{"id":7662,"date":"2021-06-08T21:57:12","date_gmt":"2021-06-08T21:57:12","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/shifting-equilibria-le-chateliers-principle\/"},"modified":"2021-10-07T20:34:36","modified_gmt":"2021-10-07T20:34:36","slug":"shifting-equilibria-le-chateliers-principle","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/shifting-equilibria-le-chateliers-principle\/","title":{"raw":"Shifting Equilibria: Le Chatelier\u2019s Principle","rendered":"Shifting Equilibria: Le Chatelier\u2019s Principle"},"content":{"raw":"[latexpage]\r\n<div class=\"textbox textbox--learning-objectives\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Learning Objectives<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>Define <em>Le Chatelier\u2019s principle<\/em>.<\/li>\r\n \t<li>Predict the direction of shift for an equilibrium under stress.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nOnce equilibrium is established, the reaction is over, right? Not exactly. An experimenter has some ability to affect the equilibrium.\r\n\r\nChemical equilibria can be shifted by changing the conditions that the system experiences. We say that we \u201cstress\u201d the equilibrium. When we stress the equilibrium, the chemical reaction is no longer at equilibrium, and the reaction starts to move back toward equilibrium in such a way as to decrease the stress. The formal statement is called <strong>Le Chatelier\u2019s principle<\/strong>: If an equilibrium is stressed, then the reaction shifts to reduce the stress.\r\n<p class=\"page-break-after\">There are several ways to stress an equilibrium. One way is to add or remove a product or a reactant in a chemical reaction at equilibrium. When additional reactant is added, the equilibrium shifts to reduce this stress: it makes more product. When additional product is added, the equilibrium shifts to reactants to reduce the stress. If reactant or product is removed, the equilibrium shifts to make more reactant or product, respectively, to make up for the loss.<\/p>\r\n\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 13.6<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nGiven this reaction at equilibrium:\r\n<p style=\"text-align: center;\">N<sub>2<\/sub> + 3H<sub>2<\/sub> \u21c4 2NH<sub>3<\/sub><\/p>\r\nIn which direction \u2014 toward reactants or toward products \u2014 does the reaction shift if the equilibrium is stressed by each change?\r\n<ol>\r\n \t<li>H<sub>2<\/sub> is added.<\/li>\r\n \t<li>NH<sub>3<\/sub> is added.<\/li>\r\n \t<li>NH<sub>3<\/sub> is removed.<\/li>\r\n<\/ol>\r\n<em>Solution<\/em>\r\n<ol>\r\n \t<li>If H<sub>2<\/sub> is added, there is now more reactant, so the reaction will shift toward products to reduce the added H<sub>2<\/sub>.<\/li>\r\n \t<li>If NH<sub>3<\/sub> is added, there is now more product, so the reaction will shift toward reactants to reduce the added NH<sub>3<\/sub>.<\/li>\r\n \t<li>If NH<sub>3<\/sub> is removed, there is now less product, so the reaction will shift toward products to replace the product removed.<\/li>\r\n<\/ol>\r\n<em>Test Yourself<\/em>\r\nGiven this reaction at equilibrium:\r\n<p style=\"text-align: center;\">CO(g) + Br<sub>2<\/sub>(g) \u21c4 COBr<sub>2<\/sub>(g)<\/p>\r\nIn which direction \u2014 toward reactants or toward products \u2014 does the reaction shift if the equilibrium is stressed by each change?\r\n<ol>\r\n \t<li>Br<sub>2<\/sub> is removed.<\/li>\r\n \t<li>COBr<sub>2<\/sub> is added.<\/li>\r\n<\/ol>\r\n<em>Answers<\/em>\r\n<ol>\r\n \t<li>toward reactants<\/li>\r\n \t<li>toward reactants<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nIt is worth noting that when reactants or products are added or removed, <em>the value of the K<sub>eq<\/sub> does not change<\/em>. The chemical reaction simply shifts, in a predictable fashion, to reestablish concentrations so that the <em>K<\/em><sub>eq<\/sub> expression reverts to the correct value.