{"id":223239,"date":"2025-06-01T11:25:05","date_gmt":"2025-06-01T11:25:05","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223239"},"modified":"2025-06-01T11:25:08","modified_gmt":"2025-06-01T11:25:08","slug":"at-this-point-hesss-law-can-be-applied-to-determine-the-molar-heat-of-formation-of-mgo-by-using-the-data-collected-in-the-lab-and-the-literature-value-for-the-heat-of-formation-of-h2o","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/01\/at-this-point-hesss-law-can-be-applied-to-determine-the-molar-heat-of-formation-of-mgo-by-using-the-data-collected-in-the-lab-and-the-literature-value-for-the-heat-of-formation-of-h2o\/","title":{"rendered":"At this point, Hess&#8217;s law can be applied to determine the molar heat of formation of MgO by using the data collected in the lab and the literature value for the heat of formation of H2O."},"content":{"rendered":"\n<p>At this point, Hess&#8217;s law can be applied to determine the molar heat of formation of MgO by using the data collected in the lab and the literature value for the heat of formation of H2O. The equation is as follows: Mg(s) + 2HCl(aq) \u00e2\u2020\u2019 MgCl2(aq) + H2(g) + 1\/2O2(g) + H2O(l) To combine the three equations above and create the stoichiometric equation that shows the formation of MgO, we can cancel out the common species on both sides of the equations: Mg(s) + 2HCl(aq) \u00e2\u2020\u2019 MgCl2(aq) + H2(g) + 1\/2O2(g) + H2O(l) 2HCl(aq) + MgO(s) \u00e2\u2020\u2019 MgCl2(aq) + H2O(l) H2(g) + 1\/2O2(g) \u00e2\u2020\u2019 H2O(l) By adding these equations together, we get: Mg(s) + MgO(s) \u00e2\u2020\u2019 MgCl2(aq) + H2(g) + 1\/2O2(g) To determine the experimental heat of formation of MgO using Hess&#8217;s law, we need to calculate the enthalpy change for the reaction: \u00ce\u201dH_MgO = \u00ce\u201dH_1 + \u00ce\u201dH_2 + \u00ce\u201dH_3 where \u00ce\u201dH_1, \u00ce\u201dH_2, and \u00ce\u201dH_3 are the enthalpy changes for the three equations above.<br>At this point, Hess&#8217;s law can be applied to determine the molar heat of formation of MgO by using the data collected in the lab and the literature value for the heat of formation of H2O. The equation is as follows: Mg(s) + 2HCl(aq) \u00e2\u2020\u2019 MgCl2(aq) + H2(g) + 1\/2O2(g) + H2O(l) To combine the three equations above and create the stoichiometric equation that shows the formation of MgO, we can cancel out the common species on both sides of the equations: Mg(s) + 2HCl(aq) \u00e2\u2020\u2019 MgCl2(aq) + H2(g) + 1\/2O2(g) + H2O(l) 2HCl(aq) + MgO(s) \u00e2\u2020\u2019 MgCl2(aq) + H2O(l) H2(g) + 1\/2O2(g) \u00e2\u2020\u2019 H2O(l) By adding these equations together, we get: Mg(s) + MgO(s) \u00e2\u2020\u2019 MgCl2(aq) + H2(g) + 1\/2O2(g) To determine the experimental heat of formation of MgO using Hess&#8217;s law, we need to calculate the enthalpy change for the reaction: \u00ce\u201dH_MgO = \u00ce\u201dH_1 + \u00ce\u201dH_2 + \u00ce\u201dH_3 where \u00ce\u201dH_1, \u00ce\u201dH_2, and \u00ce\u201dH_3 are the enthalpy changes for the three equations above<\/p>\n\n\n\n<p><em><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/em><\/p>\n\n\n\n<p>To correctly determine the <strong>molar heat of formation (\u0394H_f\u00b0)<\/strong> of <strong>magnesium oxide (MgO)<\/strong> using <strong>Hess\u2019s Law<\/strong>, we must understand how to apply this principle to experimentally derived enthalpy data and known literature values. Below is the <strong>corrected approach<\/strong> and a <strong>comprehensive 1000-word explanation<\/strong> of the theory, methodology, chemical equations, and calculation process.