{"id":2743,"date":"2016-08-24T18:00:15","date_gmt":"2016-08-24T18:00:15","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2743"},"modified":"2016-08-24T21:23:27","modified_gmt":"2016-08-24T21:23:27","slug":"effect-of-concentration","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/effect-of-concentration\/","title":{"raw":"Effect of Concentration","rendered":"Effect of Concentration"},"content":{"raw":"<div class=\"x-ck12-data-objectives\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Describe the effects of concentration changes on the equilibrium of a reaction.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<p id=\"x-ck12-MDFmZTdmMTZmN2FhMzYyZGJjNDgyOGI5N2RlNTE4Zjg.-yw9\"><span class=\"x-ck12-img-inline\"> <img class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212953\/20140811155757094471.jpeg\" alt=\"The pH in a solution can be influenced by the addition or removal of H+ ions\" width=\"200\" \/><\/span><\/p>\r\n<p id=\"x-ck12-NzMxNTZmZGQ3MGQxZTFhN2FlM2RlMWJjZWExOTg4YmI.-ld1\">Phenolphthalein is one of those chemicals that has one structure in a high acid environment and another structure in a low acid environment. If the hydrogen ion concentration is high, the compound is colorless, but turns red if the hydrogen ion concentration is low. By adding hydrogen ions to the solution or removing them through a chemical reaction, we can vary the color of the dye.<\/p>\r\n\r\n<h3>Effect of Concentration<\/h3>\r\n<p id=\"x-ck12-MWEwZmIwYzExMGQ0MWIzNTliMzA2ZDRkNTM3NWM5NWE.-ekv\">A change in concentration of one of the substances in an equilibrium system typically involves either the addition or the removal of one of the reactants or products. Consider the Haber-Bosch process for the industrial production of ammonia from nitrogen and hydrogen gases.<\/p>\r\n<p id=\"x-ck12-gyh\"><img id=\"x-ck12-MTM2Nzg4NjE4NzQ1Mw..\" 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\/19212955\/9371e722019f73b3dcbd26ffd0e09f10.png\" alt=\"text{N}_2(g)+3text{H}_2(g) rightleftarrows 2text{NH}_3(g)\" width=\"212\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-MzEwY2Q2YzI0YmNlMWMzMDJmZDcxY2FjNzdmNjBlOTc.-s10\">If the concentration of one substance in a system is increased, the system will respond by favoring the reaction that removes that substance. When more N <sub> 2 <\/sub> is added, the forward reaction will be favored because the forward reaction uses up N <sub> 2 <\/sub> and converts it to NH <sub> 3 <\/sub> . The forward reaction speeds up temporarily as a result of the addition of a reactant. The position of equilibrium shifts as more NH <sub> 3 <\/sub> is produced. The concentration of NH <sub> 3 <\/sub> increases, while the concentrations of N <sub> 2 <\/sub> and H <sub> 2 <\/sub> decrease. After some time passes, equilibrium is reestablished with new concentrations of all three substances. As can be seen in Figure 1, if more N <sub> 2 <\/sub> is added, a new equilibrium is achieved by the system. The new concentration of NH <sub> 3 <\/sub> is higher because of the favoring of the forward reaction. The new concentration of the H <sub> 2 <\/sub> is lower. The concentration of N <sub> 2 <\/sub> is higher than in the original equilibrium, but went down slightly following the addition of the N <sub> 2 <\/sub> that disturbed the original equilibrium. By responding in this way, the value of the equilibrium constant for the reaction, <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\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> , does not change as a result of the stress to the system.<\/p>\r\n\r\n<div id=\"x-ck12-Zjc2MGU1ODhhYTc5M2NkZjJhMWYxOTgwMmVmNWVhNjY.-7jn\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img id=\"x-ck12-OTgwNDUtMTM2NTE2MTE4Ni04Ny0zNS01LjMuNi4y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212955\/20140811155757216856.png\" alt=\"Adding more nitrogen in the Haber-Bosch process generates additional ammonia\" width=\"500\" height=\"310\" longdesc=\"The%20Haber-Bosch%20process%20is%20an%20equilibrium%20between%20reactant%20N%3Csub%3E2%3C\/sub%3E%20and%20H%3Csub%3E2%3C\/sub%3E%20and%20product%20NH%3Csub%3E3%3C\/sub%3E.