\r\n\r\nHow does an equilibrium react to a change in pressure? Pressure changes do not markedly affect the solid or liquid phases. However, pressure strongly impacts the gas phase. Le Chatelier\u2019s principle implies that a pressure increase shifts an equilibrium to the side of the reaction with the fewer number of moles of gas, while a pressure decrease shifts an equilibrium to the side of the reaction with the greater number of moles of gas. If the number of moles of gas is the same on both sides of the reaction, pressure has no effect.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 13.7<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nWhat is the effect on this equilibrium if pressure is increased?\r\n<p style=\"text-align: center;\">N<sub>2<\/sub>(g) + 3H<sub>2<\/sub>(g) \u21c4 2NH<sub>3<\/sub>(g)<\/p>\r\n<em>Solution<\/em>\r\nAccording to Le Chatelier\u2019s principle, if pressure is increased, then the equilibrium shifts to the side with the fewer number of moles of gas. This particular reaction shows a total of 4 mol of gas as reactants and 2 mol of gas as products, so the reaction shifts toward the products side.\r\n\r\n<em>Test Yourself<\/em>\r\nWhat is the effect on this equilibrium if pressure is decreased?\r\n<p style=\"text-align: center;\">3O<sub>2<\/sub>(g) \u21c4 2O<sub>3<\/sub>(g)<\/p>\r\n<em>Answer<\/em>\r\nReaction shifts toward reactants.\r\n\r\n<\/div>\r\n<\/div>\r\nWhat is the effect of temperature changes on an equilibrium? It depends on whether the reaction is endothermic or exothermic. Recall that <em>endothermic<\/em> means that energy is absorbed by a chemical reaction, while <em>exothermic<\/em> means that energy is given off by the reaction. As such, energy can be thought of as a reactant or a product, respectively, of a reaction:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rrcl}\r\n\\text{endothermic: }&amp;\\text{energy}+\\text{reactants}&amp;\\rightarrow&amp;\\text{products} \\\\\r\n\\text{exothermic: }&amp;\\text{reactants}&amp;\\rightarrow&amp;\\text{products}+\\text{energy}\r\n\\end{array}[\/latex]<\/p>\r\n<p class=\"page-break-after\">Because temperature is a measure of the energy of the system, increasing temperature can be thought of as adding energy. The reaction will react as if a reactant or a product is being added and will act accordingly by shifting to the other side. For example, if the temperature is increased for an endothermic reaction, essentially a reactant is being added, so the equilibrium shifts toward products. Decreasing the temperature is equivalent to decreasing a reactant (for endothermic reactions) or a product (for exothermic reactions), and the equilibrium shifts accordingly.<\/p>\r\n\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 13.8<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nPredict the effect of increasing the temperature on this equilibrium.\r\n<p style=\"text-align: center;\">PCl<sub>3<\/sub> + Cl<sub>2<\/sub> \u21c4 PCl<sub>5<\/sub> + 60 kJ<\/p>\r\n<em>Solution<\/em>\r\nBecause energy is listed as a product, it is being produced, so the reaction is exothermic. If the temperature is increasing, a product is being added to the equilibrium, so the equilibrium shifts to minimize the addition of extra product: it shifts back toward reactants.\r\n\r\n<em>Test Yourself<\/em>\r\nPredict the effect of decreasing the temperature on this equilibrium.\r\n<p style=\"text-align: center;\">N<sub>2<\/sub>O<sub>4<\/sub> + 57 kJ \u21c4 2NO<sub>2<\/sub><\/p>\r\n<em>Answer<\/em>\r\nEquilibrium shifts toward reactants.\r\n\r\n<\/div>\r\n<\/div>\r\nIn the case of temperature, the value of the equilibrium has changed because the <em>K<\/em><sub>eq<\/sub> is dependent on temperature. That is why equilibria shift with changes in temperature.\r\n\r\nA <strong>catalyst<\/strong>\u00a0is a substance that increases the speed of a reaction. Overall, a catalyst is not a reactant and is not used up, but it still affects how fast a reaction proceeds. However, a catalyst does not affect the extent or position of a reaction at equilibrium. It helps a reaction achieve equilibrium faster.\r\n<div class=\"textbox shaded\">\r\n<h1>Chemistry Is Everywhere: Equilibria in the Garden<\/h1>\r\nHydrangeas are common flowering plants around the world. Although many hydrangeas are white, there is one common species (<em>Hydrangea macrophylla<\/em>) whose flowers can be either red or blue, as shown in the accompanying figure. How is it that a plant can have different coloured flowers like this?