{"id":8342,"date":"2026-09-18T12:00:38","date_gmt":"2026-09-18T11:00:38","guid":{"rendered":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-6-energetics\/standard-enthalpy-changes\/"},"modified":"2026-10-04T23:02:15","modified_gmt":"2026-10-04T22:02:15","slug":"standard-enthalpy-changes","status":"publish","type":"page","link":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-6-energetics\/standard-enthalpy-changes\/","title":{"rendered":"Standard Enthalpy Changes"},"content":{"rendered":"\n<section class=\"ols-revision-page ols-energetics-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: #667085;\n      --body-text: #1f2937;\n      --border: rgba(28, 36, 75, 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href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-6-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\/edexcel-international\/topic-6-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\/edexcel-international\/topic-6-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\/edexcel-international\/topic-6-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\/edexcel-international\/topic-6-energetics\/bond-enthalpies\/\">Bond Enthalpies<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/core-practicals\/cp-2-enthalpy-change-via-hess-law\/\">Core Practical 2<\/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\/edexcel-international\/topic-3-bonding-structure\/\">Bonding &amp; Structure<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-5-alkenes\/\">Alkenes<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-7-intermolecular-forces\/\">Intermolecular Forces<\/a>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-8a-redox-chemistry\/\">Redox Chemistry<\/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, 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href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-6-energetics\/\">Topic 6 Energetics<\/a> \/\n<span>Standard Enthalpy Changes<\/span>\n<\/nav>\n\n      <header class=\"ols-title-card\">\n        <h1>Standard Enthalpy Changes<\/h1>\n        <p class=\"ols-page-intro\">A concise revision guide to the standard enthalpy changes of reaction, formation, combustion and neutralisation and atomisation, with the equation that goes with each definition, for Edexcel International A Level Chemistry.<\/p>\n        <div class=\"ols-badges\">\n<div class=\"ols-badge\">Exam board: Edexcel International<\/div>\n<div class=\"ols-badge\">Unit 2: WCH12\/01<\/div>\n<div class=\"ols-badge\">Topic 6: Energetics<\/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>GCSE Recap: Burning and Neutralising<\/h2>\n<p>Before you start, check what complete combustion and neutralisation actually produce.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"771\"><\/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>Why the Wording Matters<\/h2>\n<\/div>\n<p>Each <strong>standard enthalpy change<\/strong> is defined by exactly what is formed, burned or produced, and how much of it. Mark schemes award a mark for each part of a definition, so a definition that leaves out &#8220;one mole&#8221;, &#8220;standard states&#8221; or &#8220;completely&#8221; loses marks even when the chemistry is understood.<\/p>\n<div class=\"ols-table-wrap\">\n<table class=\"ols-table\">\n<thead>\n<tr><th>Enthalpy change<\/th><th>Symbol<\/th><th>Definition<\/th><th>Sign<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><td><strong>Reaction<\/strong><\/td><td>\u0394<sub>r<\/sub>H<sup>\u29b5<\/sup><\/td><td>the enthalpy change for the molar quantities in a stated equation, under standard conditions, with all substances in their standard states<\/td><td>varies<\/td><\/tr>\n<tr><td><strong>Formation<\/strong><\/td><td>\u0394<sub>f<\/sub>H<sup>\u29b5<\/sup><\/td><td>the enthalpy change when one mole of a compound is formed from its elements under standard conditions, all substances in their standard states<\/td><td>usually negative; zero for an element<\/td><\/tr>\n<tr><td><strong>Combustion<\/strong><\/td><td>\u0394<sub>c<\/sub>H<sup>\u29b5<\/sup><\/td><td>the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions, all substances in their standard states<\/td><td>always negative<\/td><\/tr>\n<tr><td><strong>Neutralisation<\/strong><\/td><td>\u0394<sub>neut<\/sub>H<sup>\u29b5<\/sup><\/td><td>the enthalpy change when one mole of water is formed in a reaction between an acid and a base in dilute aqueous solution under standard conditions<\/td><td>always negative<\/td><\/tr>\n<tr><td><strong>Atomisation<\/strong><\/td><td>\u0394<sub>at<\/sub>H<sup>\u29b5<\/sup><\/td><td>the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state under standard conditions<\/td><td>always positive<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Exam focus:<\/strong> Learn each definition word for word, then check your version has three parts: the quantity (one mole of what), the process, and standard conditions with standard states.