{"id":755,"date":"2021-07-23T09:20:37","date_gmt":"2021-07-23T13:20:37","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/equilibrium-constants\/"},"modified":"2022-06-23T09:18:45","modified_gmt":"2022-06-23T13:18:45","slug":"equilibrium-constants","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/equilibrium-constants\/","title":{"raw":"13.2 Equilibrium Constants","rendered":"13.2 Equilibrium Constants"},"content":{"raw":"<strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em;background-color: #cbd4b6;color: #000000\">Learning Objectives<\/span><\/strong>\r\n<div class=\"textbox textbox--learning-objectives\">\r\n\r\nBy the end of this section, 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\">The status of a reversible reaction is conveniently assessed by evaluating its <strong>reaction quotient (<em data-effect=\"italics\">Q<\/em>)<\/strong>. For a reversible reaction described by<\/p>\r\n\r\n<div id=\"fs-idp5986304\" style=\"text-align: center\" data-type=\"equation\"><em>m<\/em>A + <em>n<\/em>B \u21cc <em>x<\/em>C + <em>y<\/em>D<\/div>\r\n<p id=\"fs-idp11973152\">the reaction quotient is derived directly from the stoichiometry of the balanced equation as<\/p>\r\n\r\n<div id=\"fs-idp44297776\" data-type=\"equation\"><img class=\"wp-image-1779 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2a.png\" alt=\"\" width=\"119\" height=\"58\" \/><\/div>\r\n<p id=\"fs-idm7959200\">where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:<\/p>\r\n\r\n<div id=\"fs-idp23922448\" data-type=\"equation\"><img class=\"alignnone wp-image-1780 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2b.png\" alt=\"\" width=\"119\" height=\"64\" \/><\/div>\r\n<p id=\"fs-idm352046000\">Note that the reaction quotient equations above are a simplification of more rigorous expressions that use <em data-effect=\"italics\">relative<\/em> values for concentrations and pressures rather than <em data-effect=\"italics\">absolute<\/em> values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing <em data-effect=\"italics\">Q<\/em>. In most cases, this will introduce only modest errors in calculations involving reaction quotients.<\/p>\r\n\r\n<div id=\"fs-idp168372720\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp68042176\"><strong>Writing Reaction Quotient Expressions:<\/strong><\/p>\r\nWrite the concentration-based reaction quotient expression for each of the following reactions:\r\n<p id=\"fs-idp246365472\">(a) 3O<sub>2<\/sub>(<em>g<\/em>) \u21cc 2O<sub>3<\/sub>(<em>g<\/em>)<\/p>\r\n<p id=\"fs-idp142495760\">(b) N<sub>2<\/sub>(<em>g<\/em>) + 3H<sub>2<\/sub>(<em>g<\/em>) \u21cc 2NH<sub>3<\/sub>(<em>g<\/em>)<\/p>\r\n<p id=\"fs-idp82472256\">(c) 4NH<sub>3<\/sub>(<em>g<\/em>) + 7O<sub>2<\/sub>(<em>g<\/em>) \u21cc 4NO<sub>2<\/sub>(<em>g<\/em>) + 6H<sub>2<\/sub>O(<em>g<\/em>)<\/p>\r\n<p id=\"fs-idp63878256\"><strong>Solution:<\/strong><\/p>\r\n<span style=\"text-align: initial;font-size: 1em\"><img class=\"alignnone wp-image-1781\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2c-300x246.png\" alt=\"\" width=\"185\" height=\"152\" \/><\/span>\r\n<p id=\"fs-idp17156768\"><strong>Check Your Learning:<\/strong><\/p>\r\nWrite the concentration-based reaction quotient expression for each of the following reactions:\r\n<p id=\"fs-idp35394896\">(a) 2SO<sub>2<\/sub>(<em>g<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u21cc 2SO<sub>3<\/sub>(<em>g<\/em>)<\/p>\r\n<p id=\"fs-idp63748560\">(b) C<sub>4<\/sub>H<sub>8<\/sub>(<em>g<\/em>) \u21cc 2C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>)<\/p>\r\n<p id=\"fs-idp105907744\">(c) 2C<sub>4<\/sub>H<sub>10<\/sub>(<em>g<\/em>) + 13O<sub>2<\/sub>(<em>g<\/em>) \u21cc 8CO<sub>2<\/sub>(<em>g<\/em>) + 10H<sub>2<\/sub>O(<em>g<\/em>)<\/p>\r\n<strong>Answer:<\/strong>\r\n<div id=\"fs-idp167282768\" data-type=\"note\">\r\n<div data-type=\"title\"><img class=\"alignnone wp-image-1782\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2d-300x37.png\" alt=\"\" width=\"364\" height=\"45\" \/><\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"CNX_Chem_13_02_quotient\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">Changes in concentrations and <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> for a chemical equilibrium achieved beginning with (a) a mixture of reactants only and (b) products only.<\/div>\r\n<span id=\"fs-idp134344480\" data-type=\"media\" data-alt=\"Four graphs are shown and labeled, \u201ca,\u201d \u201cb,\u201d \u201cc,\u201d and \u201cd.\u201d All four graphs have a vertical dotted line running through the middle labeled, \u201cEquilibrium is reached.\u201d The y-axis on graph a is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph a. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts high on the y-axis, ends midway down the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it starts and ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph b is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph b. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts low on the y-axis, ends midway up the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph c is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph c. This curve begins at the bottom of the y-axis and rises steeply up near the top of the y-axis, then levels off into a horizontal line. The top point of this line is labeled, \u201ck.\u201d The y-axis on graph d is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph d. This curve begins near the edge of the top of the y-axis and falls steeply toward the x-axis, then levels off into a horizontal line. The bottom point of this line is labeled, \u201ck.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_13_02_quotient-1.jpg\" alt=\"Four graphs are shown and labeled, \u201ca,\u201d \u201cb,\u201d \u201cc,\u201d and \u201cd.\u201d All four graphs have a vertical dotted line running through the middle labeled, \u201cEquilibrium is reached.\u201d The y-axis on graph a is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph a. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts high on the y-axis, ends midway down the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it starts and ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph b is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph b. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts low on the y-axis, ends midway up the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph c is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph c. This curve begins at the bottom of the y-axis and rises steeply up near the top of the y-axis, then levels off into a horizontal line. The top point of this line is labeled, \u201ck.\u201d The y-axis on graph d is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph d. This curve begins near the edge of the top of the y-axis and falls steeply toward the x-axis, then levels off into a horizontal line. The bottom point of this line is labeled, \u201ck.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idm122432\">The numerical value of <em data-effect=\"italics\">Q<\/em> varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction\u2019s status. To illustrate this point, consider the oxidation of sulfur dioxide:<\/p>\r\n\r\n<div id=\"fs-idm382376384\" style=\"text-align: center\" data-type=\"equation\">2SO<sub>2<\/sub>(<em>g<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u21cc 2SO<sub>3<\/sub>(<em>g<\/em>)<\/div>\r\n<p id=\"fs-idp138846544\">Two different experimental scenarios are depicted in <a class=\"autogenerated-content\" href=\"#CNX_Chem_13_02_quotient\">(Figure)<\/a>, one in which this reaction is initiated with a mixture of reactants only, SO<sub>2<\/sub> and O<sub>2<\/sub>, and another that begins with only product, SO<sub>3<\/sub>. For the reaction that begins with a mixture of reactants only, <em data-effect=\"italics\">Q<\/em> is initially equal to zero:<\/p>\r\n\r\n<div id=\"fs-idm329826160\" data-type=\"equation\"><img class=\" wp-image-1783 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2e-300x62.png\" alt=\"\" width=\"285\" height=\"59\" \/><\/div>\r\n<p id=\"fs-idm375739520\">As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>), product concentration increases (as does the numerator of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>.<\/p>\r\n<p id=\"fs-idm378445456\">If the reaction begins with only product present, the value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> is initially undefined (immeasurably large, or infinite):<\/p>\r\n\r\n<div id=\"fs-idm324502592\" data-type=\"equation\"><img class=\" wp-image-1784 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2f-300x65.png\" alt=\"\" width=\"263\" height=\"57\" \/><\/div>\r\n<p id=\"fs-idm374784256\">In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> decrease with time, the reactant concentrations and the denominator of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.<\/p>\r\n<p id=\"fs-idm389114784\">The constant value of <em data-effect=\"italics\">Q<\/em> exhibited by a system at equilibrium is called the <strong>equilibrium constant, <em data-effect=\"italics\">K<\/em><\/strong>:<\/p>\r\n\r\n<div id=\"fs-idp266400656\" style=\"text-align: center\" data-type=\"equation\"><em>K<\/em> =<span style=\"font-size: 1em\"><em> Q<\/em> at equilibrium<\/span><\/div>\r\n<p id=\"fs-idm328721280\">Comparison of the data plots in <a class=\"autogenerated-content\" href=\"#CNX_Chem_13_02_quotient\">(Figure)<\/a> shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the <span data-type=\"term\">law of mass action<\/span>: At a given temperature, the reaction quotient for a system at equilibrium is constant.