Dinitrogen monoxide, N\u2082O, can decompose to nitrogen and oxygen gas:
2 N2O(g) \u2192 2 N2(g) + O2(g)
Use bond dissociation enthalpies (see Table on last page) to estimate the enthalpy change for this reaction. (Hint: As in problem #6, first draw the Lewis structures of the reactants and products that obey the octet rule so you knov the types of bonds involved to determine their bond dissociation energies. N\u2082O has the following stable Lewis
structure: :N=N-0:<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To estimate the enthalpy change for the decomposition of dinitrogen monoxide (N\u2082O) into nitrogen (N\u2082) and oxygen (O\u2082) using bond dissociation enthalpies, follow these steps:<\/p>\n\n\n\n The decomposition of dinitrogen monoxide can be written as: We will use the following Lewis structures for the molecules:<\/p>\n\n\n\n To estimate the enthalpy change for the reaction, use bond dissociation enthalpies for each bond involved.<\/p>\n\n\n\n To find the total enthalpy change, we need to consider both the bonds broken (reactants) and the bonds formed (products).<\/p>\n\n\n\n The enthalpy change (\u0394H) can be calculated by subtracting the total energy required to break bonds from the energy released when new bonds are formed: Thus, the enthalpy change for the decomposition of 2 moles of N\u2082O into nitrogen and oxygen is approximately +1668 kJ\/mol<\/strong>. Since the reaction is endothermic (energy is absorbed), the positive value indicates that the decomposition requires energy input to proceed.<\/p>\n\n\n\n The enthalpy change for the reaction is positive, meaning the decomposition of N\u2082O absorbs energy, making it an endothermic process.<\/p>\n","protected":false},"excerpt":{"rendered":" Dinitrogen monoxide, N\u2082O, can decompose to nitrogen and oxygen gas:2 N2O(g) \u2192 2 N2(g) + O2(g)Use bond dissociation enthalpies (see Table on last page) to estimate the enthalpy change for this reaction. (Hint: As in problem #6, first draw the Lewis structures of the reactants and products that obey the octet rule so you knov […]<\/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-192976","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192976","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=192976"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192976\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192976"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192976"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192976"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Write the balanced reaction:<\/strong><\/h3>\n\n\n\n
[
2 \\text{N}_2\\text{O}(g) \\rightarrow 2 \\text{N}_2(g) + \\text{O}_2(g)
]<\/p>\n\n\n\n2. Draw the Lewis Structures:<\/strong><\/h3>\n\n\n\n
\n
3. Bond Dissociation Enthalpies:<\/strong><\/h3>\n\n\n\n
\n
4. Calculate the Enthalpy Change:<\/strong><\/h3>\n\n\n\n
\n
5. Enthalpy Change for the Reaction:<\/strong><\/h3>\n\n\n\n
[
\\Delta H = \\text{Bonds formed} – \\text{Bonds broken}
]
[
\\Delta H = 2388 \\, \\text{kJ\/mol} – 720 \\, \\text{kJ\/mol} = 1668 \\, \\text{kJ\/mol}
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n