216\">(c) A 2.00-L flask containing 230 g of SO<sub>3<\/sub>(g):<\/p>\n<div id=\"fs-idp294747856\" data-type=\"equation\">[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right){K}_{c}=0.230[\/latex]<\/div>\n<div id=\"fs-idp32992592\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong>(a) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 6.45 \u00d7 103, shifts right. (b) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 0.12, shifts left. (c) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 0, shifts right<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-idp168987520\">In Example 2, it was mentioned that the common practice is to omit units when evaluating reaction quotients and equilibrium constants. It should be pointed out that using concentrations in these computations is a convenient but simplified approach that sometimes leads to results that seemingly conflict with the law of mass action. For example, equilibria involving aqueous ions often exhibit equilibrium constants that vary quite significantly (are <em data-effect=\"italics\">not<\/em> constant) at high solution concentrations. This may be avoided by computing <em data-effect=\"italics\">K<sub>c<\/sub><\/em> values using the <em data-effect=\"italics\">activities<\/em> of the reactants and products in the equilibrium system instead of their concentrations. The <span data-type=\"term\">activity<\/span> of a substance is a measure of its effective concentration under specified conditions. While a detailed discussion of this important quantity is beyond the scope of an introductory text, it is necessary to be aware of a few important aspects:<\/p>\n<ul id=\"fs-idp284227328\" data-bullet-style=\"bullet\">\n<li>Activities are dimensionless (unitless) quantities and are in essence \u201cadjusted\u201d concentrations.<\/li>\n<li>For relatively dilute solutions, a substance&#8217;s activity and its molar concentration are roughly equal.<\/li>\n<li>Activities for pure condensed phases (solids and liquids) are equal to 1.<\/li>\n<\/ul>\n<p id=\"fs-idp43748544\">As a consequence of this last consideration, <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> and <em data-effect=\"italics\">K<sub>c<\/sub><\/em> expressions do not contain terms for solids or liquids (being numerically equal to 1, these terms have no effect on the expression&#8217;s value)<em data-effect=\"italics\">.<\/em> Several examples of equilibria yielding such expressions will be encountered below.<\/p>\n<section id=\"fs-idp86160096\" data-depth=\"1\">\n<h2 data-type=\"title\">Homogeneous Equilibria<\/h2>\n<p id=\"fs-idm54871216\">A <span data-type=\"term\">homogeneous equilibrium<\/span> is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). In this chapter, we will concentrate on the two most common types of homogeneous equilibria: those occurring in liquid-phase solutions and those involving exclusively gaseous species. Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both. Several examples are provided below.<\/p>\n<div id=\"fs-idp38922960\" data-type=\"equation\">[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){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]<\/div>\n<div id=\"fs-idp25144000\" data-type=\"equation\">[latex]{\\text{I}}_{2}\\left(aq\\right)+{\\text{I}}^{-}\\left(aq\\right)\\rightleftharpoons{\\text{I}}_{3}{}^{-}\\left(aq\\right){K}_{c}=\\frac{\\left[{\\text{I}}_{3}{}^{-}\\right]}{\\left[{\\text{I}}_{2}\\right]\\left[{\\text{I}}^{-}\\right]}[\/latex]<\/div>\n<div id=\"fs-idm76151744\" data-type=\"equation\">[latex]{\\text{Hg}}_{2}{}^{2+}\\left(aq\\right)+{\\text{NO}}_{3}{}^{-}\\left(aq\\right)+3{\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)\\rightleftharpoons2{\\text{Hg}}^{2+}\\left(aq\\right)+{\\text{HNO}}_{2}\\left(aq\\right)+4{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/div>\n<div id=\"fs-idp122601856\" data-type=\"equation\">[latex]{K}_{c}=\\frac{{\\left[{\\text{Hg}}^{2+}\\right]}^{2}\\left[{\\text{HNO}}_{2}\\right]}{\\left[{\\text{Hg}}_{2}{}^{2+}\\right]\\left[{\\text{NO}}_{3}{}^{-}\\right]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{3}}[\/latex]<\/div>\n<div id=\"fs-idp101805696\" data-type=\"equation\">[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){K}_{c}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{F}}^{-}\\right]}{\\left[\\text{HF}\\right]}[\/latex]<\/div>\n<div id=\"fs-idp15332976\" data-type=\"equation\">[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){K}_{c}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}[\/latex]<\/div>\n<p id=\"fs-idp11719264\">In each of these examples, the equilibrium system is an aqueous solution, as denoted by the <em data-effect=\"italics\">aq<\/em> annotations on the solute formulas. Since H<sub>2<\/sub>O(<em data-effect=\"italics\">l<\/em>) is the solvent for these solutions, its concentration does not appear as a term in the <em data-effect=\"italics\">K<sub>c<\/sub><\/em> expression, as discussed above, even though it may also appear as a reactant or product in the chemical equation.<\/p>\n<p id=\"fs-idp11805840\">Reactions in which all reactants and products are gases represent a second class of homogeneous equilibria. We use molar concentrations in the following examples, but we will see shortly that partial pressures of the gases may be used as well.