{"id":1338,"date":"2020-06-23T17:10:06","date_gmt":"2020-06-23T21:10:06","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chbe220\/?post_type=chapter&#038;p=1338"},"modified":"2020-08-11T18:13:58","modified_gmt":"2020-08-11T22:13:58","slug":"equilibrium-reactions","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chbe220\/chapter\/equilibrium-reactions\/","title":{"raw":"Equilibrium Reactions","rendered":"Equilibrium Reactions"},"content":{"raw":"<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\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\r\nBy the end of this section, you should be able to:\r\n\r\n<strong>Analyze<\/strong>\u00a0equilibrium reaction equations\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\nWhen <strong>the concentrations of reactants and products become constant<\/strong>, the system is said to have reached a point of equilibrium. The equilibrium state is <strong>dynamic<\/strong>. At equilibrium, the reactions are <strong>not stopped<\/strong>, but rather <strong>the rates of opposing reactions have become equal<\/strong> such that concentrations of chemical species do not change.\r\n\r\n<span style=\"font-size: 1em;text-align: initial\">Laboratory kinetic studies may be performed on reactions far from equilibrium where products are in low concentration and the reverse reactions are not important (this is what we considered previously).<\/span>\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">[latex]CO_{(g)} + H_{2}O_{(g)} \u2192 CO_{2(g)} + H_{2(g)}[\/latex]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n\r\nCloser to equilibrium however we must take these reverse reactions into account.\r\n\r\n[latex]CO_{(g)} + H_{2}O_{(g)} \u21cc CO_{2(g)} + H_{2(g)}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"prompt input_prompt\"><\/div>\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h2 id=\"Equilibrium-Constants\">Equilibrium Constant<\/h2>\r\n<\/div>\r\n<h3>Definition of the Equilibrium Constant<\/h3>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 51.3859%;height: 54px\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]K\\;\\; or\\;\\; K_{eq}=\\prod_{i} a_{i,eq}^{vi}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nUnpack the notation:\r\n<ul>\r\n \t<li><span style=\"text-align: initial;font-size: 1em\">[latex]K[\/latex] or [latex]K_{eq}[\/latex] is the equilibrium constant<\/span><\/li>\r\n \t<li>[latex]\\prod_{i}[\/latex] is the product notation (similar to [latex]\\sum_{i}[\/latex], but terms are multiplied rather than added)<\/li>\r\n \t<li>[latex]a_{i,eq}[\/latex] is the activity of compound [latex]i[\/latex] at equilibrium, for now we will assume concentration divided by a unit concentration term. This will give us the concentration of the substace in a dimensionless form, eg. [latex]\\frac{[A]}{c^\\theta}[\/latex], where [latex][A][\/latex] and [latex]{c^\\theta}[\/latex] are in units of mol\/L, and [latex]c^\\theta[\/latex] has magnitude of 1.<\/li>\r\n \t<li>[latex]vi[\/latex] is the stoichiometric coefficient of compound [latex]i[\/latex] we saw previously. It is positive for products and negative for reactants.