<\/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\/t17x-s3definitions.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\/t17x-s3definitions.jpg\" alt=\"Five cards giving the definitions, symbols and signs of reaction, formation, combustion, neutralisation and atomisation enthalpy changes.\" data-fullscreen-src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/09\/t17x-s3definitions.jpg\">\n<\/a>\n<\/div>\n<div class=\"ols-figure-caption\"><p>Each standard enthalpy change fixes one mole of something different, and a full definition names the quantity, the process and standard conditions.<\/p><\/div>\n<\/div>\n<\/article>\n<article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">2<\/div>\n<h2>Enthalpy Change of Reaction<\/h2>\n<\/div>\n<p>The <strong>enthalpy change of reaction<\/strong> refers to the amounts written in a particular equation. If the equation is doubled, the enthalpy change doubles; if it is reversed, the sign changes. For example:<\/p>\n<p>N\u2082(g) + 3H\u2082(g) \u2192 2NH\u2083(g) &nbsp; \u0394<sub>r<\/sub>H<sup>\u29b5<\/sup> = \u221292 kJ mol\u207b\u00b9<\/p>\n<p>\u00bdN\u2082(g) + 1\u00bdH\u2082(g) \u2192 NH\u2083(g) &nbsp; \u0394<sub>r<\/sub>H<sup>\u29b5<\/sup> = \u221246 kJ mol\u207b\u00b9<\/p>\n<p>2NH\u2083(g) \u2192 N\u2082(g) + 3H\u2082(g) &nbsp; \u0394<sub>r<\/sub>H<sup>\u29b5<\/sup> = +92 kJ mol\u207b\u00b9<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Key idea:<\/strong> kJ mol\u207b\u00b9 here means &#8220;per mole of the equation as written&#8221;, so always quote the equation beside the 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: Change the Equation, Change the Value<\/h2>\n<p>Five quick questions on what happens to a value when the equation is halved, doubled or reversed.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-772\" class=\"h5p-iframe\" data-content-id=\"772\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Standard Enthalpy Changes Quick Fire: Doubling, Halving and Reversing an Equation\"><\/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>Enthalpy Change of Formation<\/h2>\n<\/div>\n<p>The <strong>enthalpy change of formation<\/strong> is for one mole of a compound made from its <strong>elements in their standard states<\/strong>. The equation must show exactly one mole of product, which often means fractions in front of the elements:<\/p>\n<p>2C(s) + 3H\u2082(g) + \u00bdO\u2082(g) \u2192 C\u2082H\u2085OH(l) &nbsp; \u0394<sub>f<\/sub>H<sup>\u29b5<\/sup> = \u2212277 kJ mol\u207b\u00b9<\/p>\n<p>Because nothing changes when an element is &#8220;formed&#8221; from itself, the enthalpy change of formation of any element in its standard state is <strong>zero<\/strong>.<\/p><p>The <strong>standard state<\/strong> is the state the element is in at 298 K and 100 kPa, so settle the state symbol from those conditions before writing the equation.<\/p><p>Many formation reactions cannot be carried out directly, because the elements would react to give other products. Their values are found indirectly using Hess&#8217;s law.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Common mistake:<\/strong> Writing 4C(s) + 6H\u2082(g) + O\u2082(g) \u2192 2C\u2082H\u2085OH(l). That equation balances, but it makes two moles of ethanol. The definition fixes one mole of product, so halve that equation to give 2C(s) + 3H\u2082(g) + \u00bdO\u2082(g) \u2192 C\u2082H\u2085OH(l).<\/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: Write the Formation Equation<\/h2>\n<p>Drag the missing amounts and formulae into four equations the page has not shown you.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-553\" class=\"h5p-iframe\" data-content-id=\"553\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Standard Enthalpy Changes Drag: Building Four Formation Equations\"><\/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>Enthalpy Change of Combustion<\/h2>\n<\/div>\n<p>The <strong>enthalpy change of combustion<\/strong> is for one mole of a substance <strong>completely<\/strong> burned in oxygen. Complete combustion of a hydrocarbon or alcohol gives carbon dioxide and water only; incomplete combustion also forms carbon monoxide or soot and releases less energy.<\/p>\n<p>C\u2083H\u2088(g) + 5O\u2082(g) \u2192 3CO\u2082(g) + 4H\u2082O(l) &nbsp; \u0394<sub>c<\/sub>H<sup>\u29b5<\/sup> = \u22122220 kJ mol\u207b\u00b9<\/p>\n<p>Combustion is always exothermic, so these values are always negative. The equation must show one mole of the fuel, which again may need a fractional amount of oxygen: C\u2082H\u2086(g) + 3\u00bdO\u2082(g) \u2192 2CO\u2082(g) + 3H\u2082O(l).