{"id":12337,"date":"2026-09-30T19:38:35","date_gmt":"2026-09-30T18:38:35","guid":{"rendered":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-equations-orders-and-the-rate-constant\/"},"modified":"2026-10-04T23:04:12","modified_gmt":"2026-10-04T22:04:12","slug":"rate-equations-orders-and-the-rate-constant","status":"publish","type":"page","link":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-equations-orders-and-the-rate-constant\/","title":{"rendered":"Rate Equations, Orders and the Rate Constant"},"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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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>Rate Equations, Orders and the Rate Constant<\/span>\n<\/nav>\n\n      <header class=\"ols-title-card\">\n        <h1>Rate Equations, Orders and the Rate Constant<\/h1>\n        <p class=\"ols-page-intro\">A concise revision guide to the rate equation rate = k[A]\u1d50[B]\u207f: what order with respect to a substance and overall order mean, why orders come only from experiment, what the rate constant is, and how to work out its units for any overall order.<\/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      <section class=\"ols-h5p-card ols-h5p-inline ols-h5p-recap\">\n<span class=\"ols-h5p-kicker\">Before you start<\/span>\n<h2>AS Recap: Rates and Collision Theory<\/h2>\n<p>Three quick questions on what you already know: rate as a change in concentration per unit time, why collisions need energy above the activation energy, and how a rate is read from a gradient.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1075\"><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">1<\/div>\n<h2>What the Rate Equation Says<\/h2>\n<\/div>\n<p>At AS you learned that a higher concentration usually gives a faster reaction. The <strong>rate equation<\/strong> makes that relationship exact.<\/p><p>For a reaction between A and B it is written rate = k[A]\u1d50[B]\u207f, and each symbol has a fixed meaning.<\/p><div class=\"ols-table-wrap\"><table class=\"ols-table\"><thead><tr><th>Symbol<\/th><th>Meaning<\/th><\/tr><\/thead><tbody><tr><td><strong>rate<\/strong><\/td><td>The change in concentration of a reactant or product per unit time, usually in mol dm\u207b\u00b3 s\u207b\u00b9<\/td><\/tr><tr><td><strong>[A]<\/strong><\/td><td>The square brackets mean the concentration of A in mol dm\u207b\u00b3<\/td><\/tr><tr><td><strong>m<\/strong> and <strong>n<\/strong><\/td><td>The powers are the orders of reaction with respect to A and B<\/td><\/tr><tr><td><strong>k<\/strong><\/td><td>The rate constant, a number that links the concentrations to the rate at a particular temperature<\/td><\/tr><\/tbody><\/table><\/div>\n<p>The equation is a statement about how the rate responds when a concentration changes. If m is 1, doubling [A] doubles the rate; if m is 2, doubling [A] quadruples it; if m is 0, changing [A] does nothing at all.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Definition:<\/strong> The rate equation, rate = k[A]\u1d50[B]\u207f, gives the rate of reaction in terms of the concentrations of the species that affect it. k is the rate constant and m and n are the orders with respect to A and B; each is 0, 1 or 2 at A Level.<\/p>\n<\/div>\n<\/article>\n<article class=\"ols-note-card soft\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">2<\/div>\n<h2>Order With Respect to a Substance<\/h2>\n<\/div>\n<p>The <strong>order with respect to a substance<\/strong> is the power to which its concentration is raised in the rate equation. Three values are met at A Level.<\/p><div class=\"ols-table-wrap\"><table class=\"ols-table\"><thead><tr><th>Order<\/th><th>Proportionality<\/th><th>Effect of changing [A]<\/th><\/tr><\/thead><tbody><tr><td><strong>Zero order<\/strong><\/td><td>Rate \u221d [A]\u2070, so the rate is independent of [A]<\/td><td>Doubling the concentration leaves the rate unchanged, and the substance does not appear in the rate equation at all, because [A]\u2070 = 1.<\/td><\/tr><tr><td><strong>First order<\/strong><\/td><td>Rate \u221d [A]<\/td><td>Doubling [A] doubles the rate, trebling it trebles the rate.<\/td><\/tr><tr><td><strong>Second order<\/strong><\/td><td>Rate \u221d [A]\u00b2<\/td><td>Doubling [A] multiplies the rate by 2\u00b2, which is 4, and trebling it multiplies the rate by 9.<\/td><\/tr><\/tbody><\/table><\/div>\n<p>The quickest way to find an order from data is the <strong>doubling test<\/strong>: double one concentration while keeping every other concentration the same, and see what happens to the rate. Rate \u00d7 1 means zero order, rate \u00d7 2 means first order, rate \u00d7 4 means second order.<\/p>\n<p>Orders are <strong>found by experiment<\/strong>, and they have nothing to do with the balancing numbers in the equation.<\/p><p>The reaction 2NO + O\u2082 \u2192 2NO\u2082 happens to have the rate equation rate = k[NO]\u00b2[O\u2082].<\/p><p>But the reaction between hydrogen peroxide and iodide ions, H\u2082O\u2082 + 2I\u207b + 2H\u207a \u2192 I\u2082 + 2H\u2082O, is first order in H\u2082O\u2082, first order in I\u207b and zero order in H\u207a even though the equation shows two of each.<\/p><p>The orders reflect the mechanism (page 6), which the balanced equation cannot show.