{"id":8373,"date":"2026-09-18T12:01:39","date_gmt":"2026-09-18T11:01:39","guid":{"rendered":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/hess-law-and-enthalpy-cycles\/"},"modified":"2026-09-24T14:58:49","modified_gmt":"2026-09-24T13:58:49","slug":"hess-law-and-enthalpy-cycles","status":"publish","type":"page","link":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/hess-law-and-enthalpy-cycles\/","title":{"rendered":"Hess&#8217;s Law and Enthalpy Cycles"},"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      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.ols-figure-placeholder .ols-figure-caption p { margin: 10px 0 0; font-size: 14px; color: #667085; text-align: center; }\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 5 Chemical Energetics<\/h4>\n\n    <ul class=\"ols-topic-list\">\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/\">Topic 5 Overview<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/enthalpy-changes-and-standard-conditions\/\">Enthalpy Changes and Standard Conditions<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/standard-enthalpy-changes\/\">Standard Enthalpy Changes<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/calorimetry-measuring-enthalpy-changes\/\">Calorimetry: Measuring Enthalpy Changes<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/hess-law-and-enthalpy-cycles\/\">Hess's Law and Enthalpy Cycles<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/bond-energies\/\">Bond Energies<\/a>\n      <\/li>\n    <\/ul>\n  <\/div>\n\n  <div class=\"ols-topic-group\">\n    <h4>Other Topics<\/h4>\n\n    <ul class=\"ols-topic-list\">\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/\">Topic 2 Atoms, Molecules and Stoichiometry<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-3-chemical-bonding\/\">Topic 3 Chemical Bonding<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-4-states-of-matter\/\">Topic 4 States of Matter<\/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 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href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/\">Cambridge International (CIE)<\/a> \/\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/\">Topic 5 Chemical Energetics<\/a> \/\n<span>Hess&#8217;s Law and Enthalpy Cycles<\/span>\n<\/nav>\n\n      <header class=\"ols-title-card\">\n        <h1>Hess&#8217;s Law and Enthalpy Cycles<\/h1>\n        <p class=\"ols-page-intro\">A concise revision guide to Hess&#8217;s law: building enthalpy cycles, calculating enthalpy changes from formation and combustion data, and using unfamiliar cycles, for Cambridge International A Level Chemistry.<\/p>\n\n        <div class=\"ols-badges\">\n<div class=\"ols-badge\">AS Level<\/div>\n<div class=\"ols-badge\">Topic 5: Chemical Energetics<\/div>\n<div class=\"ols-badge\">9701 Papers 1 and 2<\/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      <section class=\"ols-h5p-card ols-h5p-inline ols-h5p-recap\">\n<span class=\"ols-h5p-kicker\">Before you start<\/span>\n<h2>GCSE Recap: Energy Is Conserved<\/h2>\n<p>Before you start, check three GCSE ideas that every enthalpy cycle depends on.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"769\"><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card soft\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">1<\/div>\n<h2>Hess&#8217;s Law<\/h2>\n<\/div>\n<p><strong>Hess&#8217;s law<\/strong> states that the total enthalpy change for a reaction is independent of the route taken, provided the starting and finishing conditions are the same. It follows from the conservation of energy: if two routes between the same reactants and products gave different energy changes, energy could be created by going round the cycle.<\/p>\n<p>This allows enthalpy changes that cannot be measured directly, such as the enthalpy change of formation of methane or the hydration of an anhydrous salt, to be calculated from values that can be measured.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Definition:<\/strong> Hess&#8217;s law: the enthalpy change for a reaction is independent of the route taken from reactants to products, provided the starting and finishing conditions are the same.