<\/p>\r\n\r\n<div id=\"fs-idp54423808\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp23994768\"><strong>Evaluating a Reaction Quotient:<\/strong><\/p>\r\nGaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:\r\n<div id=\"fs-idp64726272\" style=\"text-align: center\" data-type=\"equation\">2NO<sub>2<\/sub>(<em>g<\/em>) \u21cc N<sub>2<\/sub>O<sub>4<\/sub>(<em>g<\/em>)<\/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<p id=\"fs-idp179491728\"><strong>Solution:<\/strong><\/p>\r\nAs for all equilibrium calculations in this text, use the simplified equations for <em data-effect=\"italics\">Q<\/em> and <em data-effect=\"italics\">K<\/em> and disregard any concentration or pressure units, as noted previously in this section.\r\n<p id=\"fs-idm358571008\">(a) Before any product is formed,<\/p>\r\n<img class=\"alignnone wp-image-1785\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2g-300x45.png\" alt=\"\" width=\"240\" height=\"36\" \/>\r\n\r\nand [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\"><img class=\"alignnone wp-image-1786\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2h-300x81.png\" alt=\"\" width=\"226\" height=\"61\" \/><\/div>\r\n<p id=\"fs-idp110962640\">(b) At equilibrium,<\/p>\r\n<img class=\"alignnone size-medium wp-image-1787\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2i-300x42.png\" alt=\"\" width=\"300\" height=\"42\" \/>\r\n<p id=\"fs-idp284528080\"><strong>Check Your Learning:<\/strong><\/p>\r\nFor the reaction 2SO<sub>2<\/sub>(<em>g<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u21cc 2SO<sub>3<\/sub>(<em>g<\/em>), 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\r\n&nbsp;\r\n<div id=\"fs-idp10573648\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp130245328\"><em data-effect=\"italics\">K<sub>c<\/sub> =<\/em> 4.3<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm80100288\">By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large <em data-effect=\"italics\">K<\/em> will reach equilibrium when most of the reactant has been converted to product, whereas a small <em data-effect=\"italics\">K<\/em> indicates the reaction achieves equilibrium after very little reactant has been converted. It\u2019s important to keep in mind that the magnitude of <em data-effect=\"italics\">K<\/em> does <em data-effect=\"italics\">not<\/em> indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.<\/p>\r\n<p id=\"fs-idm377606544\">The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and\/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing <em data-effect=\"italics\">Q<\/em> to <em data-effect=\"italics\">K<\/em> for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.<\/p>\r\n<p id=\"fs-idm323833328\">To further illustrate this important point, consider the reversible reaction shown below:<\/p>\r\n\r\n<div id=\"fs-idp63923872\" style=\"text-align: center\" data-type=\"equation\">CO(<em>g<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>) \u21cc CO<sub>2<\/sub>(<em>g<\/em>) + H<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c <\/sub><\/em>= 0.640\u00a0 \u00a0at\u00a0 <em>T <\/em>= 800\u00b0C<\/div>\r\n<p id=\"fs-idm373405968\">The bar charts in <a class=\"autogenerated-content\" href=\"#CNX_Chem_13_02_mixtures\">(Figure)<\/a> represent changes in\u00a0 reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction\u2019s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_13_02_mixtures\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">Compositions of three mixtures before (Q<sub>c<\/sub> \u2260 K<sub>c<\/sub>) and after (Q<sub>c<\/sub> = K<sub>c<\/sub>) equilibrium is established for the reaction CO(<em>g<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>) \u21cc CO<sub>2<\/sub>(<em>g<\/em>) + H<sub>2<\/sub>(<em>g<\/em>)<\/div>\r\n<span id=\"fs-idp62217296\" data-type=\"media\" data-alt=\"Two sets of bar graphs are shown. The left is labeled, \u201cBefore reaction,\u201d and the right is labeled, \u201cAt equilibrium.\u201d Both graphs have y-axes labeled, \u201cConcentration ( M ),\u201d and three bars on the x-axes labeled, \u201cMixture 1,\u201d \u201cMixture 2,\u201d and \u201cMixture 3.\u201d The y-axis has a scale beginning at 0.00 and ending at 0.10, with measurement increments of 0.02. The bars on the graphs are color coded and a key is provided with a legend. Red is labeled, \u201cC O;\u201d blue is labeled, \u201cH subscript 2 O;\u201d green is labeled, \u201cC O subscript 2,\u201d and yellow is labeled, \u201cH subscript 2.\u201d The graph on the left shows the red bar for mixture one just above 0.02, labeled \u201c0.0243,\u201d and the blue bar near 0.05, labeled \u201c0.0243.\u201d For mixture two, the green bar is near 0.05, labeled \u201c0.0468,\u201d and the yellow bar is near 0.09, labeled \u201c0.0468.\u201d For mixture 3, the red bar is near 0.01, labeled \u201c0.0330,\u201d the blue bar is slightly above that, labeled \u201c0.190,\u201d with green and yellow topping it off at 0.02. Green is labeled \u201c0.00175\u201d and yellow is labeled \u201c0.00160.\u201d On the right graph, the bar for mixture one shows the red bar slightly above 0.01, labeled \u201c0.0135,\u201d the blue bar stacked on it rising slightly above 0.02, labeled \u201c0.0135,\u201d the green rising near 0.04, labeled \u201c0.0108,\u201d and the yellow bar reaching near 0.05, labeled \u201c0.0108.\u201d A label above this bar reads, \u201cQ equals 0.640.\u201d The bar for mixture two shows the red bar slightly above 0.02, labeled \u201c0.0260,\u201d the blue bar stacked on it rising near 0.05, labeled \u201c0.0260,\u201d the green rising near 0.07, labeled \u201c0.0208,\u201d and the yellow bar reaching near 0.10, labeled \u201c0.0208.\u201d A label above this bar reads \u201cQ equals 0.640.\u201d The bar for mixture three shows the red bar near 0.01, labeled \u201c0.0231,\u201d the blue bar stacked on it rising slightly above 0.01, labeled \u201c0.00909,\u201d the green rising near 0.02, labeled \u201c0.0115,\u201d and the yellow bar reaching 0.02, labeled \u201c0.0117.\u201d A label above this bar reads \u201cQ equals 0.640\u201d.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_13_02_mixtures-1.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, labeled \u201c0.0243,\u201d and the blue bar near 0.05, labeled \u201c0.0243.\u201d For mixture two, the green bar is near 0.05, labeled \u201c0.0468,\u201d and the yellow bar is near 0.09, labeled \u201c0.0468.\u201d For mixture 3, the red bar is near 0.01, labeled \u201c0.0330,\u201d the blue bar is slightly above that, labeled \u201c0.190,\u201d with green and yellow topping it off at 0.02. Green is labeled \u201c0.00175\u201d and yellow is labeled \u201c0.00160.\u201d On the right graph, the bar for mixture one shows the red bar slightly above 0.01, labeled \u201c0.0135,\u201d the blue bar stacked on it rising slightly above 0.02, labeled \u201c0.0135,\u201d the green rising near 0.04, labeled \u201c0.0108,\u201d and the yellow bar reaching near 0.05, labeled \u201c0.0108.\u201d A label above this bar reads, \u201cQ equals 0.640.\u201d The bar for mixture two shows the red bar slightly above 0.02, labeled \u201c0.0260,\u201d the blue bar stacked on it rising near 0.05, labeled \u201c0.0260,\u201d the green rising near 0.07, labeled \u201c0.0208,\u201d and the yellow bar reaching near 0.10, labeled \u201c0.0208.\u201d A label above this bar reads \u201cQ equals 0.640.\u201d The bar for mixture three shows the red bar near 0.01, labeled \u201c0.0231,\u201d the blue bar stacked on it rising slightly above 0.01, labeled \u201c0.00909,\u201d the green rising near 0.02, labeled \u201c0.0115,\u201d and the yellow bar reaching 0.02, labeled \u201c0.0117.\u201d A label above this bar reads \u201cQ equals 0.640\u201d.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-idp174376240\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp60883568\"><span data-type=\"title\">Predicting the Direction of Reaction<\/span> Given here are the starting concentrations of reactants and products for three experiments involving this reaction:<\/p>\r\n\r\n<div id=\"fs-idm31469104\" style=\"text-align: center\" data-type=\"equation\">CO(<em>g<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>) \u21cc CO<sub>2<\/sub>(<em>g<\/em>) + H<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 <em>K<sub>c<\/sub><\/em> = 0.64<\/div>\r\n<div data-type=\"equation\"><\/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=\"middle\">\r\n<th data-align=\"left\">Reactants\/Products<\/th>\r\n<th data-align=\"left\">Experiment 1<\/th>\r\n<th data-align=\"left\">Experiment 2<\/th>\r\n<th data-align=\"left\">Experiment 3<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">[CO]<sub>i<\/sub><\/td>\r\n<td data-align=\"left\">0.020 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td data-align=\"left\">0.011 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td data-align=\"left\">0.0094 <em data-effect=\"italics\">M<\/em><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">[H<sub>2<\/sub>O]<sub>i<\/sub><\/td>\r\n<td data-align=\"left\">0.020 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td data-align=\"left\">0.0011 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td data-align=\"left\">0.0025 <em data-effect=\"italics\">M<\/em><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">[CO<sub>2<\/sub>]<sub>i<\/sub><\/td>\r\n<td data-align=\"left\">0.0040 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td data-align=\"left\">0.037 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td data-align=\"left\">0.0015 <em data-effect=\"italics\">M<\/em><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">[H<sub>2<\/sub>]<sub>i<\/sub><\/td>\r\n<td data-align=\"left\">0.0040 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td data-align=\"left\">0.046 <em data-effect=\"italics\">M<\/em><\/td>\r\n<td data-align=\"left\">0.0076 <em data-effect=\"italics\">M<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp3711408\"><span data-type=\"title\">Solution<\/span> Experiment 1:<\/p>\r\n\r\n<div id=\"fs-idp24553712\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1789\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2j-300x41.png\" alt=\"\" width=\"300\" height=\"41\" \/><\/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> (0.040 &lt; 0.64)<\/p>\r\n<p id=\"fs-idp267578784\">The reaction will proceed in the forward direction.