{"id":930,"date":"2017-12-14T21:51:27","date_gmt":"2017-12-14T21:51:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/activation-energy-and-the-arrhenius-equation\/"},"modified":"2017-12-14T21:51:27","modified_gmt":"2017-12-14T21:51:27","slug":"activation-energy-and-the-arrhenius-equation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/chapter\/activation-energy-and-the-arrhenius-equation\/","title":{"raw":"Activation Energy and the Arrhenius Equation","rendered":"Activation Energy and the Arrhenius Equation"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul><li>To gain an understanding of activation energy.<\/li>\n\t<li>To determine activation energy graphically or algebraically.<\/li>\n<\/ul><\/div>\n<span class=\"Apple-style-span\">Figure 17.11. Svante Arrhenius<\/span>\n\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/05\/Arrhenius2.jpg\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14215124\/Arrhenius2-239x300-1.jpg\" alt=\"Figure 17.11. Svante Arrhenius\" class=\"alignnone wp-image-2428 size-medium\" height=\"300\" width=\"239\"\/><\/a>\n\n<span class=\"Apple-style-span\">Swedish scientist Svante Arrhenius<\/span>[footnote]Svante Arrhenius\/Public Domain[\/footnote]\n\nEarlier in the chapter, reactions were discussed in terms of effective collision frequency and molecule energy levels. In 1889, a Swedish scientist named Svante Arrhenius proposed an equation that\u00a0relates these concepts with the rate constant:\n\n$$\\textit{k } = \\textit{A}e^{-E_a\/RT}\\textit{}\\$$\n\nwhere <em>k<\/em> represents the rate constant, <em>E<\/em><sub>a<\/sub> is the activation energy, <em>R<\/em> is the gas constant (8.3145 J\/K mol), and <em>T<\/em> is the temperature expressed in Kelvin. <em>A<\/em> is known as the <b>frequency factor<\/b>, having units of L mol<sup>-1<\/sup> s<sup>-1<\/sup>, and takes into account the frequency of reactions and likelihood of correct molecular orientation.\n\nThe Arrhenius equation allows us to calculate activation energies if the rate constant is known, or vice versa. As well, it mathematically expresses the relationships we established earlier: as activation energy term <em>E<\/em><sub>a<\/sub> increases, the rate constant <i>k<\/i> decreases and therefore the rate of reaction decreases.\n<h2>Determining the Activation Energy<\/h2>\n<h3>Graphically<\/h3>\nWe can graphically determine the activation energy by manipulating the Arrhenius equation to put it into the form of a straight line. Taking the natural logarithm of both sides gives us:\n\nln$$\\textit{k} = -\\frac{E_a}{RT} + ln \\textit{A} \\$$\n\nA slight rearrangement of this equation then gives us a straight line plot (<em>y<\/em> = <em>mx<\/em> + <em>b<\/em>) for ln <i>k<\/i> versus 1\/<em>T<\/em>, where the slope is \u2013<em>E<\/em><sub>a<\/sub>\/<em>R<\/em>:\n\nln $$\\textit{k} = - \\frac{E_a}{R}\\left(\\frac{1}{t}\\right)\\ + ln \\textit{A}\\$$\n<div class=\"textbox shaded\">\n\n<strong>Example 7<\/strong>\n\nUsing the data from the following table, determine the activation energy of the reaction:\n<table border=\"1\" cellpadding=\"0\" style=\"border-spacing: 0px;\"><tbody><tr><td valign=\"top\" style=\"width: 168px;\"><strong>Temperature (K)<\/strong><\/td>\n<td valign=\"top\" style=\"width: 151px;\"><strong>Rate Constant,<i>\u00a0k<\/i> (s<sup>-1<\/sup>)<\/strong><\/td>\n<\/tr><tr><td valign=\"top\" style=\"width: 168px;\">375<\/td>\n<td valign=\"top\" style=\"width: 151px;\">1.68 x 10<sup>-5<\/sup><\/td>\n<\/tr><tr><td valign=\"top\" style=\"width: 168px;\">400<\/td>\n<td valign=\"top\" style=\"width: 151px;\">3.5 x 10<sup>-5<\/sup><\/td>\n<\/tr><tr><td valign=\"top\" style=\"width: 168px;\">500<\/td>\n<td valign=\"top\" style=\"width: 151px;\">4.2 x 10<sup>-4<\/sup><\/td>\n<\/tr><tr><td valign=\"top\" style=\"width: 168px;\">600<\/td>\n<td valign=\"top\" style=\"width: 151px;\">2.11 x 10<sup>-3<\/sup><\/td>\n<\/tr><\/tbody><\/table><address>\u00a0<\/address>Solution\n\nWe can obtain the activation energy by plotting ln <em>k<\/em> versus 1\/<em>T<\/em>, knowing that the slope will be equal to \u2013(<i>E<\/i><sub>a<\/sub>\/<em>R<\/em>).