\r\n\r\n[caption id=\"attachment_3258\" align=\"aligncenter\" width=\"374\"]<img class=\"wp-image-728 \" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/5996174498_eb660e46c2_b-e1412019701873-1.jpg\" alt=\"Blue hydrangeas.\" width=\"374\" height=\"299\" \/> Figure 13.1 \"Garden Equilibria.\" This species of hydrangea has flowers that can be either red or blue. Why the colour difference?[\/caption]\r\n\r\nInterestingly, the colour of the flowers is due to the acidity of the soil that the hydrangea is planted in. An astute gardener can adjust the pH of the soil and actually change the colour of the flowers. However, it is not the H<sup>+<\/sup> or OH<sup>\u2212<\/sup> ions that affect the colour of the flowers. Rather, it is the presence of aluminum that causes the colour change.\r\n\r\nThe solubility of aluminum in soil \u2014 and thus the ability of plants to absorb it \u2014 is dependent on the acidity of the soil. If the soil is relatively acidic, the aluminum is more soluble, and plants can absorb it more easily. Under these conditions, hydrangea flowers are blue as Al ions interact with anthocyanin pigments in the plant. In more basic soils, aluminum is less soluble, and under these conditions the hydrangea flowers are red. Gardeners who change the pH of their soils to change the colour of their hydrangea flowers are therefore employing Le Chatelier\u2019s principle: the amount of acid in the soil changes the equilibrium of aluminum solubility, which in turn affects the colour of the flowers.\r\n\r\n<\/div>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Takeaways<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ul>\r\n \t<li>Le Chatelier\u2019s principle addresses how an equilibrium shifts when the conditions of an equilibrium are changed.<\/li>\r\n \t<li>The direction of shift can be predicted for changes in concentrations, temperature, or pressure.<\/li>\r\n \t<li>Catalysts do not affect the position of an equilibrium; they help reactions achieve equilibrium faster.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercises<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<h1>Questions<\/h1>\r\n<ol>\r\n \t<li>Define <em>Le Chatelier\u2019s principle<\/em>.<\/li>\r\n \t<li>What is meant by a stress? What are some of the ways an equilibrium can be stressed?<\/li>\r\n \t<li>Given this equilibrium, predict the direction of shift for each stress.\r\n<p style=\"text-align: center;\">H<sub>2<\/sub>(g) + I<sub>2<\/sub>(s) + 53 kJ \u21c4 2HI(g)<\/p>\r\n\r\n<ol type=\"a\">\r\n \t<li>decreased temperature<\/li>\r\n \t<li>increased pressure<\/li>\r\n \t<li>removal of HI<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Given this equilibrium, predict the direction of shift for each stress.\r\n<p style=\"text-align: center;\">H<sub>2<\/sub>(g)\u00a0+\u00a0F<sub>2<\/sub>(g) \u21c4 2HF(g) + 546 kJ<\/p>\r\n\r\n<ol type=\"a\">\r\n \t<li>increased temperature<\/li>\r\n \t<li>addition of H<sub>2<\/sub><\/li>\r\n \t<li>decreased pressure<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Given this equilibrium, predict the direction of shift for each stress.\r\n<p style=\"text-align: center;\">2SO<sub>2<\/sub>(g) + O<sub>2<\/sub>(g) \u21c4 2SO<sub>3<\/sub>(g) + 196 kJ<\/p>\r\n\r\n<ol type=\"a\">\r\n \t<li>removal of SO<sub>3<\/sub><\/li>\r\n \t<li>addition of O<sub>2<\/sub><\/li>\r\n \t<li>decreased temperature<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Given this equilibrium, predict the direction of shift for each stress listed.\r\n<p style=\"text-align: center;\">CO<sub>2<\/sub>(g) + C(s) + 171 kJ \u21c4 2CO(g)<\/p>\r\n\r\n<ol type=\"a\">\r\n \t<li>addition of CO<\/li>\r\n \t<li>increased pressure<\/li>\r\n \t<li>addition of a catalyst<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The synthesis of NH<sub>3<\/sub> uses this chemical reaction.