<\/p>\n<div id=\"fs-idp17340352\" data-type=\"equation\">[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){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]<\/div>\n<div id=\"fs-idp100807584\" data-type=\"equation\">[latex]3{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{O}}_{3}\\left(g\\right){K}_{c}=\\frac{{\\left[{\\text{O}}_{3}\\right]}^{2}}{{\\left[{\\text{O}}_{2}\\right]}^{3}}[\/latex]<\/div>\n<div id=\"fs-idp68583712\" data-type=\"equation\">[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right){K}_{c}=\\frac{{\\left[{\\text{NH}}_{3}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}[\/latex]<\/div>\n<div id=\"fs-idp86498496\" data-type=\"equation\">[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){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]<\/div>\n<p id=\"fs-idp58545648\">Note that the concentration of H<sub>2<\/sub>O(<em data-effect=\"italics\">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 id=\"fs-idp165025968\">Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the reaction quotient because the partial pressure of a gas is directly proportional to its concentration at constant temperature. This relationship can be derived from the ideal gas equation, where <em data-effect=\"italics\">M<\/em> is the molar concentration of gas, [latex]\\frac{n}{V}\\text{.}[\/latex]<\/p>\n<div id=\"fs-idp111709808\" data-type=\"equation\">[latex]PV=nRT[\/latex]<\/div>\n<div id=\"fs-idm75268224\" data-type=\"equation\">[latex]P=\\left(\\frac{n}{V}\\right)RT[\/latex]<\/div>\n<div id=\"fs-idp44758960\" data-type=\"equation\">[latex]=MRT[\/latex]<\/div>\n<p id=\"fs-idp123497808\">Thus, at constant temperature, the pressure of a gas is directly proportional to its concentration.<\/p>\n<p id=\"fs-idp43798512\">Using the partial pressures of the gases, we can write the reaction quotient for the system [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] by following the same guidelines for deriving concentration-based expressions:<\/p>\n<div id=\"fs-idp279715456\" data-type=\"equation\">[latex]{Q}_{P}=\\frac{{P}_{{\\text{C}}_{2}{\\text{H}}_{4}}{P}_{{\\text{H}}_{2}}}{{P}_{{\\text{C}}_{2}{\\text{H}}_{6}}}[\/latex]<\/div>\n<p id=\"fs-idp180873632\">In this equation we use <em data-effect=\"italics\">Q<sub>P<\/sub><\/em> to indicate a reaction quotient written with partial pressures: [latex]{P}_{{\\text{C}}_{2}{\\text{H}}_{6}}[\/latex] is the partial pressure of C<sub>2<\/sub>H<sub>6<\/sub>; [latex]{P}_{{\\text{H}}_{2}}[\/latex], the partial pressure of H<sub>2<\/sub>; and [latex]{P}_{{\\text{C}}_{2}{\\text{H}}_{6}}[\/latex], the partial pressure of C<sub>2<\/sub>H<sub>4<\/sub>. At equilibrium:<\/p>\n<div id=\"fs-idp116676256\" data-type=\"equation\">[latex]{K}_{P}={Q}_{P}=\\frac{{P}_{{\\text{C}}_{2}{\\text{H}}_{4}}{P}_{{\\text{H}}_{2}}}{{P}_{{\\text{C}}_{2}{\\text{H}}_{6}}}[\/latex]<\/div>\n<p id=\"fs-idp155025056\">The subscript <em data-effect=\"italics\">P<\/em> in the symbol <span data-type=\"term\"><em data-effect=\"italics\">K<sub>P<\/sub><\/em><\/span> designates an equilibrium constant derived using partial pressures instead of concentrations. The equilibrium constant, <em data-effect=\"italics\">K<sub>P<\/sub><\/em>, is still a constant, but its numeric value may differ from the equilibrium constant found for the same reaction by using concentrations.<\/p>\n<p id=\"fs-idp122706016\">Conversion between a value for <span data-type=\"term\"><em data-effect=\"italics\">K<sub>c<\/sub><\/em><\/span>, an equilibrium constant expressed in terms of concentrations, and a value for <em data-effect=\"italics\">K<sub>P<\/sub><\/em>, an equilibrium constant expressed in terms of pressures, is straightforward (a <em data-effect=\"italics\">K<\/em> or <em data-effect=\"italics\">Q<\/em> without a subscript could be either concentration or pressure).<\/p>\n<p id=\"fs-idp160302688\">The equation relating <em data-effect=\"italics\">K<sub>c<\/sub><\/em> and <em data-effect=\"italics\">K<sub>P<\/sub><\/em> is derived as follows. For the gas-phase reaction [latex]m\\text{A}+n\\text{B}\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex]:<\/p>\n<div id=\"fs-idp171727376\" data-type=\"equation\">[latex]{K}_{P}=\\frac{{\\left({P}_{C}\\right)}^{x}{\\left({P}_{D}\\right)}^{y}}{{\\left({P}_{A}\\right)}^{m}{\\left({P}_{B}\\right)}^{n}}[\/latex]<\/div>\n<div id=\"fs-idp89891888\" data-type=\"equation\">[latex]=\\frac{{\\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}}[\/latex]<\/div>\n<div id=\"fs-idp72794512\" data-type=\"equation\">[latex]=\\frac{{\\left[\\text{C}\\right]}^{x}{\\left[\\text{D}\\right]}^{y}}{{\\left[\\text{A}\\right]}^{m}{\\left[\\text{B}\\right]}^{n}}\\times \\frac{{\\left(RT\\right)}^{x+y}}{{\\left(RT\\right)}^{m+n}}[\/latex]<\/div>\n<div id=\"fs-idp96527968\" data-type=\"equation\">[latex]={K}_{c}{\\left(RT\\right)}^{\\left(x+y\\right)-\\left(m+n\\right)}[\/latex]<\/div>\n<div id=\"fs-idp66307536\" data-type=\"equation\">[latex]={K}_{c}{\\left(RT\\right)}^{\\Delta n}[\/latex]<\/div>\n<p id=\"fs-idp93962000\">The relationship between <em data-effect=\"italics\">K<sub>c<\/sub><\/em> and <em data-effect=\"italics\">K<sub>P<\/sub><\/em> is<\/p>\n<div id=\"fs-idp142684400\" data-type=\"equation\">[latex]{K}_{P}={K}_{c}{\\left(RT\\right)}^{\\Delta n}[\/latex]<\/div>\n<p