Given data:<\/h3>\n\n\n\n\n- Standard enthalpy of formation of NO\u2082:
N2(g)+O2(g)\u21922\u2009NO2(g)\u0394Hf=33.9\u2009kJ\/molN2\u200b(g)+O2\u200b(g)\u21922NO2\u200b(g)\u0394H<\/em>f<\/em>\u200b=33.9kJ\/mol<\/li>\n<\/ul>\n\n\n\n1. Find \u0394H\u0394H<\/em> for:<\/h3>\n\n\n\nN2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/p>\n\n\n\n
Solution:<\/strong><\/p>\n\n\n\nThe given formation reaction involves 1 mol of N\u2082 and 1 mol of O\u2082 producing 2 mols of NO\u2082.
The target reaction involves 1 mol of N\u2082 and 2 mols of O\u2082 producing 2 mols of NO\u2082.<\/p>\n\n\n\n
Since the formation reaction produces 2 mols of NO\u2082 from 1 mol N\u2082 and 1 mol O\u2082, to produce 2 mols of NO\u2082 with 1 mol N\u2082 and 2 mols O\u2082, we can think of the second O\u2082 as an additional reactant that does not participate in the formation directly, but we can write the reaction as:<\/p>\n\n\n\n
N2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/p>\n\n\n\n
Applying Hess’s Law:
\u0394H=\u0394Hformation of 2 mols NO2+additional reaction involving O2\u0394H<\/em>=\u0394H<\/em>formation of 2 mols NO2\u200b\u200b+additional reaction involving O2\u200b<\/p>\n\n\n\nBut since the original formation reaction accounts for 2 mols NO\u2082 from 1 mol N\u2082 and 1 mol O\u2082, and the second O\u2082 is just an extra reactant, the enthalpy change for adding an extra O\u2082 molecule (which is just the standard enthalpy of O\u2082, which is zero as an element in its standard state), the total enthalpy change for this reaction is:<\/p>\n\n\n\n
\u0394H=33.9\u2009kJ\u0394H<\/em>=33.9kJ\u200b<\/p>\n\n\n\nAnswer:<\/strong><\/p>\n\n\n\n\u0394H=33.9\u2009kJ\u0394H<\/em>=33.9kJ\u200b<\/p>\n\n\n\n
\n\n\n\n2. Find \u0394H\u0394H<\/em> for:<\/h3>\n\n\n\n2\u2009NO2(g)\u2192N2(g)+2\u2009O2(g)2NO2\u200b(g)\u2192N2\u200b(g)+2O2\u200b(g)<\/p>\n\n\n\n
Solution:<\/strong><\/p>\n\n\n\nThis is the reverse of the formation reaction:<\/p>\n\n\n\n
N2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/p>\n\n\n\n
The enthalpy change for the reverse reaction is simply the negative of the forward reaction:<\/p>\n\n\n\n
\u0394H=\u221233.9\u2009kJ\u0394H<\/em>=\u221233.9kJ\u200b<\/p>\n\n\n\n
\n\n\n\n3. Find \u0394H\u0394H<\/em> for:<\/h3>\n\n\n\n2\u2009N2(g)+4\u2009O2(g)\u21924\u2009NO(g)2N2\u200b(g)+4O2\u200b(g)\u21924NO(g)<\/p>\n\n\n\n
Solution:<\/strong><\/p>\n\n\n\nThis reaction involves formation of nitric oxide (NO), not NO\u2082. To find this, we need the standard enthalpy of formation of NO:<\/p>\n\n\n\n
\n- N2(g)+O2(g)\u21922\u2009NO(g)N2\u200b(g)+O2\u200b(g)\u21922NO(g)
\u0394Hf\u224890.3\u2009kJ\/mol\u0394H<\/em>f<\/em>\u200b\u224890.3kJ\/mol (from standard data)<\/li>\n<\/ul>\n\n\n\nSince the reaction involves 2 mol of N\u2082 and 4 mol of O\u2082 producing 4 mols of NO, the enthalpy change is:<\/p>\n\n\n\n
\u0394H=2\u00d790.3\u2009kJ=180.6\u2009kJ\u0394H<\/em>=2\u00d790.3kJ=180.6kJ<\/p>\n\n\n\nAnswer:<\/strong><\/p>\n\n\n\n\u0394H=180.6\u2009kJ\u0394H<\/em>=180.6kJ\u200b<\/p>\n\n\n\n
\n\n\n\nSummary:<\/h2>\n\n\n\nReaction<\/th> \u0394H\u0394H<\/em> (kJ)<\/th><\/tr><\/thead>N2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/td> 33.9<\/strong><\/td><\/tr>2\u2009NO2(g)\u2192N2(g)+2\u2009O2(g)2NO2\u200b(g)\u2192N2\u200b(g)+2O2\u200b(g)<\/td> -33.9<\/strong><\/td><\/tr>2\u2009N2(g)+4\u2009O2(g)\u21924\u2009NO(g)2N2\u200b(g)+4O2\u200b(g)\u21924NO(g)<\/td> 180.6<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\nExplanation:<\/h2>\n\n\n\n
