{"id":12488,"date":"2026-10-03T08:32:53","date_gmt":"2026-10-03T07:32:53","guid":{"rendered":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/calculating-entropy-changes\/"},"modified":"2026-10-04T23:04:53","modified_gmt":"2026-10-04T22:04:53","slug":"calculating-entropy-changes","status":"publish","type":"page","link":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/calculating-entropy-changes\/","title":{"rendered":"Calculating Entropy Changes"},"content":{"rendered":"\n<section class=\"ols-revision-page ols-nature-of-covalent-bonding-9ch0-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      --soft-orange: #fff7ed;\n      --grey-text: #667085;\n      --body-text: #1f2937;\n      --border: 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26px !important; position: relative !important; z-index: 2 !important; }\n    .ols-zoom-card img.ols-zoomable-img { display: block !important; max-width: 100% !important; height: auto !important; border-radius: 18px !important; cursor: zoom-in !important; pointer-events: auto !important; user-select: none !important; -webkit-user-drag: none !important; transform: translateZ(0) scale(1) !important; transform-origin: center center !important; transition: transform 0.32s ease, box-shadow 0.32s ease, filter 0.32s ease !important; position: relative !important; z-index: 2 !important; }\n    @media (hover: hover) and (pointer: fine) { .ols-zoom-card img.ols-zoomable-img:hover { transform: translateZ(0) scale(1.35) !important; box-shadow: 0 28px 70px rgba(28, 36, 75, 0.34) !important; filter: saturate(1.02) contrast(1.01) !important; z-index: 100 !important; } }\n    .ols-zoom-card-caption { padding: 18px 22px 20px !important; background: linear-gradient(135deg, #ffffff, #f8fbff) !important; border-bottom-left-radius: 26px !important; border-bottom-right-radius: 26px !important; position: relative !important; z-index: 1 !important; }\n    .ols-zoom-card-caption p { margin: 0 !important; color: #5f6b85 !important; font-size: 15px !important; line-height: 1.65 !important; font-weight: 300 !important; font-style: italic !important; font-family: Poppins, Arial, sans-serif !important; }\n    .ols-image-lightbox { position: fixed; inset: 0; z-index: 999999; display: none; align-items: center; justify-content: center; padding: 34px; background: rgba(10, 15, 35, 0.86); backdrop-filter: blur(8px); -webkit-backdrop-filter: blur(8px); }\n    .ols-image-lightbox.is-open { display: flex; }\n    .ols-image-lightbox-inner { position: relative; width: min(96vw, 1500px); max-height: 92vh; display: flex; align-items: center; justify-content: center; }\n    .ols-image-lightbox-img { display: block; max-width: 100%; max-height: 92vh; height: auto; width: auto; border-radius: 22px; background: #ffffff; box-shadow: 0 32px 90px rgba(0, 0, 0, 0.45); object-fit: contain; }\n    .ols-image-lightbox-close { position: absolute; top: -18px; right: -18px; width: 46px; height: 46px; border: 0; border-radius: 50%; background: #ffffff; color: var(--navy); font-family: Poppins, Arial, sans-serif; font-size: 28px; line-height: 1; font-weight: 700; cursor: pointer; box-shadow: 0 16px 34px rgba(0, 0, 0, 0.28); display: flex; align-items: center; justify-content: center; transition: transform 0.2s ease, background 0.2s ease, color 0.2s ease; }\n    .ols-image-lightbox-close:hover { transform: scale(1.08); background: var(--blue); color: #ffffff; }\n    @media (max-width: 760px) { .ols-zoom-card { border-radius: 22px !important; overflow: hidden !important; } .ols-zoom-card-image { min-height: auto !important; padding: 12px !important; overflow: hidden !important; border-top-left-radius: 22px !important; border-top-right-radius: 22px !important; } .ols-zoom-card img.ols-zoomable-img, 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line-height: 1.45; color: var(--body-text); vertical-align: top; overflow-wrap: anywhere; }\n    .ols-table tr:last-child td { border-bottom: none; }\n\n    .ols-h5p-card { background: linear-gradient(135deg, #ffffff 0%, #f7f0ff 100%); }\n    .ols-h5p-frame { margin-top: 22px; padding: 18px; border-radius: 24px; background: #ffffff; border: 1px solid var(--border); box-shadow: inset 0 0 0 1px rgba(28, 36, 75, 0.03); }\n\n    .ols-faq-list { display: grid; gap: 14px; margin-top: 