\n\nFirst determine the values of ln <em>k<\/em> and 1\/<em>T<\/em>, and plot them in a graph:\n<table style=\"width: 171px; border-spacing: 0px;\" border=\"1\" cellpadding=\"0\"><tbody><tr><td valign=\"bottom\" style=\"white-space: nowrap; width: 77px;\"><strong>1\/<em>T<\/em><\/strong><\/td>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 95px;\"><strong>ln <em>k<\/em><\/strong><\/td>\n<\/tr><tr><td valign=\"bottom\" style=\"white-space: nowrap; width: 77px;\">\n<p style=\"text-align: right;\">0.002667<\/p>\n<\/td>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 95px;\">\n<p style=\"text-align: right;\">-10.9941<\/p>\n<\/td>\n<\/tr><tr><td valign=\"bottom\" style=\"white-space: nowrap; width: 77px;\">\n<p style=\"text-align: right;\">0.0025<\/p>\n<\/td>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 95px;\">\n<p style=\"text-align: right;\">-10.2602<\/p>\n<\/td>\n<\/tr><tr><td valign=\"bottom\" style=\"white-space: nowrap; width: 77px;\">\n<p style=\"text-align: right;\">0.002<\/p>\n<\/td>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 95px;\">\n<p style=\"text-align: right;\">-7.77526<\/p>\n<\/td>\n<\/tr><tr><td valign=\"bottom\" style=\"white-space: nowrap; width: 77px;\">\n<p style=\"text-align: right;\">0.001667<\/p>\n<\/td>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 95px;\">\n<p style=\"text-align: right;\">-6.16107<\/p>\n<\/td>\n<\/tr><\/tbody><\/table>\n[caption id=\"attachment_1268\" align=\"alignnone\" width=\"481\"]<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/05\/17.5-graphical-determination-of-Ea-example-plot.jpg\"><img class=\"wp-image-1268 size-full\" alt=\"Graphical determination of Ea example plot\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14215126\/17.5-graphical-determination-of-Ea-example-plot-1.jpg\" height=\"289\" width=\"481\"\/><\/a> Graphical determination of <em>E<\/em><sub>a <\/sub>example plot[\/caption]\n\n\u00a0\n\nSlope = - $$\\frac{E_a}{R}\\$$\n\n-4865 K = - $$\\frac{E_a}{8.3145\\ J\\ K^{-1}{mol}^{-1}}\\$$\n\n<i>E<\/i><sub>a <\/sub>= 4.0 x 10<sup>4<\/sup> J\/mol\n\n<\/div>\n\u00a0\n<h3>Algebraically<\/h3>\nThe activation energy can also be calculated algebraically if <i>k <\/i>is known at two different temperatures:\n\nAt temperature 1: ln $$\\textit{k}_{1}\\$$= -$$\\frac{E_a}{RT_1} + ln \\textit{A} \\$$\n\nAt temperature 2: ln $$\\textit{k}_{2}\\$$ = - $$\\frac{E_a}{RT_2} + ln \\textit{A} \\$$\n\nWe can subtract one of these equations from the other:\n\nln $$\\textit{k}_{1} - ln \\textit{k}_{2}\\$$ = $$\\left({\\rm -}{\\rm \\ }\\frac{E_a}{RT_1}{\\rm \\ +\\ ln\\ }A{\\rm \\ }\\right) - \\left({\\rm -}{\\rm \\ }\\frac{E_a}{RT_2}{\\rm \\ +\\ ln\\ }A\\right)\\$$\n\nThis equation can then be further simplified to:\n\nln $$\\frac{k_1}{k_2}\\$$ = $$\\frac{E_a}{R}\\left({\\rm \\ }\\frac{1}{T_2}-\\frac{1}{T_1}{\\rm \\ }\\right)\\$$\n<div class=\"textbox shaded\">\n\n<strong>Example 8<\/strong>\n\nDetermine the value of <i>E<\/i><sub>a<\/sub> given the following values of<i> k<\/i> at the temperatures indicated:\n\n600 K: <i>k<\/i> = 2.75 x 10<sup>-8<\/sup> L mol<sup>-1<\/sup> s<sup>-1<\/sup>\n\n800 