The standard enthalpy of combustion for naphthalene, C10H8(s), is<\/p>\n\n\n\n
-5156.8 kJ\/mol. Using this data and the standard enthalpies of formation, \u2206H: H2O(l) = -285.9 kJ mol; CO2(g)= -393.5 kJ mol, calculate the standard enthalpy of formation of C10H8(s), in kJ mol.
a) +78.2kj
b) +935.9 kJ
C) -1065.4 kJ
d) +3619.7 kJ
e) -10235.4 kj<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the standard enthalpy of formation ((\\Delta H_f^\\circ)) for naphthalene ((C_{10}H_8(s))), we will use the following formula derived from Hess’s law:<\/p>\n\n\n\n [ The combustion of naphthalene can be represented by the following balanced chemical equation:<\/p>\n\n\n\n [ We are given:<\/p>\n\n\n\n Using the balanced combustion equation, we can calculate the enthalpy change for the products:<\/p>\n\n\n\n [ [ [ Now, using Hess’s law, the standard enthalpy of formation of naphthalene is:<\/p>\n\n\n\n [ [ [ Thus, the standard enthalpy of formation of naphthalene is (-78.2 \\, \\text{kJ\/mol}), which corresponds to the answer a)<\/strong>.<\/p>\n\n\n\n The standard enthalpy of formation of naphthalene is (-78.2 \\, \\text{kJ\/mol}), making the correct answer a)<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" The standard enthalpy of combustion for naphthalene, C10H8(s), is -5156.8 kJ\/mol. Using this data and the standard enthalpies of formation, \u2206H: H2O(l) = -285.9 kJ mol; CO2(g)= -393.5 kJ mol, calculate the standard enthalpy of formation of C10H8(s), in kJ mol.a) +78.2kjb) +935.9 kJC) -1065.4 kJd) +3619.7 kJe) -10235.4 kj The correct answer and explanation […]<\/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-197998","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197998","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=197998"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197998\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197998"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197998"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197998"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\Delta H_{\\text{combustion}}^\\circ = \\sum (\\Delta H_f^\\circ \\, \\text{of products}) – \\sum (\\Delta H_f^\\circ \\, \\text{of reactants})
]<\/p>\n\n\n\nStep 1: Write the combustion reaction for naphthalene<\/h3>\n\n\n\n
C_{10}H_8(s) + 12 O_2(g) \\rightarrow 10 CO_2(g) + 4 H_2O(l)
]<\/p>\n\n\n\nStep 2: Apply the given data<\/h3>\n\n\n\n
\n
Step 3: Calculate the sum of the standard enthalpies of formation of the products<\/h3>\n\n\n\n
\\Delta H_f^\\circ \\, (\\text{products}) = 10 \\times \\Delta H_f^\\circ \\, \\text{CO}_2(g) + 4 \\times \\Delta H_f^\\circ \\, \\text{H}_2\\text{O(l)}
]<\/p>\n\n\n\n
\\Delta H_f^\\circ \\, (\\text{products}) = 10 \\times (-393.5) + 4 \\times (-285.9)
]<\/p>\n\n\n\n
\\Delta H_f^\\circ \\, (\\text{products}) = -3935 + (-1143.6) = -5078.6 \\, \\text{kJ\/mol}
]<\/p>\n\n\n\nStep 4: Apply Hess’s law to find the standard enthalpy of formation of naphthalene<\/h3>\n\n\n\n
\\Delta H_f^\\circ \\, (C_{10}H_8(s)) = \\Delta H_{\\text{combustion}}^\\circ – \\Delta H_f^\\circ \\, (\\text{products})
]<\/p>\n\n\n\n
\\Delta H_f^\\circ \\, (C_{10}H_8(s)) = -5156.8 – (-5078.6)
]<\/p>\n\n\n\n
\\Delta H_f^\\circ \\, (C_{10}H_8(s)) = -5156.8 + 5078.6 = -78.2 \\, \\text{kJ\/mol}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n