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">\u2705 <strong>Correct Approach Using Hess\u2019s Law<\/strong><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Understand the Goal<\/strong><\/h3>\n\n\n\n<p>We want to determine the <strong>enthalpy of formation of MgO (\u0394H_f\u00b0 for MgO)<\/strong>: Target&nbsp;equation:&nbsp;Mg(s)&nbsp;+&nbsp;12O2(g)\u2192MgO(s)\\text{Target equation: } \\text{Mg(s) + } \\frac{1}{2}\\text{O}_2(g) \\rightarrow \\text{MgO(s)}<\/p>\n\n\n\n<p>This reaction forms <strong>1 mole of solid MgO from its elements in their standard states<\/strong>. This is the <strong>definition of enthalpy of formation<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Identify the Experimental Reactions<\/strong><\/h3>\n\n\n\n<p>In the lab, we perform two reactions:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Reaction 1 (R\u2081): Magnesium metal in hydrochloric acid<\/strong><\/h4>\n\n\n\n<p>Mg(s)&nbsp;+&nbsp;2HCl(aq)&nbsp;\u2192&nbsp;MgCl2(aq)+H2(g)(\u0394H1)\\text{Mg(s) + 2HCl(aq) } \\rightarrow \\text{ MgCl}_2(aq) + H_2(g) \\quad (\\Delta H_1)<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Reaction 2 (R\u2082): Magnesium oxide in hydrochloric acid<\/strong><\/h4>\n\n\n\n<p>MgO(s)&nbsp;+&nbsp;2HCl(aq)&nbsp;\u2192&nbsp;MgCl2(aq)+H2O(l)(\u0394H2)\\text{MgO(s) + 2HCl(aq) } \\rightarrow \\text{ MgCl}_2(aq) + H_2O(l) \\quad (\\Delta H_2)<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Reaction 3 (R\u2083): Hydrogen gas combusting to form water (known from literature)<\/strong><\/h4>\n\n\n\n<p>H2(g)+12O2(g)\u2192H2O(l)(\u0394H3=\u2212285.8&nbsp;kJ\/mol)\\text{H}_2(g) + \\frac{1}{2}O_2(g) \\rightarrow \\text{H}_2O(l) \\quad (\\Delta H_3 = -285.8 \\text{ kJ\/mol})<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Combine the Reactions Using Hess\u2019s Law<\/strong><\/h3>\n\n\n\n<p>To derive the formation equation for MgO, manipulate and combine the above reactions so that their sum gives the target equation:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step-by-step addition:<\/strong><\/h4>\n\n\n\n<p>From R\u2081: Mg(s)&nbsp;+&nbsp;2HCl(aq)\u2192MgCl2(aq)+H2(g)\\text{Mg(s) + 2HCl(aq)} \\rightarrow \\text{MgCl}_2(aq) + H_2(g)<\/p>\n\n\n\n<p>From R\u2082 (reverse this reaction): MgCl2(aq)+H2O(l)\u2192MgO(s)&nbsp;+&nbsp;2HCl(aq)(\u2212\u0394H2)\\text{MgCl}_2(aq) + H_2O(l) \\rightarrow \\text{MgO(s) + 2HCl(aq)} \\quad (-\\Delta H_2)<\/p>\n\n\n\n<p>From R\u2083: H2(g)+12O2(g)\u2192H2O(l)\\text{H}_2(g) + \\frac{1}{2}O_2(g) \\rightarrow \\text{H}_2O(l)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Add these equations:<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>R\u2081:<\/strong> Mg(s) + 2HCl(aq) \u2192 MgCl\u2082(aq) + H\u2082(g)<\/li>\n\n\n\n<li><strong>Reverse of R\u2082:<\/strong> MgCl\u2082(aq) + H\u2082O(l) \u2192 MgO(s) + 2HCl(aq)<\/li>\n\n\n\n<li><strong>R\u2083:<\/strong> H\u2082(g) + \u00bdO\u2082(g) \u2192 H\u2082O(l)<\/li>\n<\/ol>\n\n\n\n<p><strong>Add them together:<\/strong><\/p>\n\n\n\n<p><strong>Left side:<\/strong><br>Mg(s) + 2HCl(aq) + MgCl\u2082(aq) + H\u2082O(l) + H\u2082(g) + \u00bdO\u2082(g)<\/p>\n\n\n\n<p><strong>Right side:<\/strong><br>MgCl\u2082(aq) + H\u2082(g) + MgO(s) + 2HCl(aq) + H\u2082O(l)<\/p>\n\n\n\n<p>Cancel species common on both sides:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2HCl(aq), MgCl\u2082(aq), H\u2082(g), H\u2082O(l)<\/li>\n<\/ul>\n\n\n\n<p><strong>Final net reaction:<\/strong> Mg(s)&nbsp;+&nbsp;12O2(g)\u2192MgO(s)\\text{Mg(s) + } \\frac{1}{2}\\text{O}_2(g) \\rightarrow \\text{MgO(s)}<\/p>\n\n\n\n<p>This is our <strong>target equation<\/strong>. Thus, by Hess\u2019s Law: \u0394Hf\u2218(MgO)=\u0394H1\u2212\u0394H2+\u0394H3\\Delta H_{f}^{\\circ}(\\text{MgO}) = \\Delta H_1 &#8211; \\Delta H_2 + \\Delta H_3<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">\ud83e\uddea <strong>Enthalpy Calculations<\/strong><\/h2>\n\n\n\n<p>You will have measured <strong>\u0394H\u2081 and \u0394H\u2082 experimentally<\/strong> in the lab (likely using calorimetry), and \u0394H\u2083 is known from literature:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394H\u2083 = <strong>\u2013285.8 kJ\/mol<\/strong> (combustion of 1 mole of hydrogen gas)<\/li>\n<\/ul>\n\n\n\n<p>Suppose (example):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394H\u2081 (Mg + HCl) = \u2013445.6 kJ\/mol (measured)<\/li>\n\n\n\n<li>\u0394H\u2082 (MgO + HCl) = \u201374.4 kJ\/mol (measured)<\/li>\n<\/ul>\n\n\n\n<p>Then: \u0394Hf\u2218(MgO)=(\u2212445.6&nbsp;kJ\/mol)\u2212(\u221274.4&nbsp;kJ\/mol)+(\u2212285.8&nbsp;kJ\/mol)=\u2212445.6+74.4\u2212285.8=\u2212656.9&nbsp;kJ\/mol\\Delta H_f^{\\circ}(\\text{MgO}) = (-445.6 \\text{ kJ\/mol}) &#8211; (-74.4 \\text{ kJ\/mol}) + (-285.8 \\text{ kJ\/mol}) \\\\ = -445.6 + 74.4 &#8211; 285.8 = \\boxed{-656.9 \\text{ kJ\/mol}}<\/p>\n\n\n\n<p>This is close to the accepted value of <strong>\u2013601.6 kJ\/mol<\/strong>, and any deviation can be discussed as experimental error.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">\ud83d\udcd8 <strong>Theoretical Explanation <\/strong><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>What is Hess\u2019s Law?<\/strong><\/h3>\n\n\n\n<p>Hess&#8217;s Law states that the total enthalpy change of a reaction is <strong>independent of the pathway taken<\/strong>. If a chemical equation can be expressed as the sum of two or more other reactions, then the enthalpy change of that reaction is the <strong>sum of the enthalpy changes<\/strong> of the constituent reactions.<\/p>\n\n\n\n<p>Mathematically: \u0394Hnet=\u2211\u0394Hsteps\\Delta H_{\\text{net}} = \\sum \\Delta H_{\\text{steps}}<\/p>\n\n\n\n<p>This is possible because <strong>enthalpy is a state function<\/strong>\u2014its change depends only on the initial and final states of the system, not on how it gets from one to the other.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Why Can\u2019t We Measure \u0394H_f of MgO Directly?<\/strong><\/h3>\n\n\n\n<p>The reaction: Mg(s)&nbsp;+&nbsp;12O2(g)\u2192MgO(s)\\text{Mg(s) + } \\frac{1}{2}O_2(g) \\rightarrow \\text{MgO(s)}<\/p>\n\n\n\n<p>is highly exothermic and <strong>difficult to measure directly<\/strong> because:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>It happens too quickly.<\/li>\n\n\n\n<li>The heat released may be lost to the environment.<\/li>\n\n\n\n<li>The reaction may be incomplete or involve side reactions (e.g., formation of Mg(OH)\u2082).<\/li>\n<\/ul>\n\n\n\n<p>Instead, we use <strong>indirect methods<\/strong>, like <strong>calorimetry<\/strong>, and apply <strong>Hess\u2019s Law<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Purpose of Each Reaction<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>R\u2081 (Mg + HCl):<\/strong> Easy to carry out in a lab calorimeter. We measure temperature change to find \u0394H\u2081.<\/li>\n\n\n\n<li><strong>R\u2082 (MgO + HCl):<\/strong> Allows us to indirectly calculate the energy required to break down MgO, again via temperature change.<\/li>\n\n\n\n<li><strong>R\u2083 (H\u2082 combustion):<\/strong> Already known accurately from literature.