K: <i>k<\/i> = 1.95 x 10<sup>-7<\/sup> L mol<sup>-1<\/sup> s<sup>-1<\/sup>\n\nSolution\n\nSubstitute the values stated into the algebraic method equation:\n\nln $$\\frac{k_1}{k_2}\\$$ = $$\\frac{E_a}{R}\\left({\\rm \\ }\\frac{1}{T_2}-\\frac{1}{T_1}{\\rm \\ }\\right)\\$$\n\nln $$\\frac{{{\\rm 2.75\\ x\\ 10}}^{{\\rm -}{\\rm 8}{\\rm \\ }}{\\rm L\\ }{{\\rm mol}}^{{\\rm -}{\\rm 1}}{\\rm \\ }{{\\rm s}}^{{\\rm -}{\\rm 1}}}{{{\\rm 1.95\\ x\\ 10}}^{{\\rm -}{\\rm 7}}{\\rm \\ L}{{\\rm \\ mol}}^{{\\rm -}{\\rm 1}}{\\rm \\ }{{\\rm s}}^{{\\rm -}{\\rm 1}}}\\$$ = $$\\frac{E_a}{8.3145\\ J\\ K^{-1}{mol}^{-1}}\\left({\\rm \\ }\\frac{1}{{\\rm 800\\ K}}-\\frac{1}{{\\rm 600\\ K}}{\\rm \\ }\\right)\\$$\n\n$$\\-1.96\\$$ = $$\\frac{E_a}{8.3145\\ J\\ K^{-1}{mol}^{-1}}\\left({\\rm -}{\\rm 4.16\\ x}{10}^{-4}{\\rm \\ }{{\\rm K}}^{{\\rm -}{\\rm 1\\ }}\\right)\\$$\n\n$$\\ 4.704\\ x\\ 10{}^{-3}{}^{ }{{\\rm K}}^{{\\rm -}{\\rm 1\\ }} \\$$= $$\\frac{E_a}{8.3145\\ J\\ K^{-1}{mol}^{-1}}\\$$\n\n<i>E<\/i><sub>a<\/sub><i> =\u00a0<\/i>3.92 x 10<sup>4<\/sup> J\/mol\n\n<\/div>\n\u00a0\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul><li>The activation energy can be graphically determined by manipulating the Arrhenius equation.<\/li>\n\t<li>The activation energy can also be calculated algebraically if\u00a0<i>k\u00a0<\/i>is known at two different temperatures.<\/li>\n<\/ul><\/div>\n\u00a0","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>To gain an understanding of activation energy.<\/li>\n<li>To determine activation energy graphically or algebraically.<\/li>\n<\/ul>\n<\/div>\n<p><span class=\"Apple-style-span\">Figure 17.11. Svante Arrhenius<\/span><\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/05\/Arrhenius2.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14215124\/Arrhenius2-239x300-1.jpg\" alt=\"Figure 17.11. Svante Arrhenius\" class=\"alignnone wp-image-2428 size-medium\" height=\"300\" width=\"239\" \/><\/a><\/p>\n<p><span class=\"Apple-style-span\">Swedish scientist Svante Arrhenius<\/span><a class=\"footnote\" title=\"Svante Arrhenius\/Public Domain\" id=\"return-footnote-930-1\" href=\"#footnote-930-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a><\/p>\n<p>Earlier in the chapter, reactions were discussed in terms of effective collision frequency and molecule energy levels. In 1889, a Swedish scientist named Svante Arrhenius proposed an equation that\u00a0relates these concepts with the rate constant:<\/p>\n<p>$$\\textit{k } = \\textit{A}e^{-E_a\/RT}\\textit{}\\$$<\/p>\n<p>where <em>k<\/em> represents the rate constant, <em>E<\/em><sub>a<\/sub> is the activation energy, <em>R<\/em> is the gas constant (8.3145 J\/K mol), and <em>T<\/em> is the temperature expressed in Kelvin. <em>A<\/em> is known as the <b>frequency factor<\/b>, having units of L mol<sup>-1<\/sup> s<sup>-1<\/sup>, and takes into account the frequency of reactions and likelihood of correct molecular orientation.<\/p>\n<p>The Arrhenius equation allows us to calculate activation energies if the rate constant is known, or vice versa. As well, it mathematically expresses the relationships we established earlier: as activation energy term <em>E<\/em><sub>a<\/sub> increases, the rate constant <i>k<\/i> decreases and therefore the rate of reaction decreases.