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n\r\n&nbsp;\r\n<p style=\"text-align: left\">Consider a system with the following reaction:<\/p>\r\n<p style=\"text-align: center\">[latex]A\u21ccB[\/latex]<\/p>\r\nLet's say the system is composed of 2 first-order reactions:\r\n<blockquote>Forward reaction: [latex]A\u2192B[\/latex], where the reaction rate is expressed by: [latex]r=k_{r}\u00d7[A][\/latex]\r\nReverse reaction: [latex]B\u2192A[\/latex], where the reaction rate is expressed by: [latex]r'=k'_{r}\u00d7[B][\/latex]<\/blockquote>\r\nNet rate of change in [latex][A][\/latex] considering both reactions is:\r\n<p style=\"text-align: center\">[latex]\\frac{d[A]}{dt}=r'-r=k'_{r}\u00d7[B]-k_{r}\u00d7[A][\/latex]<\/p>\r\nIf the initial concentration of A is [latex][A]_{0}[\/latex] and no B is present, then at all times: [latex][A]+[B]=[A]_{0}[\/latex]. Using this in the equation above:\r\n\r\n\\begin{align*}\r\n\\frac{d[A]}{dt} &amp;=k'_{r}[B]-k_{r}[A]\\\\\r\n&amp; =k'_{r}([A]_{0}-[A])-k_{r}[A]\\\\\r\n&amp; =k'_{r}[A]_{0}-(k_{r}+k'_{r})[A]\r\n\\end{align*}\r\n\r\nThis is what is known as an ordinary differential equation, as [latex][A][\/latex] is both a differential and a separate term. You may not know how to solve these equations and do not need to know this for this course. However, if we were to solve this ordinary differential equation, we would get the following:\r\n<p style=\"text-align: center\">[latex][A]=\\frac{k'_{r}+k_{r}e^{-(k'_{r}+k{r})t}}{k'_{r}+k_{r}}[A]_{0}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"prompt input_prompt\"><\/div>\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n\r\nUsing the expression for [latex][A][\/latex] above, [latex][B]=[A]_{0}-[A][\/latex], taking arbitrary values of [latex]k_{r}=0.693[\/latex] and [latex]k'_{r}=0.262[\/latex], the following graph is produced:\r\n\r\n<img class=\"wp-image-1096 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Capture-concentration-change-2-first-orders-300x191.png\" alt=\"\" width=\"484\" height=\"308\" \/>As you can see from the graph, the concentrations change less over time. After a long time, they should reach a state of equilibrium. The length of time to reach equilibrium (or close to it) will depend on the reaction rates.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n\r\nAs [latex]t\u2192\\infty[\/latex], we approach equilibrium and we get the following:\r\n<p style=\"text-align: center\">[latex][A]_{eq}=\\frac{k'_{r}+k_{r}e^{-(k'_{r}+k{r})t}}{k'_{r}+k_{r}}[A]_{0}=\\frac{k'_{r}}{k'_{r}+k_{r}}[A]_{0}[\/latex]<\/p>\r\n<p style=\"text-align: center\">and\r\n[latex][B]_{eq}=[A]_{0}-[A]_{eq}=\\frac{k_{r}}{k'_{r}+k_{r}}[A]_{0}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<p style=\"text-align: left\">The equilibrium constant in this example is defined as:<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 80.1005%;height: 67px\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]K=\\frac{[B]_{eq}\/c^\\theta}{[A]_{eq}\/c^\\theta}=\\frac{[B]_{eq}}{[A]_{eq}}=\\frac{k_{r}}{k'_{r}}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: left\">At equilibrium, the rates of forward and reverse reactions should be the same (meaning no change in overall concentrations), so it makes sense that when we rearrange this we get:<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 38.2083%;height: 91px\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]k'_{r}[B]_{eq}=k_{r}[A]_{eq}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"prompt input_prompt\"><\/div>\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n\r\n&nbsp;\r\n<p style=\"text-align: left\"><strong>Consider another system with the following reaction:<\/strong><\/p>\r\n<p style=\"text-align: center\">[latex]A+B\u21ccC[\/latex]<\/p>\r\n\r\n<blockquote>Forward reaction: [latex]A+B\u2192C[\/latex], where the forward reaction rate is expressed by: [latex]r=k_{r}[A][B][\/latex]\r\n\r\nReverse reaction: [latex]C\u2192A+B[\/latex], where the reverse reaction rate is expressed by: [latex]r'=k'_{r}[C][\/latex]<\/blockquote>\r\nHere, in equilibrium we have:\r\n<p style=\"text-align: center\">[latex]r=r'=k_{r}[A]_{eq}[B]_{eq}=k'_{r}[C]_{eq}[\/latex]<\/p>\r\nThe equilibrium constant is defined as dimensionless, so here is where the [latex]c^\\theta[\/latex] terms make a difference. The equilibrium constant can be espressed as:\r\n<p style=\"text-align: center\">[latex]K=\\frac{[C]_{eq}\/c^\\theta}{[A]_{eq}\/c^\\theta\u00d7[B]_{eq}\/c^\\theta}=\\Big(\\frac{[C]}{[A][B]}\\Big)_{eq}=\\frac{k_{r}}{k'_{r}}c^\\theta[\/latex]<\/p>\r\nRecall that [latex]k_{r}[\/latex] and [latex]k'_{r}[\/latex] here will have different units, so the [latex]c^\\theta[\/latex] term ensures that these become dimensionless when one is divided by the other.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"prompt input_prompt\"><\/div>\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Equilibrium Constant<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n1. For the balanced chemical reaction below, write an equation for the equilibrium constant(K).[latex]^{[1]}[\/latex]\r\n<p style=\"text-align: center\">[latex]2H_{2(g)}+N_{2(g)} \u21ccN_{2}H_{4(g)}[\/latex]<\/p>\r\n2. Consider the following reaction system:\r\n<p style=\"text-align: center\">[latex]A\u21ccB+C[\/latex]<\/p>\r\nFrom initial rate and equilibrium experiments we know the following: <span style=\"font-size: 1em\"><span style=\"text-align: initial;font-size: 1em\">[latex][A]_{eq} = 0.5 mol\/L[\/latex]<\/span><\/span><span style=\"text-align: initial;font-size: 1em\">, [latex][B]_{eq} = 0.1 mol\/L[\/latex], [latex][C]_{eq} =0.2 mol\/L[\/latex], [latex]k_{r} = 0.1 min^{-1}[\/latex].[latex]^{[1]}[\/latex]<\/span>\r\n\r\nWhat are the values of the equilibrium constant([latex]K[\/latex]) and reverse reaction constant([latex]k'_{r}[\/latex])?\r\n\r\n&nbsp;\r\n\r\n[latex]K\\; or\\; K_{eq}=\\prod_{i} a_{i,eq}^{vi}[\/latex], at equilibrium [latex]r=r'[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n<div class=\"textbox\">\r\n<h3>Solution<\/h3>\r\n1. [latex]K=\\frac{[N_{2}H_{4}]}{[H_{2}]^2[N_{2}]}[\/latex]\r\n\r\n2. Start with expressing K in forms of concentrations and [latex]k_{r}[\/latex]:\r\n\r\n\\begin{align*}\r\nK &amp; =\\frac{[B]_{eq}\/c^\\theta*[C]_{eq}\/c^\\theta}{[A]_{eq}\/c^\\theta}\\\\\r\n&amp; = (\\frac{[B][C]}{[A]})_{eq}\\frac{1}{c^\\theta}\\\\\r\n&amp; = \\frac{k_{r}}{k'_{r}*c^\\theta}\r\n\\end{align*}\r\n\r\nCalculate K from the expression with concentrations:\r\n\r\n\\begin{align*}\r\nK &amp; = (\\frac{[B][C]}{[A]})_{eq}\\frac{1}{c^\\theta}\\\\\r\n&amp; = (\\frac{0.1\\frac{mol}{L}*0.2\\frac{mol}{L}}{0.5\\frac{mol}{L}})_{eq}\\frac{1}{c^\\theta}\\\\\r\n&amp; = 0.04\r\n\\end{align*}\r\n\r\nSubsitude the K value into the expression with [latex]k_{r}[\/latex]:\r\n\r\n\\begin{align*}\r\nK &amp; = \\frac{k_{r}}{k'_{r}*c^\\theta}\\\\\r\nk'_{r}&amp; = \\frac{k_{r}}{K*c^\\theta}\\\\\r\n&amp; = \\frac{0.1min^{-1}}{0.04*1\\frac{mol}{L}}\\\\\r\n&amp; = 2.5 \\frac{L}{mol*min}\r\n\\end{align*}\r\n\r\nBack substitute [latex]k_{r}[\/latex] into the rate equations to check this:\r\n<p style=\"text-align: center\">[latex]r=k_{r}[A]_{eq}=0.1min^{-1}*0.5\\frac{mol}{L}=0.05\\frac{mol}{L*min}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]r'=k'_{r}[B]_{eq}[C]_{eq}=2.5 \\frac{L}{mol*min}*0.1\\frac{mol}{L}*0.2\\frac{mol}{L}=0.05\\frac{mol}{L*min}[\/latex]<\/p>\r\nForward and reverse reaction rates are the same at [latex]0.05\\frac{mol}{L*min}[\/latex] , as we would expect at equilibrium.