{"id":12343,"date":"2026-09-30T19:38:45","date_gmt":"2026-09-30T18:38:45","guid":{"rendered":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/activation-energy-and-the-arrhenius-equation\/"},"modified":"2026-10-04T23:04:21","modified_gmt":"2026-10-04T22:04:21","slug":"activation-energy-and-the-arrhenius-equation","status":"publish","type":"page","link":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/activation-energy-and-the-arrhenius-equation\/","title":{"rendered":"Activation Energy and the Arrhenius Equation"},"content":{"rendered":"\n<section class=\"ols-revision-page ols-kinetics-page\">\n  <style>\n    .ols-revision-page {\n      --navy: #1C244B;\n      --blue: #2563eb;\n      --soft-blue: #eef4ff;\n      --soft-red: #fff7f7;\n      --soft-purple: #f7f0ff;\n      --soft-green: #f0f7f1;\n      --gold: #c9973a;\n      --grey-text: 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var(--inner-shadow);\n    }\n.ols-faq-item h3 {\n      margin: 0 0 8px;\n      font-size: 20px;\n      line-height: 1.3;\n      color: var(--navy);\n    }\n.ols-faq-item p {\n      margin: 0;\n      font-size: 16px;\n      line-height: 1.65;\n      color: var(--body-text);\n    }\n.ols-faq-card,\n      .ols-quicksnap-card,\n      .ols-attribution-card {\n        padding: 24px 18px;\n        border-radius: 22px;\n      }\n  <\/style>\n\n    <aside class=\"ols-sidebar\">\n  <div class=\"ols-sidebar-header\">\n    <h3>Revision Notes<\/h3>\n    <p>A Level Chemistry<\/p>\n  <\/div>\n\n  <div class=\"ols-topic-group\">\n    <h4>Topic 11 Kinetics<\/h4>\n\n    <ul class=\"ols-topic-list\">\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/\">Topic 11 Kinetics Overview<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-equations-orders-and-the-rate-constant\/\">Rate Equations, Orders and the Rate Constant<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/techniques-for-measuring-rates\/\">Techniques for Measuring Rates<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/concentration-time-graphs-and-half-life\/\">Concentration\u2013Time Graphs and Half-Life<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-concentration-graphs-and-the-initial-rates-method\/\">Rate\u2013Concentration Graphs and the Initial-Rates Method<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/the-iodine-propanone-reaction\/\">The Iodine\u2013Propanone Reaction<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-determining-step-and-reaction-mechanisms\/\">Rate-Determining Step and Reaction Mechanisms<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/activation-energy-and-the-arrhenius-equation\/\">Activation Energy and the Arrhenius Equation<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/heterogeneous-and-homogeneous-catalysis\/\">Heterogeneous and Homogeneous Catalysis<\/a>\n      <\/li>\n    <\/ul>\n  <\/div>\n\n  <div class=\"ols-topic-group\">\n    <h4>Other Sections<\/h4>\n\n    <ul class=\"ols-topic-list\">\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-9-kinetics-and-equilibria\/9a-kinetics\/\">Topic 9A Kinetics (AS)<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-10-alcohols-halogenoalkanes-and-spectra\/\">Topic 10 Alcohols, Halogenoalkanes &amp; Spectra<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-6-energetics\/\">Topic 6 Energetics<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/core-practicals\/\">Core Practicals<\/a>\n      <\/li>\n    <\/ul>\n  <\/div>\n<\/aside>\n\n<script>\r\n(function() {\r\n  function normalisePath(path) {\r\n    return String(path || '')\r\n      .split('?')