id=\"fs-idp38982608\">In this equation, \u0394<em data-effect=\"italics\">n<\/em> is the difference between the sum of the coefficients of the <em data-effect=\"italics\">gaseous<\/em> products and the sum of the coefficients of the <em data-effect=\"italics\">gaseous<\/em> reactants in the reaction (the change in moles of gas between the reactants and the products). For the gas-phase reaction [latex]m\\text{A}+n\\text{B}\\rightleftharpoons x\\text{C}+y\\text{D}[\/latex], we have<\/p>\n<div id=\"fs-idp126597184\" data-type=\"equation\">[latex]\\Delta n=\\left(x+y\\right)-\\left(m+n\\right)[\/latex]<\/div>\n<div id=\"fs-idp74619616\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 4<\/h3>\n<h4 id=\"fs-idp74619872\"><span data-type=\"title\">Calculation of <em data-effect=\"italics\">K<sub>P<\/sub><\/em><\/span><\/h4>\n<p>Write the equations for the conversion of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> to <em data-effect=\"italics\">K<sub>P<\/sub><\/em> for each of the following reactions:<\/p>\n<p id=\"fs-idp61114160\">(a) [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]<\/p>\n<p id=\"fs-idp127890016\">(b) [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]<\/p>\n<p id=\"fs-idp98159632\">(c) [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 id=\"fs-idp175509632\">(d) <em data-effect=\"italics\">K<sub>c<\/sub><\/em> is equal to 0.28 for the following reaction at 900 \u00b0C:<\/p>\n<div id=\"fs-idp277886816\" data-type=\"equation\">[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]<\/div>\n<p id=\"fs-idp237503488\">What is <em data-effect=\"italics\">K<sub>P<\/sub><\/em> at this temperature?<\/p>\n<h4 id=\"fs-idp22632784\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>(a) \u0394<em data-effect=\"italics\">n<\/em> = (2) &#8211; (1) = 1 <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>\u0394<em data-effect=\"italics\">n<\/em><\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>1<\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<\/p>\n<p id=\"fs-idp15495888\">(b) \u0394<em data-effect=\"italics\">n<\/em> = (2) &#8211; (2) = 0 <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>\u0394<em data-effect=\"italics\">n<\/em><\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>0<\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em><\/p>\n<p id=\"fs-idp64674800\">(c) \u0394<em data-effect=\"italics\">n<\/em> = (2) &#8211; (1 + 3) = -2 <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>\u0394<em data-effect=\"italics\">n<\/em><\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (<em data-effect=\"italics\">RT<\/em>)<sup>-2<\/sup> = [latex]\\frac{{K}_{c}}{{\\left(RT\\right)}^{2}}[\/latex]<\/p>\n<p id=\"fs-idp100458064\">(d) <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (RT) <sup>\u0394<em data-effect=\"italics\">n<\/em><\/sup> = (0.28)[(0.0821)(1173)]<sup>-2<\/sup> = 3.0 \u00d7 10<sup>-5<\/sup><\/p>\n<h4 id=\"fs-idp112448864\"><span data-type=\"title\">Check Your Learning<\/span><\/h4>\n<p>Write the equations for the conversion of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> to <em data-effect=\"italics\">K<sub>P<\/sub><\/em> for each of the following reactions, which occur in the gas phase:<\/p>\n<p id=\"fs-idp181427520\">(a) [latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right)[\/latex]<\/p>\n<p id=\"fs-idp162294032\">(b) [latex]{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)\\rightleftharpoons2{\\text{NO}}_{2}\\left(g\\right)[\/latex]<\/p>\n<p id=\"fs-idp103215280\">(c) [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]<\/p>\n<p id=\"fs-idp64364800\">(d) At 227 \u00b0C, the following reaction has <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = 0.0952:<\/p>\n<div id=\"fs-idp23547856\" data-type=\"equation\">[latex]{\\text{CH}}_{3}\\text{OH}\\left(g\\right)\\rightleftharpoons\\text{CO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/div>\n<p id=\"fs-idp131624768\">What would be the value of <em data-effect=\"italics\">K<sub>P<\/sub><\/em> at this temperature?<\/p>\n<div id=\"fs-idp33474544\" data-type=\"note\">\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong>(a) <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>)<sup>-1<\/sup>; (b) <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>); (c) <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>); (d) 160 or 1.6 \u00d7 10<sup>2<\/sup><\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-idp112997072\" data-depth=\"1\">\n<h2 data-type=\"title\">Heterogeneous Equilibria<\/h2>\n<p id=\"fs-idp27217312\">A <span data-type=\"term\">heterogeneous equilibrium<\/span> is a system in which reactants and products are found in two or more phases. The phases may be any combination of solid, liquid, or gas phases, and solutions. When dealing with these equilibria, remember that solids and pure liquids do not appear in equilibrium constant expressions (the activities of pure solids, pure liquids, and solvents are 1).