<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Remember:<\/strong> Water is H\u2082O(l) under standard conditions. Writing H\u2082O(g) changes the 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: Which Equation Is the Right One?<\/h2>\n<p>Only one of these four equations matches the definition exactly.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-551\" class=\"h5p-iframe\" data-content-id=\"551\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Standard Enthalpy Changes MCQ: The Combustion Equation for Ethanal\"><\/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>Enthalpy Change of Neutralisation<\/h2>\n<\/div>\n<p>The <strong>enthalpy change of neutralisation<\/strong> is for the formation of one mole of water, H\u207a(aq) + OH\u207b(aq) \u2192 H\u2082O(l). For any strong acid with any strong base the value is about \u221257 kJ mol\u207b\u00b9, because the only reaction taking place is the same combination of hydrogen and hydroxide ions.<\/p>\n<p>A weak acid such as ethanoic acid gives a slightly less exothermic value, because some energy is used to complete its ionisation as the reaction proceeds.<\/p><p>Sulfuric acid is diprotic, so one mole of H\u2082SO\u2084 forms two moles of water. The enthalpy change for the equation is doubled, but the enthalpy change of neutralisation is still quoted per mole of water.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Exam trap:<\/strong> The definition is per mole of water, not per mole of acid. With sulfuric acid, divide the energy by the moles of water formed.<\/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: Pick the Accurate Statement<\/h2>\n<p>In each round, choose the one statement that is accurate.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-552\" class=\"h5p-iframe\" data-content-id=\"552\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Standard Enthalpy Changes Summary: Per Mole of Water, Not Per Mole of Acid\"><\/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>Enthalpy Change of Atomisation<\/h2>\n<\/div>\n<p>The <strong>enthalpy change of atomisation<\/strong> is for the formation of one mole of <strong>gaseous atoms<\/strong> from an element in its standard state. Bonds or forces between particles are always broken, so atomisation is always endothermic.<\/p>\n<p>\u00bdCl\u2082(g) \u2192 Cl(g) &nbsp; \u0394<sub>at<\/sub>H<sup>\u29b5<\/sup> = +122 kJ mol\u207b\u00b9<\/p>\n<p>Na(s) \u2192 Na(g) &nbsp; \u0394<sub>at<\/sub>H<sup>\u29b5<\/sup> = +107 kJ mol\u207b\u00b9<\/p>\n<p>Notice the \u00bd in front of Cl\u2082: the definition fixes one mole of atoms, not one mole of molecules. For a diatomic <strong>gas<\/strong> the atomisation enthalpy is therefore half the bond enthalpy.<\/p>\n<p>It is not half the bond enthalpy for bromine, whose standard state is a liquid. The equation is \u00bdBr\u2082(l) \u2192 Br(g), so the energy needed to vaporise the liquid is included as well.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Key idea:<\/strong> Atomisation always produces gaseous atoms, so check the state symbol of the product is (g) and the amount is one mole of atoms.<\/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: Atomising a Solid Element<\/h2>\n<p>Apply the definition to an element the page has not used.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-773\" class=\"h5p-iframe\" data-content-id=\"773\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Standard Enthalpy Changes MCQ: The Atomisation Equation for Iodine\"><\/iframe><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card purple\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">7<\/div>\n<h2>Writing the Equation for a Definition<\/h2>\n<\/div>\n<p>Most questions on definitions ask for an equation as well as words. Work backwards from what the definition fixes.<\/p>\n<div class=\"ols-table-wrap\">\n<table class=\"ols-table\">\n<thead>\n<tr><th>Enthalpy change<\/th><th>Fix one mole of<\/th><th>Example equation<\/th><\/tr>\n<\/thead>\n<tbody>\n<tr><td><strong>Formation<\/strong><\/td><td>the compound formed<\/td><td>Na(s) + \u00bdCl\u2082(g) \u2192 NaCl(s)<\/td><\/tr>\n<tr><td><strong>Combustion<\/strong><\/td><td>the substance burned<\/td><td>CH\u2083OH(l) + 1\u00bdO\u2082(g) \u2192 CO\u2082(g) + 2H\u2082O(l)<\/td><\/tr>\n<tr><td><strong>Neutralisation<\/strong><\/td><td>water formed<\/td><td>\u00bdH\u2082SO\u2084(aq) + NaOH(aq) \u2192 \u00bdNa\u2082SO\u2084(aq) + H\u2082O(l)<\/td><\/tr>\n<tr><td><strong>Atomisation<\/strong><\/td><td>gaseous atoms formed<\/td><td>\u00bdBr\u2082(l) \u2192 Br(g)<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\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-2.jpg\" alt=\"summary card of the four standard enthalpy changes (reaction, formation, combustion, neutralisation), each with its one-mole quantity highlighted and an example equation\"><\/div>\n<div class=\"ols-figure-caption\"><p>Each definition fixes one mole of something different; the highlighted species in each equation is the one that must be one mole.