\r\n<p style=\"text-align: center;\">N<sub>2<\/sub>(g) + 3H<sub>2<\/sub>(g) \u21c4 2NH<sub>3<\/sub>(g) + 92 kJ<\/p>\r\nIdentify three stresses that can be imposed on the equilibrium to maximize the amount of NH<sub>3<\/sub>.<\/li>\r\n \t<li>The synthesis of CaCO<sub>3<\/sub> uses this chemical reaction.\r\n<p style=\"text-align: center;\">CaO(s)\u00a0+\u00a0CO<sub>2<\/sub>(g) \u21c4 CaCO<sub>3<\/sub>(s)\u00a0+\u00a0180\u00a0kJ<\/p>\r\nIdentify three stresses that can be imposed on the equilibrium to maximize the amount of CaCO<sub>3<\/sub>.<\/li>\r\n<\/ol>\r\n<h1>Answers<\/h1>\r\n<ol>\r\n \t<li>When an equilibrium is stressed, the equilibrium shifts to minimize that stress.<\/li>\r\n<\/ol>\r\n<ol start=\"3\">\r\n \t<li>\r\n<ol type=\"a\">\r\n \t<li>toward reactants<\/li>\r\n \t<li>toward reactants<\/li>\r\n \t<li>toward products<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol start=\"5\">\r\n \t<li>\r\n<ol type=\"a\">\r\n \t<li>toward products<\/li>\r\n \t<li>toward products<\/li>\r\n \t<li>toward products<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<ol start=\"7\">\r\n \t<li>increased pressure, decreased temperature, removal of NH<sub>3<\/sub><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h3>Media Attributions<\/h3>\r\n<ul>\r\n \t<li><a href=\"https:\/\/www.flickr.com\/photos\/jannem\/5996174498\/\">\u201cHydrangea\u201d<\/a> \u00a9 <a href=\"https:\/\/www.flickr.com\/photos\/jannem\/\">2011 by Janne Moren<\/a> is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/2.0\/\">CC BY-NC-SA (Attribution-NonCommercial-ShareAlike)<\/a> license<\/li>\r\n<\/ul>","rendered":"<div class=\"textbox textbox--learning-objectives\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Learning Objectives<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>Define <em>Le Chatelier\u2019s principle<\/em>.<\/li>\n<li>Predict the direction of shift for an equilibrium under stress.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p>Once equilibrium is established, the reaction is over, right? Not exactly. An experimenter has some ability to affect the equilibrium.<\/p>\n<p>Chemical equilibria can be shifted by changing the conditions that the system experiences. We say that we \u201cstress\u201d the equilibrium. When we stress the equilibrium, the chemical reaction is no longer at equilibrium, and the reaction starts to move back toward equilibrium in such a way as to decrease the stress. The formal statement is called <strong>Le Chatelier\u2019s principle<\/strong>: If an equilibrium is stressed, then the reaction shifts to reduce the stress.<\/p>\n<p class=\"page-break-after\">There are several ways to stress an equilibrium. One way is to add or remove a product or a reactant in a chemical reaction at equilibrium. When additional reactant is added, the equilibrium shifts to reduce this stress: it makes more product. When additional product is added, the equilibrium shifts to reactants to reduce the stress. If reactant or product is removed, the equilibrium shifts to make more reactant or product, respectively, to make up for the loss.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 13.6<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Given this reaction at equilibrium:<\/p>\n<p style=\"text-align: center;\">N<sub>2<\/sub> + 3H<sub>2<\/sub> \u21c4 2NH<sub>3<\/sub><\/p>\n<p>In which direction \u2014 toward reactants or toward products \u2014 does the reaction shift if the equilibrium is stressed by each change?<\/p>\n<ol>\n<li>H<sub>2<\/sub> is added.<\/li>\n<li>NH<sub>3<\/sub> is added.<\/li>\n<li>NH<sub>3<\/sub> is removed.<\/li>\n<\/ol>\n<p><em>Solution<\/em><\/p>\n<ol>\n<li>If H<sub>2<\/sub> is added, there is now more reactant, so the reaction will shift toward products to reduce the added H<sub>2<\/sub>.<\/li>\n<li>If NH<sub>3<\/sub> is added, there is now more product, so the reaction will shift toward reactants to reduce the added NH<sub>3<\/sub>.<\/li>\n<li>If NH<sub>3<\/sub> is removed, there is now less product, so the reaction will shift toward products to replace the product removed.