{"id":222325,"date":"2025-05-31T08:40:29","date_gmt":"2025-05-31T08:40:29","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222325"},"modified":"2025-05-31T08:41:19","modified_gmt":"2025-05-31T08:41:19","slug":"calculate-the-standard-enthalpy-change-in-kj-mol-1-for-the-reaction-between-carbon-disulfide-cs2-and-oxygen-shown-in-the-following-equation","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/calculate-the-standard-enthalpy-change-in-kj-mol-1-for-the-reaction-between-carbon-disulfide-cs2-and-oxygen-shown-in-the-following-equation\/","title":{"rendered":"Calculate the standard enthalpy change, in kJ mol\u20131, for the reaction between carbon disulfide, CS2, and oxygen shown in the following equation."},"content":{"rendered":"\n

Calculate the standard enthalpy change, in kJ mol\u20131, for the reaction between carbon disulfide, CS2, and oxygen shown in the following equation.<\/p>\n\n\n\n

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

Sure! First, we need the balanced chemical equation<\/strong> for the combustion of carbon disulfide (CS\u2082) in oxygen: CS2(l)+3O2(g)\u2192CO2(g)+2SO2(g)\\text{CS}_2(l) + 3\\text{O}_2(g) \\rightarrow \\text{CO}_2(g) + 2\\text{SO}_2(g)<\/p>\n\n\n\n

Step 1: Use Standard Enthalpies of Formation (\u0394H\u2070f)<\/h3>\n\n\n\n

To calculate the standard enthalpy change of reaction (\u0394H\u2070rxn)<\/strong>, use the formula: \u0394Hrxn\u2218=\u2211\u0394Hf\u2218(products)\u2212\u2211\u0394Hf\u2218(reactants)\\Delta H^\\circ_{\\text{rxn}} = \\sum \\Delta H^\\circ_f(\\text{products}) – \\sum \\Delta H^\\circ_f(\\text{reactants})<\/p>\n\n\n\n

Standard enthalpies of formation (in kJ\/mol):<\/p>\n\n\n\n

    \n
  • \u0394H\u2070f [CS\u2082(l)] = +116.9<\/strong><\/li>\n\n\n\n
  • \u0394H\u2070f [O\u2082(g)] = 0<\/strong> (element in standard state)<\/li>\n\n\n\n
  • \u0394H\u2070f [CO\u2082(g)] = \u2013393.5<\/strong><\/li>\n\n\n\n
  • \u0394H\u2070f [SO\u2082(g)] = \u2013296.8<\/strong><\/li>\n<\/ul>\n\n\n\n

    Step 2: Plug into the Formula<\/h3>\n\n\n\n

    \u0394Hrxn\u2218=[1(\u2212393.5)+2(\u2212296.8)]\u2212[1(116.9)+3(0)]\\Delta H^\\circ_{\\text{rxn}} = [1(-393.5) + 2(-296.8)] – [1(116.9) + 3(0)] =[\u2212393.5\u2212593.6]\u2212[116.9]= [-393.5 – 593.6] – [116.9] =\u2212987.1\u2212116.9=\u22121104.0 kJ mol\u22121= -987.1 – 116.9 = \\boxed{-1104.0\\ \\text{kJ mol}^{-1}}<\/p>\n\n\n\n

    \n\n\n\n

    Explanation<\/h3>\n\n\n\n

    The standard enthalpy change of a chemical reaction, denoted as \u0394H\u2070rxn, represents the heat energy released or absorbed when a reaction occurs under standard conditions (298 K and 1 atm pressure), with all reactants and products in their standard states. In this case, we examine the combustion of carbon disulfide (CS\u2082), a flammable liquid, in the presence of oxygen to form carbon dioxide (CO\u2082) and sulfur dioxide (SO\u2082).<\/p>\n\n\n\n

    To determine the enthalpy change, we apply Hess\u2019s Law<\/strong>, which states that the total enthalpy change of a reaction depends only on the enthalpy of the products and the reactants\u2014not the path taken. Therefore, we can use tabulated standard enthalpies of formation (\u0394H\u2070f), which reflect the enthalpy change when one mole of a compound forms from its elements in their standard states.<\/p>\n\n\n\n

    The formula used is: \u0394Hrxn\u2218=\u2211\u0394Hf\u2218(products)\u2212\u2211\u0394Hf\u2218(reactants)\\Delta H^\\circ_{\\text{rxn}} = \\sum \\Delta H^\\circ_f(\\text{products}) – \\sum \\Delta H^\\circ_f(\\text{reactants})<\/p>\n\n\n\n

    Here, CS\u2082 reacts with three moles of O\u2082 to produce one mole of CO\u2082 and two moles of SO\u2082. Using standard values, we substitute and calculate that the reaction has an enthalpy change of \u20131104.0 kJ\/mol<\/strong>, indicating it is strongly exothermic<\/strong>. This makes sense because combustion reactions typically release energy as bonds in reactants are broken and more stable bonds form in the products.<\/p>\n\n\n\n

    Understanding \u0394H\u2070rxn is crucial in thermodynamics, environmental science, and industrial applications, where energy balances and heat generation are important for safety, efficiency, and sustainability.<\/p>\n\n\n\n

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

    Calculate the standard enthalpy change, in kJ mol\u20131, for the reaction between carbon disulfide, CS2, and oxygen shown in the following equation. The Correct Answer and Explanation is: Sure! First, we need the balanced chemical equation for the combustion of carbon disulfide (CS\u2082) in oxygen: CS2(l)+3O2(g)\u2192CO2(g)+2SO2(g)\\text{CS}_2(l) + 3\\text{O}_2(g) \\rightarrow \\text{CO}_2(g) + 2\\text{SO}_2(g) Step 1: Use […]<\/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-222325","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222325","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=222325"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222325\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222325"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222325"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}