{"id":195441,"date":"2025-02-28T09:09:22","date_gmt":"2025-02-28T09:09:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=195441"},"modified":"2025-02-28T09:09:25","modified_gmt":"2025-02-28T09:09:25","slug":"the-combustion-of-heptane-c7h16","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/28\/the-combustion-of-heptane-c7h16\/","title":{"rendered":"The combustion of heptane, C7H16"},"content":{"rendered":"\n<p>The combustion of heptane, C7H16, occurs via the reaction<br>C7H16(g)+1102(g)\u21927CO2(g) + 8H2O(g)<br>with heat of formation values given by the following table:<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1635.png\" alt=\"\" class=\"wp-image-195442\"\/><\/figure>\n\n\n\n<p>Calculate the enthalpy for the combustion of 1 mole of heptane.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1636.png\" alt=\"\" class=\"wp-image-195443\"\/><\/figure>\n\n\n\n<p>To calculate the enthalpy change (\u0394H) for the combustion of 1 mole of heptane (C\u2087H\u2081\u2086), we use the standard enthalpies of formation (\u0394H_f\u00b0) of the reactants and products. The balanced chemical equation for the combustion is:<\/p>\n\n\n\n<p>C\u2087H\u2081\u2086(g) + 11O\u2082(g) \u2192 7CO\u2082(g) + 8H\u2082O(g)<\/p>\n\n\n\n<p>The standard enthalpy change of the reaction can be determined using Hess&#8217;s law, which states:<\/p>\n\n\n\n<p>\u0394H\u00b0_reaction = \u03a3(\u0394H_f\u00b0 of products) &#8211; \u03a3(\u0394H_f\u00b0 of reactants)<\/p>\n\n\n\n<p>Given the standard enthalpies of formation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394H_f\u00b0[C\u2087H\u2081\u2086(g)] = -187.9 kJ\/mol<\/li>\n\n\n\n<li>\u0394H_f\u00b0[CO\u2082(g)] = -393.5 kJ\/mol<\/li>\n\n\n\n<li>\u0394H_f\u00b0[H\u2082O(g)] = -241.8 kJ\/mol<\/li>\n\n\n\n<li>\u0394H_f\u00b0[O\u2082(g)] = 0 kJ\/mol (standard state)<\/li>\n<\/ul>\n\n\n\n<p>Calculating the total enthalpy of the products:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>7 moles of CO\u2082: 7 \u00d7 (-393.5 kJ\/mol) = -2,754.5 kJ<\/li>\n\n\n\n<li>8 moles of H\u2082O: 8 \u00d7 (-241.8 kJ\/mol) = -1,934.4 kJ<\/li>\n<\/ul>\n\n\n\n<p>Total enthalpy of products = -2,754.5 kJ + -1,934.4 kJ = -4,688.9 kJ<\/p>\n\n\n\n<p>Calculating the total enthalpy of the reactants:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 mole of C\u2087H\u2081\u2086: 1 \u00d7 (-187.9 kJ\/mol) = -187.9 kJ<\/li>\n\n\n\n<li>11 moles of O\u2082: 11 \u00d7 0 kJ\/mol = 0 kJ<\/li>\n<\/ul>\n\n\n\n<p>Total enthalpy of reactants = -187.9 kJ<\/p>\n\n\n\n<p>Applying Hess&#8217;s law:<br>\u0394H\u00b0_reaction = (-4,688.9 kJ) &#8211; (-187.9 kJ) = -4,501 kJ<\/p>\n\n\n\n<p>Therefore, the enthalpy change for the combustion of 1 mole of heptane is -4,501 kJ.<\/p>\n\n\n\n<p>This negative value indicates that the combustion of heptane is an exothermic process, releasing 4,501 kJ of energy per mole. Combustion reactions of hydrocarbons like heptane are highly exothermic due to the formation of strong bonds in the products (CO\u2082 and H\u2082O) compared to the reactants. The energy released during such reactions is harnessed in various applications, including internal combustion engines and heating systems.<\/p>\n\n\n\n<p>It&#8217;s important to note that the enthalpy change calculated here assumes standard conditions (298 K and 1 atm pressure) and that all reactants and products are in their standard states. Variations in temperature, pressure, or physical states can affect the enthalpy change of the reaction.<\/p>\n\n\n\n<p>For a visual representation of the balanced combustion equation and further explanation, you may refer to the following video:<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The combustion of heptane, C7H16, occurs via the reactionC7H16(g)+1102(g)\u21927CO2(g) + 8H2O(g)with heat of formation values given by the following table: Calculate the enthalpy for the combustion of 1 mole of heptane. The Correct Answer and Explanation is : To calculate the enthalpy change (\u0394H) for the combustion of 1 mole of heptane (C\u2087H\u2081\u2086), we use [&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-195441","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195441","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=195441"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195441\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195441"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195441"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195441"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}