This problem demonstrates the application of Hess’s Law and thermochemical properties. The key idea is that enthalpy changes are state functions, meaning the total enthalpy change depends only on initial and final states, not the pathway taken. The formation reaction for NO\u2082 provides a baseline, from which other reactions can be derived by reversing or scaling the equations. For reactions involving the formation or decomposition of compounds, the enthalpy of formation is central. Reversing a reaction changes the sign of \u0394H\u0394H<\/em>, and scaling the coefficients scales the enthalpy accordingly. For reactions involving different compounds, standard enthalpies of formation from tables are used.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-841.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Consider the standard enthalpy of formation of nitrogen dioxide: % N2(g) + O2(g) \u00e2\u2020\u2019 2 NO2(g) (\u00ce\u201dHf = 33.9 kJ\/mol) Use the properties of thermochemical equations to find \u00ce\u201dH for each related reaction (3 pts): N2(g) + 2 O2(g) \u00e2\u2020\u2019 2 NO2(g) \u00ce\u201dH = ? 2 NO2(g) \u00e2\u2020\u2019 N2(g) + 2 O2(g) \u00ce\u201dH = ? […]<\/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-231002","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231002","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=231002"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231002\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231002"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231002"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231002"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
N2(g)+O2(g)\u21922\u2009NO2(g)\u0394Hf=33.9\u2009kJ\/molN2\u200b(g)+O2\u200b(g)\u21922NO2\u200b(g)\u0394H<\/em>f<\/em>\u200b=33.9kJ\/mol<\/li>\n<\/ul>\n\n\n\n
1. Find \u0394H\u0394H<\/em> for:<\/h3>\n\n\n\nN2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/p>\n\n\n\n
Solution:<\/strong><\/p>\n\n\n\nThe given formation reaction involves 1 mol of N\u2082 and 1 mol of O\u2082 producing 2 mols of NO\u2082.
The target reaction involves 1 mol of N\u2082 and 2 mols of O\u2082 producing 2 mols of NO\u2082.<\/p>\n\n\n\n
Since the formation reaction produces 2 mols of NO\u2082 from 1 mol N\u2082 and 1 mol O\u2082, to produce 2 mols of NO\u2082 with 1 mol N\u2082 and 2 mols O\u2082, we can think of the second O\u2082 as an additional reactant that does not participate in the formation directly, but we can write the reaction as:<\/p>\n\n\n\n
N2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/p>\n\n\n\n
Applying Hess’s Law:
\u0394H=\u0394Hformation of 2 mols NO2+additional reaction involving O2\u0394H<\/em>=\u0394H<\/em>formation of 2 mols NO2\u200b\u200b+additional reaction involving O2\u200b<\/p>\n\n\n\nBut since the original formation reaction accounts for 2 mols NO\u2082 from 1 mol N\u2082 and 1 mol O\u2082, and the second O\u2082 is just an extra reactant, the enthalpy change for adding an extra O\u2082 molecule (which is just the standard enthalpy of O\u2082, which is zero as an element in its standard state), the total enthalpy change for this reaction is:<\/p>\n\n\n\n
\u0394H=33.9\u2009kJ\u0394H<\/em>=33.9kJ\u200b<\/p>\n\n\n\nAnswer:<\/strong><\/p>\n\n\n\n\u0394H=33.9\u2009kJ\u0394H<\/em>=33.9kJ\u200b<\/p>\n\n\n\n