\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h2>References<\/h2>\r\n<span style=\"font-size: 1em\">[1] Chemistry LibreTexts. 2020. <\/span><i style=\"font-size: 1em\">Characteristics of the Equilibrium State.<\/i><span style=\"font-size: 1em\"> [online] Available at: &lt;<a href=\"https:\/\/chem.libretexts.org\/Bookshelves\/Physical_and_Theoretical_Chemistry_Textbook_Maps\/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)\/Equilibria\/Chemical_Equilibria\/Principles_of_Chemical_Equilibria\/Principles_of_Chemical_Equilibrium\/Characteristics_Of_The_Equilibrium_State\">https:\/\/chem.libretexts.org\/Bookshelves\/Physical_and_Theoretical_Chemistry_Textbook_Maps\/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)\/Equilibria\/Chemical_Equilibria\/Principles_of_Chemical_Equilibria\/Principles_of_Chemical_Equilibrium\/Characteristics_Of_The_Equilibrium_State<\/a>&gt; [Accessed 24 April, 2020].<\/span>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<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<p>By the end of this section, you should be able to:<\/p>\n<p><strong>Analyze<\/strong>\u00a0equilibrium reaction equations<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>When <strong>the concentrations of reactants and products become constant<\/strong>, the system is said to have reached a point of equilibrium. The equilibrium state is <strong>dynamic<\/strong>. At equilibrium, the reactions are <strong>not stopped<\/strong>, but rather <strong>the rates of opposing reactions have become equal<\/strong> such that concentrations of chemical species do not change.<\/p>\n<p><span style=\"font-size: 1em;text-align: initial\">Laboratory kinetic studies may be performed on reactions far from equilibrium where products are in low concentration and the reverse reactions are not important (this is what we considered previously).<\/span><\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">[latex]CO_{(g)} + H_{2}O_{(g)} \u2192 CO_{2(g)} + H_{2(g)}[\/latex]<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<p>Closer to equilibrium however we must take these reverse reactions into account.<\/p>\n<p>[latex]CO_{(g)} + H_{2}O_{(g)} \u21cc CO_{2(g)} + H_{2(g)}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"prompt input_prompt\"><\/div>\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h2 id=\"Equilibrium-Constants\">Equilibrium Constant<\/h2>\n<\/div>\n<h3>Definition of the Equilibrium Constant<\/h3>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 51.3859%;height: 54px\">\n<tbody>\n<tr>\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]K\\;\\; or\\;\\; K_{eq}=\\prod_{i} a_{i,eq}^{vi}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Unpack the notation:<\/p>\n<ul>\n<li><span style=\"text-align: initial;font-size: 1em\">[latex]K[\/latex] or [latex]K_{eq}[\/latex] is the equilibrium constant<\/span><\/li>\n<li>[latex]\\prod_{i}[\/latex] is the product notation (similar to [latex]\\sum_{i}[\/latex], but terms are multiplied rather than added)<\/li>\n<li>[latex]a_{i,eq}[\/latex] is the activity of compound [latex]i[\/latex] at equilibrium, for now we will assume concentration divided by a unit concentration term. This will give us the concentration of the substace in a dimensionless form, eg. [latex]\\frac{[A]}{c^\\theta}[\/latex], where [latex][A][\/latex] and [latex]{c^\\theta}[\/latex] are in units of mol\/L, and [latex]c^\\theta[\/latex] has magnitude of 1.<\/li>\n<li>[latex]vi[\/latex] is the stoichiometric coefficient of compound [latex]i[\/latex] we saw previously. It is positive for products and negative for reactants.