[0]\r\n      .split('#')[0]\r\n      .replace(\/\\\/+$\/, '')\r\n      .toLowerCase();\r\n  }\r\n\r\n  function highlightActive() {\r\n    var sidebar = document.querySelector('.ols-sidebar');\r\n    if (!sidebar) return false;\r\n\r\n    var currentPath = normalisePath(window.location.pathname);\r\n    var links = sidebar.querySelectorAll('a[href]');\r\n    var matched = null;\r\n    var matchedLength = 0;\r\n\r\n    sidebar.querySelectorAll('.active, .active-main, .parent-active').forEach(function(item) {\r\n      item.classList.remove('active', 'active-main', 'parent-active');\r\n    });\r\n\r\n    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'1');\r\n        }\r\n      }\r\n    }\r\n\r\n    return true;\r\n  }\r\n\r\n  if (!highlightActive()) {\r\n    document.addEventListener('DOMContentLoaded', highlightActive);\r\n  }\r\n})();\r\n<\/script>\r\n\n\n    <main class=\"ols-main\">\n\n      <!-- BREADCRUMBS - updated with full path and correct links -->\n      <nav class=\"ols-breadcrumbs\" aria-label=\"Breadcrumb\">\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/\">Revision Notes<\/a> \/\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/\">A Level Chemistry<\/a> \/\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/\">Edexcel International<\/a> \/\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/\">Topic 11 Kinetics<\/a> \/\n<span>Activation Energy and the Arrhenius Equation<\/span>\n<\/nav>\n\n      <header class=\"ols-title-card\">\n        <h1>Activation Energy and the Arrhenius Equation<\/h1>\n        <p class=\"ols-page-intro\">A concise revision guide to how temperature changes the rate constant: the Arrhenius equation k = Ae^(\u2212E\u2090\/RT), the straight-line form ln k = \u2212E\u2090\/RT + ln A, finding the activation energy from the gradient of a graph of ln k against 1\/T, and calculations with the equation.<\/p>\n        <div class=\"ols-badges\">\n<div class=\"ols-badge\">Exam board: Edexcel International<\/div>\n<div class=\"ols-badge\">Unit 4: WCH14\/01<\/div>\n<div class=\"ols-badge\">Topic 11: Kinetics<\/div>\n<\/div>\n        <div class=\"ols-author\">\n\n    <img decoding=\"async\"\n      class=\"ols-author-avatar-img\"\n      src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Author-Profile.jpeg\"\n      alt=\"Dr. Mohammed Al-Fatah\"\n    >\n\n    <div class=\"ols-author-content\">\n\n      <h2 class=\"ols-author-title\">\n        Written by: Dr. Mohammed Al-Fatah\n      <\/h2>\n\n      <p class=\"ols-author-description\">\n        Chemistry specialist revision notes for A Level Chemistry.\n      <\/p>\n\n      <a class=\"ols-linkedin-pill\" href=\"https:\/\/www.linkedin.com\/in\/doctormohammedfatah\/\" target=\"_blank\" rel=\"noopener noreferrer\">\n        <svg class=\"ols-linkedin-icon\" viewBox=\"0 0 24 24\" fill=\"currentColor\" aria-hidden=\"true\">\n          <path d=\"M4.98 3.5C4.98 4.88 3.86 6 2.48 6S0 4.88 0 3.5 1.12 1 2.48 1s2.5 1.12 2.5 2.5zM.5 8h4V24h-4V8zm7 0h3.8v2.2h.1c.5-.9 1.8-2.2 3.9-2.2 4.2 0 5 2.8 5 6.4V24h-4v-7.6c0-1.8 0-4.2-2.6-4.2s-3 2-3 4v7.8h-4V8z\"\/>\n        <\/svg>\n        View LinkedIn Profile\n      <\/a>\n\n    <\/div>\n\n  <\/div>\n      <\/header>\n\n      <article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">1<\/div>\n<h2>Temperature and the Rate Constant<\/h2>\n<\/div>\n<p>Concentration changes affect the rate through the rate equation, but they leave k alone. Temperature is different: the <strong>rate constant itself increases with temperature<\/strong>, and for many reactions a rise of about 10 \u00b0C roughly doubles it.<\/p><p>Collision theory explains why. Warmer particles collide only slightly more often, but the <strong>fraction of collisions with energy of at least the activation energy<\/strong>, E\u2090, rises steeply.<\/p><p>This is because the high-energy tail of the Boltzmann distribution grows much faster than the average. The rate constant is the measure of how likely a collision is to succeed, so it grows with that fraction.