{"id":2739,"date":"2016-08-24T17:52:27","date_gmt":"2016-08-24T17:52:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2739"},"modified":"2016-08-24T21:20:55","modified_gmt":"2016-08-24T21:20:55","slug":"equilibrium-constant","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/equilibrium-constant\/","title":{"raw":"Equilibrium Constant","rendered":"Equilibrium Constant"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Define equilibrium constant.<\/li>\r\n \t<li>Write the equation for the general equilibrium constant.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3><strong>What is carbon monoxide poisoning? <\/strong><\/h3>\r\n<p id=\"x-ck12-YWJmOWI5YzI0NmYwYjU3N2ZhMWFkYjFiMGE5YjVjZWU.-qxk\">Red blood cells transport oxygen to the tissues so they can function. In the absence of oxygen, cells cannot carry out their biochemical responsibilities. Oxygen moves to the cells attached to hemoglobin, a protein found in the red cells. In cases of carbon monoxide poisoning, CO binds much more strongly to the hemoglobin, blocking oxygen attachment and lowering the amount of oxygen reaching the cells. Treatment involves the patient breathing pure oxygen to displace the carbon monoxide. The equilibrium reaction shown below illustrates the shift toward the right when excess oxygen is added to the system:<\/p>\r\n<p id=\"x-ck12-74d\"><img class=\"x-ck12-block-math aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212935\/2d4b591497a669859120c166e0ec19bf.png\" alt=\"text{Hb}(text{CO})_4(aq) + 4text{O}_2 (g) leftrightharpoons text{Hb}(text{O}_2)_4(aq) + 4text{CO} (g)\" width=\"384\" height=\"18\" \/><\/p>\r\n<img class=\"size-full wp-image-2996 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/515\/2016\/08\/24211944\/20140811155756616388-e1472073601445.jpeg\" alt=\"Adding excess oxygen can help displace carbon monoxide from red blood cells\" width=\"500\" height=\"402\" \/>\r\n\r\n<\/div>\r\n<h3>Equilibrium Constant<\/h3>\r\n<p id=\"x-ck12-N2Q2MTZjNjAwNjA5NTdmZmM2N2JmM2QwOGEzZjIzOGQ.-gkn\">Consider the hypothetical reversible reaction in which reactants<em> A<\/em>\u00a0and<em> B<\/em>\u00a0react to form products\u00a0<em>C<\/em>\u00a0and <em>D<\/em>. This equilibrium can be shown below, where the lower case letters represent the coefficients of each substance.<\/p>\r\n<p id=\"x-ck12-uro\"><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\/19212937\/1615882ca20053a92feb06d594725961.png\" alt=\"aA + bB rightleftarrows cC+dD\" width=\"163\" height=\"14\" \/><\/p>\r\n<p id=\"x-ck12-YzNkZDQ3YjU2ZTQ3NjdkMjFhOTI4OTE3YmNiYWI2ZTI.-rsv\">As we have established, the rates of the forward and reverse reactions are the same at equilibrium, and so the concentrations of all of the substances are constant. Since that is the case, it stands to reason that a ratio of the concentrations for any given reaction at equilibrium maintains a constant value. The <strong> equilibrium constant (<em>K<sub>eq<\/sub><\/em>)<\/strong>\u00a0is the ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. For the general reaction above, the equilibrium constant expression is written as follows:<\/p>\r\n<p id=\"x-ck12-gqc\"><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\/19212938\/fe01eb8f755a384aa40fa7b40217490d.png\" alt=\"K_{eq}=frac{left [ Cright ]^c left [ Dright ]^d}{left [ Aright ]^a left [B right ]^b}\" width=\"121\" height=\"50\" \/><\/p>\r\n<p id=\"x-ck12-MmU5OTkxNmNmZTU2ODBkMWE5ZDNlZTIzYTQyZTcxNTM.-w5j\">The concentrations of each substance, indicated by the square brackets around the formula, are measured in molarity units (mol\/L).<\/p>\r\n<p id=\"x-ck12-YjU3ZDk3MTEwN2E2NWM2ODYwNDAxZWI1YzY3OWRlYjM.-fj4\">The value of the equilibrium constant for any reaction is only determined by experiment. As detailed in the above section, the position of equilibrium for a given reaction does not depend on the starting concentrations and so the value of the equilibrium constant is truly constant. It does, however, depend on the temperature of the reaction. This is because equilibrium is defined as a condition resulting from the rates of forward and reverse reactions being equal. If the temperature changes, the corresponding change in those reaction rates will alter the equilibrium constant. For any reaction in which a\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0\u00a0is given, the temperature should be specified.