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<p>&nbsp;<\/p>\n<p style=\"text-align: left\">Consider a system with the following reaction:<\/p>\n<p style=\"text-align: center\">[latex]A\u21ccB[\/latex]<\/p>\n<p>Let&#8217;s say the system is composed of 2 first-order reactions:<\/p>\n<blockquote><p>Forward reaction: [latex]A\u2192B[\/latex], where the reaction rate is expressed by: [latex]r=k_{r}\u00d7[A][\/latex]<br \/>\nReverse reaction: [latex]B\u2192A[\/latex], where the reaction rate is expressed by: [latex]r'=k'_{r}\u00d7[B][\/latex]<\/p><\/blockquote>\n<p>Net rate of change in [latex][A][\/latex] considering both reactions is:<\/p>\n<p style=\"text-align: center\">[latex]\\frac{d[A]}{dt}=r'-r=k'_{r}\u00d7[B]-k_{r}\u00d7[A][\/latex]<\/p>\n<p>If the initial concentration of A is [latex][A]_{0}[\/latex] and no B is present, then at all times: [latex][A]+[B]=[A]_{0}[\/latex]. Using this in the equation above:<\/p>\n<p>\\begin{align*}<br \/>\n\\frac{d[A]}{dt} &amp;=k&#8217;_{r}[B]-k_{r}[A]\\\\<br \/>\n&amp; =k&#8217;_{r}([A]_{0}-[A])-k_{r}[A]\\\\<br \/>\n&amp; =k&#8217;_{r}[A]_{0}-(k_{r}+k&#8217;_{r})[A]<br \/>\n\\end{align*}<\/p>\n<p>This is what is known as an ordinary differential equation, as [latex][A][\/latex] is both a differential and a separate term. You may not know how to solve these equations and do not need to know this for this course. However, if we were to solve this ordinary differential equation, we would get the following:<\/p>\n<p style=\"text-align: center\">[latex][A]=\\frac{k'_{r}+k_{r}e^{-(k'_{r}+k{r})t}}{k'_{r}+k_{r}}[A]_{0}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"prompt input_prompt\"><\/div>\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<p>Using the expression for [latex][A][\/latex] above, [latex][B]=[A]_{0}-[A][\/latex], taking arbitrary values of [latex]k_{r}=0.693[\/latex] and [latex]k'_{r}=0.262[\/latex], the following graph is produced:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1096 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Capture-concentration-change-2-first-orders-300x191.png\" alt=\"\" width=\"484\" height=\"308\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Capture-concentration-change-2-first-orders-300x191.png 300w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Capture-concentration-change-2-first-orders-768x490.png 768w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Capture-concentration-change-2-first-orders-65x41.png 65w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Capture-concentration-change-2-first-orders-225x144.png 225w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Capture-concentration-change-2-first-orders-350x223.png 350w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/Capture-concentration-change-2-first-orders.png 964w\" sizes=\"auto, (max-width: 484px) 100vw, 484px\" \/>As you can see from the graph, the concentrations change less over time. After a long time, they should reach a state of equilibrium. The length of time to reach equilibrium (or close to it) will depend on the reaction rates.