Given that AH; CI(g)| = 121.3 kJ-mol- AH; C(g)] = 716.7 kJ-mol- AH; (CC,(g) = -95.7 kJ-mol-! calculate the average molar bond enthalpy of the carbon-chlorine bond in a CCl, molecule. None O 16 6 Given that Y H N AH; [CI(g)] = 121.3 kJ.mol-\u00c2\u00b9 AH; [C(g)] = 716.7 kJ\u00c2\u00b7mol\u00c2\u00af\u00c2\u00b9 AH; [CC14(g)] = \u00e2\u02c6\u201995.7 kJ\u00c2\u00b7mol\u00e2\u02c6\u2019\u00c2\u00b9 calculate the average molar bond enthalpy of the carbon-chlorine bond in a CCl4 molecule. AHC-Cl = 17 & 7 J f8 3 e 8 U UT about us K fg careers privacy policy terms of use W 9 alt hp O L f10 V O F11 P ctri ? contact us help { f12 \u00e2\u0153\u02c6 [ 44 11 ins Question Source: MRG – General Chemistry | Publisher: Universit prt sc ] pause delete backspace kJ.mol-1 8 home enter num lock T shift 10: 7\/21 end 7 home 4<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the average molar bond enthalpy of the carbon-chlorine (C-Cl) bond in a CCl\u2084 molecule, we need to use the following given information:<\/p>\n\n\n\n We can apply Hess’s Law, which states that the 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 The formation of CCl\u2084 from its elements in their standard states can be written as:<\/p>\n\n\n\n [ \\text{C}(s) + 2\\text{Cl}_2(g) \\rightarrow \\text{CCl}_4(g) ]<\/p>\n\n\n\n We can break this into several steps:<\/p>\n\n\n\n Now, we can sum these steps:<\/p>\n\n\n\n [ This value represents the energy required to form CCl\u2084 in its gaseous form from the elements. Since CCl\u2084 has 4 C-Cl bonds, we divide the total enthalpy change by 4 to find the bond enthalpy for one C-Cl bond:<\/p>\n\n\n\n [ The average molar bond enthalpy of the carbon-chlorine bond in a CCl\u2084 molecule is approximately 215.9 kJ\/mol<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" Given that AH; CI(g)| = 121.3 kJ-mol- AH; C(g)] = 716.7 kJ-mol- AH; (CC,(g) = -95.7 kJ-mol-! calculate the average molar bond enthalpy of the carbon-chlorine bond in a CCl, molecule. None O 16 6 Given that Y H N AH; [CI(g)] = 121.3 kJ.mol-\u00c2\u00b9 AH; [C(g)] = 716.7 kJ\u00c2\u00b7mol\u00c2\u00af\u00c2\u00b9 AH; [CC14(g)] = \u00e2\u02c6\u201995.7 kJ\u00c2\u00b7mol\u00e2\u02c6\u2019\u00c2\u00b9 […]<\/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-186464","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186464","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=186464"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186464\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186464"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186464"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186464"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Step 1: Use Hess’s Law<\/h3>\n\n\n\n
\n
[ \\text{C}(s) \\rightarrow \\text{C}(g) \\quad \\Delta H = 716.7 \\, \\text{kJ\/mol} ]<\/li>\n\n\n\n
[ \\text{Cl}_2(g) \\rightarrow 2\\text{Cl}(g) \\quad \\Delta H = 2 \\times 121.3 = 242.6 \\, \\text{kJ\/mol} ]<\/li>\n\n\n\n
[ \\text{C}(g) + 2\\text{Cl}_2(g) \\rightarrow \\text{CCl}_4(g) \\quad \\Delta H = -95.7 \\, \\text{kJ\/mol} ]<\/li>\n<\/ol>\n\n\n\nStep 2: Calculate the enthalpy change for breaking the C-Cl bonds<\/h3>\n\n\n\n
\\Delta H = 716.7 + 242.6 – 95.7 = 863.6 \\, \\text{kJ\/mol}
]<\/p>\n\n\n\n
\\text{Average bond enthalpy of C-Cl} = \\frac{863.6}{4} = 215.9 \\, \\text{kJ\/mol}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n