<\/p>\n<p id=\"fs-idp134404832\">Some heterogeneous equilibria involve chemical changes; for example:<\/p>\n<div id=\"fs-idp129666976\" data-type=\"equation\">[latex]{\\text{PbCl}}_{2}\\left(s\\right)\\rightleftharpoons{\\text{Pb}}^{2+}\\left(aq\\right)+2{\\text{Cl}}^{-}\\left(aq\\right){K}_{c}=\\left[{\\text{Pb}}^{2+}\\right]{\\left[{\\text{Cl}}^{-}\\right]}^{2}[\/latex]<\/div>\n<div id=\"fs-idp89064704\" data-type=\"equation\">[latex]\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CaCO}}_{3}\\left(s\\right){K}_{c}=\\frac{1}{\\left[{\\text{CO}}_{2}\\right]}[\/latex]<\/div>\n<div id=\"fs-idp172202816\" data-type=\"equation\">[latex]\\text{C}\\left(s\\right)+2\\text{S}\\left(g\\right)\\rightleftharpoons{\\text{CS}}_{2}\\left(g\\right){K}_{c}=\\frac{\\left[{\\text{CS}}_{2}\\right]}{{\\left[\\text{S}\\right]}^{2}}[\/latex]<\/div>\n<p id=\"fs-idp84722304\">Other heterogeneous equilibria involve phase changes, for example, the evaporation of liquid bromine, as shown in the following equation:<\/p>\n<div id=\"fs-idp64021520\" data-type=\"equation\">[latex]{\\text{Br}}_{2}\\left(l\\right)\\rightleftharpoons{\\text{Br}}_{2}\\left(g\\right){K}_{c}=\\left[{\\text{Br}}_{2}\\right][\/latex]<\/div>\n<p id=\"fs-idp8357744\">We can write equations for reaction quotients of heterogeneous equilibria that involve gases, using partial pressures instead of concentrations. Two examples are:<\/p>\n<div id=\"fs-idp11158544\" data-type=\"equation\">[latex]\\text{CaO}\\left(s\\right)+{\\text{CO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CaCO}}_{3}\\left(s\\right){K}_{P}=\\frac{1}{{P}_{{\\text{CO}}_{2}}}[\/latex]<\/div>\n<div id=\"fs-idp269174800\" data-type=\"equation\">[latex]\\text{C}\\left(s\\right)+2\\text{S}\\left(g\\right)\\rightleftharpoons{\\text{CS}}_{2}\\left(g\\right){K}_{P}=\\frac{{P}_{C{S}_{2}}}{{\\left({P}_{S}\\right)}^{2}}[\/latex]<\/div>\n<\/section>\n<section id=\"fs-idp284371824\" class=\"summary\" data-depth=\"1\">\n<div class=\"bcc-box bcc-success\">\n<h2>Key Concepts and Summary<\/h2>\n<p id=\"fs-idp165520928\">For any reaction that is at equilibrium, the reaction quotient <em data-effect=\"italics\">Q<\/em> is equal to the equilibrium constant <em data-effect=\"italics\">K<\/em> for the reaction. If a reactant or product is a pure solid, a pure liquid, or the solvent in a dilute solution, the concentration of this component does not appear in the expression for the equilibrium constant. At equilibrium, the values of the concentrations of the reactants and products are constant. Their particular values may vary depending on conditions, but the value of the reaction quotient will always equal <em data-effect=\"italics\">K<\/em> (<em data-effect=\"italics\">K<sub>c<\/sub><\/em> when using concentrations or <em data-effect=\"italics\">K<sub>P<\/sub><\/em> when using partial pressures).<\/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. We can decide whether a reaction is at equilibrium by comparing the reaction quotient with the equilibrium constant for the reaction.<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idp116522752\" class=\"key-equations\" data-depth=\"1\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Equations<\/h3>\n<ul id=\"fs-idp16702192\" data-bullet-style=\"bullet\">\n<li>[latex]Q=\\frac{{\\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}=\\frac{{\\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><em data-effect=\"italics\">P<\/em> = <em data-effect=\"italics\">MRT<\/em><\/li>\n<li><em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>)<sup>\u0394<em data-effect=\"italics\">n<\/em><\/sup><\/li>\n<\/ul>\n<\/div>\n<\/section>\n<section id=\"fs-idp16341104\" class=\"exercises\" data-depth=\"1\">\n<div class=\"bcc-box bcc-info\">\n<h3>Chemistry End of Chapter Exercises<\/h3>\n<div id=\"fs-idp109971312\" data-type=\"exercise\">\n<div id=\"fs-idp60916480\" data-type=\"problem\">\n<ol>\n<li>Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.<\/li>\n<li>Explain why an equilibrium between Br<sub>2<\/sub>(<em data-effect=\"italics\">l<\/em>) and Br<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) would not be established if the container were not a closed vessel shown in Figure 3.\n<div id=\"attachment_2253\" style=\"width: 310px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2253\" class=\"wp-image-2253\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212357\/CNX_Chem_13_01_bromine1.jpg\" alt=\"A glass container is shown that is filled with an orange-brown gas and a small amount of dark orange liquid.\" width=\"300\" height=\"354\" \/><\/p>\n<p id=\"caption-attachment-2253\" class=\"wp-caption-text\">Figure 3. A sample of liquid bromine at equilibrium with bromine vapor in a closed container. (credit: http:\/\/images-of-elements.com\/bromine.php)<\/p>\n<\/div>\n<\/li>\n<li>If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO<sub>2<\/sub> or with pure N<sub>2<\/sub>O<sub>4<\/sub>?<br \/>\n[latex]2{\\text{NO}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{4}\\left(g\\right)[\/latex]<\/li>\n<li>Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg<sub>2<\/sub>Cl<sub>2<\/sub>, AgCl, PbCl<sub>2<\/sub>, and CuCl.\n<ol>\n<li>Write the expression for the equilibrium constant for the reaction represented by the equation [latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons{\\text{Ag}}^{+}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)\\text{.}[\/latex] Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481? Explain your answer.<\/li>\n<li>Write the expression for the equilibrium constant for the reaction represented by the equation [latex]{\\text{Pb}}^{2+}\\left(aq\\right)+2{\\text{Cl}}^{-}\\left(aq\\right)\\rightleftharpoons{\\text{PbCl}}_{2}\\left(s\\right)\\text{.