{"id":245817,"date":"2025-07-06T13:26:11","date_gmt":"2025-07-06T13:26:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=245817"},"modified":"2025-07-06T13:26:14","modified_gmt":"2025-07-06T13:26:14","slug":"the-thermochemical-equation-for-the-combustion-of-methanol-is-given-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/06\/the-thermochemical-equation-for-the-combustion-of-methanol-is-given-below\/","title":{"rendered":"The thermochemical equation for the combustion of methanol is given below."},"content":{"rendered":"\n
The thermochemical equation for the combustion of methanol is given below.<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
The thermochemical equation for the combustion of methanol is:<\/p>\n\n\n\n
CH3OH(l) + 3\/2 O2(g) \u2192 CO2(g) + 2 H2O(g)
\u0394H\u00b0 = -638.7 kJ\/mol<\/p>\n\n\n\n
We are asked to find the enthalpy change for the combustion of 8.59 g of CH3OH.<\/p>\n\n\n\n
Step-by-step solution:<\/h3>\n\n\n\n
    \n
  1. Determine the molar mass of CH3OH:<\/strong>
    • Carbon (C): 12.01 g\/mol<\/li>
    • Hydrogen (H): 1.008 g\/mol (x4 for the 4 hydrogens)<\/li>
    • Oxygen (O): 16.00 g\/mol<\/li><\/ul>Molar mass of CH3OH = 12.01 + (4 x 1.008) + 16.00 = 32.04 g\/mol<\/li>\n\n\n\n
    • Convert the mass of CH3OH to moles:<\/strong>
      We are given 8.59 g of CH3OH. To convert this to moles: moles\u00a0of\u00a0CH3OH=8.59\u2009g32.04\u2009g\/mol=0.268\u2009mol\\text{moles of CH3OH} = \\frac{8.59 \\, \\text{g}}{32.04 \\, \\text{g\/mol}} = 0.268 \\, \\text{mol}moles\u00a0of\u00a0CH3OH=32.04g\/mol8.59g\u200b=0.268mol<\/li>\n\n\n\n
    • Calculate the enthalpy change:<\/strong>
      From the equation, we know that 1 mole of CH3OH releases 638.7 kJ of energy when it combusts. So, for 0.268 moles of CH3OH: enthalpy\u00a0change=0.268\u2009mol\u00d7(\u2212638.7\u2009kJ\/mol)=\u2212171.3\u2009kJ\\text{enthalpy change} = 0.268 \\, \\text{mol} \\times (-638.7 \\, \\text{kJ\/mol}) = -171.3 \\, \\text{kJ}enthalpy\u00a0change=0.268mol\u00d7(\u2212638.7kJ\/mol)=\u2212171.3kJ<\/li>\n<\/ol>\n\n\n\n
      Final answer:<\/h3>\n\n\n\n
      The enthalpy change for the combustion of 8.59 g of CH3OH is -171.3 kJ<\/strong>.<\/p>\n\n\n\n
      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
      The thermochemical equation for the combustion of methanol is given below. The Correct Answer and Explanation is: The thermochemical equation for the combustion of methanol is: CH3OH(l) + 3\/2 O2(g) \u2192 CO2(g) + 2 H2O(g)\u0394H\u00b0 = -638.7 kJ\/mol We are asked to find the enthalpy change for the combustion of 8.59 g of CH3OH. Step-by-step […]<\/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-245817","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245817","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=245817"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245817\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245817"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245817"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245817"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}