<\/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>Quick Check: Which Values Need Hess&#039;s Law?<\/h2>\n<p>Click every compound that cannot be made cleanly from its elements in one reaction.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-561\" class=\"h5p-iframe\" data-content-id=\"561\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Hess&#039;s Law Mark the Words: Which Formation Enthalpies Cannot Be Measured\"><\/iframe><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">2<\/div>\n<h2>Building an Enthalpy Cycle<\/h2>\n<\/div>\n<p>An <strong>enthalpy cycle<\/strong> links the reaction you want to a set of reactions whose enthalpy changes are known. Draw the target reaction across the top, then an alternative route through a common intermediate.<\/p>\n<ol>\n<li>Write the balanced target equation across the top with an arrow labelled with the unknown \u0394H.<\/li>\n<li>Choose the connecting species: the <strong>elements<\/strong> when you have enthalpies of formation, or the <strong>combustion products<\/strong> when you have enthalpies of combustion.<\/li>\n<li>Draw arrows from each side to the connecting box and label them with the known values, multiplied by the number of moles in the equation.<\/li>\n<li>Take any route you like from the reactants to the products: follow an arrow and add its value, go against an arrow and subtract it.<\/li>\n<\/ol>\n<div class=\"ols-figure-card\">\n<div class=\"ols-figure-image\"><img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/energetics-3-fixed.jpg\" alt=\"a formation cycle (elements in a box at the bottom, arrows up) beside a combustion cycle (CO\u2082 and H\u2082O in a box at the bottom, arrows down) for the same reaction\"><\/div>\n<div class=\"ols-figure-caption\"><p>Formation arrows point up from the elements; combustion arrows point down to the combustion products. The direction decides whether each value is added or subtracted.<\/p><\/div>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Exam focus:<\/strong> Show the cycle and the working. A correct final answer with no cycle scores less than a cycle with one arithmetic slip.<\/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>Quick Check: Follow the Arrows<\/h2>\n<p>Drag the words and numbers into place to complete a cycle you have not seen before.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-558\" class=\"h5p-iframe\" data-content-id=\"558\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Hess&#039;s Law Drag: Reading the Arrows on a Zinc Oxide Cycle\"><\/iframe><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card purple\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">3<\/div>\n<h2>Using Enthalpy Changes of Formation<\/h2>\n<\/div>\n<p>When the data are enthalpy changes of formation, the cycle reduces to one rule:<\/p>\n<p><strong>\u0394H<sub>r<\/sub><sup>\u29b5<\/sup> = \u03a3\u0394H<sub>f<\/sub><sup>\u29b5<\/sup>(products) \u2212 \u03a3\u0394H<sub>f<\/sub><sup>\u29b5<\/sup>(reactants)<\/strong><\/p>\n<p><strong>Worked example.<\/strong> Calculate the enthalpy change for the thermal decomposition of sodium hydrogencarbonate.<\/p>\n<p>2NaHCO\u2083(s) \u2192 Na\u2082CO\u2083(s) + H\u2082O(l) + CO\u2082(g)<\/p>\n<div class=\"ols-table-wrap\">\n<table class=\"ols-table\">\n<thead>\n<tr><th>Substance<\/th><th>\u0394H<sub>f<\/sub><sup>\u29b5<\/sup> \/ kJ mol\u207b\u00b9<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><td><strong>NaHCO\u2083(s)<\/strong><\/td><td>\u2212951<\/td><\/tr>\n<tr><td><strong>Na\u2082CO\u2083(s)<\/strong><\/td><td>\u22121131<\/td><\/tr>\n<tr><td><strong>H\u2082O(l)<\/strong><\/td><td>\u2212286<\/td><\/tr>\n<tr><td><strong>CO\u2082(g)<\/strong><\/td><td>\u2212394<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<ol>\n<li>\u03a3\u0394H<sub>f<\/sub><sup>\u29b5<\/sup>(products) = \u22121131 + (\u2212286) + (\u2212394) = \u22121811 kJ mol\u207b\u00b9<\/li>\n<li>\u03a3\u0394H<sub>f<\/sub><sup>\u29b5<\/sup>(reactants) = 2 \u00d7 (\u2212951) = \u22121902 kJ mol\u207b\u00b9<\/li>\n<li>\u0394H<sub>r<\/sub><sup>\u29b5<\/sup> = \u22121811 \u2212 (\u22121902) = <strong>+91 kJ mol\u207b\u00b9<\/strong><\/li>\n<\/ol>\n<div class=\"ols-key-box\">\n<p><strong>Common mistake:<\/strong> Forgetting to multiply by the coefficient (2 \u00d7 NaHCO\u2083) or giving elements a non-zero value. Elements in their standard states have an enthalpy change of formation of zero. Check the <strong>state symbols<\/strong> too: the value taken from the data must be for the state written in the equation, because H\u2082O(l) and H\u2082O(g) do not have the same value.