<\/p>\n<p>A substance can be made to behave as zero order by using it in a <strong>large excess<\/strong>. Its concentration then barely changes during the reaction, so it has no measurable effect on the rate and is described as <strong>pseudo-zero order<\/strong>.<\/p><p>This is how experiments isolate the order with respect to one reactant at a time.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Key idea:<\/strong> Order 0: rate does not change. Order 1: rate \u221d [A]. Order 2: rate \u221d [A]\u00b2. The orders come from experiment, never from the balanced equation.<\/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-doublingtest.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-doublingtest.jpg\" alt=\"Decision poster linking rate \u00d7 1, \u00d7 2 and \u00d7 4 to orders 0, 1 and 2, with pseudo-zero order from a large excess.\" data-fullscreen-src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/t11x-doublingtest.jpg\">\n<\/a>\n<\/div>\n<div class=\"ols-figure-caption\"><p>Double one concentration and watch the rate: unchanged means zero order, doubled means first order and quadrupled means second order, whatever the balanced equation shows.<\/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: What an Order Means<\/h2>\n<p>Decide how the rate changes when a concentration is changed for orders and reactions not used on this page.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-1076\" class=\"h5p-iframe\" data-content-id=\"1076\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Kinetics Summary: What an Order Means\"><\/iframe><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">3<\/div>\n<h2>Overall Order and Writing Rate Equations<\/h2>\n<\/div>\n<p>The <strong>overall order<\/strong> is the sum of the individual orders, m + n.<\/p><p>A reaction that is first order in A and second order in B has the rate equation rate = k[A][B]\u00b2 and is third order overall.<\/p><p>A reaction that is first order in A and zero order in B has rate = k[A][B]\u2070, which is written simply as rate = k[A], and is first order overall. Powers of 1 are not written, and a zero-order species is left out.<\/p>\n<p>Two features of rate equations surprise students. First, a reactant in the balanced equation can be <strong>absent<\/strong> from the rate equation, as H\u207a is absent from the hydrogen peroxide and iodide example above.<\/p>\n<p>Second, a species that is not in the balanced equation at all can <strong>appear<\/strong> in the rate equation, most often a catalyst.<\/p>\n<p>The acid-catalysed reaction between iodine and propanone (page 5) has the rate equation rate = k[CH\u2083COCH\u2083][H\u207a], with the catalyst H\u207a in it and the reactant iodine left out.<\/p>\n<div class=\"ols-key-box\"><p><strong>Key idea:<\/strong> Both follow from the rate equation describing the slowest step of the mechanism.<\/p><\/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\/kinetics-t11-01-rate-equation.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-t11-01-rate-equation.jpg\" alt=\"The rate equation rate = k[A]\u1d50[B]\u207f with each term labelled, graphs of rate against concentration for orders 0, 1 and 2, and the units of k for overall orders 1 to 3\" data-fullscreen-src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/kinetics-t11-01-rate-equation.jpg\">\n<\/a>\n<\/div>\n<div class=\"ols-figure-caption\"><p>The rate equation with every symbol labelled, the three orders as rate\u2013concentration sketches, and the units of k for overall orders 1, 2 and 3.<\/p><\/div>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Exam wording:<\/strong> To write a rate equation from given orders, write k, then each species with a non-zero order raised to its order, and leave out any zero-order species: &#8220;first order in A, second order in B&#8221; becomes rate = k[A][B]\u00b2.<\/p>\n<\/div>\n<\/article>\n<article class=\"ols-note-card soft\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">4<\/div>\n<h2>The Rate Constant and Its Units<\/h2>\n<\/div>\n<p>The <strong>rate constant<\/strong>, k, is the proportionality constant in the rate equation. At a fixed temperature it has a fixed value for a given reaction, however the concentrations change and however far the reaction has gone.<\/p><p>It is not a universal constant: <strong>k increases when the temperature rises<\/strong>, which is the real reason reactions speed up on heating.<\/p><p>It also changes if a catalyst is added because the mechanism changes (page 7 puts numbers on the temperature effect).<\/p>\n<p>The <strong>units of k<\/strong> depend on the overall order, so they must be worked out for each rate equation.<\/p><p>The method is always the same: rearrange to make k the subject, substitute the units of rate and of each concentration, and cancel.<\/p><p>For rate = k[A][B]\u00b2, k = rate \u00f7 ([A][B]\u00b2) = mol dm\u207b\u00b3 s\u207b\u00b9 \u00f7 (mol dm\u207b\u00b3 \u00d7 mol\u00b2 dm\u207b\u2076) = mol dm\u207b\u00b3 s\u207b\u00b9 \u00f7 mol\u00b3 dm\u207b\u2079 = mol\u207b\u00b2 dm\u2076 s\u207b\u00b9.