<\/p>\r\n<p id=\"fs-idp56047184\"><img class=\"alignnone size-medium wp-image-1790\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2k-300x40.png\" alt=\"\" width=\"300\" height=\"40\" \/><\/p>\r\n<em data-effect=\"italics\">Q<sub>c<\/sub><\/em> &gt; <em data-effect=\"italics\">K<sub>c<\/sub><\/em> (140 &gt; 0.64)\r\n<p id=\"fs-idp68828672\">The reaction will proceed in the reverse direction.<\/p>\r\n<p id=\"fs-idp101164320\">Experiment 3:<\/p>\r\n\r\n<div id=\"fs-idp74212144\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1791\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2l-300x42.png\" alt=\"\" width=\"300\" height=\"42\" \/><\/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 proceed in the forward direction.<\/p>\r\n<p id=\"fs-idm76258576\"><strong>Check Your Learning:<\/strong><\/p>\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(<em>g<\/em>), 0.0155 mol of Cl2(<em>g<\/em>), and 0.500 mol of NOCl(<em>g<\/em>):<\/p>\r\n\r\n<div id=\"fs-idp56406272\" style=\"text-align: center\" data-type=\"equation\">2NO(<em>g<\/em>) + Cl<sub>2<\/sub>(<em>g<\/em>) \u21cc 2NOCl(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c<\/sub><\/em> = 4.6 \u00d7 10<sup>4<\/sup><\/div>\r\n<div data-type=\"equation\"><\/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\" style=\"text-align: center\" data-type=\"equation\">N<sub>2<\/sub>(<em>g<\/em>) + 3H<sub>2<\/sub>(<em>g<\/em>) \u21cc 2NH<sub>3<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c <\/sub><\/em>= 0.060<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp110625216\">(c) A 2.00-L flask containing 230 g of SO<sub>3<\/sub>(<em>g<\/em>):<\/p>\r\n\r\n<div id=\"fs-idp294747856\" style=\"text-align: center\" data-type=\"equation\">2SO<sub>3<\/sub>(<em>g<\/em>) \u21cc 2SO<sub>2<\/sub>(<em>g<\/em>) + O<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0Kc = 0.230<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idp32992592\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp96400256\">(a) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 6.45 \u00d7 10<sup>3<\/sup>, forward. (b) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 0.23, reverse. (c) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 0, forward.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp86160096\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Homogeneous Equilibria<\/strong><\/h3>\r\n<p id=\"fs-idm54871216\">A <strong>homogeneous equilibrium<\/strong> is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in <em data-effect=\"italics\">solutions<\/em>. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:<\/p>\r\n\r\n<div id=\"fs-idp38922960\" data-type=\"equation\"><img class=\"wp-image-1794 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2m-300x88.png\" alt=\"\" width=\"460\" height=\"135\" \/><\/div>\r\n<p id=\"fs-idp11719264\">These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is <em data-effect=\"italics\">not<\/em> included in the reaction quotient. The reason for this omission is related to the more rigorous form of the <em data-effect=\"italics\">Q<\/em> (or <em data-effect=\"italics\">K<\/em>) expression mentioned previously in this chapter, in which <em data-effect=\"italics\">relative concentrations for liquids and solids are equal to 1 and needn\u2019t be included<\/em>. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.<\/p>\r\n<p id=\"fs-idp11805840\">The equilibria below all involve gas-phase solutions:<\/p>\r\n\r\n<div id=\"fs-idp17340352\" data-type=\"equation\"><img class=\"wp-image-1795 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2n-300x102.png\" alt=\"\" width=\"462\" height=\"157\" \/><\/div>\r\n<p id=\"fs-idp58545648\">For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (<em data-effect=\"italics\">K<sub>c<\/sub><\/em>) or partial pressures (<em data-effect=\"italics\">K<sub>p<\/sub><\/em>) of the reactants and products. Note that the numerical values of <em>K<sub>c<\/sub><\/em> and <em>K<sub>p<\/sub><\/em> are usually different; be sure you note the difference.<\/p>\r\n&nbsp;\r\n\r\n<strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em\">Heterogeneous Equilibria<\/span><\/strong>\r\n\r\n<\/div>\r\n<div id=\"fs-idp112997072\" class=\"bc-section section\" data-depth=\"1\">\r\n<p id=\"fs-idp27217312\">A <strong>heterogeneous equilibrium<\/strong> involves reactants and products in two or more different phases, as illustrated by the following examples:<\/p>\r\n\r\n<div id=\"fs-idp129666976\" data-type=\"equation\"><img class=\"wp-image-1796 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2p-300x72.png\" alt=\"\" width=\"446\" height=\"107\" \/><\/div>\r\n<p id=\"fs-idp8357744\">Again, note that concentration terms are only included for gaseous and solute species, as discussed previously.<\/p>\r\n<p id=\"fs-idm328525392\">Two of the above examples include terms for gaseous species only in their equilibrium constants, and so <em data-effect=\"italics\">K<sub>p<\/sub><\/em> expressions may also be written:<\/p>\r\n\r\n<div id=\"fs-idp11158544\" data-type=\"equation\"><img class=\" wp-image-1797 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2q-300x57.png\" alt=\"\" width=\"364\" height=\"69\" \/><\/div>\r\n<\/div>\r\n<div id=\"fs-idm377214672\" class=\"bc-section section\" data-depth=\"1\">\r\n\r\n&nbsp;\r\n\r\n<strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em\">Coupled Equilibria<\/span><\/strong>\r\n<p id=\"fs-idm374944688\">The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more <em data-effect=\"italics\">coupled<\/em> equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.<\/p>\r\n<p id=\"fs-idm328298096\">1. Changing the direction of a chemical equation essentially swaps the identities of \u201creactants\u201d and \u201cproducts,\u201d and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.<\/p>\r\n\r\n<div id=\"fs-idm514156544\" data-type=\"equation\"><img class=\"wp-image-1798 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2r-300x211.png\" alt=\"\" width=\"203\" height=\"143\" \/><\/div>\r\n<p id=\"fs-idm336251504\">2. Changing the stoichiometric coefficients in an equation by some factor <em data-effect=\"italics\">x<\/em> results in an exponential change in the equilibrium constant by that same factor:<\/p>\r\n\r\n<div id=\"fs-idm357343392\" data-type=\"equation\"><img class=\"wp-image-1799 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2s-300x158.png\" alt=\"\" width=\"270\" height=\"142\" \/><\/div>\r\n<p id=\"fs-idm351795744\">3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction\u2019s K values:<\/p>\r\n\r\n<div id=\"fs-idm351795360\" data-type=\"equation\"><img class=\" wp-image-1800 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2t-300x109.png\" alt=\"\" width=\"234\" height=\"85\" \/><\/div>\r\n<p id=\"fs-idm374676944\">The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:<\/p>\r\n\r\n<div id=\"fs-idm345491344\" data-type=\"equation\"><img class=\"wp-image-1801 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2u-300x129.png\" alt=\"\" width=\"242\" height=\"104\" \/><\/div>\r\n<p id=\"fs-idm380030416\">Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:<\/p>\r\n\r\n<div id=\"fs-idm371974992\" data-type=\"equation\"><img class=\"size-medium wp-image-1802 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2v-300x107.png\" alt=\"\" width=\"300\" height=\"107\" \/><\/div>\r\n<p id=\"fs-idm329039744\"><a class=\"autogenerated-content\" href=\"#fs-idm384162720\">(Figure)<\/a> demonstrates the use of this strategy in describing coupled equilibrium processes.<\/p>\r\n\r\n<div id=\"fs-idm384162720\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm384162464\"><strong>Equilibrium Constants for Coupled Reactions:<\/strong><\/p>\r\nA mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 \u00b0C:\r\n<div id=\"fs-idm329006080\" style=\"text-align: center\" data-type=\"equation\">2NH<sub>3<\/sub>(<em>g<\/em>) + 3I<sub>2<\/sub>(<em>g<\/em>) \u21cc N<sub>2<\/sub>(<em>g<\/em>) + 6HI(<em>g<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm493175904\">Use the information below to calculate Kc for this reaction.