The standard molar enthalpy of formation of NH3(g) is \u201345.9 kJ\/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g) react to produce NH3(g)? <\/p>\n\n\n\n
A. \u201310.3 kJ <\/p>\n\n\n\n
B. \u201320.7 kJ <\/p>\n\n\n\n
C. \u201329.8 kJ <\/p>\n\n\n\n
D. \u201343.7 kJ <\/p>\n\n\n\n
E. \u201365.6 kJ<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the enthalpy change when 9.51 g of N\u2082(g) and 1.96 g of H\u2082(g) react to produce NH\u2083(g), we need to follow these steps:<\/p>\n\n\n\n The balanced equation for the formation of ammonia is:<\/p>\n\n\n\n [ We start by calculating the moles of N\u2082 and H\u2082 based on their given masses.<\/p>\n\n\n\n For N\u2082: For H\u2082: From the balanced equation, we see that 1 mol of N\u2082 reacts with 3 mol of H\u2082. Therefore, we need 3 moles of H\u2082 for every mole of N\u2082. Let’s check if we have enough H\u2082 to react with all the N\u2082:<\/p>\n\n\n\n Since we only have 0.98 mol of H\u2082, which is less than the required 1.017 mol, H\u2082 is the limiting reactant.<\/p>\n\n\n\n Now, we know that 1 mol of N\u2082 reacts with 3 mol of H\u2082 to produce 2 mol of NH\u2083. The standard molar enthalpy of formation of NH\u2083(g) is \u201345.9 kJ\/mol. The enthalpy change for the reaction is proportional to the amount of limiting reactant, H\u2082.<\/p>\n\n\n\n For every 3 moles of H\u2082, 2 moles of NH\u2083 are formed. The enthalpy change for 2 moles of NH\u2083 is:<\/p>\n\n\n\n [ Now, calculate the enthalpy change for the 0.98 mol of H\u2082:<\/p>\n\n\n\n [ Thus, the enthalpy change for the reaction is \u201329.8 kJ<\/strong>.<\/p>\n\n\n\n The standard molar enthalpy of formation of NH3(g) is \u201345.9 kJ\/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g) react to produce NH3(g)? A. \u201310.3 kJ B. \u201320.7 kJ C. \u201329.8 kJ D. \u201343.7 kJ E. \u201365.6 kJ The Correct Answer and Explanation is : To determine the enthalpy change […]<\/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-189822","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189822","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=189822"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189822\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189822"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189822"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189822"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Write the balanced chemical equation<\/h3>\n\n\n\n
N_2(g) + 3H_2(g) \\rightarrow 2NH_3(g)
]<\/p>\n\n\n\nStep 2: Calculate the moles of N\u2082 and H\u2082<\/h3>\n\n\n\n
\n
[
\\text{moles of N}_2 = \\frac{\\text{mass of N}_2}{\\text{molar mass of N}_2} = \\frac{9.51 \\, \\text{g}}{28.0 \\, \\text{g\/mol}} = 0.339 \\, \\text{mol}
]<\/p>\n\n\n\n
[
\\text{moles of H}_2 = \\frac{\\text{mass of H}_2}{\\text{molar mass of H}_2} = \\frac{1.96 \\, \\text{g}}{2.0 \\, \\text{g\/mol}} = 0.98 \\, \\text{mol}
]<\/p>\n\n\n\nStep 3: Determine the limiting reactant<\/h3>\n\n\n\n
\n
[
0.339 \\, \\text{mol N}_2 \\times 3 = 1.017 \\, \\text{mol H}_2
]<\/li>\n<\/ul>\n\n\n\nStep 4: Calculate the enthalpy change<\/h3>\n\n\n\n
\\Delta H = 2 \\times (-45.9 \\, \\text{kJ\/mol}) = -91.8 \\, \\text{kJ}
]<\/p>\n\n\n\n
\\Delta H_{\\text{reaction}} = \\frac{0.98}{3} \\times (-91.8 \\, \\text{kJ}) = -29.8 \\, \\text{kJ}
]<\/p>\n\n\n\nCorrect Answer: C. \u201329.8 kJ<\/strong><\/h3>\n","protected":false},"excerpt":{"rendered":"