\" \/> Figure 1. The Haber-Bosch process is an equilibrium between reactant N<sub>2<\/sub> and H<sub>2<\/sub> and product NH<sub>3<\/sub> .[\/caption]\r\n\r\n<\/div>\r\n<p id=\"x-ck12-MjhmMDc2OTYxMDUzN2VhODEyZjczOTViYzZlMzkyN2U.-8n7\">If more NH <sub> 3 <\/sub> were added, the reverse reaction would be favored. This \u201cfavoring\u201d of a reaction means temporarily speeding up the reaction in that direction until equilibrium is reestablished. Recall that once equilibrium is reestablished, the rates of the forward and reverse reactions are again equal. The addition of NH <sub> 3 <\/sub> would result in increased formation of the reactants, N <sub> 2 <\/sub> and H <sub> 2 <\/sub> .<\/p>\r\n<p id=\"x-ck12-MmExODA3Yzc1NDlhMzg1ZTE4ZmU5Y2Y3OTIwNTY3Njc.-3th\">An equilibrium can also be disrupted by the removal of one of the substances. If the concentration of a substance is decreased, the system will respond by favoring the reaction that replaces that substance. In the industrial Haber-Bosch process, NH <sub> 3 <\/sub> is removed from the equilibrium system as the reaction proceeds. As a result, the forward reaction is favored so that more NH <sub> 3 <\/sub> will be produced. The concentrations of N <sub> 2 <\/sub> and H <sub> 2 <\/sub> decrease. Continued removal of NH <sub> 3 <\/sub> will eventually force the reaction to go to completion until all of the reactants are used up. If either N <sub> 2 <\/sub> or H <sub> 2 <\/sub> were removed from the equilibrium system, the reverse reaction would be favored and the concentration of NH <sub> 3 <\/sub> would decrease.<\/p>\r\n<p id=\"x-ck12-NTVkZTJlOWNhMTFjOGI4OGQwMTg0M2YwNjI1YTJhMzU.-z87\">The effect of changes in concentration on an equilibrium system according to Le Ch\u00e2telier\u2019s Principle is summarized in the <strong> Table <\/strong> below.<\/p>\r\n\r\n<table id=\"x-ck12-MTM2NjcyMDAzOTEyMw..\" 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>addition of reactant<\/td>\r\n<td>forward reaction favored<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>addition of product<\/td>\r\n<td>reverse reaction favored<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>removal of reactant<\/td>\r\n<td>reverse reaction favored<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>removal of product<\/td>\r\n<td>forward reaction favored<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox examples\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-ZmIxNmUzMzk4ZTRmOTk3OTk2Y2E1ZTVhZTBjZTZkM2Q.-bq9\">\r\n \t<li>The effects of concentration changes on an equilibrium are 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-YmMzM2NmMmUyYzQ1NmUxNTQ2YTNmYjA2MWZlY2FkODA.-ydq\">Watch the video below and answer the following questions:<\/p>\r\nhttps:\/\/youtu.be\/hfaC_ksuJ1k\r\n<ol id=\"x-ck12-MmY0MTZkNGJiMDM1YmRmMGIwZTZhNmQ1MGI2Y2E2NzI.-zeg\">\r\n \t<li>What is stress in an equilibrium reactant?<\/li>\r\n \t<li>What happens if more reactants are added?<\/li>\r\n \t<li>What happens if you remove product?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-MGRjMjE2N2EzOGY2YjI0MzM2MDIxYzc5ZjE1NzMzZjE.-qlr\">\r\n \t<li>In the Haber process, what happens if you add more hydrogen gas?<\/li>\r\n \t<li>You miscalculate and add too little nitrogen gas. Which way will the equilibrium shift?<\/li>\r\n \t<li>A mislabeled tank pumps in extra ammonia. What happens to the equilibrium?<\/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>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Phenolphthalein-at-pH-9.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Phenolphthalein-at-pH-9.jpg <\/a>.<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"x-ck12-data-objectives\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Describe the effects of concentration changes on the equilibrium of a reaction.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p id=\"x-ck12-MDFmZTdmMTZmN2FhMzYyZGJjNDgyOGI5N2RlNTE4Zjg.