<\/p>\n<p>The same idea explains a catalyst in terms of k: by providing a route with a lower E\u2090, the catalyst increases the fraction of collisions that can react at a given temperature, so k for the catalysed reaction is larger.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Key idea:<\/strong> Concentration changes the rate; temperature (and a catalyst) changes the rate constant. A higher temperature means a larger k because a larger fraction of collisions have E \u2265 E\u2090.<\/p>\n<\/div>\n<\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Check: Temperature and k<\/h2>\n<p>Qualitative questions on why k rises with temperature and what happens to k, and to the rate, when conditions change in reactions not mentioned on this page.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1093\"><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card soft\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">2<\/div>\n<h2>The Arrhenius Equation<\/h2>\n<\/div>\n<p>The relationship between k and temperature is the <strong>Arrhenius equation<\/strong>:<\/p>\n<p style=\"text-align:center\"><strong>k = Ae^(\u2212E\u2090\/RT)<\/strong><\/p>\n<div class=\"ols-table-wrap\"><table class=\"ols-table\"><thead><tr><th>Symbol<\/th><th>Meaning<\/th><\/tr><\/thead><tbody><tr><td><strong>A<\/strong><\/td><td>The <strong>pre-exponential factor<\/strong> (also called the Arrhenius constant or frequency factor), a constant for the reaction that has the <strong>same units as k<\/strong><\/td><\/tr><tr><td><strong>E\u2090<\/strong><\/td><td>The activation energy in J mol\u207b\u00b9<\/td><\/tr><tr><td><strong>R<\/strong><\/td><td>The gas constant, 8.31 J K\u207b\u00b9 mol\u207b\u00b9<\/td><\/tr><tr><td><strong>T<\/strong><\/td><td>The temperature in <strong>kelvin<\/strong><\/td><\/tr><\/tbody><\/table><\/div><p>Here the exponential term, e^(\u2212E\u2090\/RT), is the <strong>fraction of collisions with energy of at least E\u2090<\/strong>; it is always between 0 and 1 and it rises towards 1 as T rises.<\/p><p>A represents the rate constant the reaction would have if every collision were successful. It includes the collision frequency and the fraction of collisions with the right orientation.<\/p>\n<p>The equation shows the two things that make a reaction fast: a large A (frequent, well-oriented collisions) and a small E\u2090.<\/p><p>It also shows why the effect of temperature is so large: because E\u2090\/RT sits in an exponent, a small change in T makes a big change in e^(\u2212E\u2090\/RT).<\/p><div class=\"ols-key-box\"><p><strong>Remember:<\/strong> the equation is given on the data sheet if it is needed; what has to be known is what each symbol means and how to use it.<\/p><\/div>\n<div class=\"ols-key-box\">\n<p><strong>Definition:<\/strong> k = Ae^(\u2212E\u2090\/RT): A, the pre-exponential factor with the units of k; E\u2090, the activation energy in J mol\u207b\u00b9; R = 8.31 J K\u207b\u00b9 mol\u207b\u00b9; T in K.<\/p>\n<\/div>\n<div class=\"ols-figure-card ols-zoom-pop\">\n<div class=\"ols-figure-image\">\n<a class=\"ols-image-fullscreen-link\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/t11x-arrsymbols.jpg\" target=\"_blank\" rel=\"noopener\" aria-label=\"Open image fullscreen\">\n<img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/t11x-arrsymbols.jpg\" alt=\"Labelled Arrhenius equation with the meaning and unit of each symbol and a table of the energy fraction at three temperatures.\" data-fullscreen-src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/t11x-arrsymbols.jpg\">\n<\/a>\n<\/div>\n<div class=\"ols-figure-caption\"><p>Each symbol in k = Ae^(\u2212E\u2090\/RT) has a fixed meaning and unit, and the exponential term is the fraction of collisions with enough energy, which roughly doubles for a 10 K rise.