{"id":392,"date":"2017-08-21T06:47:51","date_gmt":"2017-08-21T06:47:51","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=392"},"modified":"2017-09-20T18:22:46","modified_gmt":"2017-09-20T18:22:46","slug":"le-chateliers-principle-and-equilibrium-constants","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/le-chateliers-principle-and-equilibrium-constants\/","title":{"rendered":"Le Ch\u00e2telier\u2019s Principle and Equilibrium Constants"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/hesss-law-and-gibbs-free-energy\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/review-4\">Next Lesson\u00a0\u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Reactions and Reactivity: Le Ch\u00e2telier\u2019s Principle and Equilibrium Constants<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will use Le Ch\u00e2telier\u2019s principle to determine changes in concentration that will occur at\u00a0equilibrium. We will also look at equilibrium constants of reactions.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>Determining the enthalpy and entropy of a system using Hess\u2019s law<\/li>\n<li>Using Gibbs Free Energy to calculate the following:\n<ul>\n<li>Entropy<\/li>\n<li>Enthalpy<\/li>\n<li>Solving for missing values<\/li>\n<li>Determining spontaneity<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<section>\n<h3>Le Ch\u00e2telier\u2019s Principle<\/h3>\n<p>Le Ch\u00e2telier\u2019s principle states that if a\u00a0change (or stress) is applied to a system at equilibrium, the position of equilibrium will shift in a direction so as to reduce that change. Therefore:<\/p>\n<ul>\n<li>If\u00a0a reactant is added to a reaction at\u00a0equilibrium, the reaction will produce more products.<\/li>\n<li>If\u00a0a reactant is removed from a reaction at\u00a0equilibrium, the reaction will shift in the direction which forms more of that reactant.<\/li>\n<li>If\u00a0a product is removed from a reaction at\u00a0equilibrium, the reaction will shift to form more of that product.<\/li>\n<li>If\u00a0a product is added to a reaction at\u00a0equilibrium, the reaction will shift to make more of the reactants.<\/li>\n<li>An\u00a0increase of temperature for an endothermic reaction shifts equilibrium towards the products.<\/li>\n<li>An\u00a0increase of temperature for an exothermic\u00a0reaction shifts equilibrium towards the reactants.<\/li>\n<li>When only gases are present, a decrease in volume will cause the equilibrium\u00a0to shift toward the smaller total number of moles of gases. Remember that to decrease the volume, the pressure must be\u00a0increased.<\/li>\n<\/ul>\n<p>As an example, let\u2019s look at the following\u00a0reaction:<\/p>\n<p class=\"center\">N<sub>2<\/sub>(<em>g<\/em>)+ 3H<sub>2<\/sub>(<em>g<\/em>)\u00a0\u2194\u00a02NH<sub>3<\/sub>(<em>g<\/em>)+ energy<\/p>\n<p align=\"left\">The changes shown below will result in the corresponding shift.<\/p>\n<table class=\"gas_law_table\">\n<tbody>\n<tr>\n<td valign=\"top\" width=\"231\">\n<div align=\"left\"><strong>Change <\/strong><\/div>\n<\/td>\n<td valign=\"top\" width=\"227\"><strong>Shift <\/strong><\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"left\">Addition of N<sub>2<\/sub>(<em>g<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">Towards the products (right)<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"left\">Removal of N<sub>2<\/sub>(<em>g<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">Towards the reactants (left)<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"left\">Addition of H<sub>2<\/sub> (<em>g<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">Towards the products (right)<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"left\">Removal of H<sub>2<\/sub> (<em>g<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">Towards the reactants (left)<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"left\">Addition of NH<sub>3<\/sub>(<em>g<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">Towards the reactants (left)<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"left\">Removal of NH<sub>3<\/sub> (<em>g<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">Towards the products (right)<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"left\">Addition of energy by heating<\/div>\n<\/td>\n<td valign=\"top\">Towards the reactants (left)<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"left\">Removal of energy by cooling<\/div>\n<\/td>\n<td valign=\"top\">Towards the products (right)<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"left\">Decrease the volume<\/div>\n<\/td>\n<td valign=\"top\">Toward the products (right)<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"left\">Increase the volume<\/div>\n<\/td>\n<td valign=\"top\">Toward the reactants (left)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Determine the shift in equilibrium when more carbon dioxide is added to the following reaction.