{"id":12340,"date":"2026-09-30T19:38:40","date_gmt":"2026-09-30T18:38:40","guid":{"rendered":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-concentration-graphs-and-the-initial-rates-method\/"},"modified":"2026-10-04T23:04:17","modified_gmt":"2026-10-04T22:04:17","slug":"rate-concentration-graphs-and-the-initial-rates-method","status":"publish","type":"page","link":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-concentration-graphs-and-the-initial-rates-method\/","title":{"rendered":"Rate\u2013Concentration Graphs and the Initial-Rates Method"},"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   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border-radius: 18px;\n      padding: 18px 20px;\n      box-shadow: 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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<header class=\"ols-title-card\">\n        <h1>Rate\u2013Concentration Graphs and the Initial-Rates Method<\/h1>\n        <p class=\"ols-page-intro\">A concise revision guide to deducing orders from rate\u2013concentration graphs and from initial-rates data: comparing experiments where one concentration changes, handling two changes at once, writing the rate equation and calculating the rate constant with its units.<\/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>Rate\u2013Concentration Graphs<\/h2>\n<\/div>\n<p>A <strong>rate\u2013concentration graph<\/strong> plots the rate of reaction (usually the initial rate) against the concentration of one reactant. Its shape gives the order with respect to that reactant directly.<\/p><div class=\"ols-table-wrap\"><table class=\"ols-table\"><thead><tr><th>Order<\/th><th>Shape of graph<\/th><th>Why<\/th><\/tr><\/thead><tbody><tr><td>For a <strong>zero-order<\/strong> reactant<\/td><td>A <strong>horizontal line<\/strong><\/td><td>The rate is the same at every concentration, and the height of the line is k.<\/td><\/tr><tr><td>For a <strong>first-order<\/strong> reactant<\/td><td>A <strong>straight line through the origin<\/strong><\/td><td>Because rate = k[A] is the equation of a straight line with gradient k, so k can be read from the gradient.<\/td><\/tr><tr><td>For a <strong>second-order<\/strong> reactant<\/td><td>A <strong>curve<\/strong> that rises ever more steeply<\/td><td>Since rate = k[A]\u00b2; plotting the rate against [A]\u00b2 instead gives a straight line through the origin, which confirms second order.<\/td><\/tr><\/tbody><\/table><\/div>\n<p>The points on the graph come from a set of experiments in which only the concentration of that one reactant is changed.<\/p><p>Every other concentration, and the temperature, must be kept constant, otherwise the change in rate cannot be attributed to the reactant being studied.<\/p><p>Each rate is either a gradient at t = 0 from a concentration\u2013time curve or a 1\/t value from a clock reaction, as page 2 described.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Key idea:<\/strong> Rate against [A]: horizontal for zero order, straight through the origin for first order (gradient = k), curving upwards for second order (rate against [A]\u00b2 is straight).<\/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: Reading a Rate\u2013Concentration Graph<\/h2>\n<p>Deduce the order, and where possible k, from rate\u2013concentration graphs for reactions not drawn on this page.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-1085\" class=\"h5p-iframe\" data-content-id=\"1085\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Kinetics Summary: Reading Rate\u2013Concentration Graphs\"><\/iframe><\/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>Deducing Orders From Initial-Rates Data<\/h2>\n<\/div>\n<p>Most initial-rates questions give a table rather than a graph. The method is to <strong>compare pairs of experiments<\/strong> in which only one concentration changes and see what the rate does.<\/p><p>Doubling a concentration and finding the rate unchanged means zero order; the rate doubling means first order; the rate quadrupling means second order.<\/p><p>The same logic works for any factor: if a concentration is trebled and the rate goes up nine times, the order is 2, because 3\u00b2 = 9.<\/p>\n<p>The table below is for the reaction A + B \u2192 products.<\/p><p><strong>Step 1:<\/strong> Comparing experiments 1 and 2, [B] is unchanged while [A] doubles, and the rate doubles from 2.0 \u00d7 10\u207b\u2074 to 4.0 \u00d7 10\u207b\u2074 mol dm\u207b\u00b3 s\u207b\u00b9, so the reaction is <strong>first order in A<\/strong>.<\/p><p><strong>Step 2:<\/strong> Comparing experiments 1 and 3, [A] is unchanged while [B] doubles, and the rate rises four times, from 2.0 \u00d7 10\u207b\u2074 to 8.0 \u00d7 10\u207b\u2074 mol dm\u207b\u00b3 s\u207b\u00b9, so the reaction is <strong>second order in B<\/strong>.<\/p><p><strong>Answer:<\/strong> The rate equation is rate = k[A][B]\u00b2, third order overall.