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<p>As [latex]t\u2192\\infty[\/latex], we approach equilibrium and we get the following:<\/p>\n<p style=\"text-align: center\">[latex][A]_{eq}=\\frac{k'_{r}+k_{r}e^{-(k'_{r}+k{r})t}}{k'_{r}+k_{r}}[A]_{0}=\\frac{k'_{r}}{k'_{r}+k_{r}}[A]_{0}[\/latex]<\/p>\n<p style=\"text-align: center\">and<br \/>\n[latex][B]_{eq}=[A]_{0}-[A]_{eq}=\\frac{k_{r}}{k'_{r}+k_{r}}[A]_{0}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<p style=\"text-align: left\">The equilibrium constant in this example is defined as:<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 80.1005%;height: 67px\">\n<tbody>\n<tr>\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]K=\\frac{[B]_{eq}\/c^\\theta}{[A]_{eq}\/c^\\theta}=\\frac{[B]_{eq}}{[A]_{eq}}=\\frac{k_{r}}{k'_{r}}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: left\">At equilibrium, the rates of forward and reverse reactions should be the same (meaning no change in overall concentrations), so it makes sense that when we rearrange this we get:<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 38.2083%;height: 91px\">\n<tbody>\n<tr>\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]k'_{r}[B]_{eq}=k_{r}[A]_{eq}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"prompt input_prompt\"><\/div>\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<p>&nbsp;<\/p>\n<p style=\"text-align: left\"><strong>Consider another system with the following reaction:<\/strong><\/p>\n<p style=\"text-align: center\">[latex]A+B\u21ccC[\/latex]<\/p>\n<blockquote><p>Forward reaction: [latex]A+B\u2192C[\/latex], where the forward reaction rate is expressed by: [latex]r=k_{r}[A][B][\/latex]<\/p>\n<p>Reverse reaction: [latex]C\u2192A+B[\/latex], where the reverse reaction rate is expressed by: [latex]r'=k'_{r}[C][\/latex]<\/p><\/blockquote>\n<p>Here, in equilibrium we have:<\/p>\n<p style=\"text-align: center\">[latex]r=r'=k_{r}[A]_{eq}[B]_{eq}=k'_{r}[C]_{eq}[\/latex]<\/p>\n<p>The equilibrium constant is defined as dimensionless, so here is where the [latex]c^\\theta[\/latex] terms make a difference. The equilibrium constant can be espressed as:<\/p>\n<p style=\"text-align: center\">[latex]K=\\frac{[C]_{eq}\/c^\\theta}{[A]_{eq}\/c^\\theta\u00d7[B]_{eq}\/c^\\theta}=\\Big(\\frac{[C]}{[A][B]}\\Big)_{eq}=\\frac{k_{r}}{k'_{r}}c^\\theta[\/latex]<\/p>\n<p>Recall that [latex]k_{r}[\/latex] and [latex]k'_{r}[\/latex] here will have different units, so the [latex]c^\\theta[\/latex] term ensures that these become dimensionless when one is divided by the other.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"prompt input_prompt\"><\/div>\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Equilibrium Constant<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>1. For the balanced chemical reaction below, write an equation for the equilibrium constant(K).[latex]^{[1]}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]2H_{2(g)}+N_{2(g)} \u21ccN_{2}H_{4(g)}[\/latex]<\/p>\n<p>2. Consider the following reaction system:<\/p>\n<p style=\"text-align: center\">[latex]A\u21ccB+C[\/latex]<\/p>\n<p>From initial rate and equilibrium experiments we know the following: <span style=\"font-size: 1em\"><span style=\"text-align: initial;font-size: 1em\">[latex][A]_{eq} = 0.5 mol\/L[\/latex]<\/span><\/span><span style=\"text-align: initial;font-size: 1em\">, [latex][B]_{eq} = 0.1 mol\/L[\/latex], [latex][C]_{eq} =0.2 mol\/L[\/latex], [latex]k_{r} = 0.1 min^{-1}[\/latex].[latex]^{[1]}[\/latex]<\/span><\/p>\n<p>What are the values of the equilibrium constant([latex]K[\/latex]) and reverse reaction constant([latex]k'_{r}[\/latex])?