18px; }\n    .ols-faq-item { background: #ffffff; border: 1px solid var(--border); border-radius: 18px; padding: 18px 20px; box-shadow: var(--inner-shadow); }\n    .ols-faq-item h3 { margin: 0 0 8px; font-size: 20px; line-height: 1.3; color: var(--navy); }\n    .ols-faq-item p { margin: 0; font-size: 16px; line-height: 1.65; color: var(--body-text); }\n\n    .ols-related-grid { display: grid; grid-template-columns: repeat(3, minmax(0, 1fr)); gap: 16px; margin-top: 18px; }\n    .ols-related-item { border: 1px 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}\n    .ols-course-cta-kicker { display: inline-flex; align-items: center; padding: 8px 14px; border-radius: 999px; background: #ffffff; border: 1px solid rgba(37, 99, 235, 0.18); color: var(--blue); font-size: 14px; line-height: 1.2; font-weight: 700; box-shadow: var(--inner-shadow); }\n    .ols-course-cta-covalent h2 { margin: 0 0 14px; font-size: clamp(28px, 3.5vw, 42px); line-height: 1.15; font-weight: 800; letter-spacing: -0.03em; color: #111827; }\n    .ols-course-cta-intro { margin: 0 0 24px; color: var(--body-text); font-size: clamp(16px, 1.4vw, 18px); line-height: 1.7; font-weight: 300; }\n    .ols-course-cta-features { display: grid; grid-template-columns: repeat(2, minmax(0, 1fr)); gap: 14px; margin: 0 0 30px; }\n    .ols-course-feature { background: rgba(255, 255, 255, 0.78); border: 1px solid var(--border); border-radius: 18px; padding: 16px 18px; }\n    .ols-course-feature h3 { margin: 0 0 6px; color: var(--navy); font-size: 18px; line-height: 1.3; font-weight: 800; }\n   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760px) { .ols-h5p-card.ols-h5p-inline { padding: 20px 16px; } }\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 13 Energetics II<\/h4>\n\n    <ul class=\"ols-topic-list\">\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/\">Topic 13 Energetics II Overview<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/lattice-energy-and-born-haber-cycles\/\">Lattice Energy and Born\u2013Haber Cycles<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/lattice-energy-trends-and-covalent-character\/\">Lattice Energy Trends and Covalent Character<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/enthalpy-of-solution-and-hydration\/\">Enthalpy of Solution and Hydration<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/entropy-and-the-direction-of-change\/\">Entropy and the Direction of Change<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/calculating-entropy-changes\/\">Calculating Entropy Changes<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/feasibility-gibbs-energy-and-temperature\/\">Feasibility, Gibbs Energy and Temperature<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/thermodynamic-and-kinetic-stability\/\">Thermodynamic and Kinetic Stability<\/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\/topic-8-energetics-i\/\">Topic 8 Energetics I<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-16-kinetics-ii\/\">Topic 16 Kinetics II<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-4-inorganic-chemistry-and-the-periodic-table\/\">Topic 4 Inorganic Chemistry<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/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    sidebar.querySelectorAll('a[data-ols-disabled-parent=\"1\"], 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document.addEventListener('DOMContentLoaded', highlightActive);\r\n  }\r\n})();\r\n<\/script>\n\n  <main class=\"ols-main\">\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\/\">Edexcel<\/a> \/\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/\">Topic 13 Energetics II<\/a> \/\n<span>Calculating Entropy Changes<\/span>\n<\/nav>\n\n      <header class=\"ols-title-card\">\n        <h1>Calculating Entropy Changes<\/h1>\n        <p class=\"ols-page-intro\">A concise revision guide to calculating the entropy change of a reaction from standard entropies, the entropy change of the surroundings from \u0394Ssurroundings = \u2212\u0394H\/T, and the total entropy change that decides whether a reaction is feasible.