<\/p>\r\n\r\n<div id=\"fs-idm515104288\" style=\"text-align: center\" data-type=\"equation\">N<sub>2<\/sub>(<em>g<\/em>) + 3H<sub>2<\/sub>(<em>g<\/em>) \u21cc 2NH<sub>3<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c1 <\/sub><\/em>= 0.50\u00a0 at\u00a0 400\u00b0C<\/div>\r\n<div style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>(<em>g<\/em>) + I<sub>2<\/sub>(<em>g<\/em>) \u21cc 2HI(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c2 <\/sub><\/em>= 50\u00a0 at\u00a0 400\u00b0C<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm375742144\"><strong>Solution:<\/strong><\/p>\r\nThe equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.\r\n<p id=\"fs-idm385085648\">Reverse the first coupled reaction equation:<\/p>\r\n\r\n<div id=\"fs-idm385085264\" data-type=\"equation\"><img class=\"alignnone wp-image-1803 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2w-300x31.png\" alt=\"\" width=\"348\" height=\"36\" \/><\/div>\r\n<p id=\"fs-idm339773312\">Multiply the second coupled reaction by 3:<\/p>\r\n\r\n<div id=\"fs-idm508785744\" style=\"text-align: center\" data-type=\"equation\">3H<sub>2<\/sub>(<em>g<\/em>) + 3I<sub>2<\/sub>(<em>g<\/em>) \u21cc 6HI(<em>g<\/em>)\u00a0 \u00a0 \u00a0 <em>K<sub>c2'<\/sub><\/em> = <em>K<sub>c2<\/sub><\/em><sup>3 <\/sup>= 50<sup>3 <\/sup>= 1.2 \u00d7 10<sup>5<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm372296992\">Finally, add the two revised equations:<\/p>\r\n\r\n<div id=\"fs-idm372296608\" style=\"text-align: center\" data-type=\"equation\">2NH<sub>3<\/sub>(<em>g<\/em>) + <del>3H<sub>2<\/sub>(<em>g<\/em>)<\/del> + 3I<sub><em>2<\/em><\/sub>(<em>g<\/em>) \u21cc N<sub>2<\/sub>(<em>g<\/em>) + <del>3H<sub>2<\/sub>(<em>g<\/em>)<\/del> + 6HI(<em>g<\/em>)<\/div>\r\n<div style=\"text-align: center\" data-type=\"equation\">2NH<sub>3<\/sub>(<em>g<\/em>) + 3I<sub><em>2<\/em><\/sub>(<em>g<\/em>) \u21cc N<sub>2<\/sub>(<em>g<\/em>) + 6HI(<em>g<\/em>)<\/div>\r\n<div style=\"text-align: center\" data-type=\"equation\"><em>K<sub>c<\/sub> <\/em>= <em>K<sub>c1'<\/sub>K<sub>c2'<\/sub><\/em> = (2.0)(1.2 \u00d7 10<sup>5<\/sup>) = 2.5 \u00d7 10<sup>5<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm379751344\"><strong>Check Your Learning:<\/strong><\/p>\r\nUse the provided information to calculate <em>K<sub>c<\/sub><\/em> for the following reaction at 550 \u00b0C:\r\n<div id=\"fs-idm379750528\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>(<em>g<\/em>) + CO<sub>2<\/sub>(<em>g<\/em>) \u21cc CO(<em>g<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>)\u00a0 \u00a0 \u00a0Kc=?<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div style=\"text-align: center\" data-type=\"equation\">Given:\u00a0 CoO(<em>s<\/em>) + CO(<em>g<\/em>) \u21cc Co(<em>s<\/em>) + CO<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c1<\/sub><\/em> = 490<\/div>\r\n<div style=\"text-align: center\" data-type=\"equation\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 CoO(<em>s<\/em>) + H<sub>2<\/sub>(<em>g<\/em>) \u21cc Co(<em>s<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c2<\/sub><\/em> = 67<\/div>\r\n<div id=\"fs-idm384024512\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm500608128\">K<sub>c<\/sub> = 0.14<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp284371824\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\r\n<p id=\"fs-idp165520928\">The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, <em data-effect=\"italics\">Q<\/em>. For a reaction at equilibrium, the composition is constant, and <em data-effect=\"italics\">Q<\/em> is called the equilibrium constant, <em data-effect=\"italics\">K<\/em>.<\/p>\r\n<p id=\"fs-idp116522496\">A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp116522752\" class=\"key-equations\" data-depth=\"1\"><\/div>\r\n<div id=\"fs-idp16341104\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idp72435440\" data-type=\"exercise\">\r\n<div id=\"fs-idp164607296\" data-type=\"problem\">\r\n<p id=\"fs-idp164607552\"><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"glossary\">\r\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\r\n<dl id=\"fs-idp63345136\">\r\n \t<dt>equilibrium constant (<em data-effect=\"italics\">K<\/em>)<\/dt>\r\n \t<dd id=\"fs-idp101767312\">value of the reaction quotient for a system at equilibrium; may be expressed using concentrations (<em data-effect=\"italics\">K<sub>c<\/sub><\/em>) or partial pressures (<em data-effect=\"italics\">K<sub>p<\/sub><\/em>)<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102632368\">\r\n \t<dt>heterogeneous equilibria<\/dt>\r\n \t<dd id=\"fs-idp35066448\">equilibria in which reactants and products occupy two or more different phases<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp35066960\">\r\n \t<dt>homogeneous equilibria<\/dt>\r\n \t<dd id=\"fs-idp55285008\">equilibria in which all reactants and products occupy the same phase<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp74361936\">\r\n \t<dt>reaction quotient (<em data-effect=\"italics\">Q<\/em>)<\/dt>\r\n \t<dd id=\"fs-idp97709856\">mathematical function describing the relative amounts of reactants and products in a reaction mixture; may be expressed in terms of concentrations (<em data-effect=\"italics\">Q<sub>c<\/sub><\/em>) or pressures (<em data-effect=\"italics\">Q<sub>p<\/sub><\/em>)<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<p><strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em;background-color: #cbd4b6;color: #000000\">Learning Objectives<\/span><\/strong><\/p>\n<div class=\"textbox textbox--learning-objectives\">\n<p>By the end of this section, 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\">The status of a reversible reaction is conveniently assessed by evaluating its <strong>reaction quotient (<em data-effect=\"italics\">Q<\/em>)<\/strong>. For a reversible reaction described by<\/p>\n<div id=\"fs-idp5986304\" style=\"text-align: center\" data-type=\"equation\"><em>m<\/em>A + <em>n<\/em>B \u21cc <em>x<\/em>C + <em>y<\/em>D<\/div>\n<p id=\"fs-idp11973152\">the reaction quotient is derived directly from the stoichiometry of the balanced equation as<\/p>\n<div id=\"fs-idp44297776\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1779 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2a.png\" alt=\"\" width=\"119\" height=\"58\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2a.png 245w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2a-65x32.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2a-225x109.png 225w\" sizes=\"auto, (max-width: 119px) 100vw, 119px\" \/><\/div>\n<p id=\"fs-idm7959200\">where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:<\/p>\n<div id=\"fs-idp23922448\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1780 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2b.png\" alt=\"\" width=\"119\" height=\"64\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2b.png 229w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2b-65x35.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2b-225x121.png 225w\" sizes=\"auto, (max-width: 119px) 100vw, 119px\" \/><\/div>\n<p id=\"fs-idm352046000\">Note that the reaction quotient equations above are a simplification of more rigorous expressions that use <em data-effect=\"italics\">relative<\/em> values for concentrations and pressures rather than <em data-effect=\"italics\">absolute<\/em> values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing <em data-effect=\"italics\">Q<\/em>. In most cases, this will introduce only modest errors in calculations involving reaction quotients.<\/p>\n<div id=\"fs-idp168372720\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp68042176\"><strong>Writing Reaction Quotient Expressions:<\/strong><\/p>\n<p>Write the concentration-based reaction quotient expression for each of the following reactions:<\/p>\n<p id=\"fs-idp246365472\">(a) 3O<sub>2<\/sub>(<em>g<\/em>) \u21cc 2O<sub>3<\/sub>(<em>g<\/em>)<\/p>\n<p id=\"fs-idp142495760\">(b) N<sub>2<\/sub>(<em>g<\/em>) + 3H<sub>2<\/sub>(<em>g<\/em>) \u21cc 2NH<sub>3<\/sub>(<em>g<\/em>)<\/p>\n<p id=\"fs-idp82472256\">(c) 4NH<sub>3<\/sub>(<em>g<\/em>) + 7O<sub>2<\/sub>(<em>g<\/em>) \u21cc 4NO<sub>2<\/sub>(<em>g<\/em>) + 6H<sub>2<\/sub>O(<em>g<\/em>)<\/p>\n<p id=\"fs-idp63878256\"><strong>Solution:<\/strong><\/p>\n<p><span style=\"text-align: initial;font-size: 1em\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1781\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2c-300x246.png\" alt=\"\" width=\"185\" height=\"152\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2c-300x246.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2c-65x53.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2c-225x185.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2c.png 313w\" sizes=\"auto, (max-width: 185px) 100vw, 185px\" \/><\/span><\/p>\n<p id=\"fs-idp17156768\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Write the concentration-based reaction quotient expression for each of the following reactions:<\/p>\n<p id=\"fs-idp35394896\">(a) 2SO<sub>2<\/sub>(<em>g<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u21cc 2SO<sub>3<\/sub>(<em>g<\/em>)<\/p>\n<p id=\"fs-idp63748560\">(b) C<sub>4<\/sub>H<sub>8<\/sub>(<em>g<\/em>) \u21cc 2C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>)<\/p>\n<p id=\"fs-idp105907744\">(c) 2C<sub>4<\/sub>H<sub>10<\/sub>(<em>g<\/em>) + 13O<sub>2<\/sub>(<em>g<\/em>) \u21cc 8CO<sub>2<\/sub>(<em>g<\/em>) + 10H<sub>2<\/sub>O(<em>g<\/em>)<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<div id=\"fs-idp167282768\" data-type=\"note\">\n<div data-type=\"title\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1782\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2d-300x37.png\" alt=\"\" width=\"364\" height=\"45\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2d-300x37.