{"id":2745,"date":"2016-08-24T18:08:37","date_gmt":"2016-08-24T18:08:37","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2745"},"modified":"2016-08-24T21:25:24","modified_gmt":"2016-08-24T21:25:24","slug":"effect-of-pressure","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/effect-of-pressure\/","title":{"raw":"Effect of Pressure","rendered":"Effect of Pressure"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Describe the effect of pressure on an equilibrium reaction.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3><strong>A dual purpose <\/strong><\/h3>\r\n<p id=\"x-ck12-YzBiYTlkNmYyZjk1YmZiYTU2ZDA1ZjdhNjgyNzE4NWI.-jtq\">The ammonia storage tank in the picture above does two things, One it stores ammonia at high pressure to minimize the reverse reaction that would lead to less ammonia and more nitrogen and hydrogen. Secondly, it sends an important message. Ammonia is used to make methamphetamine, a dangerous drug of abuse. Locks and other safety mechanisms built into the tanks help stop the theft of ammonia to be used in this illicit activity.<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213003\/20140811155757808851.jpeg\" alt=\"Ammonia storage tanks are under high pressure to prevent the ammonia from decomposing\" width=\"400\" \/>\r\n\r\n<\/div>\r\n<h2>Effect of Pressure<\/h2>\r\n<p id=\"x-ck12-YTA5NWU5ZGY1MjUwN2Q0M2I3MDRiYTY3YzgxY2YzMzM.-uly\">Changing the pressure of an equilibrium system in which gases are involved is also a stress to the system. A change in the pressure on a liquid or a solid has a negligible effect. We will return again to the equilibrium for the Haber-Bosch process. Imagine the gases are contained in a closed system in which the volume of the system is controlled by an adjustable piston as shown in <strong>Figure 1<\/strong>.<\/p>\r\n\r\n<div id=\"x-ck12-YjMyNzE1ZGQ1OGFkZThiMzI0ZDc2NjIzOTI1ZWUxYTU.-z2k\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img id=\"x-ck12-OTgwNDUtMTM2NTIyNzgwMS03OC0yOC01LjUuOC4y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213004\/20140811155757937264.png\" alt=\"Compressing a mixture of nitrogen and hydrogen creates more ammonia\" width=\"500\" height=\"309\" longdesc=\"Effect%20of%20pressure%20on%20ammonia%20formation.\" \/> Figure 1.\u00a0Effect of pressure on ammonia formation.[\/caption]\r\n\r\n<\/div>\r\n<p id=\"x-ck12-ZjNiMmZkZDZkNDg4YTllMzA0ZDlkYjI3ZDRiNWYwZWM.-neh\">On the far left, the reaction system contains primarily N <sub> 2 <\/sub> and H <sub> 2 <\/sub> , with only one molecule of NH <sub> 3 <\/sub> present. As the piston is pushed inwards, the pressure of the system increases according to Boyle\u2019s Law. This is a stress to the equilibrium. In the middle image, the same number of molecules is now confined in a smaller space and so the pressure has increased. According to LeCh\u00e2telier\u2019s principle, the system responds in order to relieve the stress. In the image on the right, the forward reaction has been favored, in which one molecule of N <sub> 2 <\/sub> combines with three molecules of H <sub> 2 <\/sub> to form two molecules of NH <sub> 3 <\/sub> . The overall result is a decrease in the number of gas molecules in the entire system. This in turn decreases the pressure and provides a relief to the original stress of a pressure increase. An increase in pressure on an equilibrium system favors the reaction which produces fewer total moles of gas. In this case, it is the forward reaction that is favored.<\/p>\r\n<p id=\"x-ck12-NDM5MDYzYzMzYWFhMWFmMjcwZGY1Nzg4OWYwYzkzMjk.-bvi\">A decrease in pressure on the above system could be achieved by pulling the piston outward, increasing the container volume. The equilibrium would respond by favoring the reverse reaction in which NH <sub> 3 <\/sub> decomposes to N <sub> 2 <\/sub> and H <sub> 2 <\/sub> . This is because the overall number of gas molecules would increases and so would the pressure. A decrease in pressure on an equilibrium system favors the reaction which produces more total moles of gas. This is summarized in the <strong> Table <\/strong> below.