<\/p>\n\n        <div class=\"ols-badges\">\n<div class=\"ols-badge\">Paper 1 and 3<\/div>\n<div class=\"ols-badge\">Topic 13: Energetics II<\/div>\n<div class=\"ols-badge\">9CH0\/01<\/div>\n<\/div>\n\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:<br><span>Dr. Mohammed Al-Fatah<\/span>\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>\u0394S of the System From Standard Entropies<\/h2>\n<\/div>\n<p>The <strong>entropy change of the system<\/strong>, \u0394S_system, is the difference between the entropies of the products and the reactants.<\/p>\n<p>It is found from tabulated standard entropies, in the same way that an enthalpy change is found from enthalpies of formation:<\/p>\n<p>\u0394S_system = \u03a3S\u29b5(products) \u2212 \u03a3S\u29b5(reactants)<\/p>\n<ul class=\"ols-list\">\n<li>Each entropy is multiplied by the number of moles of that substance in the equation.<\/li>\n<li>The answer has the same units as the data, <strong>J K\u207b\u00b9 mol\u207b\u00b9<\/strong>.<\/li>\n<li>There is one difference from enthalpy calculations: elements have non-zero entropies, so an element in the equation must be included with its tabulated value, not treated as zero.<\/li>\n<\/ul>\n<h3>Worked example 1: a gas is made<\/h3>\n<p>The thermal decomposition of calcium carbonate is CaCO\u2083(s) \u2192 CaO(s) + CO\u2082(g).<\/p>\n<div class=\"ols-table-wrap\">\n<table class=\"ols-table\">\n<thead>\n<tr><th>Substance<\/th><th>S\u29b5 \/ J K\u207b\u00b9 mol\u207b\u00b9<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><td><strong>CaCO\u2083(s)<\/strong><\/td><td>92.9<\/td><\/tr>\n<tr><td><strong>CaO(s)<\/strong><\/td><td>39.7<\/td><\/tr>\n<tr><td><strong>CO\u2082(g)<\/strong><\/td><td>213.6<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p><strong>Step 1.<\/strong> Add the entropies of the products: 39.7 + 213.6 = 253.3 J K\u207b\u00b9 mol\u207b\u00b9.<\/p>\n<p><strong>Step 2.<\/strong> Subtract the entropy of the reactant: \u0394S_system = (39.7 + 213.6) \u2212 92.9.<\/p>\n<p><strong>Answer.<\/strong> \u0394S_system = <strong>+160.4 J K\u207b\u00b9 mol\u207b\u00b9<\/strong>.<\/p>\n<ul class=\"ols-list\">\n<li>The sign is positive, as page 1 predicted for a reaction that makes a gas from a solid.<\/li>\n<li>The size is typical of a reaction that produces one mole of gas: roughly 150 to 200 J K\u207b\u00b9 mol\u207b\u00b9 per mole of gas gained.<\/li>\n<\/ul>\n<h3>Worked example 2: gas moles fall<\/h3>\n<p>For N\u2082(g) + 3H\u2082(g) \u2192 2NH\u2083(g), S\u29b5 = 192 (N\u2082), 131 (H\u2082) and 193 (NH\u2083) J K\u207b\u00b9 mol\u207b\u00b9.<\/p>\n<p><strong>Step 1.<\/strong> Products: 2 \u00d7 193 = 386 J K\u207b\u00b9 mol\u207b\u00b9.<\/p>\n<p><strong>Step 2.<\/strong> Reactants: 192 + (3 \u00d7 131) = 585 J K\u207b\u00b9 mol\u207b\u00b9.<\/p>\n<p><strong>Answer.<\/strong> \u0394S_system = 386 \u2212 585 = \u2212199 J K\u207b\u00b9 mol\u207b\u00b9. Four moles of gas become two, so the entropy of the system falls.<\/p>\n<div class=\"ols-zoom-card\">\n<div class=\"ols-zoom-card-image\">\n<a class=\"ols-lightbox-link\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/entropy-t12-03-scalc.jpg\" aria-label=\"Open image full screen\">\n<img decoding=\"async\" class=\"ols-zoomable-img ols-lightbox-target\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/entropy-t12-03-scalc.jpg\" alt=\"Worked card calculating \u0394S of the system for the thermal decomposition of calcium car\">\n<\/a>\n<\/div>\n<div class=\"ols-zoom-card-caption\"><p>The calcium carbonate calculation laid out in full: standard entropies, \u0394S of the system, then the surroundings and total entropy at 298 K and at 1200 K.<\/p><\/div>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Worked example:<\/strong> CaCO\u2083(s) \u2192 CaO(s) + CO\u2082(g): \u0394S_system = \u03a3S(products) \u2212 \u03a3S(reactants) = (39.7 + 213.6) \u2212 92.9 = +160.4 J K\u207b\u00b9 mol\u207b\u00b9. Multiply each S by its balancing number, and include the elements.<\/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: \u0394S From Standard Entropies<\/h2>\n<p>Calculate the entropy change of the system with its sign and units for reactions not used on this page.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1102\"><\/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>\u0394S of the Surroundings<\/h2>\n<\/div>\n<p>The reaction also changes the entropy of everything around it.