\n\n\n\n2. Find \u0394H\u0394H<\/em> for:<\/h3>\n\n\n\n2\u2009NO2(g)\u2192N2(g)+2\u2009O2(g)2NO2\u200b(g)\u2192N2\u200b(g)+2O2\u200b(g)<\/p>\n\n\n\n
Solution:<\/strong><\/p>\n\n\n\nThis is the reverse of the formation reaction:<\/p>\n\n\n\n
N2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/p>\n\n\n\n
The enthalpy change for the reverse reaction is simply the negative of the forward reaction:<\/p>\n\n\n\n
\u0394H=\u221233.9\u2009kJ\u0394H<\/em>=\u221233.9kJ\u200b<\/p>\n\n\n\n
\n\n\n\n3. Find \u0394H\u0394H<\/em> for:<\/h3>\n\n\n\n2\u2009N2(g)+4\u2009O2(g)\u21924\u2009NO(g)2N2\u200b(g)+4O2\u200b(g)\u21924NO(g)<\/p>\n\n\n\n
Solution:<\/strong><\/p>\n\n\n\nThis reaction involves formation of nitric oxide (NO), not NO\u2082. To find this, we need the standard enthalpy of formation of NO:<\/p>\n\n\n\n
\n- N2(g)+O2(g)\u21922\u2009NO(g)N2\u200b(g)+O2\u200b(g)\u21922NO(g)
\u0394Hf\u224890.3\u2009kJ\/mol\u0394H<\/em>f<\/em>\u200b\u224890.3kJ\/mol (from standard data)<\/li>\n<\/ul>\n\n\n\nSince the reaction involves 2 mol of N\u2082 and 4 mol of O\u2082 producing 4 mols of NO, the enthalpy change is:<\/p>\n\n\n\n
\u0394H=2\u00d790.3\u2009kJ=180.6\u2009kJ\u0394H<\/em>=2\u00d790.3kJ=180.6kJ<\/p>\n\n\n\nAnswer:<\/strong><\/p>\n\n\n\n\u0394H=180.6\u2009kJ\u0394H<\/em>=180.6kJ\u200b<\/p>\n\n\n\n
\n\n\n\nSummary:<\/h2>\n\n\n\nReaction<\/th> \u0394H\u0394H<\/em> (kJ)<\/th><\/tr><\/thead>N2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/td> 33.9<\/strong><\/td><\/tr>2\u2009NO2(g)\u2192N2(g)+2\u2009O2(g)2NO2\u200b(g)\u2192N2\u200b(g)+2O2\u200b(g)<\/td> -33.9<\/strong><\/td><\/tr>2\u2009N2(g)+4\u2009O2(g)\u21924\u2009NO(g)2N2\u200b(g)+4O2\u200b(g)\u21924NO(g)<\/td> 180.6<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\nExplanation:<\/h2>\n\n\n\n
This problem demonstrates the application of Hess’s Law and thermochemical properties. The key idea is that enthalpy changes are state functions, meaning the total enthalpy change depends only on initial and final states, not the pathway taken. The formation reaction for NO\u2082 provides a baseline, from which other reactions can be derived by reversing or scaling the equations. For reactions involving the formation or decomposition of compounds, the enthalpy of formation is central. Reversing a reaction changes the sign of \u0394H\u0394H<\/em>, and scaling the coefficients scales the enthalpy accordingly. For reactions involving different compounds, standard enthalpies of formation from tables are used.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Consider the standard enthalpy of formation of nitrogen dioxide: % N2(g) + O2(g) \u00e2\u2020\u2019 2 NO2(g) (\u00ce\u201dHf = 33.9 kJ\/mol) Use the properties of thermochemical equations to find \u00ce\u201dH for each related reaction (3 pts): N2(g) + 2 O2(g) \u00e2\u2020\u2019 2 NO2(g) \u00ce\u201dH = ? 2 NO2(g) \u00e2\u2020\u2019 N2(g) + 2 O2(g) \u00ce\u201dH = ? […]<\/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-231002","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231002","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=231002"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231002\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231002"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231002"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231002"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
The given formation reaction involves 1 mol of N\u2082 and 1 mol of O\u2082 producing 2 mols of NO\u2082.