<\/p>\r\n\r\n<table id=\"x-ck12-MTM2Njc4MzAzMTc4Nw..\" class=\"x-ck12-nofloat\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td><strong> <span class=\"x-ck12-underline\"> Stress <\/span> <\/strong><\/td>\r\n<td><strong> <span class=\"x-ck12-underline\"> Response <\/span> <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>pressure increase<\/td>\r\n<td>reaction produces fewer gas molecules<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>pressure decrease<\/td>\r\n<td>reaction produces more gas molecules<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"x-ck12-YmZhYzQyMTYxNzdkNzQxOWIyNDYzYThmNjM5ODRmMjM.-3qh\">Like changes in concentration, the\u00a0 <img id=\"x-ck12-MTM2Nzk0NjYzMzEyNw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> value for a given reaction is unchanged by a change in pressure.<\/p>\r\n<p id=\"x-ck12-M2Y1ZTExODZlMmVmMjhmN2Q3N2ZlYWFiZmY1NzlkMTE.-unk\">It is important to remember when analyzing the effect of a pressure change on equilibrium that only gases are affected. If a certain reaction involves liquids or solids, they should be ignored. Calcium carbonate decomposes according to the equilibrium reaction:<\/p>\r\n<p id=\"x-ck12-f6t\"><img id=\"x-ck12-MTM2Nzk0NjYzMzEyOA..\" 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\/19213006\/f5cc0b2847ccec167b773674d7760165.png\" alt=\"text{CaCO}_3(s) rightleftarrows text{CaO}(s)+text{O}_2(g)\" width=\"230\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-MmViNzJmYTYwMGUxOGU2OTU4YjUwYTM2NGYwZDIwMGQ.-lor\">Oxygen is the only gas in the system. An increase in the pressure of the system slows the rate of decomposition of CaCO <sub> 3 <\/sub> because the reverse reaction is favored. When a system contains equal moles of gas on both sides of the equation, pressure has no effect on the equilibrium position, as in the formation of HCl from H <sub> 2 <\/sub> and Cl <sub> 2 <\/sub> .<\/p>\r\n<p id=\"x-ck12-cdu\"><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\/19213008\/fb54dff81091c7942141ec0671448512.png\" alt=\"text{H}_2(g)+text{Cl}_2(g) rightleftarrows 2text{HCl}(g)\" width=\"205\" height=\"18\" \/><\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-NGQyMjk4NmNjMzI4NDU3NDAwNDlkNzgyN2ZjOTEzM2M.-sry\">\r\n \t<li>The influence of pressure on an equilibrium system 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-YmE3YjU5Yzg1ZmZhOTYzZmMxOTc4Y2I1NTMwYzc4ZGE.-vhp\">Read the material on pressure at <a href=\"http:\/\/www.chemguide.co.uk\/physical\/equilibria\/change.html\" target=\"_blank\">ChemGuide.co.uk<\/a>\u00a0and answer the following questions:<\/p>\r\n\r\n<ol id=\"x-ck12-YmVkYTk0MjEzYzJiNTE1ZGIwMTk4NDU0MWM4NGQ5NGM.-3vq\">\r\n \t<li>How is pressure directly related to an equilibrium constant?<\/li>\r\n \t<li>What is done to related pressures to moles?<\/li>\r\n \t<li>If you decrease P, what has to happen to the mole fractions of C and D?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-ZDcwZGYxNjY4NmQyYWI1OGYwZjFjNTBlNDViZmQ4MTA.-qc0\">\r\n \t<li>Why does the pressure increase in the middle picture?<\/li>\r\n \t<li>How is the stress due to increased pressure relieved?<\/li>\r\n \t<li>Why does an increase in pressure slow the rate of CaCO<sub>3 <\/sub> decomposition?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"x-ck12-data-problem-set\">\r\n\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>Bill Whittaker (User:Billwhittaker\/Wikipedia).<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Meth_ammonia_tank_Otley_iowa.JPG\">http:\/\/commons.wikimedia.org\/wiki\/File:Meth_ammonia_tank_Otley_iowa.JPG <\/a>.<\/li>\r\n \t<li>CK-12 Foundation - Zachary Wilson.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"x-ck12-data-vocabulary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Describe the effect of pressure on an equilibrium reaction.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3><strong>A dual purpose <\/strong><\/h3>\n<p id=\"x-ck12-YzBiYTlkNmYyZjk1YmZiYTU2ZDA1ZjdhNjgyNzE4NWI.