<\/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>Quick Check: Coefficients and Elements<\/h2>\n<p>Use the formation data given in the question to find the enthalpy change for the equation.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-559\" class=\"h5p-iframe\" data-content-id=\"559\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Hess&#039;s Law MCQ: Burning Ammonia from Formation Data\"><\/iframe><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">4<\/div>\n<h2>Using Enthalpy Changes of Combustion<\/h2>\n<\/div>\n<p>When the data are enthalpy changes of combustion, the arrows point the other way, so the rule is reversed:<\/p>\n<p><strong>\u0394H<sub>r<\/sub><sup>\u29b5<\/sup> = \u03a3\u0394H<sub>c<\/sub><sup>\u29b5<\/sup>(reactants) \u2212 \u03a3\u0394H<sub>c<\/sub><sup>\u29b5<\/sup>(products)<\/strong><\/p>\n<p><strong>Worked example.<\/strong> Calculate the enthalpy change of formation of propane from the enthalpies of combustion of carbon (\u2212394), hydrogen (\u2212286) and propane (\u22122220 kJ mol\u207b\u00b9).<\/p>\n<p>3C(s) + 4H\u2082(g) \u2192 C\u2083H\u2088(g)<\/p>\n<ol>\n<li>\u03a3\u0394H<sub>c<\/sub><sup>\u29b5<\/sup>(reactants) = 3 \u00d7 (\u2212394) + 4 \u00d7 (\u2212286) = \u22121182 \u2212 1144 = \u22122326 kJ mol\u207b\u00b9<\/li>\n<li>\u03a3\u0394H<sub>c<\/sub><sup>\u29b5<\/sup>(products) = \u22122220 kJ mol\u207b\u00b9<\/li>\n<li>\u0394H<sub>f<\/sub><sup>\u29b5<\/sup>(C\u2083H\u2088) = \u22122326 \u2212 (\u22122220) = <strong>\u2212106 kJ mol\u207b\u00b9<\/strong><\/li>\n<\/ol>\n<p>This is exactly the kind of value that cannot be measured directly, because carbon and hydrogen do not react to give propane alone.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Key idea:<\/strong> Formation: products minus reactants. Combustion: reactants minus products. If you draw the cycle, you never need to remember which.<\/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>Quick Check: Order the Route<\/h2>\n<p>Drag the six steps of this calculation into the order you would carry them out.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-560\" class=\"h5p-iframe\" data-content-id=\"560\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Hess&#039;s Law Sort: The Route to the Formation Enthalpy of Methanol\"><\/iframe><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card soft\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">5<\/div>\n<h2>Unfamiliar Cycles and Practical Hess<\/h2>\n<\/div>\n<p>Exam questions often give a cycle you have not seen before. The method does not change: identify the route, follow the arrows, reverse the sign of any step you travel against, and multiply by the moles.<\/p>\n<p>A classic practical use is finding the enthalpy change for the hydration of anhydrous copper(II) sulfate, CuSO\u2084(s) + 5H\u2082O(l) \u2192 CuSO\u2084\u00b75H\u2082O(s), which cannot be measured directly because the solid cannot be given exactly five moles of water without dissolving. Instead, the enthalpy changes of solution of the anhydrous and hydrated salts are measured separately, and Hess&#8217;s law combines them. Both dissolve to give the same solution, so that solution is the common intermediate. Measuring gives \u0394H of solution of CuSO\u2084(s) as \u221266.5 kJ mol\u207b\u00b9 and of CuSO\u2084\u00b75H\u2082O(s) as +11.7 kJ mol\u207b\u00b9. Travelling with the first arrow and against the second gives \u0394H = \u221266.5 \u2212 (+11.7) = <strong>\u221278.2 kJ mol\u207b\u00b9<\/strong> for the hydration.