<\/p><\/div>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Exam sentence:<\/strong> The standard enthalpy change of formation of ethanol is the enthalpy change when one mole of ethanol is formed from its elements in their standard states under standard conditions.<\/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: Does the Equation Fit?<\/h2>\n<p>Decide whether each equation represents the standard enthalpy change it claims to.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"1067\"><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">8<\/div>\n<h2>Common Exam Points<\/h2>\n<\/div>\n<h3>Define the standard enthalpy change of combustion<\/h3><p>The enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions, all substances in their standard states.<\/p>\n<h3>Write the equation for the enthalpy change of formation<\/h3><p>One mole of product, elements in their standard states, fractions allowed on the left.<\/p>\n<h3>State the enthalpy change of formation of oxygen<\/h3><p>Zero, because it is an element in its standard state.<\/p>\n<h3>Do not say<\/h3><p>&#8220;One mole of reactants&#8221;; &#8220;burned in air&#8221; instead of &#8220;completely burned in oxygen&#8221;; &#8220;per mole of acid&#8221; for neutralisation.<\/p>\n<\/article>\n\n<section class=\"ols-faq-card\">\n<h2>FAQs<\/h2>\n<p>Use these quick answers to check the definitions Edexcel International expects word for word.<\/p>\n\n<div class=\"ols-faq-list\">\n<div class=\"ols-faq-item\">\n<h3>Why does the definition say &#8220;one mole&#8221;?<\/h3>\n<p>Each standard enthalpy change is fixed per mole of a particular species, so the equation has to show one mole of it, even if that needs fractions elsewhere.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why is the enthalpy change of formation of an element zero?<\/h3>\n<p>Forming an element in its standard state from itself involves no change, so no energy is transferred.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>What is the difference between combustion in oxygen and burning in air?<\/h3>\n<p>The definition requires complete combustion in oxygen, which gives only carbon dioxide and water from a hydrocarbon. Burning in air can be incomplete, which releases less energy.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why is the enthalpy change of neutralisation about the same for all strong acids and bases?<\/h3>\n<p>The reaction in each case is H\u207a(aq) + OH\u207b(aq) \u2192 H\u2082O(l), so the energy released per mole of water is the same, about \u221257 kJ mol\u207b\u00b9.<\/p>\n<\/div>\n\n<div class=\"ols-faq-item\">\n<h3>Why is the enthalpy change of atomisation always positive?<\/h3>\n<p>Forming gaseous atoms always involves breaking bonds or overcoming forces between particles, which needs energy.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"ols-related-card\">\n<h2>Related Topic 6 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\/edexcel-international\/topic-6-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\/edexcel-international\/topic-6-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\/edexcel-international\/topic-6-energetics\/hess-law-and-enthalpy-cycles\/\">Hess&#8217;s Law and Enthalpy Cycles<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-6-energetics\/bond-enthalpies\/\">Bond Enthalpies<\/a>\n<a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/core-practicals\/cp-2-enthalpy-change-via-hess-law\/\">CP2 Enthalpy Change via Hess&#8217;s Law<\/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-6-energetics\/standard-enthalpy-changes\/#webpage\",\n      \"url\": \"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/edexcel-international\/topic-6-energetics\/standard-enthalpy-changes\/\",\n      \"name\": \"Standard Enthalpy Changes | Topic 6 Energetics | Online Learning System\",\n      \"description\": \"Standard enthalpy changes for Edexcel International A Level Chemistry: definitions of reaction, formation, combustion and neutralisation and atomisation with example equations.\",\n      \"isPartOf\": {\n        \"@id\": 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