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2d-768x95.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2d-65x8.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2d-225x28.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2d-350x43.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2d.png 844w\" sizes=\"auto, (max-width: 364px) 100vw, 364px\" \/><\/div>\n<\/div>\n<\/div>\n<div id=\"CNX_Chem_13_02_quotient\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">Changes in concentrations and <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> for a chemical equilibrium achieved beginning with (a) a mixture of reactants only and (b) products only.<\/div>\n<p><span id=\"fs-idp134344480\" data-type=\"media\" data-alt=\"Four graphs are shown and labeled, \u201ca,\u201d \u201cb,\u201d \u201cc,\u201d and \u201cd.\u201d All four graphs have a vertical dotted line running through the middle labeled, \u201cEquilibrium is reached.\u201d The y-axis on graph a is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph a. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts high on the y-axis, ends midway down the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it starts and ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph b is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph b. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts low on the y-axis, ends midway up the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph c is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph c. This curve begins at the bottom of the y-axis and rises steeply up near the top of the y-axis, then levels off into a horizontal line. The top point of this line is labeled, \u201ck.\u201d The y-axis on graph d is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph d. This curve begins near the edge of the top of the y-axis and falls steeply toward the x-axis, then levels off into a horizontal line. The bottom point of this line is labeled, \u201ck.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_13_02_quotient-1.jpg\" alt=\"Four graphs are shown and labeled, \u201ca,\u201d \u201cb,\u201d \u201cc,\u201d and \u201cd.\u201d All four graphs have a vertical dotted line running through the middle labeled, \u201cEquilibrium is reached.\u201d The y-axis on graph a is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph a. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts high on the y-axis, ends midway down the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it starts and ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph b is labeled, \u201cConcentration,\u201d and the x-axis is labeled, \u201cTime.\u201d Three curves are plotted on graph b. The first is labeled, \u201c[ S O subscript 2 ];\u201d this line starts low on the y-axis, ends midway up the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, \u201c[ O subscript 2 ];\u201d this line mimics the first except that it ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, \u201c[ S O subscript 3 ].\u201d The y-axis on graph c is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph c. This curve begins at the bottom of the y-axis and rises steeply up near the top of the y-axis, then levels off into a horizontal line. The top point of this line is labeled, \u201ck.\u201d The y-axis on graph d is labeled, \u201cReaction Quotient,\u201d and the x-axis is labeled, \u201cTime.\u201d A single curve is plotted on graph d. This curve begins near the edge of the top of the y-axis and falls steeply toward the x-axis, then levels off into a horizontal line. The bottom point of this line is labeled, \u201ck.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idm122432\">The numerical value of <em data-effect=\"italics\">Q<\/em> varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction\u2019s status. To illustrate this point, consider the oxidation of sulfur dioxide:<\/p>\n<div id=\"fs-idm382376384\" style=\"text-align: center\" data-type=\"equation\">2SO<sub>2<\/sub>(<em>g<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u21cc 2SO<sub>3<\/sub>(<em>g<\/em>)<\/div>\n<p id=\"fs-idp138846544\">Two different experimental scenarios are depicted in <a class=\"autogenerated-content\" href=\"#CNX_Chem_13_02_quotient\">(Figure)<\/a>, one in which this reaction is initiated with a mixture of reactants only, SO<sub>2<\/sub> and O<sub>2<\/sub>, and another that begins with only product, SO<sub>3<\/sub>. For the reaction that begins with a mixture of reactants only, <em data-effect=\"italics\">Q<\/em> is initially equal to zero:<\/p>\n<div id=\"fs-idm329826160\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1783 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2e-300x62.png\" alt=\"\" width=\"285\" height=\"59\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2e-300x62.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2e-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2e-225x46.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2e-350x72.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2e.png 571w\" sizes=\"auto, (max-width: 285px) 100vw, 285px\" \/><\/div>\n<p id=\"fs-idm375739520\">As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>), product concentration increases (as does the numerator of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em>.<\/p>\n<p id=\"fs-idm378445456\">If the reaction begins with only product present, the value of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> is initially undefined (immeasurably large, or infinite):<\/p>\n<div id=\"fs-idm324502592\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1784 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2f-300x65.png\" alt=\"\" width=\"263\" height=\"57\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2f-300x65.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2f-65x14.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2f-225x49.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2f-350x76.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2f.png 522w\" sizes=\"auto, (max-width: 263px) 100vw, 263px\" \/><\/div>\n<p id=\"fs-idm374784256\">In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> decrease with time, the reactant concentrations and the denominator of <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.<\/p>\n<p id=\"fs-idm389114784\">The constant value of <em data-effect=\"italics\">Q<\/em> exhibited by a system at equilibrium is called the <strong>equilibrium constant, <em data-effect=\"italics\">K<\/em><\/strong>:<\/p>\n<div id=\"fs-idp266400656\" style=\"text-align: center\" data-type=\"equation\"><em>K<\/em> =<span style=\"font-size: 1em\"><em> Q<\/em> at equilibrium<\/span><\/div>\n<p id=\"fs-idm328721280\">Comparison of the data plots in <a class=\"autogenerated-content\" href=\"#CNX_Chem_13_02_quotient\">(Figure)<\/a> shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the <span data-type=\"term\">law of mass action<\/span>: At a given temperature, the reaction quotient for a system at equilibrium is constant.<\/p>\n<div id=\"fs-idp54423808\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp23994768\"><strong>Evaluating a Reaction Quotient:<\/strong><\/p>\n<p>Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:<\/p>\n<div id=\"fs-idp64726272\" style=\"text-align: center\" data-type=\"equation\">2NO<sub>2<\/sub>(<em>g<\/em>) \u21cc N<sub>2<\/sub>O<sub>4<\/sub>(<em>g<\/em>)<\/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<p id=\"fs-idp179491728\"><strong>Solution:<\/strong><\/p>\n<p>As for all equilibrium calculations in this text, use the simplified equations for <em data-effect=\"italics\">Q<\/em> and <em data-effect=\"italics\">K<\/em> and disregard any concentration or pressure units, as noted previously in this section.<\/p>\n<p id=\"fs-idm358571008\">(a) Before any product is formed,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1785\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2g-300x45.png\" alt=\"\" width=\"240\" height=\"36\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2g-300x45.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2g-65x10.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2g-225x34.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2g-350x53.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2g.png 409w\" sizes=\"auto, (max-width: 240px) 100vw, 240px\" \/><\/p>\n<p>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\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1786\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2h-300x81.png\" alt=\"\" width=\"226\" height=\"61\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2h-300x81.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2h-65x18.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2h-225x61.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2h-350x95.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2h.png 429w\" sizes=\"auto, (max-width: 226px) 100vw, 226px\" \/><\/div>\n<p id=\"fs-idp110962640\">(b) At equilibrium,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1787\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2i-300x42.png\" alt=\"\" width=\"300\" height=\"42\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2i-300x42.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2i-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2i-225x31.