}[\/latex] Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481? Explain your answer.<\/li>\n<\/ol>\n<\/li>\n<li>Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates\u2014except those of the ammonium ion and the alkali metals\u2014are insoluble.\n<ol>\n<li>Write the expression for the equilibrium constant for the reaction represented by the equation [latex]{\\text{CaCO}}_{3}\\left(s\\right)\\rightleftharpoons{\\text{Ca}}^{2+}\\left(aq\\right)+{\\text{CO}}_{3}{}^{-}\\left(aq\\right)\\text{.}[\/latex] Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481? Explain your answer.<\/li>\n<li>Write the expression for the equilibrium constant for the reaction represented by the equation [latex]3{\\text{Ba}}^{2+}\\left(aq\\right)+2{\\text{PO}}_{4}{}^{3-}\\left(aq\\right)\\rightleftharpoons{\\text{Ba}}_{3}{\\left({\\text{PO}}_{4}\\right)}_{2}\\left(s\\right)\\text{.}[\/latex] Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481? Explain your answer.<\/li>\n<\/ol>\n<\/li>\n<li>Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: [latex]3{\\text{C}}_{2}{\\text{H}}_{2}\\left(g\\right)\\rightarrow{\\text{C}}_{6}{\\text{H}}_{6}\\left(g\\right)\\text{.}[\/latex] Which value of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> would make this reaction most useful commercially? <em data-effect=\"italics\">K<sub>c<\/sub><\/em> \u2248 0.01, <em data-effect=\"italics\">K<sub>c<\/sub><\/em> \u2248 1, or <em data-effect=\"italics\">K<sub>c<\/sub><\/em> \u2248 10. Explain your answer.<\/li>\n<li>Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation [latex]\\text{KI}\\left(aq\\right)+{\\text{I}}_{2}\\left(aq\\right)\\rightleftharpoons{\\text{KI}}_{3}\\left(aq\\right)[\/latex] give the same expression for the reaction quotient. KI<sub>3<\/sub> is composed of the ions K<sup>+<\/sup> and [latex]{\\text{I}}_{3}{}^{\\text{-}}\\text{.}[\/latex]<br \/>\nThe reaction quotient for the complete form of the equation is [latex]{Q}_{c}=\\frac{\\left[{\\text{KI}}_{3}\\right]}{\\left[\\text{KI}\\right]\\left[{\\text{I}}_{2}\\right]}\\text{.}[\/latex]<\/li>\n<li>For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481 for a titration reaction?<\/li>\n<li>For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is <em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1, &lt;1, or \u22481 for a useful precipitation reaction?<\/li>\n<li>Write the mathematical expression for the reaction quotient, <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>, for each of the following reactions:\n<ol>\n<li>[latex]{\\text{CH}}_{4}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CH}}_{3}\\text{Cl}\\left(g\\right)+\\text{HCl}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{N}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NO}\\left(g\\right)[\/latex]<\/li>\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{BaSO}}_{3}\\left(s\\right)\\rightleftharpoons\\text{BaO}\\left(s\\right)+{\\text{SO}}_{2}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{P}}_{4}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{P}}_{4}{\\text{O}}_{10}\\left(s\\right)[\/latex]<\/li>\n<li>[latex]{\\text{Br}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{Br}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{CH}}_{4}\\left(g\\right)+2{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons{\\text{CO}}_{2}\\left(g\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/li>\n<li>[latex]{\\text{CuSO}}_{4}\\cdot 5{\\text{H}}_{2}\\text{O}\\left(s\\right)\\rightleftharpoons{\\text{CuSO}}_{4}\\left(s\\right)+5{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Write the mathematical expression for the reaction quotient, <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>, for each of the following reactions:\n<ol>\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)+5{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons4\\text{NO}\\left(g\\right)+6{\\text{H}}_{2}\\text{O}\\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{CO}}_{2}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons\\text{CO}\\left(g\\right)+{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{NH}}_{4}\\text{Cl}\\left(s\\right)\\rightleftharpoons{\\text{NH}}_{3}\\left(g\\right)+\\text{HCl}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]2\\text{Pb}{\\left({\\text{NO}}_{3}\\right)}_{2}\\left(s\\right)\\rightleftharpoons2\\text{PbO}\\left(s\\right)+4{\\text{NO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]2{\\text{H}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/li>\n<li>[latex]{\\text{S}}_{8}\\left(g\\right)\\rightleftharpoons8\\text{S}\\left(g\\right)[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.