<\/p><\/div>\n<\/div>\n<\/article>\n<article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">3<\/div>\n<h2>The Straight-Line Form<\/h2>\n<\/div>\n<p>Taking natural logarithms of both sides turns the exponential into something that can be plotted as a straight line:<\/p>\n<p style=\"text-align:center\"><strong>ln k = \u2212E\u2090\/R \u00d7 1\/T + ln A<\/strong><\/p>\n<p>Compare this with y = mx + c. If <strong>ln k is plotted on the y-axis against 1\/T on the x-axis<\/strong>, the points lie on a straight line with <strong>gradient \u2212E\u2090\/R<\/strong> and <strong>intercept ln A<\/strong>.<\/p><p>The gradient is negative because k falls as 1\/T rises (that is, as T falls).<\/p><p>Multiplying the gradient by \u2212R gives E\u2090; taking the exponential of the intercept gives A.<\/p>\n<p>In a real experiment k is rarely measured directly. Instead a reaction is timed at several temperatures with the same starting concentrations.<\/p><p>At fixed concentrations the rate is proportional to k, and in a clock reaction the rate is proportional to 1\/t, so <strong>ln(rate) or ln(1\/t) can be plotted instead of ln k<\/strong>.<\/p><p>The intercept changes, because the constant of proportionality is absorbed into it, but the gradient, and therefore E\u2090, is the same.<\/p>\n<div class=\"ols-figure-card ols-zoom-pop\">\n<div class=\"ols-figure-image\">\n<a class=\"ols-image-fullscreen-link\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/kinetics-t11-07-arrhenius-graph.jpg\" target=\"_blank\" rel=\"noopener\" aria-label=\"Open image fullscreen\">\n<img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/kinetics-t11-07-arrhenius-graph.jpg\" alt=\"Arrhenius plot of ln k against 1\/T with a gradient of \u22128300 K, giving an activation energy of 69 kJ mol\u207b\u00b9\" data-fullscreen-src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/kinetics-t11-07-arrhenius-graph.jpg\">\n<\/a>\n<\/div>\n<div class=\"ols-figure-caption\"><p>The Arrhenius plot: ln k against 1\/T is a straight line of gradient \u2212E\u2090\/R, with the data and the working for E\u2090.<\/p><\/div>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Exam wording:<\/strong> &#8220;A graph of ln k against 1\/T is a straight line with gradient \u2212E\u2090\/R and intercept ln A, so E\u2090 = \u2212gradient \u00d7 R.&#8221;<\/p>\n<\/div>\n<\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Check: The Arrhenius Graph<\/h2>\n<p>What is plotted on each axis, what the gradient and the intercept give, and why ln(1\/t) can stand in for ln k, for data sets not shown on this page.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1094\"><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card soft\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">4<\/div>\n<h2>Finding E\u2090 From a Graph<\/h2>\n<\/div>\n<p>The figure above gives k at four temperatures. The steps are always the same:<\/p>\n<ol><li>Convert each temperature to kelvin and calculate <strong>1\/T<\/strong> (typically 2.5 \u00d7 10\u207b\u00b3 to 3.5 \u00d7 10\u207b\u00b3 K\u207b\u00b9).<\/li><li>Calculate <strong>ln k<\/strong> (or ln(1\/t)) for each run; the values are negative for small k.<\/li><li>Plot ln k against 1\/T and draw the <strong>best-fit straight line<\/strong>.<\/li><li>Measure the gradient using a <strong>large triangle<\/strong> with points far apart on the line, not two of the data points.<\/li><li>E\u2090 = \u2212gradient \u00d7 R, in J mol\u207b\u00b9; divide by 1000 for kJ mol\u207b\u00b9.<\/li><\/ol>\n<p><strong>Step 1:<\/strong> From the plotted data the gradient is \u22128300 K (the unit of the gradient is K, because ln k has no unit and 1\/T is in K\u207b\u00b9).