{"id":2747,"date":"2016-08-24T18:15:02","date_gmt":"2016-08-24T18:15:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2747"},"modified":"2016-08-24T21:31:27","modified_gmt":"2016-08-24T21:31:27","slug":"lechateliers-principle-and-the-equilibrium-constant","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/lechateliers-principle-and-the-equilibrium-constant\/","title":{"raw":"LeCh\u00e2telier's Principle and the Equilibrium Constant","rendered":"LeCh\u00e2telier&#8217;s Principle and the Equilibrium Constant"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Describe the relationship between Le Ch\u00e2telier\u2019s principle and the equilibrium constant.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>How much is in savings this month?<\/h3>\r\n<p id=\"x-ck12-Yzk3YmE4ZmRmNjJkYWFjNzA4NzkxYTVlZjA1YjUwYmQ.-wix\">With online banking, management of your personal finances can become less complicated in some ways. You can automatically deposit paychecks, pay bills, and designate how much goes into savings or other special accounts each month. If you want to maintain 10% of your bank account in savings, you can set up a program that moves money in and out of the account when you get a paycheck or pay bills. The amount of money in savings will change as the money comes in and out of the bank, but the ratio of savings to checking will always be constant.<\/p>\r\n\r\n<\/div>\r\n<h2>LeCh\u00e2telier\u2019s Principle and the Equilibrium Constant<\/h2>\r\n<p id=\"x-ck12-MDliYmY0NzMyYzYzYjQ3N2FmMTA2NjAxNzM5Y2RhMWI.-pkk\">Occasionally, when students apply LeChatelier\u2019s principle to an equilibrium problem involving a change in concentration, they assume that <em>K<sub>eq<\/sub><\/em>\u00a0must change. This seems logical since we talk about \u201cshifting\u201d the equilibrium in one direction or the other. However,\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0is a constant, for a given equilibrium at a given temperature, so it must not change. Here is an example of how this works. Consider the simplified equilibrium below:<\/p>\r\n<p id=\"x-ck12-odx\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213017\/72e74f69fa133df5bab97520defd6e5a.png\" alt=\"A rightleftarrows B\" width=\"56\" height=\"13\" \/><\/p>\r\n<p id=\"x-ck12-OWEwYjYwNjIyOWM1MDgwYzc0OTlkMDk0NThkM2Y3MTk.-wkb\">Let\u2019s say we have a 1.0 liter container. At equilibrium the following amounts are measured.<\/p>\r\n<p id=\"x-ck12-m7m\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213017\/090753703c616b1009336feae1bce8d1.png\" alt=\"A &amp;=0.50 text{ mol} \\B &amp;=1.0 text{ mol}\" width=\"105\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-NGQyNjkxYmViZTk0ZTVmZWExYjVkYTZjMWUxZmRmNWU.-ycw\">The value of\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0is given by:<\/p>\r\n<p id=\"x-ck12-ixn\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213018\/e4a661eb9faf0712ed0d3f12e3116eec.png\" alt=\"K_{eq}= frac{left [B right ]}{left [A right ]}=frac{1.0 text{ M}}{0.50 text{ M}}=2.0\" width=\"209\" height=\"43\" \/><\/p>\r\n<p id=\"x-ck12-OWIyOWQ4MWM4Yzg2NjYxMTI4ZjYxNDA0MTJmNDMyOGI.-evb\">Now we will disturb the equilibrium by adding 0.50 mole of\u00a0<em>A<\/em>\u00a0to the mixture. The equilibrium will shift towards the right, forming more\u00a0<em>B<\/em>. Immediately after the addition of\u00a0<em>A<\/em>\u00a0and before any response, we now have 1.0 mol of\u00a0<em>A<\/em>\u00a0and 1.0 mol of <em>B<\/em>. The equilibrium then shifts in the forward direction. We will introduce a variable (<em>x<\/em>), which will represent the change in concentrations as the reaction proceeds. Since the mole ratio of <em>A<\/em> :\u00a0<em>B<\/em>\u00a0is 1:1, as [<em>A<\/em>]\u00a0decreases by the amount <em>x<\/em>, the [<em>B<\/em>]\u00a0increases by the amount <em>x<\/em>. We set up an analysis called <em> ICE<\/em>, which stands for Initial, Change, and Equilibrium. The values in the table represent molar concentrations.