<br \/>\nEnergy + CaCO<sub>3<\/sub>(<em>s<\/em>)\u00a0\u2194\u00a0CaO(<em>s<\/em>) + CO<sub>2<\/sub>(<em>g<\/em>)<\/p>\n<ol>\n<li>The reaction will shift toward the products<\/li>\n<li>The reaction will shift toward the reactants<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is B.\u00a0When more carbon dioxide is added there are more products and the\u00a0change in equilibrium will shift toward the reactants<\/p>\n<\/section>\n<h3>Dynamic Equilibrium<\/h3>\n<p>When a reaction has the same\u00a0rate going forward (producing\u00a0products) as going in reverse (the products reverting to the reactants)\u00a0the\u00a0reaction is said to be in equilibrium.<\/p>\n<p>We can calculate the\u00a0equilibrium constant of a reaction\u00a0given the concentrations of products and reactants at equilibrium.<\/p>\n<p>First we use the generic equation to set up the\u00a0equilibrium expression:<\/p>\n<p class=\"center\"><em>a<\/em>A + <em>b<\/em>B\u00a0\u2194\u00a0<em>c<\/em>C + <em>d<\/em>D<\/p>\n<p>The lower case letters represent the coefficients used to balance the equation and the capital letters the substances themselves.<\/p>\n<p>The equilibrium expression is:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_002.gif\" width=\"92\" height=\"46\" \/><\/center>The values that are used for the substances are their\u00a0concentrations\u00a0measured in <abbr title=\"Moles of solute per liter of solution\">molarity<\/abbr>.<\/p>\n<h4>Sample Problem<\/h4>\n<p>Calculate the equilibrium constant under the following conditions:<\/p>\n<p class=\"center\">N<sub>2<\/sub>(<em>g<\/em>) + 3Cl<sub>2<\/sub>(<em>g<\/em>) \u2194 2NCl<sub>3<\/sub>(<em>g<\/em>)<\/p>\n<p>The equilibrium concentrations were found to be NCl<sub>3<\/sub>(<em>g<\/em>) = 0.23 M, N<sub>2<\/sub>(<em>g<\/em>) = 0.12 M, and Cl<sub>2<\/sub>(<em>g<\/em>) = 0.13 M<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_003.gif\" alt=\"placeholder.png\" width=\"277\" height=\"48\" \/><\/center><\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>At a particular temperature, NO gas reacts with chlorine gas to produce gaseous NOCl. When the reaction reaches\u00a0equilibrium the concentrations of substances present are NO = 0.56M; Cl<sub>2<\/sub> = 0.34M; and NOCl = 0.012 M.\u00a0Calculate the value of the equilibrium constant at this temperature.<\/p>\n<ol>\n<li>1.35 \u00d7 10<sup>-3<\/sup> \/ M<\/li>\n<li>6.40 \u00d7 10<sup>-2<\/sup> \/ M<\/li>\n<li>1.59 \u00d7 10<sup>1<\/sup> \/ M<\/li>\n<li>7.40 \u00d7 10<sup>2<\/sup> \/ M<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is A.<\/p>\n<p class=\"q-reveal center\">2NO (g) + Cl<sub>2<\/sub>(g) \u2194 2NOCl(g)<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_004.gif\" width=\"306\" height=\"48\" \/><\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>At a particular temperature, hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas. Find the value\u00a0of the equilibrium constant if the following concentrations are found at equilibrium. H<sub>2<\/sub> = 0.0123 M; F<sub>2<\/sub> = 0.0342 M and HF = 0.00291 M.<\/p>\n<ol>\n<li>2.12 \u00d7 10<sup>-2<\/sup><\/li>\n<li>1.37 \u00d7 10<sup>-1<\/sup><\/li>\n<li>4.71 \u00d7 10<sup>1<\/sup><\/li>\n<li>7.27 \u00d7 10<sup>1<\/sup><\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is A.<\/p>\n<p class=\"center q-reveal\">H<sub>2<\/sub> (<em>g<\/em>) + F<sub>2<\/sub>(<em>g<\/em>) \u2194 2HF(<em>g<\/em>)<\/p>\n<p class=\"center q-reveal\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_005.gif\" width=\"317\" height=\"48\" \/><\/p>\n<\/section>\n<p>Another\u00a0way to find the equilibrium constant of\u00a0a reaction is to use the equation<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_006.gif\" width=\"116\" height=\"28\" \/><\/center><em>e<\/em> = 2.718, \u0394<em>G <\/em>is free energy, <em>R<\/em> is the ideal gas law constant 8.314 J\/mol<strong>\u00b7<\/strong>K (or 0.008314 kJ\/mol<strong>\u00b7<\/strong>K), and <em>T<\/em> is temperature in Kelvin.