For a particular isomer of C8H18,C8H18, the combustion reaction produces 5093.7 kJ5093.7 kJ of heat per mole of C8H18(g)C8H18(g) consumed, under standard conditions.<\/p>\n\n\n\n
C8H18(g)+252O2(g)?8CO2(g)+9H2O(g) ?H\u00b0rxn=-5093.7 kJ\/mol<\/p>\n\n\n\n
What is the standard enthalpy of formation of this isomer of C8H18(g)?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s solve your question carefully.<\/p>\n\n\n\n You are given: You are asked to find the standard enthalpy of formation<\/strong> ((\\Delta H^\\circ_f)) of the isomer of (C_8H_{18}(g)).<\/p>\n\n\n\n The standard enthalpy change of a reaction is: For the combustion:<\/p>\n\n\n\n Note:<\/em> The standard enthalpy of formation for (O_2(g)) is zero<\/strong>.<\/p>\n\n\n\n From standard tables:<\/p>\n\n\n\n [ First, calculate the products’ side: Now plug into the equation: Rearranging: [ In thermochemistry, the standard enthalpy of formation<\/strong> ((\\Delta H^\\circ_f)) of a compound is the heat change when one mole<\/strong> of the substance forms from its elements in their standard states. For hydrocarbons like octane ((C_8H_{18})), combustion is a convenient method to calculate it because the combustion enthalpy ((\\Delta H^\\circ_{\\text{rxn}})) is typically measured experimentally.<\/p>\n\n\n\n The given reaction shows that when 1 mole of gaseous (C_8H_{18}) combusts, it releases (-5093.7\\ \\text{kJ}) of heat. In combustion, the products are typically (CO_2(g)) and (H_2O(g)), both of which have well-known standard enthalpies of formation.<\/p>\n\n\n\n The general formula we apply is: Oxygen ((O_2)) being an element in its standard state has (\\Delta H^\\circ_f = 0), simplifying the calculation. We multiply the known (\\Delta H^\\circ_f) values by their stoichiometric coefficients (8 for (CO_2), 9 for (H_2O)) and add them. This gives the total enthalpy change for the products.<\/p>\n\n\n\n By plugging in and rearranging the formula, we isolate the unknown (\\Delta H^\\circ_f) for octane. The negative value (-230.5\\ \\text{kJ\/mol}) indicates that the formation of this isomer of octane from its elements is an exothermic<\/strong> process (it releases energy).<\/p>\n\n\n\n Understanding these relationships helps chemists predict reaction behavior and design fuels with desirable energy properties.<\/p>\n\n\n\n For a particular isomer of C8H18,C8H18, the combustion reaction produces 5093.7 kJ5093.7 kJ of heat per mole of C8H18(g)C8H18(g) consumed, under standard conditions. C8H18(g)+252O2(g)?8CO2(g)+9H2O(g) ?H\u00b0rxn=-5093.7 kJ\/mol What is the standard enthalpy of formation of this isomer of C8H18(g)? The correct answer and explanation is : Let’s solve your question carefully. You are given:[\\text{C}8\\text{H}{18}(g) + \\frac{25}{2}\\text{O}2(g) […]<\/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-209275","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209275","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=209275"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209275\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209275"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209275"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209275"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
[
\\text{C}8\\text{H}<\/em>{18}(g) + \\frac{25}{2}\\text{O}2(g) \\rightarrow 8\\text{CO}_2(g) + 9\\text{H}_2\\text{O}(g) \\quad \\Delta H^\\circ<\/em>{\\text{rxn}} = -5093.7\\ \\text{kJ\/mol}
]<\/p>\n\n\n\nStep 1: Recall the Enthalpy of Reaction Formula<\/h3>\n\n\n\n
[
\\Delta H^\\circ_{\\text{rxn}} = \\sum \\Delta H^\\circ_f\\ (\\text{products}) – \\sum \\Delta H^\\circ_f\\ (\\text{reactants})
]<\/p>\n\n\n\n\n
Step 2: Known Values<\/h3>\n\n\n\n
\n
Step 3: Write the Equation<\/h3>\n\n\n\n
-5093.7\\ \\text{kJ} = [8 \\times (-393.5) + 9 \\times (-241.8)] – [\\Delta H^\\circ_f\\ \\text{of}\\ \\text{C}8\\text{H}<\/em>{18}]
]<\/p>\n\n\n\n
[
8 \\times (-393.5) = -3148.0\\ \\text{kJ}
]
[
9 \\times (-241.8) = -2176.2\\ \\text{kJ}
]
Adding them:
[
-3148.0 + (-2176.2) = -5324.2\\ \\text{kJ}
]<\/p>\n\n\n\n
[
-5093.7 = (-5324.2) – (\\Delta H^\\circ_f\\ \\text{of}\\ \\text{C}8\\text{H}<\/em>{18})
]<\/p>\n\n\n\nStep 4: Solve for (\\Delta H^\\circ_f)<\/h3>\n\n\n\n
[
\\Delta H^\\circ_f\\ \\text{of}\\ \\text{C}8\\text{H}<\/em>{18} = -5324.2 + 5093.7
]
[
\\Delta H^\\circ_f\\ \\text{of}\\ \\text{C}8\\text{H}<\/em>{18} = -230.5\\ \\text{kJ\/mol}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\\boxed{-230.5\\ \\text{kJ\/mol}}
]<\/p>\n\n\n\n
\n\n\n\nExplanation (~300 words)<\/h3>\n\n\n\n
[
\\Delta H^\\circ_{\\text{rxn}} = \\text{Sum of} (\\Delta H^\\circ_f\\ \\text{of products}) – \\text{Sum of} (\\Delta H^\\circ_f\\ \\text{of reactants})
]<\/p>\n\n\n\n
\n","protected":false},"excerpt":{"rendered":"