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2i-350x49.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2i.png 597w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p id=\"fs-idp284528080\"><strong>Check Your Learning:<\/strong><\/p>\n<p>For the reaction 2SO<sub>2<\/sub>(<em>g<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u21cc 2SO<sub>3<\/sub>(<em>g<\/em>), 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<p>&nbsp;<\/p>\n<div id=\"fs-idp10573648\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp130245328\"><em data-effect=\"italics\">K<sub>c<\/sub> =<\/em> 4.3<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm80100288\">By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large <em data-effect=\"italics\">K<\/em> will reach equilibrium when most of the reactant has been converted to product, whereas a small <em data-effect=\"italics\">K<\/em> indicates the reaction achieves equilibrium after very little reactant has been converted. It\u2019s important to keep in mind that the magnitude of <em data-effect=\"italics\">K<\/em> does <em data-effect=\"italics\">not<\/em> indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.<\/p>\n<p id=\"fs-idm377606544\">The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and\/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing <em data-effect=\"italics\">Q<\/em> to <em data-effect=\"italics\">K<\/em> for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.<\/p>\n<p id=\"fs-idm323833328\">To further illustrate this important point, consider the reversible reaction shown below:<\/p>\n<div id=\"fs-idp63923872\" style=\"text-align: center\" data-type=\"equation\">CO(<em>g<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>) \u21cc CO<sub>2<\/sub>(<em>g<\/em>) + H<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c <\/sub><\/em>= 0.640\u00a0 \u00a0at\u00a0 <em>T <\/em>= 800\u00b0C<\/div>\n<p id=\"fs-idm373405968\">The bar charts in <a class=\"autogenerated-content\" href=\"#CNX_Chem_13_02_mixtures\">(Figure)<\/a> represent changes in\u00a0 reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction\u2019s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_13_02_mixtures\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">Compositions of three mixtures before (Q<sub>c<\/sub> \u2260 K<sub>c<\/sub>) and after (Q<sub>c<\/sub> = K<sub>c<\/sub>) equilibrium is established for the reaction CO(<em>g<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>) \u21cc CO<sub>2<\/sub>(<em>g<\/em>) + H<sub>2<\/sub>(<em>g<\/em>)<\/div>\n<p><span id=\"fs-idp62217296\" data-type=\"media\" data-alt=\"Two sets of bar graphs are shown. The left is labeled, \u201cBefore reaction,\u201d and the right is labeled, \u201cAt equilibrium.\u201d Both graphs have y-axes labeled, \u201cConcentration ( M ),\u201d and three bars on the x-axes labeled, \u201cMixture 1,\u201d \u201cMixture 2,\u201d and \u201cMixture 3.\u201d The y-axis has a scale beginning at 0.00 and ending at 0.10, with measurement increments of 0.02. The bars on the graphs are color coded and a key is provided with a legend. Red is labeled, \u201cC O;\u201d blue is labeled, \u201cH subscript 2 O;\u201d green is labeled, \u201cC O subscript 2,\u201d and yellow is labeled, \u201cH subscript 2.\u201d The graph on the left shows the red bar for mixture one just above 0.02, labeled \u201c0.0243,\u201d and the blue bar near 0.05, labeled \u201c0.0243.\u201d For mixture two, the green bar is near 0.05, labeled \u201c0.0468,\u201d and the yellow bar is near 0.09, labeled \u201c0.0468.\u201d For mixture 3, the red bar is near 0.01, labeled \u201c0.0330,\u201d the blue bar is slightly above that, labeled \u201c0.190,\u201d with green and yellow topping it off at 0.02. Green is labeled \u201c0.00175\u201d and yellow is labeled \u201c0.00160.\u201d On the right graph, the bar for mixture one shows the red bar slightly above 0.01, labeled \u201c0.0135,\u201d the blue bar stacked on it rising slightly above 0.02, labeled \u201c0.0135,\u201d the green rising near 0.04, labeled \u201c0.0108,\u201d and the yellow bar reaching near 0.05, labeled \u201c0.0108.\u201d A label above this bar reads, \u201cQ equals 0.640.\u201d The bar for mixture two shows the red bar slightly above 0.02, labeled \u201c0.0260,\u201d the blue bar stacked on it rising near 0.05, labeled \u201c0.0260,\u201d the green rising near 0.07, labeled \u201c0.0208,\u201d and the yellow bar reaching near 0.10, labeled \u201c0.0208.\u201d A label above this bar reads \u201cQ equals 0.640.\u201d The bar for mixture three shows the red bar near 0.01, labeled \u201c0.0231,\u201d the blue bar stacked on it rising slightly above 0.01, labeled \u201c0.00909,\u201d the green rising near 0.02, labeled \u201c0.0115,\u201d and the yellow bar reaching 0.02, labeled \u201c0.0117.\u201d A label above this bar reads \u201cQ equals 0.640\u201d.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_13_02_mixtures-1.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, labeled \u201c0.0243,\u201d and the blue bar near 0.05, labeled \u201c0.0243.\u201d For mixture two, the green bar is near 0.05, labeled \u201c0.0468,\u201d and the yellow bar is near 0.09, labeled \u201c0.0468.\u201d For mixture 3, the red bar is near 0.01, labeled \u201c0.0330,\u201d the blue bar is slightly above that, labeled \u201c0.190,\u201d with green and yellow topping it off at 0.02. Green is labeled \u201c0.00175\u201d and yellow is labeled \u201c0.00160.\u201d On the right graph, the bar for mixture one shows the red bar slightly above 0.01, labeled \u201c0.0135,\u201d the blue bar stacked on it rising slightly above 0.02, labeled \u201c0.0135,\u201d the green rising near 0.04, labeled \u201c0.0108,\u201d and the yellow bar reaching near 0.05, labeled \u201c0.0108.\u201d A label above this bar reads, \u201cQ equals 0.640.\u201d The bar for mixture two shows the red bar slightly above 0.02, labeled \u201c0.0260,\u201d the blue bar stacked on it rising near 0.05, labeled \u201c0.0260,\u201d the green rising near 0.07, labeled \u201c0.0208,\u201d and the yellow bar reaching near 0.10, labeled \u201c0.0208.\u201d A label above this bar reads \u201cQ equals 0.640.\u201d The bar for mixture three shows the red bar near 0.01, labeled \u201c0.0231,\u201d the blue bar stacked on it rising slightly above 0.01, labeled \u201c0.00909,\u201d the green rising near 0.02, labeled \u201c0.0115,\u201d and the yellow bar reaching 0.02, labeled \u201c0.0117.\u201d A label above this bar reads \u201cQ equals 0.640\u201d.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-idp174376240\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp60883568\"><span data-type=\"title\">Predicting the Direction of Reaction<\/span> Given here are the starting concentrations of reactants and products for three experiments involving this reaction:<\/p>\n<div id=\"fs-idm31469104\" style=\"text-align: center\" data-type=\"equation\">CO(<em>g<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>) \u21cc CO<sub>2<\/sub>(<em>g<\/em>) + H<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 <em>K<sub>c<\/sub><\/em> = 0.64<\/div>\n<div data-type=\"equation\"><\/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=\"middle\">\n<th data-align=\"left\">Reactants\/Products<\/th>\n<th data-align=\"left\">Experiment 1<\/th>\n<th data-align=\"left\">Experiment 2<\/th>\n<th data-align=\"left\">Experiment 3<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td data-align=\"left\">[CO]<sub>i<\/sub><\/td>\n<td data-align=\"left\">0.020 <em data-effect=\"italics\">M<\/em><\/td>\n<td data-align=\"left\">0.011 <em data-effect=\"italics\">M<\/em><\/td>\n<td data-align=\"left\">0.0094 <em data-effect=\"italics\">M<\/em><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">[H<sub>2<\/sub>O]<sub>i<\/sub><\/td>\n<td data-align=\"left\">0.020 <em data-effect=\"italics\">M<\/em><\/td>\n<td data-align=\"left\">0.0011 <em data-effect=\"italics\">M<\/em><\/td>\n<td data-align=\"left\">0.0025 <em data-effect=\"italics\">M<\/em><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">[CO<sub>2<\/sub>]<sub>i<\/sub><\/td>\n<td data-align=\"left\">0.0040 <em data-effect=\"italics\">M<\/em><\/td>\n<td data-align=\"left\">0.037 <em data-effect=\"italics\">M<\/em><\/td>\n<td data-align=\"left\">0.0015 <em data-effect=\"italics\">M<\/em><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">[H<sub>2<\/sub>]<sub>i<\/sub><\/td>\n<td data-align=\"left\">0.0040 <em data-effect=\"italics\">M<\/em><\/td>\n<td data-align=\"left\">0.046 <em data-effect=\"italics\">M<\/em><\/td>\n<td data-align=\"left\">0.0076 <em data-effect=\"italics\">M<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp3711408\"><span data-type=\"title\">Solution<\/span> Experiment 1:<\/p>\n<div id=\"fs-idp24553712\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1789\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2j-300x41.png\" alt=\"\" width=\"300\" height=\"41\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2j-300x41.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2j-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2j-225x30.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2j-350x47.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2j.png 731w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/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> (0.040 &lt; 0.64)<\/p>\n<p id=\"fs-idp267578784\">The reaction will proceed in the forward direction.