<\/p>\n<h4>Sample Problem<\/h4>\n<p>Using the equation above, solve for the equilibrium constant when the\u00a0value for free energy is 1.5 \u00d7 10<sup>2<\/sup> kJ at a temperature of 38\u00b0C for the reaction:<\/p>\n<p class=\"center\">N<sub>2<\/sub>(<em>g<\/em>) + 3Cl<sub>2<\/sub>(<em>g<\/em>) \u2194 2NCl<sub>3<\/sub>(<em>g<\/em>)<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_006.gif\" width=\"106\" height=\"28\" \/><\/center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_009.gif\" width=\"201\" height=\"42\" \/><\/p>\n<p class=\"center\">K<sub>eq<\/sub> = 6.57 \u00d7 10<sup>-26<\/sup><\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Determine the equilibrium\u00a0constant for the following\u00a0reaction when the value for free energy is\u00a0 2.3\u00d710<sup>3<\/sup> kJ and is\u00a0at a temperature of 1432 K.<\/p>\n<p class=\"center\">2NO (<em>g<\/em>) + Cl<sub>2<\/sub> (<em>g<\/em>) \u2194 2NOCl (<em>g<\/em>)<\/p>\n<ol>\n<li>1.26 \u00d7 10<sup> -84<\/sup><\/li>\n<li>1.93 \u00d7 10<sup> -1<\/sup><\/li>\n<li>8.24 \u00d7 10<sup> -1<\/sup><\/li>\n<li>1.93 \u00d7 10<sup> -58<\/sup><\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The\u00a0correct answer is A. Option C is arrived at by neglecting to put <em>R<\/em> into\u00a0kJ. Choice D, does not raise <em>e<\/em> to the specified value and does\u00a0not convert <em>R<\/em> into kJ. Option B does not raise the value of <em>e<\/em> to the\u00a0power, but is merely the power itself.<\/p>\n<p class=\"center q-reveal\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_006.gif\" width=\"116\" height=\"28\" \/><br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_012.gif\" width=\"202\" height=\"42\" align=\"absmiddle\" \/><br \/>\nK<sub>eq<\/sub>=<em>e<\/em><sup>\u2014193.186<\/sup><br \/>\nK<sub>eq<\/sub> = 1.26 x 10<sup>-84<\/sup><\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Determine the equilibrium constant for the following\u00a0reaction that has free\u00a0energy of -1.34\u00d710<sup>2<\/sup> kJ at a\u00a0temperature of 357 K.<\/p>\n<p class=\"center\">H<sub>2<\/sub>(<em>g<\/em>) + F<sub>2<\/sub>(<em>g<\/em>) \u2194 2HF(<em>g<\/em>)<\/p>\n<ol>\n<li>2.47 \u00d7 10<sup>-20<\/sup><\/li>\n<li>1.05 \u00d7 10<sup>1<\/sup><\/li>\n<li>4.51 \u00d7 10<sup>2<\/sup><\/li>\n<li>4.05 \u00d7 10<sup>19<\/sup><\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is D. Option C is simply the power to which <em>e<\/em> is to be\u00a0raised. Choice A is arrived at by neglecting to make the value of \u0394<em>G<\/em> negative (since the value is\u00a0negative, it should become positive ). Choice B is\u00a0obtained by failing to convert the value of <em>R<\/em> into kJ.<\/p>\n<p class=\"center q-reveal\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_006.gif\" width=\"116\" height=\"28\" \/><br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s10_015.gif\" width=\"205\" height=\"42\" \/><br \/>\nK<sub>eq<\/sub> = <em>e<\/em><sup>45.146<br \/>\n<\/sup>K<sub>eq<\/sub> = 4.05 \u00d7 10<sup>19<\/sup><\/p>\n<\/section>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/hesss-law-and-gibbs-free-energy\">\u2b05 Previous\u00a0Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/review-4\">Next Lesson\u00a0\u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson\u00a0\u27a1 Reactions and Reactivity: Le Ch\u00e2telier\u2019s Principle and Equilibrium Constants Objective In this lesson we will use Le Ch\u00e2telier\u2019s principle to determine changes in concentration that will occur at\u00a0equilibrium. We will also look at equilibrium constants of reactions. Previously we covered&#8230; Determining the enthalpy and entropy of a system using Hess\u2019s [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-392","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/392","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=392"}],"version-history":[{"count":10,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/392\/revisions"}],"predecessor-version":[{"id":936,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/392\/revisions\/936"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=392"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}