<\/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\/Using-Hesss-law-to-find-enthalpy-changes-v2.webp\" 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\/Using-Hesss-law-to-find-enthalpy-changes-v2.webp\" alt=\"Hess\u2019s law cycle used to calculate enthalpy changes from related reactions\">\n<\/a>\n<\/div>\n<div class=\"ols-zoom-card-caption\"><p>Hess&#8217;s law applied in the laboratory: two measurable enthalpy changes of solution give the hydration enthalpy that cannot be measured directly.<\/p><\/div>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Exam sentence:<\/strong> The enthalpy change cannot be measured directly, so Hess&#8217;s law is used: the enthalpy changes of two reactions that can be measured are combined, because the total enthalpy change is independent of the route.<\/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>Quick Check: Four Cycles to Work Out<\/h2>\n<p>Work each calculation out on paper in full, then flip the card to compare your cycle and your answer.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-770\" class=\"h5p-iframe\" data-content-id=\"770\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Hess&#039;s Law Flip Cards: Four Cycles, Four Unknowns\"><\/iframe><\/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>State Hess&#8217;s law<\/h3><p>The enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.<\/p>\n<h3>Calculate an enthalpy change from formation data<\/h3><p>Products minus reactants, each multiplied by its coefficient; elements count as zero.<\/p>\n<h3>Calculate an enthalpy change from combustion data<\/h3><p>Reactants minus products, each multiplied by its coefficient.<\/p>\n<h3>Explain why a value must be found indirectly<\/h3><p>The reaction does not happen cleanly on its own, or gives other products, so it cannot be measured by experiment.<\/p>\n<\/article>\n\n<section class=\"ols-faq-card\">\n<h2>FAQs<\/h2>\n<p>Use these quick answers to check the Hess&#8217;s law methods Cambridge International questions test.<\/p>\n\n<div class=\"ols-faq-list\">\n<div class=\"ols-faq-item\">\n<h3>What does Hess&#8217;s law say?<\/h3>\n<p>The enthalpy change for a reaction is independent of the route taken, provided the starting and finishing conditions are the same.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>When do I use products minus reactants?<\/h3>\n<p>When the data are enthalpy changes of formation, because the arrows point up from the elements to both sides.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>When do I use reactants minus products?<\/h3>\n<p>When the data are enthalpy changes of combustion, because the arrows point down from both sides to carbon dioxide and water.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>What happens to the sign when I go against an arrow?<\/h3>\n<p>It is reversed. Going against a step is the reverse reaction, so its enthalpy change has the opposite sign.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why use Hess cycles at all?<\/h3>\n<p>Many enthalpy changes cannot be measured directly because the reaction does not happen cleanly, so they are calculated from values that can be measured.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"ols-related-card\">\n<h2>Related Energetics Pages<\/h2>\n<p>Use these pages to connect enthalpy definitions, measurement and calculation methods across the topic.<\/p>\n<div class=\"ols-related-grid\">\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/enthalpy-changes-and-standard-conditions\/\">Enthalpy Changes and Standard Conditions<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/standard-enthalpy-changes\/\">Standard Enthalpy Changes<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/calorimetry-measuring-enthalpy-changes\/\">Calorimetry: Measuring Enthalpy Changes<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-5-chemical-energetics\/bond-energies\/\">Bond Energies<\/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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Chemistry. 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