<\/p>\n<ul class=\"ols-list\">\n<li>An exothermic reaction gives out heat. That heat spreads among the particles of the surroundings and raises their entropy.<\/li>\n<li>An endothermic reaction takes heat in and lowers the entropy of the surroundings.<\/li>\n<\/ul>\n<p>The size of the effect depends on how much heat is transferred and on the temperature at which it happens. The two are combined in a simple expression:<\/p>\n<p>\u0394S_surroundings = \u2212\u0394H \u00f7 T<\/p>\n<div class=\"ols-table-wrap\">\n<table class=\"ols-table\">\n<thead>\n<tr><th>Part of the expression<\/th><th>What to do<\/th><th>Why<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><td><strong>The minus sign<\/strong><\/td><td>keep it<\/td><td>it makes the signs come out right: a negative \u0394H (exothermic) gives a positive \u0394S_surroundings<\/td><\/tr>\n<tr><td><strong>\u0394H<\/strong><\/td><td>convert from <strong>kJ to J<\/strong> before dividing<\/td><td>entropies are in J K\u207b\u00b9 mol\u207b\u00b9<\/td><\/tr>\n<tr><td><strong>T<\/strong><\/td><td>use kelvin<\/td><td>T is in the denominator because a given amount of heat makes a bigger difference to cold surroundings, which have little energy already, than to hot ones<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<h3>Worked example: calcium carbonate at 298 K<\/h3>\n<p><strong>Step 1.<\/strong> Convert the enthalpy change: \u0394H = +178 kJ mol\u207b\u00b9 = +178 000 J mol\u207b\u00b9.<\/p>\n<p><strong>Step 2.<\/strong> Substitute: \u0394S_surroundings = \u2212178 000 \u00f7 298.<\/p>\n<p><strong>Answer.<\/strong> \u0394S_surroundings = <strong>\u2212597 J K\u207b\u00b9 mol\u207b\u00b9<\/strong>.<\/p>\n<p>The reaction is endothermic, so it takes heat from the surroundings and lowers their entropy, by an amount that far outweighs the entropy gained by the system at this temperature.<\/p>\n<h3>Worked example: the Haber process reaction at 298 K<\/h3>\n<p><strong>Step 1.<\/strong> Convert the enthalpy change: \u0394H = \u221292 kJ mol\u207b\u00b9 = \u221292 000 J mol\u207b\u00b9.<\/p>\n<p><strong>Step 2.<\/strong> Substitute: \u0394S_surroundings = \u2212(\u221292 000) \u00f7 298.<\/p>\n<p><strong>Answer.<\/strong> \u0394S_surroundings = +309 J K\u207b\u00b9 mol\u207b\u00b9.<\/p>\n<p>The heat given out raises the entropy of the surroundings by more than the entropy of the system falls.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Exam focus:<\/strong> \u0394S_surroundings = \u2212\u0394H\/T. Convert \u0394H to joules (multiply by 1000), use T in kelvin, and keep the minus sign: exothermic reactions give a positive \u0394S_surroundings, endothermic reactions a negative one.<\/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: \u0394S of the Surroundings<\/h2>\n<p>Calculate the entropy change of the surroundings from \u0394H and T, with the right units and sign, for reactions not used on this page.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1103\"><\/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>Total Entropy Change<\/h2>\n<\/div>\n<p>The <strong>total entropy change<\/strong> of a reaction is the sum of the entropy change of the system and the entropy change of the surroundings:<\/p>\n<p>\u0394S_total = \u0394S_system + \u0394S_surroundings<\/p>\n<p>A reaction is <strong>feasible<\/strong> (it can happen of its own accord) when \u0394S_total is <strong>positive<\/strong>. It does not matter if one of the two terms is negative, as long as the other outweighs it.<\/p>\n<h3>Calcium carbonate at two temperatures<\/h3>\n<p>All values in the table are in J K\u207b\u00b9 mol\u207b\u00b9.<\/p>\n<div class=\"ols-table-wrap\">\n<table class=\"ols-table\">\n<thead>\n<tr><th>Quantity<\/th><th>At 298 K<\/th><th>At 1200 K<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><td><strong>\u0394S_system<\/strong><\/td><td>+160.4<\/td><td>+160.4<\/td><\/tr>\n<tr><td><strong>\u0394S_surroundings = \u2212\u0394H \u00f7 T<\/strong><\/td><td>\u2212178 000 \u00f7 298 = \u2212597<\/td><td>\u2212178 000 \u00f7 1200 = \u2212148<\/td><\/tr>\n<tr><td><strong>\u0394S_total<\/strong><\/td><td>+160.4 + (\u2212597) = <strong>\u2212437<\/strong><\/td><td>+160.4 + (\u2212148) = <strong>+12<\/strong><\/td><\/tr>\n<tr><td><strong>Feasible?