<\/p><p><strong>Step 2:<\/strong> So E\u2090 = 8300 \u00d7 8.31 = 69 000 J mol\u207b\u00b9.<\/p><p><strong>Answer:<\/strong> E\u2090 = <strong>69 kJ mol\u207b\u00b9<\/strong>, which is a typical value for a reaction that is conveniently slow at room temperature.<\/p><p>If the intercept is required, extending the line to 1\/T = 0 gives ln A, and A = e^(ln A).<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Exam focus:<\/strong> Three marks usually hide here: the gradient with its sign, multiplying by R (not dividing), and converting J to kJ. Quote E\u2090 as a positive number with its unit.<\/p>\n<\/div>\n<\/article>\n<article class=\"ols-note-card soft\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">5<\/div>\n<h2>Calculations With the Equation<\/h2>\n<\/div>\n<p>Questions also use the equation directly, in either form. The working is a matter of substituting carefully: T in kelvin, R = 8.31 J K\u207b\u00b9 mol\u207b\u00b9, and E\u2090 in <strong>joules<\/strong> per mole (not kilojoules) so that E\u2090\/RT is a pure number.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Worked example:<\/strong> Find A. A first-order reaction has k = 2.5 \u00d7 10\u207b\u00b3 s\u207b\u00b9 at 340 K and E\u2090 = 69 kJ mol\u207b\u00b9. E\u2090\/RT = 69 000 \u00f7 (8.31 \u00d7 340) = 24.4. Rearrange k = Ae^(\u2212E\u2090\/RT) to A = k \u00f7 e^(\u221224.4) = k \u00d7 e^(24.4) = 2.5 \u00d7 10\u207b\u00b3 \u00d7 3.9 \u00d7 10\u00b9\u2070 = 1.0 \u00d7 10\u2078 s\u207b\u00b9. A has the same units as k.<\/p>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Worked example:<\/strong> Find E\u2090. For a reaction with ln A = 18.4, k = 2.5 \u00d7 10\u207b\u00b3 s\u207b\u00b9 at 340 K. ln k = \u22125.99. From ln k = \u2212E\u2090\/RT + ln A: \u22125.99 = \u2212E\u2090\/(8.31 \u00d7 340) + 18.4, so E\u2090\/(2825) = 24.4 and E\u2090 = 24.4 \u00d7 2825 = 69 000 J mol\u207b\u00b9 = 69 kJ mol\u207b\u00b9.<\/p>\n<\/div>\n<p>Two checks catch most slips. E\u2090 must come out positive and of the order of tens to a few hundred kJ mol\u207b\u00b9.<\/p>\n<p>And because k rises with temperature, if you calculate k at two temperatures the higher temperature must give the larger value. If it does not, a sign or a unit has gone wrong.<\/p>\n<p><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/core-practicals\/cp-10-activation-energy-of-a-reaction\/\">Core Practical 10 (activation energy of a reaction)<\/a> is the practical version of this page: a clock reaction is timed at five or six temperatures, ln(1\/t) is plotted against 1\/T and E\u2090 comes from the gradient.<\/p>\n<\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Check: Gradient, Intercept and Units<\/h2>\n<p>Read a gradient and an intercept from Arrhenius data not used above, get the sign of E\u2090 right and give it in the correct unit.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1095\"><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">6<\/div>\n<h2>Common Exam Points<\/h2>\n<\/div>\n<h3>Say<\/h3><p>&#8220;The rate constant increases with temperature because a greater proportion of collisions have energy \u2265 E\u2090.&#8221; &#8220;Plot ln k against 1\/T; gradient = \u2212E\u2090\/R; E\u2090 = \u2212gradient \u00d7 R.&#8221; &#8220;T must be in kelvin.&#8221;<\/p>\n<h3>Do not say<\/h3><p>&#8220;The activation energy decreases when the temperature rises&#8221; (E\u2090 is fixed; the fraction of collisions that reach it rises). &#8220;Concentration changes k.&#8221; &#8220;E\u2090 = gradient&#8221; without the \u2212R.<\/p>\n<h3>Watch for<\/h3><p>Data given in \u00b0C, rates given as times, and E\u2090 asked for in kJ mol\u207b\u00b9 when R is in J: convert every time. The gradient of an ln k against 1\/T graph is always negative; a positive E\u2090 follows from the minus sign in \u2212E\u2090\/R.