<\/p>\n<h2>Determining the Activation Energy<\/h2>\n<h3>Graphically<\/h3>\n<p>We can graphically determine the activation energy by manipulating the Arrhenius equation to put it into the form of a straight line. Taking the natural logarithm of both sides gives us:<\/p>\n<p>ln$$\\textit{k} = -\\frac{E_a}{RT} + ln \\textit{A} \\$$<\/p>\n<p>A slight rearrangement of this equation then gives us a straight line plot (<em>y<\/em> = <em>mx<\/em> + <em>b<\/em>) for ln <i>k<\/i> versus 1\/<em>T<\/em>, where the slope is \u2013<em>E<\/em><sub>a<\/sub>\/<em>R<\/em>:<\/p>\n<p>ln $$\\textit{k} = &#8211; \\frac{E_a}{R}\\left(\\frac{1}{t}\\right)\\ + ln \\textit{A}\\$$<\/p>\n<div class=\"textbox shaded\">\n<p><strong>Example 7<\/strong><\/p>\n<p>Using the data from the following table, determine the activation energy of the reaction:<\/p>\n<table cellpadding=\"0\" style=\"border-spacing: 0px;\">\n<tbody>\n<tr>\n<td valign=\"top\" style=\"width: 168px;\"><strong>Temperature (K)<\/strong><\/td>\n<td valign=\"top\" style=\"width: 151px;\"><strong>Rate Constant,<i>\u00a0k<\/i> (s<sup>-1<\/sup>)<\/strong><\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" style=\"width: 168px;\">375<\/td>\n<td valign=\"top\" style=\"width: 151px;\">1.68 x 10<sup>-5<\/sup><\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" style=\"width: 168px;\">400<\/td>\n<td valign=\"top\" style=\"width: 151px;\">3.5 x 10<sup>-5<\/sup><\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" style=\"width: 168px;\">500<\/td>\n<td valign=\"top\" style=\"width: 151px;\">4.2 x 10<sup>-4<\/sup><\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" style=\"width: 168px;\">600<\/td>\n<td valign=\"top\" style=\"width: 151px;\">2.11 x 10<sup>-3<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<address>\u00a0<\/address>\n<p>Solution<\/p>\n<p>We can obtain the activation energy by plotting ln <em>k<\/em> versus 1\/<em>T<\/em>, knowing that the slope will be equal to \u2013(<i>E<\/i><sub>a<\/sub>\/<em>R<\/em>).<\/p>\n<p>First determine the values of ln <em>k<\/em> and 1\/<em>T<\/em>, and plot them in a graph:<\/p>\n<table style=\"width: 171px; border-spacing: 0px;\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 77px;\"><strong>1\/<em>T<\/em><\/strong><\/td>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 95px;\"><strong>ln <em>k<\/em><\/strong><\/td>\n<\/tr>\n<tr>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 77px;\">\n<p style=\"text-align: right;\">0.002667<\/p>\n<\/td>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 95px;\">\n<p style=\"text-align: right;\">-10.9941<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 77px;\">\n<p style=\"text-align: right;\">0.0025<\/p>\n<\/td>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 95px;\">\n<p style=\"text-align: right;\">-10.2602<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 77px;\">\n<p style=\"text-align: right;\">0.002<\/p>\n<\/td>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 95px;\">\n<p style=\"text-align: right;\">-7.77526<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 77px;\">\n<p style=\"text-align: right;\">0.001667<\/p>\n<\/td>\n<td valign=\"bottom\" style=\"white-space: nowrap; width: 95px;\">\n<p style=\"text-align: right;\">-6.16107<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"attachment_1268\" style=\"width: 491px\" class=\"wp-caption alignnone\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/05\/17.5-graphical-determination-of-Ea-example-plot.