-jtq\">The ammonia storage tank in the picture above does two things, One it stores ammonia at high pressure to minimize the reverse reaction that would lead to less ammonia and more nitrogen and hydrogen. Secondly, it sends an important message. Ammonia is used to make methamphetamine, a dangerous drug of abuse. Locks and other safety mechanisms built into the tanks help stop the theft of ammonia to be used in this illicit activity.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213003\/20140811155757808851.jpeg\" alt=\"Ammonia storage tanks are under high pressure to prevent the ammonia from decomposing\" width=\"400\" \/><\/p>\n<\/div>\n<h2>Effect of Pressure<\/h2>\n<p id=\"x-ck12-YTA5NWU5ZGY1MjUwN2Q0M2I3MDRiYTY3YzgxY2YzMzM.-uly\">Changing the pressure of an equilibrium system in which gases are involved is also a stress to the system. A change in the pressure on a liquid or a solid has a negligible effect. We will return again to the equilibrium for the Haber-Bosch process. Imagine the gases are contained in a closed system in which the volume of the system is controlled by an adjustable piston as shown in <strong>Figure 1<\/strong>.<\/p>\n<div id=\"x-ck12-YjMyNzE1ZGQ1OGFkZThiMzI0ZDc2NjIzOTI1ZWUxYTU.-z2k\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2NTIyNzgwMS03OC0yOC01LjUuOC4y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213004\/20140811155757937264.png\" alt=\"Compressing a mixture of nitrogen and hydrogen creates more ammonia\" width=\"500\" height=\"309\" longdesc=\"Effect%20of%20pressure%20on%20ammonia%20formation.\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1.\u00a0Effect of pressure on ammonia formation.<\/p>\n<\/div>\n<\/div>\n<p id=\"x-ck12-ZjNiMmZkZDZkNDg4YTllMzA0ZDlkYjI3ZDRiNWYwZWM.-neh\">On the far left, the reaction system contains primarily N <sub> 2 <\/sub> and H <sub> 2 <\/sub> , with only one molecule of NH <sub> 3 <\/sub> present. As the piston is pushed inwards, the pressure of the system increases according to Boyle\u2019s Law. This is a stress to the equilibrium. In the middle image, the same number of molecules is now confined in a smaller space and so the pressure has increased. According to LeCh\u00e2telier\u2019s principle, the system responds in order to relieve the stress. In the image on the right, the forward reaction has been favored, in which one molecule of N <sub> 2 <\/sub> combines with three molecules of H <sub> 2 <\/sub> to form two molecules of NH <sub> 3 <\/sub> . The overall result is a decrease in the number of gas molecules in the entire system. This in turn decreases the pressure and provides a relief to the original stress of a pressure increase. An increase in pressure on an equilibrium system favors the reaction which produces fewer total moles of gas. In this case, it is the forward reaction that is favored.<\/p>\n<p id=\"x-ck12-NDM5MDYzYzMzYWFhMWFmMjcwZGY1Nzg4OWYwYzkzMjk.-bvi\">A decrease in pressure on the above system could be achieved by pulling the piston outward, increasing the container volume. The equilibrium would respond by favoring the reverse reaction in which NH <sub> 3 <\/sub> decomposes to N <sub> 2 <\/sub> and H <sub> 2 <\/sub> . This is because the overall number of gas molecules would increases and so would the pressure. A decrease in pressure on an equilibrium system favors the reaction which produces more total moles of gas. This is summarized in the <strong> Table <\/strong> below.<\/p>\n<table id=\"x-ck12-MTM2Njc4MzAzMTc4Nw..\" class=\"x-ck12-nofloat\">\n<tbody>\n<tr>\n<td><strong> <span class=\"x-ck12-underline\"> Stress <\/span> <\/strong><\/td>\n<td><strong> <span class=\"x-ck12-underline\"> Response <\/span> <\/strong><\/td>\n<\/tr>\n<tr>\n<td>pressure increase<\/td>\n<td>reaction produces fewer gas molecules<\/td>\n<\/tr>\n<tr>\n<td>pressure decrease<\/td>\n<td>reaction produces more gas molecules<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"x-ck12-YmZhYzQyMTYxNzdkNzQxOWIyNDYzYThmNjM5ODRmMjM.