-yw9\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212953\/20140811155757094471.jpeg\" alt=\"The pH in a solution can be influenced by the addition or removal of H+ ions\" width=\"200\" \/><\/span><\/p>\n<p id=\"x-ck12-NzMxNTZmZGQ3MGQxZTFhN2FlM2RlMWJjZWExOTg4YmI.-ld1\">Phenolphthalein is one of those chemicals that has one structure in a high acid environment and another structure in a low acid environment. If the hydrogen ion concentration is high, the compound is colorless, but turns red if the hydrogen ion concentration is low. By adding hydrogen ions to the solution or removing them through a chemical reaction, we can vary the color of the dye.<\/p>\n<h3>Effect of Concentration<\/h3>\n<p id=\"x-ck12-MWEwZmIwYzExMGQ0MWIzNTliMzA2ZDRkNTM3NWM5NWE.-ekv\">A change in concentration of one of the substances in an equilibrium system typically involves either the addition or the removal of one of the reactants or products. Consider the Haber-Bosch process for the industrial production of ammonia from nitrogen and hydrogen gases.<\/p>\n<p id=\"x-ck12-gyh\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Nzg4NjE4NzQ1Mw..\" 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\/19212955\/9371e722019f73b3dcbd26ffd0e09f10.png\" alt=\"text{N}_2(g)+3text{H}_2(g) rightleftarrows 2text{NH}_3(g)\" width=\"212\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-MzEwY2Q2YzI0YmNlMWMzMDJmZDcxY2FjNzdmNjBlOTc.-s10\">If the concentration of one substance in a system is increased, the system will respond by favoring the reaction that removes that substance. When more N <sub> 2 <\/sub> is added, the forward reaction will be favored because the forward reaction uses up N <sub> 2 <\/sub> and converts it to NH <sub> 3 <\/sub> . The forward reaction speeds up temporarily as a result of the addition of a reactant. The position of equilibrium shifts as more NH <sub> 3 <\/sub> is produced. The concentration of NH <sub> 3 <\/sub> increases, while the concentrations of N <sub> 2 <\/sub> and H <sub> 2 <\/sub> decrease. After some time passes, equilibrium is reestablished with new concentrations of all three substances. As can be seen in Figure 1, if more N <sub> 2 <\/sub> is added, a new equilibrium is achieved by the system. The new concentration of NH <sub> 3 <\/sub> is higher because of the favoring of the forward reaction. The new concentration of the H <sub> 2 <\/sub> is lower. The concentration of N <sub> 2 <\/sub> is higher than in the original equilibrium, but went down slightly following the addition of the N <sub> 2 <\/sub> that disturbed the original equilibrium. By responding in this way, the value of the equilibrium constant for the reaction, <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\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> , does not change as a result of the stress to the system.<\/p>\n<div id=\"x-ck12-Zjc2MGU1ODhhYTc5M2NkZjJhMWYxOTgwMmVmNWVhNjY.-7jn\" 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-OTgwNDUtMTM2NTE2MTE4Ni04Ny0zNS01LjMuNi4y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212955\/20140811155757216856.png\" alt=\"Adding more nitrogen in the Haber-Bosch process generates additional ammonia\" width=\"500\" height=\"310\" longdesc=\"The%20Haber-Bosch%20process%20is%20an%20equilibrium%20between%20reactant%20N%3Csub%3E2%3C\/sub%3E%20and%20H%3Csub%3E2%3C\/sub%3E%20and%20product%20NH%3Csub%3E3%3C\/sub%3E.\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The Haber-Bosch process is an equilibrium between reactant N<sub>2<\/sub> and H<sub>2<\/sub> and product NH<sub>3<\/sub> .<\/p>\n<\/div>\n<\/div>\n<p id=\"x-ck12-MjhmMDc2OTYxMDUzN2VhODEyZjczOTViYzZlMzkyN2U.-8n7\">If more NH <sub> 3 <\/sub> were added, the reverse reaction would be favored. This \u201cfavoring\u201d of a reaction means temporarily speeding up the reaction in that direction until equilibrium is reestablished. Recall that once equilibrium is reestablished, the rates of the forward and reverse reactions are again equal. The addition of NH <sub> 3 <\/sub> would result in increased formation of the reactants, N <sub> 2 <\/sub> and H <sub> 2 <\/sub> .