<\/p>\n<\/article>\n<section class=\"ols-faq-card\">\n<h2>FAQs<\/h2>\n<p>Use these quick answers to check the Arrhenius ideas.<\/p>\n\n<div class=\"ols-faq-list\">\n<div class=\"ols-faq-item\">\n<h3>Why do we plot ln k against 1\/T and not k against T?<\/h3>\n<p>Because k against T is a curve, from which you cannot read E\u2090. Taking natural logarithms gives ln k = \u2212E\u2090\/RT + ln A, which has the form y = mx + c with x = 1\/T, so the graph is a straight line whose gradient is \u2212E\u2090\/R.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why is the gradient negative?<\/h3>\n<p>Because k increases with temperature, and 1\/T decreases as T increases, so ln k falls as 1\/T rises. E\u2090 is positive, so the gradient \u2212E\u2090\/R must be negative; if yours comes out positive you have plotted the axes the wrong way round.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Which units do I use?<\/h3>\n<p>T in kelvin, so 1\/T in K\u207b\u00b9, and R = 8.31 J K\u207b\u00b9 mol\u207b\u00b9. The gradient is then in K, and E\u2090 = \u2212gradient \u00d7 R comes out in J mol\u207b\u00b9; divide by 1000 to quote it in kJ mol\u207b\u00b9. A gradient of about \u22126000 K, for example, gives an E\u2090 of about 50 kJ mol\u207b\u00b9.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Do I need to know the Arrhenius equation by heart?<\/h3>\n<p>No, the equation is given, but you must be able to use it: rearrange it to find E\u2090, A or k, and take logarithms correctly. Practise finding E\u2090 from a pair of k values at two temperatures as well as from the gradient of a graph.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>What does the constant A mean?<\/h3>\n<p>The pre-exponential factor. It is related to the collision frequency and the fraction of collisions in the right orientation, and e^(\u2212E\u2090\/RT) is the fraction of collisions with energy above E\u2090. Multiplying the two gives the rate constant.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"ols-related-card\">\n<h2>Related Topic 11 Kinetics Pages<\/h2>\n<p>Use these pages to connect the ideas across Topic 11 Kinetics and the rest of the course.<\/p>\n<div class=\"ols-related-grid\">\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-equations-orders-and-the-rate-constant\/\">Rate Equations, Orders and the Rate Constant<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/techniques-for-measuring-rates\/\">Techniques for Measuring Rates<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/concentration-time-graphs-and-half-life\/\">Concentration\u2013Time Graphs and Half-Life<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-concentration-graphs-and-the-initial-rates-method\/\">Rate\u2013Concentration Graphs and the Initial-Rates Method<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/the-iodine-propanone-reaction\/\">The Iodine\u2013Propanone Reaction<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-determining-step-and-reaction-mechanisms\/\">Rate-Determining Step and Reaction Mechanisms<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/heterogeneous-and-homogeneous-catalysis\/\">Heterogeneous and Homogeneous Catalysis<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/\">Topic 11 Kinetics Overview<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/core-practicals\/cp-10-activation-energy-of-a-reaction\/\">Core Practical 10 (activation energy of a reaction)<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-9-kinetics-and-equilibria\/9a-kinetics\/\">Topic 9A Kinetics (AS)<\/a>\n<\/div>\n<\/section>\n<section class=\"ols-attribution-card\">\n        <p><strong>Copyright notice:<\/strong> This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. 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