<\/p>\n<div class=\"ols-table-wrap\">\n<table class=\"ols-table\">\n<thead>\n<tr><th>Overall order<\/th><th>Example rate equation<\/th><th>Units of k<\/th><th>Pattern<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><td><strong>1<\/strong><\/td><td>rate = k[A]<\/td><td>s\u207b\u00b9<\/td><td>mol dm\u207b\u00b3 s\u207b\u00b9 \u00f7 mol dm\u207b\u00b3<\/td><\/tr>\n<tr><td><strong>2<\/strong><\/td><td>rate = k[A][B] or k[A]\u00b2<\/td><td>mol\u207b\u00b9 dm\u00b3 s\u207b\u00b9<\/td><td>mol dm\u207b\u00b3 s\u207b\u00b9 \u00f7 (mol dm\u207b\u00b3)\u00b2<\/td><\/tr>\n<tr><td><strong>3<\/strong><\/td><td>rate = k[A][B]\u00b2 or k[A]\u00b2[B]<\/td><td>mol\u207b\u00b2 dm\u2076 s\u207b\u00b9<\/td><td>mol dm\u207b\u00b3 s\u207b\u00b9 \u00f7 (mol dm\u207b\u00b3)\u00b3<\/td><\/tr>\n<tr><td><strong>0<\/strong><\/td><td>rate = k<\/td><td>mol dm\u207b\u00b3 s\u207b\u00b9<\/td><td>the same units as rate<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Exam focus:<\/strong> Always give k a unit. Work it out from the rate equation in the question, because the examiner sets the order; write the units in the order mol, dm, s, with negative powers, for example mol\u207b\u00b9 dm\u00b3 s\u207b\u00b9.<\/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: Units of k<\/h2>\n<p>Work out the units of the rate constant for rate equations that are not the ones in the table above.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1077\"><\/div><\/div>\n<\/section>\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;The order with respect to A is the power of [A] in the rate equation.&#8221; &#8220;Doubling [A] doubles the rate, so the reaction is first order with respect to A.&#8221; &#8220;k is constant at a fixed temperature and increases as the temperature rises.&#8221;<\/p>\n<h3>Do not say<\/h3><p>&#8220;The order is 2 because there are two moles of it in the equation&#8221; (orders come only from experiment). &#8220;k is constant&#8221; without adding &#8220;at a fixed temperature&#8221;. &#8220;The rate constant has no units&#8221; (only a first-order overall reaction has the simple unit s\u207b\u00b9, and even that is a unit).<\/p>\n<h3>Watch for<\/h3><p>Questions that give a rate equation and ask for the effect of changing two concentrations at once: apply each change separately and multiply the effects.<\/p><p>Questions that ask for the overall order want a single number, the sum of the powers. If a catalyst appears in the rate equation, that is intended: it is in the rate-determining step.<\/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: Overall Order and Rate Equations<\/h2>\n<p>Write rate equations from given orders and state the overall order for reactions not used on this page.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1078\"><\/div><\/div>\n<\/section>\n<section class=\"ols-faq-card\">\n<h2>FAQs<\/h2>\n<p>Use these quick answers to check the rate equation ideas that come up most often.<\/p>\n\n<div class=\"ols-faq-list\">\n<div class=\"ols-faq-item\">\n<h3>Can the order with respect to a substance be a fraction or negative?<\/h3>\n<p>At A Level the orders you meet are 0, 1 and 2, and every question is set so the data give one of those three. Check each comparison against the doubling rules: no change, doubles, quadruples.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why is the rate constant called constant if it changes with temperature?<\/h3>\n<p>It is constant for a given reaction at a given temperature, whatever the concentrations. Change the temperature (or add a catalyst) and you get a new value of k. Concentration changes never alter k.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why can I not read the orders off the balanced equation?<\/h3>\n<p>Because the balanced equation only shows the overall stoichiometry; the rate depends on the slowest step of the mechanism, which the equation does not show. Orders come only from experiment.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>How do I work out the units of k without memorising a table?<\/h3>\n<p>Rearrange to k = rate \u00f7 (concentration terms) and cancel the units. For rate = k[A][B]\u00b2 that is mol dm\u207b\u00b3 s\u207b\u00b9 \u00f7 (mol dm\u207b\u00b3)\u00b3, which simplifies to mol\u207b\u00b2 dm\u2076 s\u207b\u00b9. Each extra order divides by another mol dm\u207b\u00b3.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>If a substance is zero order, is it still needed for the reaction?<\/h3>\n<p>Yes. It still reacts, and it still appears in the balanced equation, but changing its concentration does not change the rate because it is not involved in the rate-determining step. Its concentration term is [A]\u2070 = 1, so it is left out of the rate equation.<\/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\/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\/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<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. 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-equations-orders-and-the-rate-constant\/#webpage\",\n      \"url\": \"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-11-kinetics\/rate-equations-orders-and-the-rate-constant\/\",\n      \"name\": \"Rate Equations, Orders and the Rate Constant | Topic 11 Kinetics | Online Learning System\",\n      \"description\": \"Edexcel International A Level Chemistry revision notes on rate equations: orders of reaction, the rate constant k, overall order and working out the units of k.\",\n      \"isPartOf\": {\n        \"@id\": 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