The target reaction involves 1 mol of N\u2082 and 2 mols of O\u2082 producing 2 mols of NO\u2082.<\/p>\n\n\n\n
Since the formation reaction produces 2 mols of NO\u2082 from 1 mol N\u2082 and 1 mol O\u2082, to produce 2 mols of NO\u2082 with 1 mol N\u2082 and 2 mols O\u2082, we can think of the second O\u2082 as an additional reactant that does not participate in the formation directly, but we can write the reaction as:<\/p>\n\n\n\n
N2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/p>\n\n\n\n
Applying Hess’s Law: But since the original formation reaction accounts for 2 mols NO\u2082 from 1 mol N\u2082 and 1 mol O\u2082, and the second O\u2082 is just an extra reactant, the enthalpy change for adding an extra O\u2082 molecule (which is just the standard enthalpy of O\u2082, which is zero as an element in its standard state), the total enthalpy change for this reaction is:<\/p>\n\n\n\n \u0394H=33.9\u2009kJ\u0394H<\/em>=33.9kJ\u200b<\/p>\n\n\n\n Answer:<\/strong><\/p>\n\n\n\n \u0394H=33.9\u2009kJ\u0394H<\/em>=33.9kJ\u200b<\/p>\n\n\n\n 2\u2009NO2(g)\u2192N2(g)+2\u2009O2(g)2NO2\u200b(g)\u2192N2\u200b(g)+2O2\u200b(g)<\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n This is the reverse of the formation reaction:<\/p>\n\n\n\n N2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/p>\n\n\n\n The enthalpy change for the reverse reaction is simply the negative of the forward reaction:<\/p>\n\n\n\n \u0394H=\u221233.9\u2009kJ\u0394H<\/em>=\u221233.9kJ\u200b<\/p>\n\n\n\n 2\u2009N2(g)+4\u2009O2(g)\u21924\u2009NO(g)2N2\u200b(g)+4O2\u200b(g)\u21924NO(g)<\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n This reaction involves formation of nitric oxide (NO), not NO\u2082. To find this, we need the standard enthalpy of formation of NO:<\/p>\n\n\n\n Since the reaction involves 2 mol of N\u2082 and 4 mol of O\u2082 producing 4 mols of NO, the enthalpy change is:<\/p>\n\n\n\n \u0394H=2\u00d790.3\u2009kJ=180.6\u2009kJ\u0394H<\/em>=2\u00d790.3kJ=180.6kJ<\/p>\n\n\n\n Answer:<\/strong><\/p>\n\n\n\n \u0394H=180.6\u2009kJ\u0394H<\/em>=180.6kJ\u200b<\/p>\n\n\n\n This problem demonstrates the application of Hess’s Law and thermochemical properties. The key idea is that enthalpy changes are state functions, meaning the total enthalpy change depends only on initial and final states, not the pathway taken. The formation reaction for NO\u2082 provides a baseline, from which other reactions can be derived by reversing or scaling the equations. For reactions involving the formation or decomposition of compounds, the enthalpy of formation is central. Reversing a reaction changes the sign of \u0394H\u0394H<\/em>, and scaling the coefficients scales the enthalpy accordingly. For reactions involving different compounds, standard enthalpies of formation from tables are used.<\/p>\n\n\n\n Consider the standard enthalpy of formation of nitrogen dioxide: % N2(g) + O2(g) \u00e2\u2020\u2019 2 NO2(g) (\u00ce\u201dHf = 33.9 kJ\/mol) Use the properties of thermochemical equations to find \u00ce\u201dH for each related reaction (3 pts): N2(g) + 2 O2(g) \u00e2\u2020\u2019 2 NO2(g) \u00ce\u201dH = ? 2 NO2(g) \u00e2\u2020\u2019 N2(g) + 2 O2(g) \u00ce\u201dH = ? […]<\/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-231002","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231002","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=231002"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231002\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231002"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231002"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231002"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\u0394H=\u0394Hformation of 2 mols NO2+additional reaction involving O2\u0394H<\/em>=\u0394H<\/em>formation of 2 mols NO2\u200b\u200b+additional reaction involving O2\u200b<\/p>\n\n\n\n
\n\n\n\n2. Find \u0394H\u0394H<\/em> for:<\/h3>\n\n\n\n
\n\n\n\n3. Find \u0394H\u0394H<\/em> for:<\/h3>\n\n\n\n
\n
\u0394Hf\u224890.3\u2009kJ\/mol\u0394H<\/em>f<\/em>\u200b\u224890.3kJ\/mol (from standard data)<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nSummary:<\/h2>\n\n\n\n
Reaction<\/th> \u0394H\u0394H<\/em> (kJ)<\/th><\/tr><\/thead> N2(g)+2\u2009O2(g)\u21922\u2009NO2(g)N2\u200b(g)+2O2\u200b(g)\u21922NO2\u200b(g)<\/td> 33.9<\/strong><\/td><\/tr> 2\u2009NO2(g)\u2192N2(g)+2\u2009O2(g)2NO2\u200b(g)\u2192N2\u200b(g)+2O2\u200b(g)<\/td> -33.9<\/strong><\/td><\/tr> 2\u2009N2(g)+4\u2009O2(g)\u21924\u2009NO(g)2N2\u200b(g)+4O2\u200b(g)\u21924NO(g)<\/td> 180.6<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\nExplanation:<\/h2>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"