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-NmQzM2IzNmE0ZTVjOWUxZDdjYzdmMGU2OTViMmYyOTE.-e9j\">\r\n \t<li>The equilibrium constant for a reversible 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.-icn\">Read the material at the link below and answer the following questions:<\/p>\r\n\r\n<ol id=\"x-ck12-MGY1MjFiZmY5YTQ5NjZhMTk2ZjgwMmYyMzU0NjVmMzY.-k1e\">\r\n \t<li>What is a homogeneous equilibrium?<\/li>\r\n \t<li>What is a heterogeneous equilibrium?<\/li>\r\n \t<li>What goes in the numerator in an equilibrium expression?<\/li>\r\n \t<li>Why is it important to write your equilibrium equation out before setting up the equilibrium constant?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-ZjgwOTI4YzQwY2VmNWYzZDFlMGZiNDkxMmM4NmExYjQ.-8pf\">\r\n \t<li>What does the equilibrium constant tell us?<\/li>\r\n \t<li>What does it mean if the <em>K<sub>eq<\/sub><\/em>\u00a0is &gt; 1?<\/li>\r\n \t<li>What does it mean if the\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0is &lt; 1?<\/li>\r\n \t<li>Does the position of equilibrium depend on the starting concentrations?<\/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-ZGQ1ZTM1ZTZkNGY4ZDczMjkwNTlkYjQzOGIxNDQzMTM.-czp\">\r\n \t<li><strong> equilibrium constant (<em>K<sub>eq<\/sub><\/em>)\u00a0: <\/strong> The ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation.<\/li>\r\n<\/ul>\r\n[reveal-answer q=\"836080\"]Show References[\/reveal-answer]\r\n[hidden-answer a=\"836080\"]\r\n<h2>References<\/h2>\r\n<ol>\r\n \t<li>Courtesy of Bruce Wetzel, Harry Schaefer, National Cancer Institute.<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:SEM_blood_cells.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:SEM_blood_cells.jpg <\/a>.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Define equilibrium constant.<\/li>\n<li>Write the equation for the general equilibrium constant.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3><strong>What is carbon monoxide poisoning? <\/strong><\/h3>\n<p id=\"x-ck12-YWJmOWI5YzI0NmYwYjU3N2ZhMWFkYjFiMGE5YjVjZWU.-qxk\">Red blood cells transport oxygen to the tissues so they can function. In the absence of oxygen, cells cannot carry out their biochemical responsibilities. Oxygen moves to the cells attached to hemoglobin, a protein found in the red cells. In cases of carbon monoxide poisoning, CO binds much more strongly to the hemoglobin, blocking oxygen attachment and lowering the amount of oxygen reaching the cells. Treatment involves the patient breathing pure oxygen to displace the carbon monoxide. The equilibrium reaction shown below illustrates the shift toward the right when excess oxygen is added to the system:<\/p>\n<p id=\"x-ck12-74d\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212935\/2d4b591497a669859120c166e0ec19bf.png\" alt=\"text{Hb}(text{CO})_4(aq) + 4text{O}_2 (g) leftrightharpoons text{Hb}(text{O}_2)_4(aq) + 4text{CO} (g)\" width=\"384\" height=\"18\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-2996 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/515\/2016\/08\/24211944\/20140811155756616388-e1472073601445.jpeg\" alt=\"Adding excess oxygen can help displace carbon monoxide from red blood cells\" width=\"500\" height=\"402\" \/><\/p>\n<\/div>\n<h3>Equilibrium Constant<\/h3>\n<p id=\"x-ck12-N2Q2MTZjNjAwNjA5NTdmZmM2N2JmM2QwOGEzZjIzOGQ.-gkn\">Consider the hypothetical reversible reaction in which reactants<em> A<\/em>\u00a0and<em> B<\/em>\u00a0react to form products\u00a0<em>C<\/em>\u00a0and <em>D<\/em>. This equilibrium can be shown below, where the lower case letters represent the coefficients of each substance.<\/p>\n<p id=\"x-ck12-uro\"><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\/19212937\/1615882ca20053a92feb06d594725961.png\" alt=\"aA + bB rightleftarrows cC+dD\" width=\"163\" height=\"14\" \/><\/p>\n<p id=\"x-ck12-YzNkZDQ3YjU2ZTQ3NjdkMjFhOTI4OTE3YmNiYWI2ZTI.