To calculate the standard enthalpy of formation of methane (\u0394Hf\u2218\\Delta H_f^\\circ\u0394Hf\u2218\u200b for CH\u2084), we can use the following approach based on Hess’s Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes of the steps into which the reaction can be divided.<\/p>\n\n\n\n
Step 1: Write the balanced equation for the combustion of methane.<\/h3>\n\n\n\n
The combustion of methane involves methane reacting with oxygen to form carbon dioxide and water. The balanced chemical equation is:CH4(g)+2O2(g)\u2192CO2(g)+2H2O(l)CH_4 (g) + 2O_2 (g) \\rightarrow CO_2 (g) + 2H_2O (l)CH4\u200b(g)+2O2\u200b(g)\u2192CO2\u200b(g)+2H2\u200bO(l)<\/p>\n\n\n\n
Step 2: Understand the relationship between combustion enthalpy and formation enthalpies.<\/h3>\n\n\n\n
The enthalpy change of combustion of methane (\u0394Hcombustion\\Delta H_{\\text{combustion}}\u0394Hcombustion\u200b) is the enthalpy change for the reaction above. The standard enthalpy of formation of a substance is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states. For example, the standard enthalpy of formation of methane is the enthalpy change for the reaction:C(s)+2H2(g)\u2192CH4(g)C(s) + 2H_2(g) \\rightarrow CH_4(g)C(s)+2H2\u200b(g)\u2192CH4\u200b(g)<\/p>\n\n\n\n
The standard enthalpy of formation of carbon dioxide and water are provided as:\u0394Hf\u2218(CO2(g))=\u2212393.5\u2009kJ\/mol\\Delta H_f^\\circ (CO_2 (g)) = -393.5 \\, \\text{kJ\/mol}\u0394Hf\u2218\u200b(CO2\u200b(g))=\u2212393.5kJ\/mol\u0394Hf\u2218(H2O(l))=\u2212285.8\u2009kJ\/mol\\Delta H_f^\\circ (H_2O (l)) = -285.8 \\, \\text{kJ\/mol}\u0394Hf\u2218\u200b(H2\u200bO(l))=\u2212285.8kJ\/mol<\/p>\n\n\n\n
Step 3: Use Hess’s Law to relate the combustion reaction to formation reactions.<\/h3>\n\n\n\n
We can write the combustion reaction in terms of formation reactions:CH4(g)+2O2(g)\u2192CO2(g)+2H2O(l)CH_4 (g) + 2O_2 (g) \\rightarrow CO_2 (g) + 2H_2O (l)CH4\u200b(g)+2O2\u200b(g)\u2192CO2\u200b(g)+2H2\u200bO(l)<\/p>\n\n\n\n
The enthalpy change for this reaction is:\u0394Hcombustion=[\u0394Hf\u2218(CO2)+2\u22c5\u0394Hf\u2218(H2O)]\u2212[\u0394Hf\u2218(CH4)+2\u22c5\u0394Hf\u2218(O2)]\\Delta H_{\\text{combustion}} = [\\Delta H_f^\\circ (CO_2) + 2 \\cdot \\Delta H_f^\\circ (H_2O)] – [\\Delta H_f^\\circ (CH_4) + 2 \\cdot \\Delta H_f^\\circ (O_2)]\u0394Hcombustion\u200b=[\u0394Hf\u2218\u200b(CO2\u200b)+2\u22c5\u0394Hf\u2218\u200b(H2\u200bO)]\u2212[\u0394Hf\u2218\u200b(CH4\u200b)+2\u22c5\u0394Hf\u2218\u200b(O2\u200b)]<\/p>\n\n\n\n
Since oxygen (O2O_2O2\u200b) is in its standard state, its enthalpy of formation is zero:\u0394Hf\u2218(O2)=0\\Delta H_f^\\circ (O_2) = 0\u0394Hf\u2218\u200b(O2\u200b)=0<\/p>\n\n\n\n
Thus, the equation simplifies to:\u0394Hcombustion=[\u0394Hf\u2218(CO2)+2\u22c5\u0394Hf\u2218(H2O)]\u2212\u0394Hf\u2218(CH4)\\Delta H_{\\text{combustion}} = [\\Delta H_f^\\circ (CO_2) + 2 \\cdot \\Delta H_f^\\circ (H_2O)] – \\Delta H_f^\\circ (CH_4)\u0394Hcombustion\u200b=[\u0394Hf\u2218\u200b(CO2\u200b)+2\u22c5\u0394Hf\u2218\u200b(H2\u200bO)]\u2212\u0394Hf\u2218\u200b(CH4\u200b)<\/p>\n\n\n\n