<\/p>\n<p id=\"x-ck12-MmExODA3Yzc1NDlhMzg1ZTE4ZmU5Y2Y3OTIwNTY3Njc.-3th\">An equilibrium can also be disrupted by the removal of one of the substances. If the concentration of a substance is decreased, the system will respond by favoring the reaction that replaces that substance. In the industrial Haber-Bosch process, NH <sub> 3 <\/sub> is removed from the equilibrium system as the reaction proceeds. As a result, the forward reaction is favored so that more NH <sub> 3 <\/sub> will be produced. The concentrations of N <sub> 2 <\/sub> and H <sub> 2 <\/sub> decrease. Continued removal of NH <sub> 3 <\/sub> will eventually force the reaction to go to completion until all of the reactants are used up. If either N <sub> 2 <\/sub> or H <sub> 2 <\/sub> were removed from the equilibrium system, the reverse reaction would be favored and the concentration of NH <sub> 3 <\/sub> would decrease.<\/p>\n<p id=\"x-ck12-NTVkZTJlOWNhMTFjOGI4OGQwMTg0M2YwNjI1YTJhMzU.-z87\">The effect of changes in concentration on an equilibrium system according to Le Ch\u00e2telier\u2019s Principle is summarized in the <strong> Table <\/strong> below.<\/p>\n<table id=\"x-ck12-MTM2NjcyMDAzOTEyMw..\" 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>addition of reactant<\/td>\n<td>forward reaction favored<\/td>\n<\/tr>\n<tr>\n<td>addition of product<\/td>\n<td>reverse reaction favored<\/td>\n<\/tr>\n<tr>\n<td>removal of reactant<\/td>\n<td>reverse reaction favored<\/td>\n<\/tr>\n<tr>\n<td>removal of product<\/td>\n<td>forward reaction favored<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox examples\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-ZmIxNmUzMzk4ZTRmOTk3OTk2Y2E1ZTVhZTBjZTZkM2Q.-bq9\">\n<li>The effects of concentration changes on an equilibrium are described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-YmMzM2NmMmUyYzQ1NmUxNTQ2YTNmYjA2MWZlY2FkODA.-ydq\">Watch the video below and answer the following questions:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Chemistry Tutorial 9.05a:  Le Chatelier&#39;s Principle: Changing Concentration\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/hfaC_ksuJ1k?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<ol id=\"x-ck12-MmY0MTZkNGJiMDM1YmRmMGIwZTZhNmQ1MGI2Y2E2NzI.-zeg\">\n<li>What is stress in an equilibrium reactant?<\/li>\n<li>What happens if more reactants are added?<\/li>\n<li>What happens if you remove product?<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-MGRjMjE2N2EzOGY2YjI0MzM2MDIxYzc5ZjE1NzMzZjE.-qlr\">\n<li>In the Haber process, what happens if you add more hydrogen gas?<\/li>\n<li>You miscalculate and add too little nitrogen gas. Which way will the equilibrium shift?<\/li>\n<li>A mislabeled tank pumps in extra ammonia. What happens to the equilibrium?<\/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>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Phenolphthalein-at-pH-9.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Phenolphthalein-at-pH-9.jpg <\/a>.<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung.<\/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-2743\">\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":6,"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-2743","chapter","type-chapter","status-publish","hentry"],"part":2340,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2743","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\/2743\/revisions"}],"predecessor-version":[{"id":3001,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2743\/revisions\/3001"}],"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\/2743\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2743"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2743"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2743"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2743"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}