<\/p>\n<p id=\"fs-idp56047184\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1790\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2k-300x40.png\" alt=\"\" width=\"300\" height=\"40\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2k-300x40.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2k-768x101.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2k-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2k-225x30.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2k-350x46.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2k.png 781w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p><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 proceed in the reverse direction.<\/p>\n<p id=\"fs-idp101164320\">Experiment 3:<\/p>\n<div id=\"fs-idp74212144\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1791\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2l-300x42.png\" alt=\"\" width=\"300\" height=\"42\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2l-300x42.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2l-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2l-225x31.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2l-350x48.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2l.png 722w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/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 proceed in the forward direction.<\/p>\n<p id=\"fs-idm76258576\"><strong>Check Your Learning:<\/strong><\/p>\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(<em>g<\/em>), 0.0155 mol of Cl2(<em>g<\/em>), and 0.500 mol of NOCl(<em>g<\/em>):<\/p>\n<div id=\"fs-idp56406272\" style=\"text-align: center\" data-type=\"equation\">2NO(<em>g<\/em>) + Cl<sub>2<\/sub>(<em>g<\/em>) \u21cc 2NOCl(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c<\/sub><\/em> = 4.6 \u00d7 10<sup>4<\/sup><\/div>\n<div data-type=\"equation\"><\/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\" style=\"text-align: center\" data-type=\"equation\">N<sub>2<\/sub>(<em>g<\/em>) + 3H<sub>2<\/sub>(<em>g<\/em>) \u21cc 2NH<sub>3<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c <\/sub><\/em>= 0.060<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp110625216\">(c) A 2.00-L flask containing 230 g of SO<sub>3<\/sub>(<em>g<\/em>):<\/p>\n<div id=\"fs-idp294747856\" style=\"text-align: center\" data-type=\"equation\">2SO<sub>3<\/sub>(<em>g<\/em>) \u21cc 2SO<sub>2<\/sub>(<em>g<\/em>) + O<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0Kc = 0.230<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idp32992592\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp96400256\">(a) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 6.45 \u00d7 10<sup>3<\/sup>, forward. (b) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 0.23, reverse. (c) <em data-effect=\"italics\">Q<sub>c<\/sub><\/em> = 0, forward.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp86160096\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Homogeneous Equilibria<\/strong><\/h3>\n<p id=\"fs-idm54871216\">A <strong>homogeneous equilibrium<\/strong> is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in <em data-effect=\"italics\">solutions<\/em>. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:<\/p>\n<div id=\"fs-idp38922960\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1794 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2m-300x88.png\" alt=\"\" width=\"460\" height=\"135\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2m-300x88.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2m-1024x300.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2m-768x225.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2m-65x19.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2m-225x66.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2m-350x103.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2m.png 1133w\" sizes=\"auto, (max-width: 460px) 100vw, 460px\" \/><\/div>\n<p id=\"fs-idp11719264\">These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is <em data-effect=\"italics\">not<\/em> included in the reaction quotient. The reason for this omission is related to the more rigorous form of the <em data-effect=\"italics\">Q<\/em> (or <em data-effect=\"italics\">K<\/em>) expression mentioned previously in this chapter, in which <em data-effect=\"italics\">relative concentrations for liquids and solids are equal to 1 and needn\u2019t be included<\/em>. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.<\/p>\n<p id=\"fs-idp11805840\">The equilibria below all involve gas-phase solutions:<\/p>\n<div id=\"fs-idp17340352\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1795 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2n-300x102.png\" alt=\"\" width=\"462\" height=\"157\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2n-300x102.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2n-1024x348.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2n-768x261.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2n-65x22.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2n-225x77.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2n-350x119.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2n.png 1055w\" sizes=\"auto, (max-width: 462px) 100vw, 462px\" \/><\/div>\n<p id=\"fs-idp58545648\">For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (<em data-effect=\"italics\">K<sub>c<\/sub><\/em>) or partial pressures (<em data-effect=\"italics\">K<sub>p<\/sub><\/em>) of the reactants and products. Note that the numerical values of <em>K<sub>c<\/sub><\/em> and <em>K<sub>p<\/sub><\/em> are usually different; be sure you note the difference.<\/p>\n<p>&nbsp;<\/p>\n<p><strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em\">Heterogeneous Equilibria<\/span><\/strong><\/p>\n<\/div>\n<div id=\"fs-idp112997072\" class=\"bc-section section\" data-depth=\"1\">\n<p id=\"fs-idp27217312\">A <strong>heterogeneous equilibrium<\/strong> involves reactants and products in two or more different phases, as illustrated by the following examples:<\/p>\n<div id=\"fs-idp129666976\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1796 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2p-300x72.png\" alt=\"\" width=\"446\" height=\"107\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2p-300x72.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2p-1024x247.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2p-768x185.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2p-65x16.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2p-225x54.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2p-350x84.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2p.png 1090w\" sizes=\"auto, (max-width: 446px) 100vw, 446px\" \/><\/div>\n<p id=\"fs-idp8357744\">Again, note that concentration terms are only included for gaseous and solute species, as discussed previously.<\/p>\n<p id=\"fs-idm328525392\">Two of the above examples include terms for gaseous species only in their equilibrium constants, and so <em data-effect=\"italics\">K<sub>p<\/sub><\/em> expressions may also be written:<\/p>\n<div id=\"fs-idp11158544\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1797 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2q-300x57.png\" alt=\"\" width=\"364\" height=\"69\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2q-300x57.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2q-768x147.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2q-65x12.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2q-225x43.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2q-350x67.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2q.png 806w\" sizes=\"auto, (max-width: 364px) 100vw, 364px\" \/><\/div>\n<\/div>\n<div id=\"fs-idm377214672\" class=\"bc-section section\" data-depth=\"1\">\n<p>&nbsp;<\/p>\n<p><strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em\">Coupled Equilibria<\/span><\/strong><\/p>\n<p id=\"fs-idm374944688\">The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more <em data-effect=\"italics\">coupled<\/em> equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.<\/p>\n<p id=\"fs-idm328298096\">1. Changing the direction of a chemical equation essentially swaps the identities of \u201creactants\u201d and \u201cproducts,\u201d and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.<\/p>\n<div id=\"fs-idm514156544\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1798 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2r-300x211.png\" alt=\"\" width=\"203\" height=\"143\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2r-300x211.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2r-65x46.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2r-225x158.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2r-350x246.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2r.png 425w\" sizes=\"auto, (max-width: 203px) 100vw, 203px\" \/><\/div>\n<p id=\"fs-idm336251504\">2. Changing the stoichiometric coefficients in an equation by some factor <em data-effect=\"italics\">x<\/em> results in an exponential change in the equilibrium constant by that same factor:<\/p>\n<div id=\"fs-idm357343392\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1799 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2s-300x158.png\" alt=\"\" width=\"270\" height=\"142\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2s-300x158.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2s-65x34.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2s-225x119.