\n<ol>\n<li>[latex]2{\\text{NH}}_{3}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right){K}_{c}=17\\text{;}[\/latex] [NH<sub>3<\/sub>] = 0.20 <em data-effect=\"italics\">M<\/em>, [N<sub>2<\/sub>] = 1.00 <em data-effect=\"italics\">M<\/em>, [H<sub>2<\/sub>] = 1.00 <em data-effect=\"italics\">M<br \/>\n<\/em><\/li>\n<li>[latex]2{\\text{NH}}_{3}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right){K}_{P}=6.8\\times {10}^{4}\\text{;}[\/latex] initial pressures: NH<sub>3<\/sub> = 3.0 atm, N<sub>2<\/sub> = 2.0 atm, H<sub>2<\/sub> = 1.0 atm<\/li>\n<li>[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\text{(}g\\text{)}{K}_{c}=0.230\\text{;}[\/latex] [SO<sub>3<\/sub>] = 0.00 <em data-effect=\"italics\">M<\/em>, [SO<sub>2<\/sub>] = 1.00 <em data-effect=\"italics\">M<\/em>, [O<sub>2<\/sub>] = 1.00 <em data-effect=\"italics\">M<br \/>\n<\/em><\/li>\n<li>[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right){K}_{P}=16.5\\text{;}[\/latex] initial pressures: SO<sub>3<\/sub> = 1.00 atm, SO<sub>2<\/sub> = 1.00 atm, O<sub>2<\/sub> = 1.00 atm<\/li>\n<li>[latex]2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NOCl}\\left(g\\right){K}_{c}=4.6\\times {10}^{4}\\text{;}[\/latex] [NO] = 1.00 <em data-effect=\"italics\">M<\/em>, [Cl<sub>2<\/sub>] = 1.00 <em data-effect=\"italics\">M<\/em>, [NOCl] = 0 <em data-effect=\"italics\">M<br \/>\n<\/em><\/li>\n<li>[latex]{\\text{N}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NO}\\left(g\\right){K}_{P}=0.050\\text{;}[\/latex] initial pressures: NO = 10.0 atm, N<sub>2<\/sub> = O<sub>2<\/sub> = 5 atm<\/li>\n<\/ol>\n<\/li>\n<li>The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.\n<ol>\n<li>[latex]2{\\text{NH}}_{3}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right){K}_{c}=17\\text{;}[\/latex] [NH<sub>3<\/sub>] = 0.50 <em data-effect=\"italics\">M<\/em>, [N<sub>2<\/sub>] = 0.15 <em data-effect=\"italics\">M<\/em>, [H<sub>2<\/sub>] = 0.12 <em data-effect=\"italics\">M<br \/>\n<\/em><\/li>\n<li>[latex]2{\\text{NH}}_{3}\\left(g\\right)\\rightleftharpoons{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right){K}_{P}=6.8\\times {10}^{4}\\text{;}[\/latex] initial pressures: NH<sub>3<\/sub> = 2.00 atm, N<sub>2<\/sub> = 10.00 atm, H<sub>2<\/sub> = 10.00 atm<\/li>\n<li>[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right){K}_{c}=0.230\\text{;}[\/latex] [SO<sub>3<\/sub>] = 2.00 <em data-effect=\"italics\">M<\/em>, [SO<sub>2<\/sub>] = 2.00 <em data-effect=\"italics\">M<\/em>, [O<sub>2<\/sub>] = 2.00 <em data-effect=\"italics\">M<br \/>\n<\/em><\/li>\n<li>[latex]2{\\text{SO}}_{3}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right){K}_{P}=6.5\\text{atm;}[\/latex] initial pressures: SO<sub>2<\/sub> = 1.00 atm, O<sub>2<\/sub> = 1.130 atm, SO<sub>3<\/sub> = 0 atm<\/li>\n<li>[latex]2\\text{NO}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NOCl}\\left(g\\right){K}_{P}=2.5\\times {10}^{3}\\text{;}[\/latex] initial pressures: NO = 1.00 atm, Cl<sub>2<\/sub> = 1.00 atm, NOCl = 0 atm<\/li>\n<li>[latex]{\\text{N}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{NO}\\left(g\\right){K}_{c}=0.050\\text{;}[\/latex] [N<sub>2<\/sub>] = 0.100 <em data-effect=\"italics\">M<\/em>, [O<sub>2<\/sub>] = 0.200 <em data-effect=\"italics\">M<\/em>, [NO] = 1.00 <em data-effect=\"italics\">M<\/em><\/li>\n<\/ol>\n<\/li>\n<li>The following reaction has <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = 4.50 \u00d7 10<sup>-5<\/sup> at 720 K.<br \/>\n[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<br \/>\nIf a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? <em data-effect=\"italics\">P<\/em>(NH<sub>3<\/sub>) = 93 atm, <em data-effect=\"italics\">P<\/em>(N<sub>2<\/sub>) = 48 atm, and <em data-effect=\"italics\">P<\/em>(H<sub>2<\/sub>) = 52<\/li>\n<li>Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?<\/li>\n<li>Which of the systems described in question 10 give homogeneous equilibria? Which give heterogeneous equilibria?<\/li>\n<li>Which of the systems described in question 11\u00a0give homogeneous equilibria? Which give heterogeneous equilibria?<\/li>\n<li>For which of the reactions in question 10\u00a0does <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (calculated using concentrations) equal <em data-effect=\"italics\">K<sub>P<\/sub><\/em> (calculated using pressures)?<\/li>\n<li>For which of the reactions in question 11\u00a0does <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (calculated using concentrations) equal <em data-effect=\"italics\">K<sub>P<\/sub><\/em> (calculated using pressures)?<\/li>\n<li>Convert the values of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> to values of <em data-effect=\"italics\">K<sub>P<\/sub><\/em> or the values of <em data-effect=\"italics\">K<sub>P<\/sub><\/em> to values of <em data-effect=\"italics\">K<sub>c<\/sub><\/em>.\n<ol>\n<li>[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\\text{\\textdegree C}[\/latex]<\/li>\n<li>[latex]{\\text{H}}_{2}+{\\text{I}}_{2}\\rightleftharpoons2\\text{HI}{K}_{c}=50.2\\text{at}448\\text{\\textdegree C}[\/latex]<\/li>\n<li>[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\\text{\\textdegree C}[\/latex]<\/li>\n<li>[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right){K}_{P}=0.122\\text{at}50\\text{\\textdegree C}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Convert the values of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> to values of <em data-effect=\"italics\">K<sub>P<\/sub><\/em> or the values of <em data-effect=\"italics\">K<sub>P<\/sub><\/em> to values of <em data-effect=\"italics\">K<sub>c<\/sub><\/em>.