<\/li>\n<\/ul>\n\n\n\n<p>These reactions are strategically chosen so that when summed, the <strong>desired target reaction<\/strong> emerges.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>How to Perform the Calorimetry<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Weigh a known mass<\/strong> of magnesium or magnesium oxide.<\/li>\n\n\n\n<li>React with <strong>known concentration and volume<\/strong> of HCl.<\/li>\n\n\n\n<li>Measure the <strong>initial and final temperature<\/strong> of the solution.<\/li>\n\n\n\n<li>Use the formula:<\/li>\n<\/ol>\n\n\n\n<p>q=mc\u0394Tq = mc\\Delta T<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>qq = heat evolved<\/li>\n\n\n\n<li>mm = mass of solution (assume density of 1.0 g\/mL)<\/li>\n\n\n\n<li>cc = specific heat capacity (typically 4.18 J\/g\u00b7\u00b0C)<\/li>\n\n\n\n<li>\u0394T\\Delta T = change in temperature<\/li>\n<\/ul>\n\n\n\n<p>Convert <strong>q<\/strong> to <strong>\u0394H in kJ\/mol<\/strong> by dividing by the number of moles of magnesium or MgO used.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Sources of Experimental Error<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Heat loss to the environment<\/strong> \u2192 leads to underestimating \u0394H.<\/li>\n\n\n\n<li><strong>Incomplete reaction<\/strong> \u2192 not all Mg or MgO reacts.<\/li>\n\n\n\n<li><strong>Calorimeter not perfectly insulated.<\/strong><\/li>\n\n\n\n<li><strong>Measurement errors<\/strong> in temperature or mass.<\/li>\n<\/ul>\n\n\n\n<p>These factors explain why your calculated \u0394H_f for MgO may differ from the literature value.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Significance of \u0394H_f(MgO)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Thermodynamics:<\/strong> Knowing how much heat is released helps in designing combustion processes.<\/li>\n\n\n\n<li><strong>Material Science:<\/strong> MgO is used in refractory materials; its heat stability is key.<\/li>\n\n\n\n<li><strong>Environmental Chemistry:<\/strong> Understanding formation heats is important in modeling atmospheric and geochemical cycles.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Conclusion<\/strong><\/h3>\n\n\n\n<p>By cleverly applying Hess\u2019s Law and combining laboratory measurements with literature values, we can accurately determine the <strong>enthalpy of formation<\/strong> of magnesium oxide, even when it is <strong>not directly measurable<\/strong>. This process demonstrates the <strong>power of indirect reasoning in chemistry<\/strong> and reinforces the importance of <strong>energy conservation<\/strong> in chemical reactions.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-4.jpeg\" alt=\"\" class=\"wp-image-223240\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>At this point, Hess&#8217;s law can be applied to determine the molar heat of formation of MgO by using the data collected in the lab and the literature value for the heat of formation of H2O. The equation is as follows: Mg(s) + 2HCl(aq) \u00e2\u2020\u2019 MgCl2(aq) + H2(g) + 1\/2O2(g) + H2O(l) To combine the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-223239","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223239","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=223239"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223239\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223239"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223239"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223239"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}