<\/li>\n<\/ol>\n<p><em>Test Yourself<\/em><br \/>\nGiven this reaction at equilibrium:<\/p>\n<p style=\"text-align: center;\">CO(g) + Br<sub>2<\/sub>(g) \u21c4 COBr<sub>2<\/sub>(g)<\/p>\n<p>In which direction \u2014 toward reactants or toward products \u2014 does the reaction shift if the equilibrium is stressed by each change?<\/p>\n<ol>\n<li>Br<sub>2<\/sub> is removed.<\/li>\n<li>COBr<sub>2<\/sub> is added.<\/li>\n<\/ol>\n<p><em>Answers<\/em><\/p>\n<ol>\n<li>toward reactants<\/li>\n<li>toward reactants<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p>It is worth noting that when reactants or products are added or removed, <em>the value of the K<sub>eq<\/sub> does not change<\/em>. The chemical reaction simply shifts, in a predictable fashion, to reestablish concentrations so that the <em>K<\/em><sub>eq<\/sub> expression reverts to the correct value.<\/p>\n<p>How does an equilibrium react to a change in pressure? Pressure changes do not markedly affect the solid or liquid phases. However, pressure strongly impacts the gas phase. Le Chatelier\u2019s principle implies that a pressure increase shifts an equilibrium to the side of the reaction with the fewer number of moles of gas, while a pressure decrease shifts an equilibrium to the side of the reaction with the greater number of moles of gas. If the number of moles of gas is the same on both sides of the reaction, pressure has no effect.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 13.7<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>What is the effect on this equilibrium if pressure is increased?<\/p>\n<p style=\"text-align: center;\">N<sub>2<\/sub>(g) + 3H<sub>2<\/sub>(g) \u21c4 2NH<sub>3<\/sub>(g)<\/p>\n<p><em>Solution<\/em><br \/>\nAccording to Le Chatelier\u2019s principle, if pressure is increased, then the equilibrium shifts to the side with the fewer number of moles of gas. This particular reaction shows a total of 4 mol of gas as reactants and 2 mol of gas as products, so the reaction shifts toward the products side.<\/p>\n<p><em>Test Yourself<\/em><br \/>\nWhat is the effect on this equilibrium if pressure is decreased?<\/p>\n<p style=\"text-align: center;\">3O<sub>2<\/sub>(g) \u21c4 2O<sub>3<\/sub>(g)<\/p>\n<p><em>Answer<\/em><br \/>\nReaction shifts toward reactants.<\/p>\n<\/div>\n<\/div>\n<p>What is the effect of temperature changes on an equilibrium? It depends on whether the reaction is endothermic or exothermic. Recall that <em>endothermic<\/em> means that energy is absorbed by a chemical reaction, while <em>exothermic<\/em> means that energy is given off by the reaction. As such, energy can be thought of as a reactant or a product, respectively, of a reaction:<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/ql-cache\/quicklatex.com-8f5c50c87044d501bb1c8df2b73d8f60_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#114;&#114;&#99;&#108;&#125; &#92;&#116;&#101;&#120;&#116;&#123;&#101;&#110;&#100;&#111;&#116;&#104;&#101;&#114;&#109;&#105;&#99;&#58;&#32;&#125;&#38;&#92;&#116;&#101;&#120;&#116;&#123;&#101;&#110;&#101;&#114;&#103;&#121;&#125;&#43;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#101;&#97;&#99;&#116;&#97;&#110;&#116;&#115;&#125;&#38;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#38;&#92;&#116;&#101;&#120;&#116;&#123;&#112;&#114;&#111;&#100;&#117;&#99;&#116;&#115;&#125;&#32;&#92;&#92; &#92;&#116;&#101;&#120;&#116;&#123;&#101;&#120;&#111;&#116;&#104;&#101;&#114;&#109;&#105;&#99;&#58;&#32;&#125;&#38;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#101;&#97;&#99;&#116;&#97;&#110;&#116;&#115;&#125;&#38;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#38;&#92;&#116;&#101;&#120;&#116;&#123;&#112;&#114;&#111;&#100;&#117;&#99;&#116;&#115;&#125;&#43;&#92;&#116;&#101;&#120;&#116;&#123;&#101;&#110;&#101;&#114;&#103;&#121;&#125; &#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"458\" style=\"vertical-align: -15px;\" \/><\/p>\n<p class=\"page-break-after\">Because temperature is a measure of the energy of the system, increasing temperature can be thought of as adding energy. The reaction will react as if a reactant or a product is being added and will act accordingly by shifting to the other side. For example, if the temperature is increased for an endothermic reaction, essentially a reactant is being added, so the equilibrium shifts toward products. Decreasing the temperature is equivalent to decreasing a reactant (for endothermic reactions) or a product (for exothermic reactions), and the equilibrium shifts accordingly.