{"id":256264,"date":"2025-07-16T08:22:01","date_gmt":"2025-07-16T08:22:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=256264"},"modified":"2025-07-16T08:22:04","modified_gmt":"2025-07-16T08:22:04","slug":"the-standard-enthalpy-of-reaction-for-the-combustion-of-1-mole-of-butane-is-2888-0-kj-in-this-reaction-butane-c4h10g-reacts-with-gaseous-o2-to-produce-co2g-h2ol-and-energy-as-heat","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/16\/the-standard-enthalpy-of-reaction-for-the-combustion-of-1-mole-of-butane-is-2888-0-kj-in-this-reaction-butane-c4h10g-reacts-with-gaseous-o2-to-produce-co2g-h2ol-and-energy-as-heat\/","title":{"rendered":"The standard enthalpy of reaction for the combustion of 1 mole of butane is -2888.0 kJ. In this reaction, butane (C4H10(g)) reacts with gaseous O2 to produce CO2(g), H2O(l), and energy as heat"},"content":{"rendered":"\n

The standard enthalpy of reaction for the combustion of 1 mole of butane is -2888.0 kJ. In this reaction, butane (C4H10(g)) reacts with gaseous O2 to produce CO2(g), H2O(l), and energy as heat. (a) Write and balance the combustion reaction. (b) Calculate the standard molar enthalpy of formation of butane, knowing that the enthalpies of formation of CO2(g) and H2O(l) are -393.4 kJ\/mol and -285.8 kJ\/mol, respectively.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s break this into two parts:<\/p>\n\n\n\n

(a) Write and Balance the Combustion Reaction<\/h3>\n\n\n\n

The combustion of butane (C4H10) involves the reaction of butane with oxygen to produce carbon dioxide (CO2) and water (H2O). The general form of a combustion reaction is:Fuel (CxHy)+O2(g)\u2192CO2(g)+H2O(l)\\text{Fuel (C}_x\\text{H}_y\\text{)} + \\text{O}_2(g) \\rightarrow \\text{CO}_2(g) + \\text{H}_2\\text{O}(l)Fuel (Cx\u200bHy\u200b)+O2\u200b(g)\u2192CO2\u200b(g)+H2\u200bO(l)<\/p>\n\n\n\n

For butane, the reaction is as follows:C4H10(g)+O2(g)\u2192CO2(g)+H2O(l)\\text{C}_4\\text{H}_{10}(g) + \\text{O}_2(g) \\rightarrow \\text{CO}_2(g) + \\text{H}_2\\text{O}(l)C4\u200bH10\u200b(g)+O2\u200b(g)\u2192CO2\u200b(g)+H2\u200bO(l)<\/p>\n\n\n\n

Now, balance the reaction:<\/p>\n\n\n\n

    \n
  1. Balance Carbon<\/strong>: There are 4 carbon atoms in C4H10, so you need 4 molecules of CO2: C4H10(g)+O2(g)\u21924CO2(g)+H2O(l)\\text{C}_4\\text{H}_{10}(g) + \\text{O}_2(g) \\rightarrow 4\\text{CO}_2(g) + \\text{H}_2\\text{O}(l)C4\u200bH10\u200b(g)+O2\u200b(g)\u21924CO2\u200b(g)+H2\u200bO(l)<\/li>\n\n\n\n
  2. Balance Hydrogen<\/strong>: There are 10 hydrogen atoms in C4H10, so you need 5 molecules of H2O: C4H10(g)+O2(g)\u21924CO2(g)+5H2O(l)\\text{C}_4\\text{H}_{10}(g) + \\text{O}_2(g) \\rightarrow 4\\text{CO}_2(g) + 5\\text{H}_2\\text{O}(l)C4\u200bH10\u200b(g)+O2\u200b(g)\u21924CO2\u200b(g)+5H2\u200bO(l)<\/li>\n\n\n\n
  3. Balance Oxygen<\/strong>: On the right side, there are 4\u00d72=84 \\times 2 = 84\u00d72=8 oxygen atoms from CO2 and 5\u00d71=55 \\times 1 = 55\u00d71=5 oxygen atoms from H2O, for a total of 13 oxygen atoms. Since each O2 molecule contains 2 oxygen atoms, you need 132=6.5\\frac{13}{2} = 6.5213\u200b=6.5 O2 molecules: C4H10(g)+6.5O2(g)\u21924CO2(g)+5H2O(l)\\text{C}_4\\text{H}_{10}(g) + 6.5\\text{O}_2(g) \\rightarrow 4\\text{CO}_2(g) + 5\\text{H}_2\\text{O}(l)C4\u200bH10\u200b(g)+6.5O2\u200b(g)\u21924CO2\u200b(g)+5H2\u200bO(l)<\/li>\n\n\n\n
  4. Balance the equation<\/strong>: Since you can’t have a fraction in the coefficients, multiply the entire equation by 2 to eliminate the fraction: 2C4H10(g)+13O2(g)\u21928CO2(g)+10H2O(l)2\\text{C}_4\\text{H}_{10}(g) + 13\\text{O}_2(g) \\rightarrow 8\\text{CO}_2(g) + 10\\text{H}_2\\text{O}(l)2C4\u200bH10\u200b(g)+13O2\u200b(g)\u21928CO2\u200b(g)+10H2\u200bO(l)<\/li>\n<\/ol>\n\n\n\n