<\/p>\n<div class=\"ols-table-wrap\">\n<table class=\"ols-table\">\n<thead>\n<tr><th>Experiment<\/th><th>[A] \/ mol dm\u207b\u00b3<\/th><th>[B] \/ mol dm\u207b\u00b3<\/th><th>Initial rate \/ mol dm\u207b\u00b3 s\u207b\u00b9<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><td><strong>1<\/strong><\/td><td>0.10<\/td><td>0.10<\/td><td>2.0 \u00d7 10\u207b\u2074<\/td><\/tr>\n<tr><td><strong>2<\/strong><\/td><td>0.20<\/td><td>0.10<\/td><td>4.0 \u00d7 10\u207b\u2074<\/td><\/tr>\n<tr><td><strong>3<\/strong><\/td><td>0.10<\/td><td>0.20<\/td><td>8.0 \u00d7 10\u207b\u2074<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>Write the comparison out every time, because the reasoning carries marks: &#8220;experiments 1 and 2: [A] \u00d7 2, [B] constant, rate \u00d7 2, so first order with respect to A&#8221;. A species whose concentration change leaves the rate unaltered is zero order and is left out of the rate equation, even though it is a reactant.<\/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-t11-04-initial-rates.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-t11-04-initial-rates.jpg\" alt=\"Rate\u2013concentration graphs for orders 0, 1 and 2 and a worked initial-rates table giving rate = k[A][B]\u00b2\" data-fullscreen-src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/kinetics-t11-04-initial-rates.jpg\">\n<\/a>\n<\/div>\n<div class=\"ols-figure-caption\"><p>Rate\u2013concentration graphs for the three orders beside the worked initial-rates table, with the run-by-run comparisons that give rate = k[A][B]\u00b2.<\/p><\/div>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Exam wording:<\/strong> For each order: name the two experiments, say which concentration changed and by what factor, say what the rate did, and state the order. Then write the rate equation.<\/p>\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>When Two Concentrations Change at Once<\/h2>\n<\/div>\n<p>Sometimes no pair of experiments differs in only one concentration. The rule is that the effects <strong>multiply<\/strong>.<\/p><p><strong>Step 1:<\/strong> In one comparison [A] doubles and [B] doubles together, and the rate rises eight times.<\/p><p><strong>Step 2:<\/strong> Having already found that the reaction is first order in A (which accounts for a factor of 2), the remaining factor is 8 \u00f7 2 = 4, so the reaction must be second order in B.<\/p><p>In the same way, a run in which [A] is trebled and [B] doubled for a reaction with rate = k[A][B]\u00b2 would show the rate rising 3 \u00d7 2\u00b2 = 12 times.<\/p>\n<p>The same multiplying rule predicts the rate of any new run once the rate equation is known.<\/p><p><strong>Step 1:<\/strong> For rate = k[A][B]\u00b2, a fourth experiment with [A] = 0.30 and [B] = 0.30 mol dm\u207b\u00b3 has both concentrations three times those of experiment 1.<\/p><p><strong>Step 2:<\/strong> So its rate is 3 \u00d7 3\u00b2 = 27 times the rate of experiment 1.<\/p><p><strong>Answer:<\/strong> 27 \u00d7 2.0 \u00d7 10\u207b\u2074 = 5.4 \u00d7 10\u207b\u00b3 mol dm\u207b\u00b3 s\u207b\u00b9.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Worked example:<\/strong> [A] doubles and [B] doubles; rate \u00d7 8. First order in A accounts for \u00d7 2; the remaining \u00d7 4 comes from B, so B is second order.<\/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-multiplyfactors.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-multiplyfactors.jpg\" alt=\"Grid of rate factors for orders 0, 1 and 2, with worked examples where two concentrations change together.\" data-fullscreen-src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/t11x-multiplyfactors.jpg\">\n<\/a>\n<\/div>\n<div class=\"ols-figure-caption\"><p>When two concentrations change together, raise each factor to its order and multiply, which both reveals a hidden order and predicts the rate of a new run.<\/p><\/div>\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: Orders From a Table<\/h2>\n<p>Deduce the orders and write the rate equation from initial-rates tables that are not the one on this page.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1086\"><\/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>Calculating the Rate Constant<\/h2>\n<\/div>\n<p>Once the rate equation is written, <strong>k<\/strong> is found by rearranging it and substituting the concentrations and rate from <strong>any one experiment<\/strong>.<\/p><p><strong>Step 1:<\/strong> Using experiment 1 above: k = rate \u00f7 ([A][B]\u00b2) = 2.0 \u00d7 10\u207b\u2074 \u00f7 (0.10 \u00d7 0.10\u00b2) = 2.0 \u00d7 10\u207b\u2074 \u00f7 1.0 \u00d7 10\u207b\u00b3 = 0.20.<\/p><p><strong>Step 2:<\/strong> The units come from the same rearrangement: mol dm\u207b\u00b3 s\u207b\u00b9 \u00f7 (mol dm\u207b\u00b3 \u00d7 mol\u00b2 dm\u207b\u2076) = mol\u207b\u00b2 dm\u2076 s\u207b\u00b9.<\/p><p><strong>Answer:<\/strong> k = 0.20 mol\u207b\u00b2 dm\u2076 s\u207b\u00b9.<\/p>\n<p>Because k is a constant at that temperature, substituting the values from experiment 2 or 3 must give the same answer. Checking a second run is a good way to catch an arithmetic slip.