-rsv\">As we have established, the rates of the forward and reverse reactions are the same at equilibrium, and so the concentrations of all of the substances are constant. Since that is the case, it stands to reason that a ratio of the concentrations for any given reaction at equilibrium maintains a constant value. The <strong> equilibrium constant (<em>K<sub>eq<\/sub><\/em>)<\/strong>\u00a0is the ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. For the general reaction above, the equilibrium constant expression is written as follows:<\/p>\n<p id=\"x-ck12-gqc\"><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\/19212938\/fe01eb8f755a384aa40fa7b40217490d.png\" alt=\"K_{eq}=frac{left [ Cright ]^c left [ Dright ]^d}{left [ Aright ]^a left [B right ]^b}\" width=\"121\" height=\"50\" \/><\/p>\n<p id=\"x-ck12-MmU5OTkxNmNmZTU2ODBkMWE5ZDNlZTIzYTQyZTcxNTM.-w5j\">The concentrations of each substance, indicated by the square brackets around the formula, are measured in molarity units (mol\/L).<\/p>\n<p id=\"x-ck12-YjU3ZDk3MTEwN2E2NWM2ODYwNDAxZWI1YzY3OWRlYjM.-fj4\">The value of the equilibrium constant for any reaction is only determined by experiment. As detailed in the above section, the position of equilibrium for a given reaction does not depend on the starting concentrations and so the value of the equilibrium constant is truly constant. It does, however, depend on the temperature of the reaction. This is because equilibrium is defined as a condition resulting from the rates of forward and reverse reactions being equal. If the temperature changes, the corresponding change in those reaction rates will alter the equilibrium constant. For any reaction in which a\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0\u00a0is given, the temperature should be specified.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-NmQzM2IzNmE0ZTVjOWUxZDdjYzdmMGU2OTViMmYyOTE.-e9j\">\n<li>The equilibrium constant for a reversible reaction is described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-ZDA1NTE4MmE3ZWNhMTc1ZjZmZjlhNTc2YTQzZWI5ZTg.-icn\">Read the material at the link below and answer the following questions:<\/p>\n<ol id=\"x-ck12-MGY1MjFiZmY5YTQ5NjZhMTk2ZjgwMmYyMzU0NjVmMzY.-k1e\">\n<li>What is a homogeneous equilibrium?<\/li>\n<li>What is a heterogeneous equilibrium?<\/li>\n<li>What goes in the numerator in an equilibrium expression?<\/li>\n<li>Why is it important to write your equilibrium equation out before setting up the equilibrium constant?<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-ZjgwOTI4YzQwY2VmNWYzZDFlMGZiNDkxMmM4NmExYjQ.-8pf\">\n<li>What does the equilibrium constant tell us?<\/li>\n<li>What does it mean if the <em>K<sub>eq<\/sub><\/em>\u00a0is &gt; 1?<\/li>\n<li>What does it mean if the\u00a0<em>K<sub>eq<\/sub><\/em>\u00a0is &lt; 1?<\/li>\n<li>Does the position of equilibrium depend on the starting concentrations?<\/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-ZGQ1ZTM1ZTZkNGY4ZDczMjkwNTlkYjQzOGIxNDQzMTM.-czp\">\n<li><strong> equilibrium constant (<em>K<sub>eq<\/sub><\/em>)\u00a0: <\/strong> The ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation.<\/li>\n<\/ul>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q836080\">Show References<\/span><\/p>\n<div id=\"q836080\" class=\"hidden-answer\" style=\"display: none\">\n<h2>References<\/h2>\n<ol>\n<li>Courtesy of Bruce Wetzel, Harry Schaefer, National Cancer Institute.<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:SEM_blood_cells.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:SEM_blood_cells.jpg <\/a>.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2739\">\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":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry Concepts Intermediate\",\"author\":\"Calbreath, Baxter, et al.\",\"organization\":\"CK12.org\",\"url\":\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\",\"project\":\"\",\"license\":\"cc-by-nc\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2739","chapter","type-chapter","status-publish","hentry"],"part":2340,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2739","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2739\/revisions"}],"predecessor-version":[{"id":2998,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2739\/revisions\/2998"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2340"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2739\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2739"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2739"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2739"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2739"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}