<\/p>\r\n<p id=\"x-ck12-al1\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213020\/3f4d8eaf3ccca83b51d2e8e12b9af3c3.png\" alt=\"&amp; qquad qquad qquad underline{;; A qquad qquad B ; ; ; ; ; ; ; ;;;} \\&amp; text{Initial} qquad qquad 1.0 qquad quad 1.0 \\ &amp; text{Change}qquad quad -x qquad quad +x \\ &amp; text{Equilibrium} qquad 1.0 -x qquad 1.0 +x\" width=\"281\" height=\"96\" \/><\/p>\r\n<p id=\"x-ck12-YTBmNzUwZjA0OGQxZWU2YWU2NGY0NDI3MDU0MTRhZTc.-wna\">At the new equilibrium position, the values for\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> and\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> as a function of\u00a0<em>x\u00a0<\/em>can be set equal to the value of the\u00a0<em>K<sub>eq<\/sub><\/em>. Then, one can solve for <em>x<\/em>.<\/p>\r\n<p id=\"x-ck12-lbi\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213021\/a57d233fe81ef3ae00eab66ee1318d76.png\" alt=\"K_{eq}=2.0=frac{left [ B right ]}{left [ A right ]}=frac{1.0 + x}{1.0 - x}\" width=\"209\" height=\"43\" \/><\/p>\r\n<p id=\"x-ck12-Yzc1M2IzOWJhNWMwNGI4N2MwYmRhMWM0NmRiZjI0ZWU.-wac\">Solving for <em>x<\/em>:<\/p>\r\n<p id=\"x-ck12-26d\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213021\/bfde50174b7fd5998ffaa1d4487decd6.png\" alt=\"2.0(1.0-x) &amp;=1.0 + x \\2.0 - 2.0 x &amp;=1.0+x \\3.0 x &amp;=1.0 \\x &amp;=0.33\" width=\"171\" height=\"94\" \/><\/p>\r\n<p id=\"x-ck12-ZmQ2MmQzNzRhNDNiNDZmODJlOTE1ODYxMTVhYTVkMzI.-0n3\">This value for\u00a0<em>x<\/em>\u00a0is now plugged back in to the Equilibrium line of the table and the final concentrations of\u00a0<em>A<\/em>\u00a0and\u00a0<em>B<\/em>\u00a0after the reaction is calculated.<\/p>\r\n<p id=\"x-ck12-c2z\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213022\/717a9394752fb5eebb2a19bf94b86d9e.png\" alt=\"left [ A right ]&amp;=1.0-x=0.67 text{ M} \\left [ B right ]&amp;=1.0+x=1.33 text{ M}\" width=\"180\" height=\"44\" \/><\/p>\r\n<p id=\"x-ck12-NjlmNmIzOWFjZGEwMDkwNTJhNWZhNzZjMjcxMzYxNTY.-jld\">The value of\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0has been maintained since <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213023\/47120e44b5c4b02a55052bca72484a5a.png\" alt=\"frac{1.33}{0.67}=2.0\" width=\"72\" height=\"23\" \/> . This shows that even though a change in concentration of one of the substances in equilibrium causes a shift in the equilibrium position, the value of the equilibrium constant does not change.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-ZmE2MDczOGE1YTJjM2ZlZmI3M2RhNWU5YjllNDNmOTA.-vqn\">\r\n \t<li>Maintenance of the constant\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0for a reaction is described.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Practice<\/h3>\r\n<p id=\"x-ck12-ZDA1NTE4MmE3ZWNhMTc1ZjZmZjlhNTc2YTQzZWI5ZTg.-fdo\">Read the material at <a href=\"http:\/\/www.chem.purdue.edu\/gchelp\/howtosolveit\/Equilibrium\/ICEchart.htm\" target=\"_blank\">Chem Purdue<\/a>\u00a0and answer the following questions:<\/p>\r\n\r\n<ol id=\"x-ck12-NTEyMDNjMmYxZGUxNDAzZDdmY2RkOGNjN2Q4MzUwZjA.-doh\">\r\n \t<li>What concentration units should be used?<\/li>\r\n \t<li>What quantities should you use for equilibrium problems?<\/li>\r\n \t<li>What must the change in each quantity agree with?<\/li>\r\n \t<li>What is \u201c<em>x<\/em>\u201d?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-ZjIxODk2Njk2ZDNkMTVhY2I2Mjg1MTNmN2UzNWQ1YzY.-4ix\">\r\n \t<li>Does\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0change for a given reaction at a given temperature?<\/li>\r\n \t<li>What does ICE stand for?<\/li>\r\n \t<li>Will the equilibrium position change if materials are added to or removed from the reaction?<\/li>\r\n \t<li>How does addition or removal of materials affect the <em>K<sub>eq<\/sub><\/em>?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2 class=\"x-ck12-data-problem-set\">Glossary<\/h2>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-ZGNiMmMxODliMDBhYzk3MmVjMjE2ZTFhZDlkYWVjMzE.-4ph\">\r\n \t<li><strong> Initial Change Equilibrium (ICE): <\/strong> A calculation that looks at the initial conditions, change, and new equilibrium in a reaction.