Step 4: Substitute known values and solve for \u0394Hf\u2218(CH4)\\Delta H_f^\\circ (CH_4)\u0394Hf\u2218\u200b(CH4\u200b).<\/h3>\n\n\n\n
The given combustion enthalpy is \u2212850.4\u2009kJ\/mol-850.4 \\, \\text{kJ\/mol}\u2212850.4kJ\/mol, and the enthalpies of formation for carbon dioxide and water are \u2212393.5\u2009kJ\/mol-393.5 \\, \\text{kJ\/mol}\u2212393.5kJ\/mol and \u2212285.8\u2009kJ\/mol-285.8 \\, \\text{kJ\/mol}\u2212285.8kJ\/mol, respectively.<\/p>\n\n\n\n
Now substitute these values into the equation:\u2212850.4=[(\u2212393.5)+2(\u2212285.8)]\u2212\u0394Hf\u2218(CH4)-850.4 = [(-393.5) + 2(-285.8)] – \\Delta H_f^\\circ (CH_4)\u2212850.4=[(\u2212393.5)+2(\u2212285.8)]\u2212\u0394Hf\u2218\u200b(CH4\u200b)<\/p>\n\n\n\n
First, calculate the sum on the right-hand side:\u2212393.5+2(\u2212285.8)=\u2212393.5\u2212571.6=\u2212965.1-393.5 + 2(-285.8) = -393.5 – 571.6 = -965.1\u2212393.5+2(\u2212285.8)=\u2212393.5\u2212571.6=\u2212965.1<\/p>\n\n\n\n
Now, solve for \u0394Hf\u2218(CH4)\\Delta H_f^\\circ (CH_4)\u0394Hf\u2218\u200b(CH4\u200b):\u2212850.4=\u2212965.1\u2212\u0394Hf\u2218(CH4)-850.4 = -965.1 – \\Delta H_f^\\circ (CH_4)\u2212850.4=\u2212965.1\u2212\u0394Hf\u2218\u200b(CH4\u200b)\u0394Hf\u2218(CH4)=\u2212965.1+850.4=\u2212114.7\u2009kJ\/mol\\Delta H_f^\\circ (CH_4) = -965.1 + 850.4 = -114.7 \\, \\text{kJ\/mol}\u0394Hf\u2218\u200b(CH4\u200b)=\u2212965.1+850.4=\u2212114.7kJ\/mol<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
The standard enthalpy of formation for methane is \u0394Hf\u2218(CH4)=\u2212114.7\u2009kJ\/mol\\Delta H_f^\\circ (CH_4) = -114.7 \\, \\text{kJ\/mol}\u0394Hf\u2218\u200b(CH4\u200b)=\u2212114.7kJ\/mol.<\/p>\n\n\n\n
Explanation:<\/h3>\n\n\n\n
The standard enthalpy of formation for methane is negative because it requires energy to break the bonds of elemental carbon and hydrogen in their standard states to form methane. The negative value indicates that methane is more stable than its constituent elements under standard conditions.<\/p>\n\n\n\n If the enthalpy change for the complete combustion of methane is -850.4 kJ, calculate the standard enthalpy of formation for methane, in kJ\/mol, given that the standard enthalpies of formation for carbon dioxide and water are -393.5 and -285.8 kJ\/mol, respectively The Correct Answer and Explanation is: To calculate the standard enthalpy of formation 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-251514","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/251514","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=251514"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/251514\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=251514"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=251514"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=251514"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-59.jpeg\")
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