{"id":192065,"date":"2025-02-17T06:51:13","date_gmt":"2025-02-17T06:51:13","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=192065"},"modified":"2025-02-17T06:51:17","modified_gmt":"2025-02-17T06:51:17","slug":"the-standard-enthalpy-of-combustion-of-butane-is-10-93-kcal-per-gram-of-butane-when-the-water-isproduced-in-vapor-form","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/17\/the-standard-enthalpy-of-combustion-of-butane-is-10-93-kcal-per-gram-of-butane-when-the-water-isproduced-in-vapor-form\/","title":{"rendered":"The standard enthalpy of combustion of butane is -10.93 kcal per gram of butane when the water isproduced in vapor form"},"content":{"rendered":"\n<p>The standard enthalpy of combustion of butane is -10.93 kcal per gram of butane when the water is<br>produced in vapor form.<br>a) Convert this value into units of kJ per mole of butane.<br>b) Use this reaction enthalpy to determine the standard enthalpy of formation (Hr) of butane.<\/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<p>To solve this problem, we\u2019ll break it into two parts as requested.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">a) Convert the enthalpy of combustion of butane to kJ per mole.<\/h3>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The standard enthalpy of combustion of butane (C\u2084H\u2081\u2080) is -10.93 kcal per gram of butane.<\/li>\n\n\n\n<li>The molar mass of butane is 58.12 g\/mol (12.01 g\/mol for carbon and 1.008 g\/mol for hydrogen).<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1: Convert kcal to kJ.<\/strong><br>1 kcal = 4.184 kJ, so:<br>[<br>-10.93 \\, \\text{kcal\/g} \\times 4.184 = -45.74 \\, \\text{kJ\/g}<br>]<\/p>\n\n\n\n<p><strong>Step 2: Convert the value from kJ per gram to kJ per mole.<\/strong><br>Since the molar mass of butane is 58.12 g\/mol:<br>[<br>-45.74 \\, \\text{kJ\/g} \\times 58.12 \\, \\text{g\/mol} = -2654.46 \\, \\text{kJ\/mol}<br>]<br>Thus, the standard enthalpy of combustion of butane is approximately <strong>-2654.46 kJ\/mol<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">b) Use this reaction enthalpy to determine the standard enthalpy of formation (\u0394Hf\u00b0) of butane.<\/h3>\n\n\n\n<p>The standard enthalpy of combustion is the heat released when one mole of a substance undergoes complete combustion in excess oxygen to form CO\u2082 and H\u2082O. The reaction is:<\/p>\n\n\n\n<p>[<br>\\text{C\u2084H\u2081\u2080 (g)} + 13\/2 \\, \\text{O\u2082 (g)} \\rightarrow 4 \\, \\text{CO\u2082 (g)} + 5 \\, \\text{H\u2082O (g)}<br>]<\/p>\n\n\n\n<p>The standard enthalpy of combustion is related to the standard enthalpies of formation (\u0394Hf\u00b0) of the reactants and products:<\/p>\n\n\n\n<p>[<br>\\Delta H_{\\text{combustion}} = \\sum \\Delta H_f^\\circ \\, \\text{(products)} &#8211; \\sum \\Delta H_f^\\circ \\, \\text{(reactants)}<br>]<\/p>\n\n\n\n<p>For the combustion of butane:<br>[<br>\\Delta H_{\\text{combustion}} = \\left[4\\Delta H_f^\\circ (\\text{CO\u2082}) + 5\\Delta H_f^\\circ (\\text{H\u2082O (g)}) \\right] &#8211; \\left[\\Delta H_f^\\circ (\\text{C\u2084H\u2081\u2080}) + \\frac{13}{2} \\Delta H_f^\\circ (\\text{O\u2082}) \\right]<br>]<\/p>\n\n\n\n<p>Using standard values:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394Hf\u00b0 (CO\u2082) = -393.5 kJ\/mol<\/li>\n\n\n\n<li>\u0394Hf\u00b0 (H\u2082O (g)) = -241.8 kJ\/mol<\/li>\n\n\n\n<li>\u0394Hf\u00b0 (O\u2082) = 0 kJ\/mol (since it is an elemental form)<\/li>\n<\/ul>\n\n\n\n<p>Thus:<br>[<br>\\Delta H_{\\text{combustion}} = \\left[4(-393.5) + 5(-241.8)\\right] &#8211; \\left[\\Delta H_f^\\circ (\\text{C\u2084H\u2081\u2080}) + 0 \\right]<br>]<br>[<br>-2654.46 = [-1574 + (-1209)] &#8211; \\Delta H_f^\\circ (\\text{C\u2084H\u2081\u2080})<br>]<br>[<br>-2654.46 = -2783 &#8211; \\Delta H_f^\\circ (\\text{C\u2084H\u2081\u2080})<br>]<br>[<br>\\Delta H_f^\\circ (\\text{C\u2084H\u2081\u2080}) = -2783 + 2654.46 = -128.54 \\, \\text{kJ\/mol}<br>]<\/p>\n\n\n\n<p>Thus, the standard enthalpy of formation of butane is approximately <strong>-128.54 kJ\/mol<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Image of the combustion reaction and calculation:<\/h3>\n\n\n\n<p>I will generate an image to visually represent this calculation.<\/p>\n\n\n\n<p>Here is the image representing the combustion reaction of butane along with the standard enthalpy of combustion and formation calculations. Let me know if you need any further clarification!<\/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-740.png\" alt=\"\" class=\"wp-image-192066\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The standard enthalpy of combustion of butane is -10.93 kcal per gram of butane when the water isproduced in vapor form.a) Convert this value into units of kJ per mole of butane.b) Use this reaction enthalpy to determine the standard enthalpy of formation (Hr) of butane. The Correct Answer and Explanation is : To solve [&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-192065","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192065","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=192065"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192065\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192065"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192065"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192065"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}