    This is the balanced combustion reaction for butane.<\/p>\n\n\n\n

    (b) Calculate the Standard Molar Enthalpy of Formation of Butane<\/h3>\n\n\n\n

    The standard enthalpy of reaction (\u0394H\u00b0_reaction) for the combustion of butane is given as -2888.0 kJ. We can use Hess’s Law and the standard enthalpies of formation (\u0394H\u00b0_f) for CO2 and H2O to calculate the standard molar enthalpy of formation of butane (\u0394H\u00b0_f for C4H10).<\/p>\n\n\n\n

    The enthalpy change for a reaction is given by:\u0394Hreaction=\u2211(\u0394Hf,products)\u2212\u2211(\u0394Hf,reactants)\\Delta H_{\\text{reaction}} = \\sum (\\Delta H_{\\text{f,products}}) – \\sum (\\Delta H_{\\text{f,reactants}})\u0394Hreaction\u200b=\u2211(\u0394Hf,products\u200b)\u2212\u2211(\u0394Hf,reactants\u200b)<\/p>\n\n\n\n

    For the combustion of butane:\u0394Hreaction=[8\u0394Hf,CO2+10\u0394Hf,H2O]\u2212[2\u0394Hf,C4H10+13\u0394Hf,O2]\\Delta H_{\\text{reaction}} = [8\\Delta H_{\\text{f,CO2}} + 10\\Delta H_{\\text{f,H2O}}] – [2\\Delta H_{\\text{f,C4H10}} + 13\\Delta H_{\\text{f,O2}}]\u0394Hreaction\u200b=[8\u0394Hf,CO2\u200b+10\u0394Hf,H2O\u200b]\u2212[2\u0394Hf,C4H10\u200b+13\u0394Hf,O2\u200b]<\/p>\n\n\n\n

    Since the enthalpy of formation of oxygen (O2) is zero, the equation simplifies to:\u0394Hreaction=[8(\u2212393.4)+10(\u2212285.8)]\u22122\u0394Hf,C4H10\\Delta H_{\\text{reaction}} = [8(-393.4) + 10(-285.8)] – 2\\Delta H_{\\text{f,C4H10}}\u0394Hreaction\u200b=[8(\u2212393.4)+10(\u2212285.8)]\u22122\u0394Hf,C4H10\u200b<\/p>\n\n\n\n

    Substitute the known values:\u22122888.0=[8(\u2212393.4)+10(\u2212285.8)]\u22122\u0394Hf,C4H10-2888.0 = [8(-393.4) + 10(-285.8)] – 2\\Delta H_{\\text{f,C4H10}}\u22122888.0=[8(\u2212393.4)+10(\u2212285.8)]\u22122\u0394Hf,C4H10\u200b<\/p>\n\n\n\n

    Calculate the terms:\u22122888.0=[\u22123147.2+(\u22122858.0)]\u22122\u0394Hf,C4H10-2888.0 = [-3147.2 + (-2858.0)] – 2\\Delta H_{\\text{f,C4H10}}\u22122888.0=[\u22123147.2+(\u22122858.0)]\u22122\u0394Hf,C4H10\u200b\u22122888.0=\u22126005.2\u22122\u0394Hf,C4H10-2888.0 = -6005.2 – 2\\Delta H_{\\text{f,C4H10}}\u22122888.0=\u22126005.2\u22122\u0394Hf,C4H10\u200b<\/p>\n\n\n\n

    Solve for \u0394H\u00b0_f(C4H10):2\u0394Hf,C4H10=\u22126005.2+2888.02\\Delta H_{\\text{f,C4H10}} = -6005.2 + 2888.02\u0394Hf,C4H10\u200b=\u22126005.2+2888.02\u0394Hf,C4H10=\u22123117.22\\Delta H_{\\text{f,C4H10}} = -3117.22\u0394Hf,C4H10\u200b=\u22123117.2\u0394Hf,C4H10=\u22123117.22\\Delta H_{\\text{f,C4H10}} = \\frac{-3117.2}{2}\u0394Hf,C4H10\u200b=2\u22123117.2\u200b\u0394Hf,C4H10=\u22121558.6\u2009kJ\/mol\\Delta H_{\\text{f,C4H10}} = -1558.6 \\, \\text{kJ\/mol}\u0394Hf,C4H10\u200b=\u22121558.6kJ\/mol<\/p>\n\n\n\n

    Thus, the standard molar enthalpy of formation of butane (C4H10) is -1558.6 kJ\/mol<\/strong>.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    The standard enthalpy of reaction for the combustion of 1 mole of butane is -2888.0 kJ. In this reaction, butane (C4H10(g)) reacts with gaseous O2 to produce CO2(g), H2O(l), and energy as heat. (a) Write and balance the combustion reaction. (b) Calculate the standard molar enthalpy of formation of butane, knowing that the enthalpies of […]<\/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-256264","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256264","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=256264"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256264\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=256264"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=256264"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=256264"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}