<\/strong><\/td><td>no: \u0394S_total is negative<\/td><td>yes: \u0394S_total is positive<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<ul class=\"ols-list\">\n<li>At 298 K limestone does not decompose, however long you wait.<\/li>\n<li>Raising the temperature makes the surroundings term smaller, because the same \u0394H is divided by a larger T.<\/li>\n<li>At 1200 K, in a lime kiln, the reaction goes.<\/li>\n<li>Page 3 shows how to find the exact temperature at which the sign changes.<\/li>\n<\/ul>\n<div class=\"ols-zoom-card\">\n<div class=\"ols-zoom-card-image\">\n<a class=\"ols-lightbox-link\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/entropy-t12-12-smethod.jpg\" aria-label=\"Open image full screen\">\n<img decoding=\"async\" class=\"ols-zoomable-img ols-lightbox-target\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/entropy-t12-12-smethod.jpg\" alt=\"Flow chart poster of the method for an entropy calculation: \u0394S of the system from standard entropies, then two routes to the verdict, total entropy using \u2212\u0394H\/T or Gibbs energy using \u0394G = \u0394H \u2212 T\u0394S, with three unit traps flagged and the calcium carbonate numbers showing that both routes agree\">\n<\/a>\n<\/div>\n<div class=\"ols-zoom-card-caption\"><p>The method as a flow chart: \u0394S of the system first, then the total entropy route or the Gibbs energy route, with the unit traps marked and both routes giving the same verdict for calcium carbonate.<\/p><\/div>\n<\/div>\n<h3>The link to Gibbs energy<\/h3>\n<p>The same numbers can be expressed through the <strong>Gibbs energy change<\/strong>, \u0394G = \u0394H \u2212 T\u0394S_system, which is \u2212T \u00d7 \u0394S_total in kJ.<\/p>\n<p>You will meet it as the standard way of writing the feasibility condition on page 3: a positive \u0394S_total is the same statement as a negative \u0394G.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Exam wording:<\/strong> &#8220;\u0394S_total = \u0394S_system + \u0394S_surroundings = +160.4 + (\u2212597) = \u2212437 J K\u207b\u00b9 mol\u207b\u00b9. \u0394S_total is negative, so the reaction is not feasible at 298 K.&#8221; Give both terms, the sum, its sign and the conclusion.<\/p>\n<\/div>\n<\/article>\n<article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">4<\/div>\n<h2>Common Exam Points<\/h2>\n<\/div>\n<h3>Say<\/h3>\n<ul class=\"ols-list\">\n<li>&#8220;\u0394S_system = \u03a3S(products) \u2212 \u03a3S(reactants), including the elements.&#8221;<\/li>\n<li>&#8220;\u0394S_surroundings = \u2212\u0394H\/T, with \u0394H in J and T in K.&#8221;<\/li>\n<li>&#8220;\u0394S_total is positive, so the reaction is feasible.&#8221;<\/li>\n<\/ul>\n<h3>Do not say<\/h3>\n<ul class=\"ols-list\">\n<li>&#8220;Elements have an entropy of zero&#8221; (only their enthalpy of formation is zero).<\/li>\n<li>&#8220;T = 25&#8221; (use kelvin: 298 K).<\/li>\n<li>&#8220;\u0394S_surroundings = \u2212178 \u00f7 298&#8221; (\u0394H must be in joules: \u2212178 000 \u00f7 298).<\/li>\n<\/ul>\n<h3>Watch for<\/h3>\n<ul class=\"ols-list\">\n<li>The most common slip in this topic is <strong>units<\/strong>: entropies are in J K\u207b\u00b9 mol\u207b\u00b9 and enthalpies in kJ mol\u207b\u00b9, so one of them must be converted before they are combined.<\/li>\n<li>Check that the sign of \u0394S_system matches the change in moles of gas.<\/li>\n<li>Check that a temperature given in \u00b0C has had 273 added.<\/li>\n<\/ul>\n<\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Check: A Full Calculation<\/h2>\n<p>Carry out a complete calculation of \u0394S_system, \u0394S_surroundings and \u0394S_total and decide whether a reaction not used on this page is feasible.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1104\"><\/div><\/div>\n<\/section>\n<section class=\"ols-faq-card\">\n<h2>FAQs<\/h2>\n<p>Use these quick answers to check the entropy calculations.<\/p>\n\n<div class=\"ols-faq-list\">\n<div class=\"ols-faq-item\">\n<h3>Why must I convert J to kJ?<\/h3>\n<p>Because standard entropies are tabulated in J K\u207b\u00b9 mol\u207b\u00b9 while enthalpy changes are in kJ mol\u207b\u00b9. Whenever the two meet in one equation they must share a unit, so either divide the entropy term by 1000 or multiply \u0394H by 1000. Mixing them is the most common reason for an answer that is out by a factor of a thousand.