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 13.8<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Predict the effect of increasing the temperature on this equilibrium.<\/p>\n<p style=\"text-align: center;\">PCl<sub>3<\/sub> + Cl<sub>2<\/sub> \u21c4 PCl<sub>5<\/sub> + 60 kJ<\/p>\n<p><em>Solution<\/em><br \/>\nBecause energy is listed as a product, it is being produced, so the reaction is exothermic. If the temperature is increasing, a product is being added to the equilibrium, so the equilibrium shifts to minimize the addition of extra product: it shifts back toward reactants.<\/p>\n<p><em>Test Yourself<\/em><br \/>\nPredict the effect of decreasing the temperature on this equilibrium.<\/p>\n<p style=\"text-align: center;\">N<sub>2<\/sub>O<sub>4<\/sub> + 57 kJ \u21c4 2NO<sub>2<\/sub><\/p>\n<p><em>Answer<\/em><br \/>\nEquilibrium shifts toward reactants.<\/p>\n<\/div>\n<\/div>\n<p>In the case of temperature, the value of the equilibrium has changed because the <em>K<\/em><sub>eq<\/sub> is dependent on temperature. That is why equilibria shift with changes in temperature.<\/p>\n<p>A <strong>catalyst<\/strong>\u00a0is a substance that increases the speed of a reaction. Overall, a catalyst is not a reactant and is not used up, but it still affects how fast a reaction proceeds. However, a catalyst does not affect the extent or position of a reaction at equilibrium. It helps a reaction achieve equilibrium faster.<\/p>\n<div class=\"textbox shaded\">\n<h1>Chemistry Is Everywhere: Equilibria in the Garden<\/h1>\n<p>Hydrangeas are common flowering plants around the world. Although many hydrangeas are white, there is one common species (<em>Hydrangea macrophylla<\/em>) whose flowers can be either red or blue, as shown in the accompanying figure. How is it that a plant can have different coloured flowers like this?<\/p>\n<figure id=\"attachment_3258\" aria-describedby=\"caption-attachment-3258\" style=\"width: 374px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-728\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/5996174498_eb660e46c2_b-e1412019701873-1.jpg\" alt=\"Blue hydrangeas.\" width=\"374\" height=\"299\" \/><figcaption id=\"caption-attachment-3258\" class=\"wp-caption-text\">Figure 13.1 &#8220;Garden Equilibria.&#8221; This species of hydrangea has flowers that can be either red or blue. Why the colour difference?<\/figcaption><\/figure>\n<p>Interestingly, the colour of the flowers is due to the acidity of the soil that the hydrangea is planted in. An astute gardener can adjust the pH of the soil and actually change the colour of the flowers. However, it is not the H<sup>+<\/sup> or OH<sup>\u2212<\/sup> ions that affect the colour of the flowers. Rather, it is the presence of aluminum that causes the colour change.<\/p>\n<p>The solubility of aluminum in soil \u2014 and thus the ability of plants to absorb it \u2014 is dependent on the acidity of the soil. If the soil is relatively acidic, the aluminum is more soluble, and plants can absorb it more easily. Under these conditions, hydrangea flowers are blue as Al ions interact with anthocyanin pigments in the plant. In more basic soils, aluminum is less soluble, and under these conditions the hydrangea flowers are red. Gardeners who change the pH of their soils to change the colour of their hydrangea flowers are therefore employing Le Chatelier\u2019s principle: the amount of acid in the soil changes the equilibrium of aluminum solubility, which in turn affects the colour of the flowers.<\/p>\n<\/div>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Takeaways<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ul>\n<li>Le Chatelier\u2019s principle addresses how an equilibrium shifts when the conditions of an equilibrium are changed.<\/li>\n<li>The direction of shift can be predicted for changes in concentrations, temperature, or pressure.