<\/p><p>Once k is known the rate equation can be used the other way round, to calculate the rate for any pair of concentrations or the concentration needed for a required rate.<\/p><p>If a question gives data at a different temperature, k will be different and must be recalculated. A larger k at a higher temperature is the quantitative version of &#8220;heating speeds the reaction up&#8221;.<\/p>\n\n<div class=\"ols-key-box\">\n<p><strong>Worked example:<\/strong> rate = k[A][B]\u00b2; from experiment 1, k = 2.0 \u00d7 10\u207b\u2074 \u00f7 (0.10 \u00d7 0.10\u00b2) = 0.20 mol\u207b\u00b2 dm\u2076 s\u207b\u00b9. Check with experiment 3: 8.0 \u00d7 10\u207b\u2074 \u00f7 (0.10 \u00d7 0.20\u00b2) = 8.0 \u00d7 10\u207b\u2074 \u00f7 4.0 \u00d7 10\u207b\u00b3 = 0.20. The same value, as it must be.<\/p>\n<\/div>\n<\/article>\n<article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">5<\/div>\n<h2>Common Exam Points<\/h2>\n<\/div>\n<h3>Say<\/h3><p>&#8220;Between experiments 1 and 2, [A] doubles and [B] is constant; the rate doubles, so the reaction is first order in A.&#8221; &#8220;k = rate \u00f7 [A][B]\u00b2, with units mol\u207b\u00b2 dm\u2076 s\u207b\u00b9.&#8221; &#8220;k is the same for every experiment at this temperature.&#8221;<\/p>\n<h3>Do not say<\/h3><p>&#8220;The rate doubles so the order is 2.&#8221; &#8220;k = 0.20&#8221; without a unit. &#8220;A is not involved because it is zero order&#8221; (it reacts, but not in the rate-determining step).<\/p>\n<h3>Watch for<\/h3><p>Tables where the rate is given in a different unit, or where the concentration is trebled or halved rather than doubled: apply the power, not the doubling rule.<\/p><p>Questions that give the rate equation and ask you to predict a rate: substitute and give the unit of rate, mol dm\u207b\u00b3 s\u207b\u00b9.<\/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: Calculating k<\/h2>\n<p>Calculate the rate constant with its units, and use it to predict a rate, for data that are not on this page.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1087\"><\/div><\/div>\n<\/section>\n<section class=\"ols-faq-card\">\n<h2>FAQs<\/h2>\n<p>Use these quick answers to check the initial-rates method.<\/p>\n\n<div class=\"ols-faq-list\">\n<div class=\"ols-faq-item\">\n<h3>What does a rate\u2013concentration graph look like for each order?<\/h3>\n<p>Zero order gives a horizontal line: rate does not change with concentration. First order gives a straight line through the origin, with gradient k. Second order gives a curve that gets steeper, and a plot of rate against [A]\u00b2 is then a straight line.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>How do I deduce an order from a table when two concentrations change at once?<\/h3>\n<p>Find the order of one substance first from a pair of experiments where only that concentration changes. Then allow for its effect in the other pair, and whatever change in rate is left over is due to the second substance.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>What if tripling a concentration multiplies the rate by nine?<\/h3>\n<p>That is second order, because 3\u00b2 = 9. Doubling would give \u00d74 and tripling gives \u00d79; the rate scales with the concentration squared.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Which experiment should I use to calculate k?<\/h3>\n<p>Any of them; k is the same for all runs at that temperature. Pick one with simple numbers, substitute rate and the concentrations into the rate equation and rearrange for k. Give the units every time.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why do the experiments need the same temperature?<\/h3>\n<p>Because the whole method depends on k being identical in every run, so that any change in rate is due only to the concentration you changed. A warmer run would have a bigger k and would look like a higher order.<\/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\/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\/activation-energy-and-the-arrhenius-equation\/\">Activation Energy and the Arrhenius Equation<\/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<\/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. It may not be copied, reproduced, redistributed or adapted without written permission.<\/p>\n      <\/section>\n    <\/main>\n  \n\n<script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@graph\": [\n    {\n      \"@type\": \"WebPage\",\n      \"@id\": \"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-concentration-graphs-and-the-initial-rates-method\/#webpage\",\n      \"url\": \"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-concentration-graphs-and-the-initial-rates-method\/\",\n      \"name\": \"Rate\u2013Concentration Graphs and the Initial-Rates Method | Topic 11 Kinetics | Online Learning System\",\n      \"description\": \"Edexcel International A Level Chemistry revision notes on the initial-rates method: rate\u2013concentration graphs, deducing orders from tables of data, writing the 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