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1268\" class=\"wp-image-1268 size-full\" alt=\"Graphical determination of Ea example plot\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14215126\/17.5-graphical-determination-of-Ea-example-plot-1.jpg\" height=\"289\" width=\"481\" \/><\/a><\/p>\n<p id=\"caption-attachment-1268\" class=\"wp-caption-text\">Graphical determination of <em>E<\/em><sub>a <\/sub>example plot<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<p>Slope = &#8211; $$\\frac{E_a}{R}\\$$<\/p>\n<p>-4865 K = &#8211; $$\\frac{E_a}{8.3145\\ J\\ K^{-1}{mol}^{-1}}\\$$<\/p>\n<p><i>E<\/i><sub>a <\/sub>= 4.0 x 10<sup>4<\/sup> J\/mol<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<h3>Algebraically<\/h3>\n<p>The activation energy can also be calculated algebraically if <i>k <\/i>is known at two different temperatures:<\/p>\n<p>At temperature 1: ln $$\\textit{k}_{1}\\$$= -$$\\frac{E_a}{RT_1} + ln \\textit{A} \\$$<\/p>\n<p>At temperature 2: ln $$\\textit{k}_{2}\\$$ = &#8211; $$\\frac{E_a}{RT_2} + ln \\textit{A} \\$$<\/p>\n<p>We can subtract one of these equations from the other:<\/p>\n<p>ln $$\\textit{k}_{1} &#8211; ln \\textit{k}_{2}\\$$ = $$\\left({\\rm -}{\\rm \\ }\\frac{E_a}{RT_1}{\\rm \\ +\\ ln\\ }A{\\rm \\ }\\right) &#8211; \\left({\\rm -}{\\rm \\ }\\frac{E_a}{RT_2}{\\rm \\ +\\ ln\\ }A\\right)\\$$<\/p>\n<p>This equation can then be further simplified to:<\/p>\n<p>ln $$\\frac{k_1}{k_2}\\$$ = $$\\frac{E_a}{R}\\left({\\rm \\ }\\frac{1}{T_2}-\\frac{1}{T_1}{\\rm \\ }\\right)\\$$<\/p>\n<div class=\"textbox shaded\">\n<p><strong>Example 8<\/strong><\/p>\n<p>Determine the value of <i>E<\/i><sub>a<\/sub> given the following values of<i> k<\/i> at the temperatures indicated:<\/p>\n<p>600 K: <i>k<\/i> = 2.75 x 10<sup>-8<\/sup> L mol<sup>-1<\/sup> s<sup>-1<\/sup><\/p>\n<p>800 K: <i>k<\/i> = 1.95 x 10<sup>-7<\/sup> L mol<sup>-1<\/sup> s<sup>-1<\/sup><\/p>\n<p>Solution<\/p>\n<p>Substitute the values stated into the algebraic method equation:<\/p>\n<p>ln $$\\frac{k_1}{k_2}\\$$ = $$\\frac{E_a}{R}\\left({\\rm \\ }\\frac{1}{T_2}-\\frac{1}{T_1}{\\rm \\ }\\right)\\$$<\/p>\n<p>ln $$\\frac{{{\\rm 2.75\\ x\\ 10}}^{{\\rm -}{\\rm 8}{\\rm \\ }}{\\rm L\\ }{{\\rm mol}}^{{\\rm -}{\\rm 1}}{\\rm \\ }{{\\rm s}}^{{\\rm -}{\\rm 1}}}{{{\\rm 1.95\\ x\\ 10}}^{{\\rm -}{\\rm 7}}{\\rm \\ L}{{\\rm \\ mol}}^{{\\rm -}{\\rm 1}}{\\rm \\ }{{\\rm s}}^{{\\rm -}{\\rm 1}}}\\$$ = $$\\frac{E_a}{8.3145\\ J\\ K^{-1}{mol}^{-1}}\\left({\\rm \\ }\\frac{1}{{\\rm 800\\ K}}-\\frac{1}{{\\rm 600\\ K}}{\\rm \\ }\\right)\\$$<\/p>\n<p>$$\\-1.96\\$$ = $$\\frac{E_a}{8.3145\\ J\\ K^{-1}{mol}^{-1}}\\left({\\rm -}{\\rm 4.16\\ x}{10}^{-4}{\\rm \\ }{{\\rm K}}^{{\\rm -}{\\rm 1\\ }}\\right)\\$$<\/p>\n<p>$$\\ 4.704\\ x\\ 10{}^{-3}{}^{ }{{\\rm K}}^{{\\rm -}{\\rm 1\\ }} \\$$= $$\\frac{E_a}{8.3145\\ J\\ K^{-1}{mol}^{-1}}\\$$<\/p>\n<p><i>E<\/i><sub>a<\/sub><i> =\u00a0<\/i>3.92 x 10<sup>4<\/sup> J\/mol<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul>\n<li>The activation energy can be graphically determined by manipulating the Arrhenius equation.<\/li>\n<li>The activation energy can also be calculated algebraically if\u00a0<i>k\u00a0<\/i>is known at two different temperatures.<\/li>\n<\/ul>\n<\/div>\n<p>\u00a0<\/p>\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-930\">\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>Introductory Chemistry- 1st Canadian Edition . <strong>Authored by<\/strong>: Jessie A. Key and David W. Ball. <strong>Provided by<\/strong>: BCCampus. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/opentextbc.ca\/introductorychemistry\/\">https:\/\/opentextbc.ca\/introductorychemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download this book for free at http:\/\/open.bccampus.ca<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-930-1\">Svante Arrhenius\/Public Domain <a href=\"#return-footnote-930-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":23485,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Introductory Chemistry- 1st Canadian Edition \",\"author\":\"Jessie A. Key and David W. 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