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2s-350x185.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2s.png 479w\" sizes=\"auto, (max-width: 270px) 100vw, 270px\" \/><\/div>\n<p id=\"fs-idm351795744\">3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction\u2019s K values:<\/p>\n<div id=\"fs-idm351795360\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1800 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2t-300x109.png\" alt=\"\" width=\"234\" height=\"85\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2t-300x109.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2t-65x24.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2t-225x82.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2t-350x127.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2t.png 433w\" sizes=\"auto, (max-width: 234px) 100vw, 234px\" \/><\/div>\n<p id=\"fs-idm374676944\">The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:<\/p>\n<div id=\"fs-idm345491344\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1801 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2u-300x129.png\" alt=\"\" width=\"242\" height=\"104\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2u-300x129.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2u-65x28.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2u-225x97.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2u-350x151.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2u.png 444w\" sizes=\"auto, (max-width: 242px) 100vw, 242px\" \/><\/div>\n<p id=\"fs-idm380030416\">Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:<\/p>\n<div id=\"fs-idm371974992\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1802 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2v-300x107.png\" alt=\"\" width=\"300\" height=\"107\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2v-300x107.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2v-65x23.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2v-225x80.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2v-350x125.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2v.png 708w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idm329039744\"><a class=\"autogenerated-content\" href=\"#fs-idm384162720\">(Figure)<\/a> demonstrates the use of this strategy in describing coupled equilibrium processes.<\/p>\n<div id=\"fs-idm384162720\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm384162464\"><strong>Equilibrium Constants for Coupled Reactions:<\/strong><\/p>\n<p>A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 \u00b0C:<\/p>\n<div id=\"fs-idm329006080\" style=\"text-align: center\" data-type=\"equation\">2NH<sub>3<\/sub>(<em>g<\/em>) + 3I<sub>2<\/sub>(<em>g<\/em>) \u21cc N<sub>2<\/sub>(<em>g<\/em>) + 6HI(<em>g<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm493175904\">Use the information below to calculate Kc for this reaction.<\/p>\n<div id=\"fs-idm515104288\" style=\"text-align: center\" data-type=\"equation\">N<sub>2<\/sub>(<em>g<\/em>) + 3H<sub>2<\/sub>(<em>g<\/em>) \u21cc 2NH<sub>3<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c1 <\/sub><\/em>= 0.50\u00a0 at\u00a0 400\u00b0C<\/div>\n<div style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>(<em>g<\/em>) + I<sub>2<\/sub>(<em>g<\/em>) \u21cc 2HI(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c2 <\/sub><\/em>= 50\u00a0 at\u00a0 400\u00b0C<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm375742144\"><strong>Solution:<\/strong><\/p>\n<p>The equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.<\/p>\n<p id=\"fs-idm385085648\">Reverse the first coupled reaction equation:<\/p>\n<div id=\"fs-idm385085264\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1803 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2w-300x31.png\" alt=\"\" width=\"348\" height=\"36\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2w-300x31.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2w-768x80.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2w-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2w-225x23.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2w-350x36.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/13.2w.png 972w\" sizes=\"auto, (max-width: 348px) 100vw, 348px\" \/><\/div>\n<p id=\"fs-idm339773312\">Multiply the second coupled reaction by 3:<\/p>\n<div id=\"fs-idm508785744\" style=\"text-align: center\" data-type=\"equation\">3H<sub>2<\/sub>(<em>g<\/em>) + 3I<sub>2<\/sub>(<em>g<\/em>) \u21cc 6HI(<em>g<\/em>)\u00a0 \u00a0 \u00a0 <em>K<sub>c2&#8242;<\/sub><\/em> = <em>K<sub>c2<\/sub><\/em><sup>3 <\/sup>= 50<sup>3 <\/sup>= 1.2 \u00d7 10<sup>5<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm372296992\">Finally, add the two revised equations:<\/p>\n<div id=\"fs-idm372296608\" style=\"text-align: center\" data-type=\"equation\">2NH<sub>3<\/sub>(<em>g<\/em>) + <del>3H<sub>2<\/sub>(<em>g<\/em>)<\/del> + 3I<sub><em>2<\/em><\/sub>(<em>g<\/em>) \u21cc N<sub>2<\/sub>(<em>g<\/em>) + <del>3H<sub>2<\/sub>(<em>g<\/em>)<\/del> + 6HI(<em>g<\/em>)<\/div>\n<div style=\"text-align: center\" data-type=\"equation\">2NH<sub>3<\/sub>(<em>g<\/em>) + 3I<sub><em>2<\/em><\/sub>(<em>g<\/em>) \u21cc N<sub>2<\/sub>(<em>g<\/em>) + 6HI(<em>g<\/em>)<\/div>\n<div style=\"text-align: center\" data-type=\"equation\"><em>K<sub>c<\/sub> <\/em>= <em>K<sub>c1&#8242;<\/sub>K<sub>c2&#8242;<\/sub><\/em> = (2.0)(1.2 \u00d7 10<sup>5<\/sup>) = 2.5 \u00d7 10<sup>5<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm379751344\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Use the provided information to calculate <em>K<sub>c<\/sub><\/em> for the following reaction at 550 \u00b0C:<\/p>\n<div id=\"fs-idm379750528\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>(<em>g<\/em>) + CO<sub>2<\/sub>(<em>g<\/em>) \u21cc CO(<em>g<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>)\u00a0 \u00a0 \u00a0Kc=?<\/div>\n<div data-type=\"equation\"><\/div>\n<div style=\"text-align: center\" data-type=\"equation\">Given:\u00a0 CoO(<em>s<\/em>) + CO(<em>g<\/em>) \u21cc Co(<em>s<\/em>) + CO<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c1<\/sub><\/em> = 490<\/div>\n<div style=\"text-align: center\" data-type=\"equation\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 CoO(<em>s<\/em>) + H<sub>2<\/sub>(<em>g<\/em>) \u21cc Co(<em>s<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>)\u00a0 \u00a0 \u00a0<em>K<sub>c2<\/sub><\/em> = 67<\/div>\n<div id=\"fs-idm384024512\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm500608128\">K<sub>c<\/sub> = 0.14<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp284371824\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\n<p id=\"fs-idp165520928\">The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, <em data-effect=\"italics\">Q<\/em>. For a reaction at equilibrium, the composition is constant, and <em data-effect=\"italics\">Q<\/em> is called the equilibrium constant, <em data-effect=\"italics\">K<\/em>.<\/p>\n<p id=\"fs-idp116522496\">A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases.<\/p>\n<\/div>\n<div id=\"fs-idp116522752\" class=\"key-equations\" data-depth=\"1\"><\/div>\n<div id=\"fs-idp16341104\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idp72435440\" data-type=\"exercise\">\n<div id=\"fs-idp164607296\" data-type=\"problem\">\n<p id=\"fs-idp164607552\">\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"glossary\">\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\n<dl id=\"fs-idp63345136\">\n<dt>equilibrium constant (<em data-effect=\"italics\">K<\/em>)<\/dt>\n<dd id=\"fs-idp101767312\">value of the reaction quotient for a system at equilibrium; may be expressed using concentrations (<em data-effect=\"italics\">K<sub>c<\/sub><\/em>) or partial pressures (<em data-effect=\"italics\">K<sub>p<\/sub><\/em>)<\/dd>\n<\/dl>\n<dl id=\"fs-idp102632368\">\n<dt>heterogeneous equilibria<\/dt>\n<dd id=\"fs-idp35066448\">equilibria in which reactants and products occupy two or more different phases<\/dd>\n<\/dl>\n<dl id=\"fs-idp35066960\">\n<dt>homogeneous equilibria<\/dt>\n<dd id=\"fs-idp55285008\">equilibria in which all reactants and products occupy the same phase<\/dd>\n<\/dl>\n<dl id=\"fs-idp74361936\">\n<dt>reaction quotient (<em data-effect=\"italics\">Q<\/em>)<\/dt>\n<dd id=\"fs-idp97709856\">mathematical function describing the relative amounts of reactants and products in a reaction mixture; may be expressed in terms of concentrations (<em data-effect=\"italics\">Q<sub>c<\/sub><\/em>) or pressures (<em data-effect=\"italics\">Q<sub>p<\/sub><\/em>)<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":1392,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-755","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":746,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/755","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/users\/1392"}],"version-history":[{"count":6,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/755\/revisions"}],"predecessor-version":[{"id":2167,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/755\/revisions\/2167"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/parts\/746"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/755\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=755"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=755"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=755"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=755"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}