To calculate the heat of formation (\u0394Hf\u2218\\Delta H_f^\\circ\u0394Hf\u2218\u200b) of hexane (C6H14\\text{C}_6\\text{H}_{14}C6\u200bH14\u200b), we can use Hess’s Law, which states that the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps, regardless of the pathway taken.<\/p>\n\n\n\n
Step 1: Write the combustion reaction for hexane.<\/h3>\n\n\n\n
The combustion reaction for hexane is:C6H14(l)+O2(g)\u21926CO2(g)+7H2O(l)\\text{C}_6\\text{H}_{14(l)} + \\text{O}_2(g) \\rightarrow 6\\text{CO}_2(g) + 7\\text{H}_2\\text{O(l)}C6\u200bH14(l)\u200b+O2\u200b(g)\u21926CO2\u200b(g)+7H2\u200bO(l)<\/p>\n\n\n\n
Step 2: Use the given data for the heat of combustion.<\/h3>\n\n\n\n
The heat of combustion of hexane (\u0394Hcomb\\Delta H_\\text{comb}\u0394Hcomb\u200b) is \u22124163\u2009kJ\/mol-4163 \\, \\text{kJ\/mol}\u22124163kJ\/mol. This represents the enthalpy change when one mole of hexane undergoes complete combustion.<\/p>\n\n\n\n
Step 3: Use the heats of formation of the products and reactants.<\/h3>\n\n\n\n
The standard heats of formation are:<\/p>\n\n\n\n
- \n
- \u0394Hf\u2218(CO2(g))=\u2212393.5\u2009kJ\/mol\\Delta H_f^\\circ (\\text{CO}_2(g)) = -393.5 \\, \\text{kJ\/mol}\u0394Hf\u2218\u200b(CO2\u200b(g))=\u2212393.5kJ\/mol<\/li>\n\n\n\n
- \u0394Hf\u2218(H2O(l))=\u2212285.8\u2009kJ\/mol\\Delta H_f^\\circ (\\text{H}_2\\text{O(l)}) = -285.8 \\, \\text{kJ\/mol}\u0394Hf\u2218\u200b(H2\u200bO(l))=\u2212285.8kJ\/mol<\/li>\n\n\n\n
- The heat of formation for O2(g)\\text{O}_2(g)O2\u200b(g) is 0\u2009kJ\/mol0 \\, \\text{kJ\/mol}0kJ\/mol because it is in its elemental form.<\/li>\n\n\n\n
- We are solving for \u0394Hf\u2218(C6H14(l))\\Delta H_f^\\circ (\\text{C}_6\\text{H}_{14(l)})\u0394Hf\u2218\u200b(C6\u200bH14(l)\u200b), which is the unknown.<\/li>\n<\/ul>\n\n\n\n
Step 4: Apply Hess’s Law.<\/h3>\n\n\n\n
Hess’s Law states that:\u0394Hcomb=\u2211\u0394Hf\u2218(products)\u2212\u2211\u0394Hf\u2218(reactants)\\Delta H_\\text{comb} = \\sum \\Delta H_f^\\circ (\\text{products}) – \\sum \\Delta H_f^\\circ (\\text{reactants})\u0394Hcomb\u200b=\u2211\u0394Hf\u2218\u200b(products)\u2212\u2211\u0394Hf\u2218\u200b(reactants)<\/p>\n\n\n\n
Substitute the known values:\u22124163=[6\u00d7(\u2212393.5)+7\u00d7(\u2212285.8)]\u2212[\u0394Hf\u2218(C6H14(l))+0]-4163 = [6 \\times (-393.5) + 7 \\times (-285.8)] – [\\Delta H_f^\\circ (\\text{C}_6\\text{H}_{14(l)}) + 0]\u22124163=[6\u00d7(\u2212393.5)+7\u00d7(\u2212285.8)]\u2212[\u0394Hf\u2218\u200b(C6\u200bH14(l)\u200b)+0]\u22124163=[\u22122361\u22122000.6]\u2212\u0394Hf\u2218(C6H14(l))-4163 = [-2361 – 2000.6] – \\Delta H_f^\\circ (\\text{C}_6\\text{H}_{14(l)})\u22124163=[\u22122361\u22122000.6]\u2212\u0394Hf\u2218\u200b(C6\u200bH14(l)\u200b)\u22124163=\u22124361.6\u2212\u0394Hf\u2218(C6H14(l))-4163 = -4361.6 – \\Delta H_f^\\circ (\\text{C}_6\\text{H}_{14(l)})\u22124163=\u22124361.6\u2212\u0394Hf\u2218\u200b(C6\u200bH14(l)\u200b)<\/p>\n\n\n\n
Step 5: Solve for \u0394Hf\u2218(C6H14(l))\\Delta H_f^\\circ (\\text{C}_6\\text{H}_{14(l)})\u0394Hf\u2218\u200b(C6\u200bH14(l)\u200b).<\/h3>\n\n\n\n
\u0394Hf\u2218(C6H14(l))=\u22124361.6+4163\\Delta H_f^\\circ (\\text{C}_6\\text{H}_{14(l)}) = -4361.6 + 4163\u0394Hf\u2218\u200b(C6\u200bH14(l)\u200b)=\u22124361.6+4163\u0394Hf\u2218(C6H14(l))=\u2212198.6\u2009kJ\/mol\\Delta H_f^\\circ (\\text{C}_6\\text{H}_{14(l)}) = -198.6 \\, \\text{kJ\/mol}\u0394Hf\u2218\u200b(C6\u200bH14(l)\u200b)=\u2212198.6kJ\/mol<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
The heat of formation of hexane is \u0394Hf\u2218(C6H14(l))=\u2212198.6\u2009kJ\/mol\\Delta H_f^\\circ (\\text{C}_6\\text{H}_{14(l)}) = -198.6 \\, \\text{kJ\/mol}\u0394Hf\u2218\u200b(C6\u200bH14(l)\u200b)=\u2212198.6kJ\/mol.<\/p>\n\n\n\n
Explanation:<\/h3>\n\n\n\n
This calculation uses Hess’s Law to break down the combustion reaction into its individual steps, comparing the heat of formation of the products and reactants. The negative sign indicates that the formation of hexane from its elements releases energy. The reaction is exothermic, as expected for combustion processes.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-658.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"The combustion reaction of hexane is: Type numbers in the boxes_ 10 points 19 02(9) 4 6 CO2(g) C6H14(l) 7 HzO() The heat of combustion of hexane is -4163 kJ\/mol. Given the following information, calculate the heat of formation of hexane_ AH? ; (CO_(g)) ~393.5 kJ\/mol AH? (H,O()) -285.8 kJ\/mol kJ\/mol The Correct Answer and […]<\/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-245553","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245553","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=245553"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245553\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245553"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245553"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245553"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}