<\/li>\r\n<\/ul>\r\n&nbsp;\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Describe the relationship between Le Ch\u00e2telier\u2019s principle and the equilibrium constant.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>How much is in savings this month?<\/h3>\n<p id=\"x-ck12-Yzk3YmE4ZmRmNjJkYWFjNzA4NzkxYTVlZjA1YjUwYmQ.-wix\">With online banking, management of your personal finances can become less complicated in some ways. You can automatically deposit paychecks, pay bills, and designate how much goes into savings or other special accounts each month. If you want to maintain 10% of your bank account in savings, you can set up a program that moves money in and out of the account when you get a paycheck or pay bills. The amount of money in savings will change as the money comes in and out of the bank, but the ratio of savings to checking will always be constant.<\/p>\n<\/div>\n<h2>LeCh\u00e2telier\u2019s Principle and the Equilibrium Constant<\/h2>\n<p id=\"x-ck12-MDliYmY0NzMyYzYzYjQ3N2FmMTA2NjAxNzM5Y2RhMWI.-pkk\">Occasionally, when students apply LeChatelier\u2019s principle to an equilibrium problem involving a change in concentration, they assume that <em>K<sub>eq<\/sub><\/em>\u00a0must change. This seems logical since we talk about \u201cshifting\u201d the equilibrium in one direction or the other. However,\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0is a constant, for a given equilibrium at a given temperature, so it must not change. Here is an example of how this works. Consider the simplified equilibrium below:<\/p>\n<p id=\"x-ck12-odx\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213017\/72e74f69fa133df5bab97520defd6e5a.png\" alt=\"A rightleftarrows B\" width=\"56\" height=\"13\" \/><\/p>\n<p id=\"x-ck12-OWEwYjYwNjIyOWM1MDgwYzc0OTlkMDk0NThkM2Y3MTk.-wkb\">Let\u2019s say we have a 1.0 liter container. At equilibrium the following amounts are measured.<\/p>\n<p id=\"x-ck12-m7m\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213017\/090753703c616b1009336feae1bce8d1.png\" alt=\"A &amp;=0.50 text{ mol} \\B &amp;=1.0 text{ mol}\" width=\"105\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-NGQyNjkxYmViZTk0ZTVmZWExYjVkYTZjMWUxZmRmNWU.-ycw\">The value of\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0is given by:<\/p>\n<p id=\"x-ck12-ixn\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213018\/e4a661eb9faf0712ed0d3f12e3116eec.png\" alt=\"K_{eq}= frac{left [B right ]}{left [A right ]}=frac{1.0 text{ M}}{0.50 text{ M}}=2.0\" width=\"209\" height=\"43\" \/><\/p>\n<p id=\"x-ck12-OWIyOWQ4MWM4Yzg2NjYxMTI4ZjYxNDA0MTJmNDMyOGI.-evb\">Now we will disturb the equilibrium by adding 0.50 mole of\u00a0<em>A<\/em>\u00a0to the mixture. The equilibrium will shift towards the right, forming more\u00a0<em>B<\/em>. Immediately after the addition of\u00a0<em>A<\/em>\u00a0and before any response, we now have 1.0 mol of\u00a0<em>A<\/em>\u00a0and 1.0 mol of <em>B<\/em>. The equilibrium then shifts in the forward direction. We will introduce a variable (<em>x<\/em>), which will represent the change in concentrations as the reaction proceeds. Since the mole ratio of <em>A<\/em> :\u00a0<em>B<\/em>\u00a0is 1:1, as [<em>A<\/em>]\u00a0decreases by the amount <em>x<\/em>, the [<em>B<\/em>]\u00a0increases by the amount <em>x<\/em>. We set up an analysis called <em> ICE<\/em>, which stands for Initial, Change, and Equilibrium. The values in the table represent molar concentrations.<\/p>\n<p id=\"x-ck12-al1\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213020\/3f4d8eaf3ccca83b51d2e8e12b9af3c3.png\" alt=\"&amp; qquad qquad qquad underline{;; A qquad qquad B ; ; ; ; ; ; ; ;;;} \\&amp; text{Initial} qquad qquad 1.0 qquad quad 1.0 \\ &amp; text{Change}qquad quad -x qquad quad +x \\ &amp; text{Equilibrium} qquad 1.0 -x qquad 1.0 +x\" width=\"281\" height=\"96\" \/><\/p>\n<p id=\"x-ck12-YTBmNzUwZjA0OGQxZWU2YWU2NGY0NDI3MDU0MTRhZTc.