\n<ol>\n<li>[latex]{\\text{Cl}}_{2}\\left(g\\right)+{\\text{Br}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{BrCl}\\left(g\\right){K}_{c}=4.7\\times {10}^{-2}\\text{at}25\\text{\\textdegree C}[\/latex]<\/li>\n<li>[latex]2{\\text{SO}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\rightleftharpoons2{\\text{SO}}_{3}\\left(g\\right){K}_{P}=48.2\\text{at}500\\text{\\textdegree C}[\/latex]<\/li>\n<li>[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\\text{\\textdegree C}[\/latex]<\/li>\n<li>[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right){K}_{P}=0.196\\text{at}60\\text{\\textdegree C}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>What is the value of the equilibrium constant expression for the change [latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex] at 30 \u00b0C?<\/li>\n<li>Write the expression of the reaction quotient for the ionization of HOCN in water.<\/li>\n<li>Write the reaction quotient expression for the ionization of NH<sub>3<\/sub> in water.<\/li>\n<li>What is the approximate value of the equilibrium constant <em data-effect=\"italics\">K<sub>P<\/sub><\/em> for the change [latex]{\\text{C}}_{2}{\\text{H}}_{5}{\\text{OC}}_{2}{\\text{H}}_{5}\\left(l\\right)\\rightleftharpoons{\\text{C}}_{2}{\\text{H}}_{5}{\\text{OC}}_{2}{\\text{H}}_{5}\\left(g\\right)[\/latex] at 25 \u00b0C. (Vapor pressure was described in the previous chapter on liquids and solids; refer back to this chapter to find the relevant information needed to solve this problem.)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"bcc-box bcc-info\">\n<h4>Selected Answers<\/h4>\n<p>2.\u00a0Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle, because the system is not closed; one of the components of the equilibrium, the Br<sub>2<\/sub> vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase.<\/p>\n<p>4. (a) <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = [Ag+][Cl-] &lt; 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 <em data-effect=\"italics\">M<\/em>;<\/p>\n<p>(b) [latex]{K}_{c}=\\frac{1}{\\left[{\\text{Pb}}^{2+}\\right]{\\left[{\\text{Cl}}^{-}\\right]}^{2}}[\/latex] &gt; 1 because PbCl<sub>2<\/sub> is insoluble and formation of the solid will reduce the concentration of ions to a low level (&lt;1 <em data-effect=\"italics\">M<\/em>).<\/p>\n<p>6.\u00a0Since [latex]{K}_{c}=\\frac{\\left[{\\text{C}}_{6}{\\text{H}}_{6}\\right]}{{\\left[{\\text{C}}_{2}{\\text{H}}_{2}\\right]}^{3}}[\/latex], a value of <em data-effect=\"italics\">K<sub>c<\/sub><\/em> \u2248 10 means that C<sub>6<\/sub>H<sub>6<\/sub> predominates over C<sub>2<\/sub>H<sub>2<\/sub>. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.<\/p>\n<p>8.\u00a0<em data-effect=\"italics\">K<sub>c<\/sub><\/em> &gt; 1; the product must be formed in overwhelmingly large proportions.<\/p>\n<p>10.\u00a0(a) [latex]{Q}_{c}=\\frac{\\left[{\\text{CH}}_{3}\\text{Cl}\\right]\\left[\\text{HCl}\\right]}{\\left[{\\text{CH}}_{4}\\right]\\left[{\\text{Cl}}_{2}\\right]}\\text{;}[\/latex] (b) [latex]{Q}_{c}=\\frac{{\\left[\\text{NO}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]}\\text{;}[\/latex] (c) [latex]{Q}_{c}=\\frac{{\\left[{\\text{SO}}_{3}\\right]}^{2}}{{\\left[{\\text{SO}}_{2}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]}\\text{;}[\/latex] (d) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = [SO<sub>2<\/sub>]; (e) [latex]{Q}_{c}=\\frac{1}{\\left[{\\text{P}}_{4}\\right]{\\left[{\\text{O}}_{2}\\right]}^{5}}\\text{;}[\/latex] (f) [latex]{Q}_{c}=\\frac{{\\left[\\text{Br}\\right]}^{2}}{\\left[{\\text{Br}}_{2}\\right]}\\text{;}[\/latex] (g) [latex]{Q}_{c}=\\frac{\\left[{\\text{CO}}_{2}\\right]}{\\left[{\\text{CH}}_{4}\\right]{\\left[{\\text{O}}_{2}\\right]}^{2}}\\text{;}[\/latex] (h) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = [H<sub>2<\/sub>O]<sup>5<\/sup><\/p>\n<p>12.\u00a0(a) [latex]{Q}_{c}=\\frac{\\left[{\\text{N}}_{2}\\right]{\\left[{\\text{H}}_{2}\\right]}^{3}}{{\\left[{\\text{NH}}_{3}\\right]}^{2}}=\\frac{\\left(1.00\\right){\\left(1.00\\right)}^{3}}{{\\left(0.20\\right)}^{2}}=25[\/latex]\u00a0<em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &lt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em>, proceeds left;<br \/>\n(b) [latex]{Q}_{P}=\\frac{{P}_{{\\text{N}}_{2}}{\\left({P}_{{\\text{H}}_{2}}\\right)}^{3}}{{\\left({P}_{{\\text{NH}}_{3}}\\right)}^{2}}=\\frac{\\left(2.0\\right){\\left(1.0\\right)}^{3}}{{\\left(3.0\\right)}^{2}}=0.22[\/latex]\u00a0<em data-effect=\"italics\">Q<sub>P<\/sub><\/em> &lt; <em data-effect=\"italics\">K<sub>P<\/sub><\/em>, proceeds right;<br \/>\n(c) [latex]{Q}_{c}=\\frac{{\\left[{\\text{SO}}_{2}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]}{{\\left[{\\text{SO}}_{3}\\right]}^{2}}=\\frac{{\\left(1.00\\right)}^{2}\\left(1.00\\right)}{\\left(0\\right)}=\\text{undefined}[\/latex]\u00a0<em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &gt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em>, proceeds left;<br \/>\n(d) [latex]{Q}_{P}=\\frac{{\\left({P}_{{\\text{SO}}_{2}}\\right)}^{2}{P}_{{\\text{O}}_{2}}}{{\\left({P}_{{\\text{SO}}_{3}}\\right)}^{2}}=\\frac{{\\left(1.00\\right)}^{2}\\left(1.00\\right)}{{\\left(1.00\\right)}^{2}}=1.00[\/latex]\u00a0<em data-effect=\"italics\">Q<sub>P<\/sub><\/em> &gt; <em data-effect=\"italics\">K<sub>P<\/sub><\/em>, proceeds right;<br \/>\n(e) [latex]{Q}_{P}=\\frac{{\\left({P}_{\\text{NOCl}}\\right)}^{2}}{{\\left({P}_{\\text{NO}}\\right)}^{2}{P}_{{\\text{Cl}}_{2}}}=\\frac{{\\left(0\\right)}^{2}}{{\\left(1.00\\right)}^{2}\\left(1.00\\right)}=0[\/latex]\u00a0<em data-effect=\"italics\">Q<sub>P<\/sub><\/em> &lt; <em data-effect=\"italics\">K<sub>P<\/sub><\/em>, proceeds right;<br \/>\n(f) [latex]{Q}_{c}=\\frac{{\\left[\\text{NO}\\right]}^{2}}{\\left[{\\text{N}}_{2}\\right]\\left[{\\text{O}}_{2}\\right]}=\\frac{{\\left(10.0\\right)}^{2}}{\\left(5.00\\right)\\left(5.00\\right)}=4[\/latex] <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &gt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em>, proceeds left<\/p>\n<p>14.