<\/li>\n<li>Catalysts do not affect the position of an equilibrium; they help reactions achieve equilibrium faster.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercises<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<h1>Questions<\/h1>\n<ol>\n<li>Define <em>Le Chatelier\u2019s principle<\/em>.<\/li>\n<li>What is meant by a stress? What are some of the ways an equilibrium can be stressed?<\/li>\n<li>Given this equilibrium, predict the direction of shift for each stress.\n<p style=\"text-align: center;\">H<sub>2<\/sub>(g) + I<sub>2<\/sub>(s) + 53 kJ \u21c4 2HI(g)<\/p>\n<ol type=\"a\">\n<li>decreased temperature<\/li>\n<li>increased pressure<\/li>\n<li>removal of HI<\/li>\n<\/ol>\n<\/li>\n<li>Given this equilibrium, predict the direction of shift for each stress.\n<p style=\"text-align: center;\">H<sub>2<\/sub>(g)\u00a0+\u00a0F<sub>2<\/sub>(g) \u21c4 2HF(g) + 546 kJ<\/p>\n<ol type=\"a\">\n<li>increased temperature<\/li>\n<li>addition of H<sub>2<\/sub><\/li>\n<li>decreased pressure<\/li>\n<\/ol>\n<\/li>\n<li>Given this equilibrium, predict the direction of shift for each stress.\n<p style=\"text-align: center;\">2SO<sub>2<\/sub>(g) + O<sub>2<\/sub>(g) \u21c4 2SO<sub>3<\/sub>(g) + 196 kJ<\/p>\n<ol type=\"a\">\n<li>removal of SO<sub>3<\/sub><\/li>\n<li>addition of O<sub>2<\/sub><\/li>\n<li>decreased temperature<\/li>\n<\/ol>\n<\/li>\n<li>Given this equilibrium, predict the direction of shift for each stress listed.\n<p style=\"text-align: center;\">CO<sub>2<\/sub>(g) + C(s) + 171 kJ \u21c4 2CO(g)<\/p>\n<ol type=\"a\">\n<li>addition of CO<\/li>\n<li>increased pressure<\/li>\n<li>addition of a catalyst<\/li>\n<\/ol>\n<\/li>\n<li>The synthesis of NH<sub>3<\/sub> uses this chemical reaction.\n<p style=\"text-align: center;\">N<sub>2<\/sub>(g) + 3H<sub>2<\/sub>(g) \u21c4 2NH<sub>3<\/sub>(g) + 92 kJ<\/p>\n<p>Identify three stresses that can be imposed on the equilibrium to maximize the amount of NH<sub>3<\/sub>.<\/li>\n<li>The synthesis of CaCO<sub>3<\/sub> uses this chemical reaction.\n<p style=\"text-align: center;\">CaO(s)\u00a0+\u00a0CO<sub>2<\/sub>(g) \u21c4 CaCO<sub>3<\/sub>(s)\u00a0+\u00a0180\u00a0kJ<\/p>\n<p>Identify three stresses that can be imposed on the equilibrium to maximize the amount of CaCO<sub>3<\/sub>.<\/li>\n<\/ol>\n<h1>Answers<\/h1>\n<ol>\n<li>When an equilibrium is stressed, the equilibrium shifts to minimize that stress.<\/li>\n<\/ol>\n<ol start=\"3\">\n<li>\n<ol type=\"a\">\n<li>toward reactants<\/li>\n<li>toward reactants<\/li>\n<li>toward products<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol start=\"5\">\n<li>\n<ol type=\"a\">\n<li>toward products<\/li>\n<li>toward products<\/li>\n<li>toward products<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol start=\"7\">\n<li>increased pressure, decreased temperature, removal of NH<sub>3<\/sub><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h3>Media Attributions<\/h3>\n<ul>\n<li><a href=\"https:\/\/www.flickr.com\/photos\/jannem\/5996174498\/\">\u201cHydrangea\u201d<\/a> \u00a9 <a href=\"https:\/\/www.flickr.com\/photos\/jannem\/\">2011 by Janne Moren<\/a> is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/2.0\/\">CC BY-NC-SA (Attribution-NonCommercial-ShareAlike)<\/a> license<\/li>\n<\/ul>\n","protected":false},"author":90,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-7662","chapter","type-chapter","status-publish","hentry"],"part":7638,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7662","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":7,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7662\/revisions"}],"predecessor-version":[{"id":8960,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7662\/revisions\/8960"}],"part":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/7638"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7662\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/media?parent=7662"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=7662"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/contributor?post=7662"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/license?post=7662"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}