-3qh\">Like changes in concentration, the\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Nzk0NjYzMzEyNw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> value for a given reaction is unchanged by a change in pressure.<\/p>\n<p id=\"x-ck12-M2Y1ZTExODZlMmVmMjhmN2Q3N2ZlYWFiZmY1NzlkMTE.-unk\">It is important to remember when analyzing the effect of a pressure change on equilibrium that only gases are affected. If a certain reaction involves liquids or solids, they should be ignored. Calcium carbonate decomposes according to the equilibrium reaction:<\/p>\n<p id=\"x-ck12-f6t\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Nzk0NjYzMzEyOA..\" 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\/19213006\/f5cc0b2847ccec167b773674d7760165.png\" alt=\"text{CaCO}_3(s) rightleftarrows text{CaO}(s)+text{O}_2(g)\" width=\"230\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-MmViNzJmYTYwMGUxOGU2OTU4YjUwYTM2NGYwZDIwMGQ.-lor\">Oxygen is the only gas in the system. An increase in the pressure of the system slows the rate of decomposition of CaCO <sub> 3 <\/sub> because the reverse reaction is favored. When a system contains equal moles of gas on both sides of the equation, pressure has no effect on the equilibrium position, as in the formation of HCl from H <sub> 2 <\/sub> and Cl <sub> 2 <\/sub> .<\/p>\n<p id=\"x-ck12-cdu\"><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\/19213008\/fb54dff81091c7942141ec0671448512.png\" alt=\"text{H}_2(g)+text{Cl}_2(g) rightleftarrows 2text{HCl}(g)\" width=\"205\" height=\"18\" \/><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-NGQyMjk4NmNjMzI4NDU3NDAwNDlkNzgyN2ZjOTEzM2M.-sry\">\n<li>The influence of pressure on an equilibrium system is described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-YmE3YjU5Yzg1ZmZhOTYzZmMxOTc4Y2I1NTMwYzc4ZGE.-vhp\">Read the material on pressure at <a href=\"http:\/\/www.chemguide.co.uk\/physical\/equilibria\/change.html\" target=\"_blank\">ChemGuide.co.uk<\/a>\u00a0and answer the following questions:<\/p>\n<ol id=\"x-ck12-YmVkYTk0MjEzYzJiNTE1ZGIwMTk4NDU0MWM4NGQ5NGM.-3vq\">\n<li>How is pressure directly related to an equilibrium constant?<\/li>\n<li>What is done to related pressures to moles?<\/li>\n<li>If you decrease P, what has to happen to the mole fractions of C and D?<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-ZDcwZGYxNjY4NmQyYWI1OGYwZjFjNTBlNDViZmQ4MTA.-qc0\">\n<li>Why does the pressure increase in the middle picture?<\/li>\n<li>How is the stress due to increased pressure relieved?<\/li>\n<li>Why does an increase in pressure slow the rate of CaCO<sub>3 <\/sub> decomposition?<\/li>\n<\/ol>\n<\/div>\n<div class=\"x-ck12-data-problem-set\">\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>Bill Whittaker (User:Billwhittaker\/Wikipedia).<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Meth_ammonia_tank_Otley_iowa.JPG\">http:\/\/commons.wikimedia.org\/wiki\/File:Meth_ammonia_tank_Otley_iowa.JPG <\/a>.<\/li>\n<li>CK-12 Foundation &#8211; Zachary Wilson.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"x-ck12-data-vocabulary\"><\/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-2745\">\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":8,"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-2745","chapter","type-chapter","status-publish","hentry"],"part":2340,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2745","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":2,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2745\/revisions"}],"predecessor-version":[{"id":3003,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2745\/revisions\/3003"}],"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\/2745\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2745"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2745"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2745"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2745"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}