\u00a0The reaction quotient expression for this problem is [latex]{Q}_{p}=\\frac{{\\left({P}_{{\\text{NH}}_{3}}\\right)}^{2}}{\\left({P}_{{\\text{N}}_{2}}\\right){\\left({P}_{{\\text{H}}_{2}}\\right)}^{3}}\\text{.}[\/latex] Plugging in the given values of partial pressures gives [latex]\\frac{{\\left(93\\right)}^{2}}{\\left(\\left(48\\right)\\times {\\left(52\\right)}^{3}\\right)}[\/latex], so <em data-effect=\"italics\">Q<sub>p<\/sub><\/em> = 1.3 \u00d7 10<sup>-3<\/sup>. Since this value is larger than <em data-effect=\"italics\">K<sub>P<\/sub><\/em> (4.50 \u00d7 10<sup>-5<\/sup>), the system will shift toward the reactants to reach equilibrium.<\/p>\n<p>16.\u00a0(a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous<\/p>\n<p>18.\u00a0When the number of gaseous components are the same on both sides of the equilibrium expression, <em data-effect=\"italics\">K<sub>c<\/sub><\/em> will equal <em data-effect=\"italics\">K<sub>P<\/sub><\/em>. This situation occurs in (a) and (b).<\/p>\n<p>20.\u00a0[latex]{K}_{P}={K}_{c}{\\left(RT\\right)}^{\\Delta n}[\/latex], where \u0394<em data-effect=\"italics\">n<\/em> is the sum of gaseous products minus the sum of gaseous reactants. (a) \u0394<em data-effect=\"italics\">n<\/em> = (2) &#8211; (1 + 3) = -2, <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = 0.50[0.08206 \u00d7 673.15]<sup>-2<\/sup> = 1.6 \u00d7 10<sup>-4<\/sup>; (b) \u0394<em data-effect=\"italics\">n<\/em> = (2) &#8211; (1 + 1) = 0, <em data-effect=\"italics\">K<sub>P<\/sub><\/em> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>)<sup>0<\/sup> = <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = 50.2; (c) \u0394<em data-effect=\"italics\">n<\/em> = (10) &#8211; (0) = 10, <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = <em data-effect=\"italics\">K<sub>P<\/sub><\/em>(<em data-effect=\"italics\">RT<\/em>)<sup>-\u0394<em data-effect=\"italics\">n<\/em><\/sup>, <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = 4.08 \u00d7 10<sup>-25<\/sup>[0.08206 \u00d7 298.15]<sup>-10<\/sup> = 5.31 \u00d7 10<sup>-39<\/sup>; (d) \u0394<em data-effect=\"italics\">n<\/em> = (1) &#8211; (0) = 1, <em data-effect=\"italics\">K<sub>c<\/sub><\/em> = 0.122(0.08206 \u00d7 323.15)<sup>-1<\/sup> = 4.60 \u00d7 10<sup>-3<\/sup><\/p>\n<p>22.\u00a0The equilibrium expression for this transformation is [latex]{K}_{P}={P}_{{\\text{H}}_{2}\\text{O}}\\text{.}[\/latex] The vapor pressure of H<sub>2<\/sub>O at 30 \u00baC is 31.8 torr. Converting to atmospheres gives [latex]3.18\\cancel{\\text{torr}}\\times \\frac{1\\text{atm}}{760\\cancel{\\text{torr}}}=0.042\\text{atm.}[\/latex] Therefore, [latex]{K}_{P}={\\text{P}}_{{\\text{H}}_{2}\\text{O}}=0.042\\text{.}[\/latex]<\/p>\n<p>24.\u00a0[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)\\text{.}[\/latex] Because the concentration of water is a constant, the term [H<sub>2<\/sub>O] is normally incorporated into the reaction quotient as well as the final equilibrium constant. [latex]{Q}_{c}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{HN}}_{3}\\right]}[\/latex]<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Glossary<\/h3>\n<p id=\"fs-idp63345136\" data-type=\"definition\"><strong><span data-type=\"term\">equilibrium constant (<em data-effect=\"italics\">K<\/em>)<br \/>\n<\/span><\/strong>value of the reaction quotient for a system at equilibrium<\/p>\n<p id=\"fs-idp102632368\" data-type=\"definition\"><strong><span data-type=\"term\">heterogeneous equilibria<br \/>\n<\/span><\/strong>equilibria between reactants and products in different phases<\/p>\n<p id=\"fs-idp35066960\" data-type=\"definition\"><strong><span data-type=\"term\">homogeneous equilibria<br \/>\n<\/span><\/strong>equilibria within a single phase<\/p>\n<p id=\"fs-idp55285520\" data-type=\"definition\"><strong><span data-type=\"term\"><em data-effect=\"italics\">K<sub>c<br \/>\n<\/sub><\/em><\/span><\/strong>equilibrium constant for reactions based on concentrations of reactants and products<\/p>\n<p id=\"fs-idp61369840\" data-type=\"definition\"><strong><span data-type=\"term\"><em data-effect=\"italics\">K<sub>P<br \/>\n<\/sub><\/em><\/span><\/strong>equilibrium constant for gas-phase reactions based on partial pressures of reactants and products<\/p>\n<p id=\"fs-idp89564720\" data-type=\"definition\"><strong><span data-type=\"term\">law of mass action<br \/>\n<\/span><\/strong>when a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant<\/p>\n<p id=\"fs-idp74361936\" data-type=\"definition\"><strong><span data-type=\"term\">reaction quotient (<em data-effect=\"italics\">Q<\/em>)<br \/>\n<\/span><\/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<\/div>\n<p>&nbsp;<\/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. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.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>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t 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