-wna\">At the new equilibrium position, the values for\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> and\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> as a function of\u00a0<em>x\u00a0<\/em>can be set equal to the value of the\u00a0<em>K<sub>eq<\/sub><\/em>. Then, one can solve for <em>x<\/em>.<\/p>\n<p id=\"x-ck12-lbi\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213021\/a57d233fe81ef3ae00eab66ee1318d76.png\" alt=\"K_{eq}=2.0=frac{left [ B right ]}{left [ A right ]}=frac{1.0 + x}{1.0 - x}\" width=\"209\" height=\"43\" \/><\/p>\n<p id=\"x-ck12-Yzc1M2IzOWJhNWMwNGI4N2MwYmRhMWM0NmRiZjI0ZWU.-wac\">Solving for <em>x<\/em>:<\/p>\n<p id=\"x-ck12-26d\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213021\/bfde50174b7fd5998ffaa1d4487decd6.png\" alt=\"2.0(1.0-x) &amp;=1.0 + x \\2.0 - 2.0 x &amp;=1.0+x \\3.0 x &amp;=1.0 \\x &amp;=0.33\" width=\"171\" height=\"94\" \/><\/p>\n<p id=\"x-ck12-ZmQ2MmQzNzRhNDNiNDZmODJlOTE1ODYxMTVhYTVkMzI.-0n3\">This value for\u00a0<em>x<\/em>\u00a0is now plugged back in to the Equilibrium line of the table and the final concentrations of\u00a0<em>A<\/em>\u00a0and\u00a0<em>B<\/em>\u00a0after the reaction is calculated.<\/p>\n<p id=\"x-ck12-c2z\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213022\/717a9394752fb5eebb2a19bf94b86d9e.png\" alt=\"left [ A right ]&amp;=1.0-x=0.67 text{ M} \\left [ B right ]&amp;=1.0+x=1.33 text{ M}\" width=\"180\" height=\"44\" \/><\/p>\n<p id=\"x-ck12-NjlmNmIzOWFjZGEwMDkwNTJhNWZhNzZjMjcxMzYxNTY.-jld\">The value of\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0has been maintained since <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213023\/47120e44b5c4b02a55052bca72484a5a.png\" alt=\"frac{1.33}{0.67}=2.0\" width=\"72\" height=\"23\" \/> . This shows that even though a change in concentration of one of the substances in equilibrium causes a shift in the equilibrium position, the value of the equilibrium constant does not change.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-ZmE2MDczOGE1YTJjM2ZlZmI3M2RhNWU5YjllNDNmOTA.-vqn\">\n<li>Maintenance of the constant\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0for a reaction is described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-ZDA1NTE4MmE3ZWNhMTc1ZjZmZjlhNTc2YTQzZWI5ZTg.-fdo\">Read the material at <a href=\"http:\/\/www.chem.purdue.edu\/gchelp\/howtosolveit\/Equilibrium\/ICEchart.htm\" target=\"_blank\">Chem Purdue<\/a>\u00a0and answer the following questions:<\/p>\n<ol id=\"x-ck12-NTEyMDNjMmYxZGUxNDAzZDdmY2RkOGNjN2Q4MzUwZjA.-doh\">\n<li>What concentration units should be used?<\/li>\n<li>What quantities should you use for equilibrium problems?<\/li>\n<li>What must the change in each quantity agree with?<\/li>\n<li>What is \u201c<em>x<\/em>\u201d?<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-ZjIxODk2Njk2ZDNkMTVhY2I2Mjg1MTNmN2UzNWQ1YzY.-4ix\">\n<li>Does\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0change for a given reaction at a given temperature?<\/li>\n<li>What does ICE stand for?<\/li>\n<li>Will the equilibrium position change if materials are added to or removed from the reaction?<\/li>\n<li>How does addition or removal of materials affect the <em>K<sub>eq<\/sub><\/em>?<\/li>\n<\/ol>\n<\/div>\n<h2 class=\"x-ck12-data-problem-set\">Glossary<\/h2>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-ZGNiMmMxODliMDBhYzk3MmVjMjE2ZTFhZDlkYWVjMzE.-4ph\">\n<li><strong> Initial Change Equilibrium (ICE): <\/strong> A calculation that looks at the initial conditions, change, and new equilibrium in a reaction.<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2747\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry Concepts Intermediate. <strong>Authored by<\/strong>: Calbreath, Baxter, et al.. <strong>Provided by<\/strong>: CK12.org. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\">http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc\/4.0\/\">CC BY-NC: Attribution-NonCommercial<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry Concepts Intermediate\",\"author\":\"Calbreath, Baxter, et 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