<\/p>\n<p>&nbsp;<\/p>\n<p>[latex]K\\; or\\; K_{eq}=\\prod_{i} a_{i,eq}^{vi}[\/latex], at equilibrium [latex]r=r'[\/latex]<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<div class=\"textbox\">\n<h3>Solution<\/h3>\n<p>1. [latex]K=\\frac{[N_{2}H_{4}]}{[H_{2}]^2[N_{2}]}[\/latex]<\/p>\n<p>2. Start with expressing K in forms of concentrations and [latex]k_{r}[\/latex]:<\/p>\n<p>\\begin{align*}<br \/>\nK &amp; =\\frac{[B]_{eq}\/c^\\theta*[C]_{eq}\/c^\\theta}{[A]_{eq}\/c^\\theta}\\\\<br \/>\n&amp; = (\\frac{[B][C]}{[A]})_{eq}\\frac{1}{c^\\theta}\\\\<br \/>\n&amp; = \\frac{k_{r}}{k&#8217;_{r}*c^\\theta}<br \/>\n\\end{align*}<\/p>\n<p>Calculate K from the expression with concentrations:<\/p>\n<p>\\begin{align*}<br \/>\nK &amp; = (\\frac{[B][C]}{[A]})_{eq}\\frac{1}{c^\\theta}\\\\<br \/>\n&amp; = (\\frac{0.1\\frac{mol}{L}*0.2\\frac{mol}{L}}{0.5\\frac{mol}{L}})_{eq}\\frac{1}{c^\\theta}\\\\<br \/>\n&amp; = 0.04<br \/>\n\\end{align*}<\/p>\n<p>Subsitude the K value into the expression with [latex]k_{r}[\/latex]:<\/p>\n<p>\\begin{align*}<br \/>\nK &amp; = \\frac{k_{r}}{k&#8217;_{r}*c^\\theta}\\\\<br \/>\nk&#8217;_{r}&amp; = \\frac{k_{r}}{K*c^\\theta}\\\\<br \/>\n&amp; = \\frac{0.1min^{-1}}{0.04*1\\frac{mol}{L}}\\\\<br \/>\n&amp; = 2.5 \\frac{L}{mol*min}<br \/>\n\\end{align*}<\/p>\n<p>Back substitute [latex]k_{r}[\/latex] into the rate equations to check this:<\/p>\n<p style=\"text-align: center\">[latex]r=k_{r}[A]_{eq}=0.1min^{-1}*0.5\\frac{mol}{L}=0.05\\frac{mol}{L*min}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]r'=k'_{r}[B]_{eq}[C]_{eq}=2.5 \\frac{L}{mol*min}*0.1\\frac{mol}{L}*0.2\\frac{mol}{L}=0.05\\frac{mol}{L*min}[\/latex]<\/p>\n<p>Forward and reverse reaction rates are the same at [latex]0.05\\frac{mol}{L*min}[\/latex] , as we would expect at equilibrium.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h2>References<\/h2>\n<p><span style=\"font-size: 1em\">[1] Chemistry LibreTexts. 2020. <\/span><i style=\"font-size: 1em\">Characteristics of the Equilibrium State.<\/i><span style=\"font-size: 1em\"> [online] Available at: &lt;<a href=\"https:\/\/chem.libretexts.org\/Bookshelves\/Physical_and_Theoretical_Chemistry_Textbook_Maps\/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)\/Equilibria\/Chemical_Equilibria\/Principles_of_Chemical_Equilibria\/Principles_of_Chemical_Equilibrium\/Characteristics_Of_The_Equilibrium_State\">https:\/\/chem.libretexts.org\/Bookshelves\/Physical_and_Theoretical_Chemistry_Textbook_Maps\/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)\/Equilibria\/Chemical_Equilibria\/Principles_of_Chemical_Equilibria\/Principles_of_Chemical_Equilibrium\/Characteristics_Of_The_Equilibrium_State<\/a>&gt; [Accessed 24 April, 2020].<\/span><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":948,"menu_order":5,"comment_status":"closed","ping_status":"closed","template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1338","chapter","type-chapter","status-publish","hentry"],"part":1286,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1338","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/users\/948"}],"replies":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/comments?post=1338"}],"version-history":[{"count":13,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1338\/revisions"}],"predecessor-version":[{"id":2646,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1338\/revisions\/2646"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/parts\/1286"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1338\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/media?parent=1338"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapter-type?post=1338"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/contributor?post=1338"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/license?post=1338"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}