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why do I have to multiply each standard entropy by the balancing number?<\/h3>\n<p>Because standard entropy is quoted per mole of substance, and the equation may involve two or three moles. In 2H\u2082(g) + O\u2082(g) \u2192 2H\u2082O(l) the hydrogen contributes 2 \u00d7 131 J K\u207b\u00b9 mol\u207b\u00b9 and the water 2 \u00d7 70 J K\u207b\u00b9 mol\u207b\u00b9. Forgetting the multiplier changes both the size and, sometimes, the sign of \u0394S.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why do I divide by T in \u0394Ssurroundings = \u2212\u0394H\/T?<\/h3>\n<p>Because the entropy change caused by adding a given amount of heat depends on how hot the surroundings already are. The same quantity of heat spreads energy among far more extra arrangements in cold surroundings than in hot ones, so the entropy gain is bigger at low temperature. Dividing by the temperature in kelvin builds that in, and the minus sign is there because heat given out by the system (negative \u0394H) is heat gained by the surroundings.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why does \u0394Ssurroundings have the opposite sign to \u0394H?<\/h3>\n<p>Because an exothermic reaction (negative \u0394H) gives heat to the surroundings, which increases their entropy, so \u0394Ssurroundings is positive. An endothermic reaction takes heat in and lowers the entropy of the surroundings. Sort out the sign first and then check that the number you get agrees with it.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>What temperature do I use if the question does not say?<\/h3>\n<p>Standard conditions, which means 298 K. Always convert Celsius to kelvin by adding 273 before substituting; an entropy calculation at 25 K instead of 298 K gives nonsense. If the question sets a different temperature, use that one throughout.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"ols-related-card\">\n<h2>Related Topic 13 Energetics II Pages<\/h2>\n<p>Use these pages to connect the ideas across Topic 13 Energetics II 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\/topic-13-energetics-ii\/lattice-energy-and-born-haber-cycles\/\">Lattice Energy and Born\u2013Haber Cycles<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/lattice-energy-trends-and-covalent-character\/\">Lattice Energy Trends and Covalent Character<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/enthalpy-of-solution-and-hydration\/\">Enthalpy of Solution and Hydration<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/entropy-and-the-direction-of-change\/\">Entropy and the Direction of Change<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/feasibility-gibbs-energy-and-temperature\/\">Feasibility, Gibbs Energy and Temperature<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/thermodynamic-and-kinetic-stability\/\">Thermodynamic and Kinetic Stability<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel\/topic-13-energetics-ii\/\">Topic 13 Energetics II Overview<\/a>\n<\/div>\n<\/section>\n<section class=\"ols-attribution-card\">\n        <p><strong>Copyright and author footprint:<\/strong> This OLS revision page was written for Online Learning System by <strong>Dr. Mohammed Al-Fatah<\/strong>. It is designed for A Level Chemistry revision and should not be copied or redistributed without permission.<\/p>\n      <\/section>\n\n      <div class=\"ols-image-lightbox\" id=\"olsImageLightboxNatureCovalentBonding9ch0\" aria-hidden=\"true\" role=\"dialog\" aria-modal=\"true\" aria-label=\"Expanded revision image\">\n        <div class=\"ols-image-lightbox-inner\">\n          <button class=\"ols-image-lightbox-close\" type=\"button\" aria-label=\"Close enlarged image\">\u00d7<\/button>\n          <img decoding=\"async\" class=\"ols-image-lightbox-img\" src=\"\" alt=\"\">\n        <\/div>\n      <\/div>\n\n      <script>\n        (function(){\n          var page = document.querySelector(\".ols-nature-of-covalent-bonding-9ch0-page\");\n          if (!page) { return; }\n          var lightbox = page.querySelector(\"#olsImageLightboxNatureCovalentBonding9ch0\");\n          if (!lightbox) { return; }\n          var lightboxImage = lightbox.querySelector(\".ols-image-lightbox-img\");\n    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