{"id":217984,"date":"2025-05-22T21:15:23","date_gmt":"2025-05-22T21:15:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217984"},"modified":"2025-05-22T21:15:25","modified_gmt":"2025-05-22T21:15:25","slug":"solid-ammonium-chloride-dissociates-according-to-nh4cls-nh3g-hclg","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/22\/solid-ammonium-chloride-dissociates-according-to-nh4cls-nh3g-hclg\/","title":{"rendered":"Solid ammonium chloride dissociates according to NH4Cl(s) &#8211; NH3(g) + HCl(g"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Solid ammonium chloride dissociates according to NH4Cl(s) &#8211; NH3(g) + HCl(g). The dissociation vapour pressure of NH4Cl at 427 \u00b0C is 608 kPa but at 459 \u00b0C it has risen to 1115 kPa. Calculate (i) the equilibrium constant, (ii) the standard reaction Gibbs energy, (iii) the standard enthalpy, (iv) the standard entropy of dissociation, all at 427 \u00b0C. Assume that the vapour behaves as a perfect gas and that ?rH\u00b0 and ?rS\u00b0 are independent of temperature in the range given.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We are given the dissociation reaction:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{NH}_4\\text{Cl}(s) \\rightleftharpoons \\text{NH}_3(g) + \\text{HCl}(g)<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">And the vapor pressures (which equal the <strong>total pressure<\/strong> of NH\u2083 and HCl) at:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>427 \u00b0C = 700.15 K \u2192 <strong>P\u2081 = 608 kPa<\/strong><\/li>\n\n\n\n<li>459 \u00b0C = 732.15 K \u2192 <strong>P\u2082 = 1115 kPa<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(i) <strong>Equilibrium constant $K_p$ at 427 \u00b0C:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">From the reaction, each mole of NH\u2084Cl produces 1 mole of NH\u2083 and 1 mole of HCl:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>K_p = P_{\\text{NH}<em>3} \\times P<\/em>{\\text{HCl}} = (P\/2)^2 = \\frac{P^2}{4}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">At 427 \u00b0C, $P = 608 \\, \\text{kPa}$:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>K_p = \\frac{608^2}{4} = \\frac{369664}{4} = \\boxed{92416 \\, \\text{kPa}^2}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(ii) <strong>Standard Gibbs energy change $\\Delta_r G^\\circ$:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Use the formula:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\Delta_r G^\\circ = -RT \\ln K_p<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$R = 8.314 \\, \\text{J mol}^{-1} \\text{K}^{-1}$<\/li>\n\n\n\n<li>$T = 700.15 \\, \\text{K}$<\/li>\n\n\n\n<li>$K_p = 92416$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\Delta_r G^\\circ = -8.314 \\times 700.15 \\times \\ln(92416)<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\ln(92416) \\approx 11.434<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\Delta_r G^\\circ = -8.314 \\times 700.15 \\times 11.434 \\approx \\boxed{-66586 \\, \\text{J mol}^{-1} \\text{ or } -66.6 \\, \\text{kJ\/mol}}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(iii) <strong>Standard Enthalpy change $\\Delta_r H^\\circ$:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Using <strong>van\u2019t Hoff equation<\/strong>:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\ln\\left(\\frac{K_2}{K_1}\\right) = -\\frac{\\Delta_r H^\\circ}{R} \\left(\\frac{1}{T_2} &#8211; \\frac{1}{T_1}\\right)<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">First calculate $K_2 = \\frac{1115^2}{4} = 310056.25 \\, \\text{kPa}^2$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Then:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\ln\\left(\\frac{K_2}{K_1}\\right) = \\ln\\left(\\frac{310056.25}{92416}\\right) = \\ln(3.354) \\approx 1.209<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\frac{1}{T_2} &#8211; \\frac{1}{T_1} = \\frac{1}{732.15} &#8211; \\frac{1}{700.15} \\approx -0.0000628 \\, \\text{K}^{-1}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\Delta_r H^\\circ = -\\frac{1.209 \\times 8.314}{-0.0000628} \\approx \\boxed{160096 \\, \\text{J\/mol} = 160.1 \\, \\text{kJ\/mol}}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(iv) <strong>Standard Entropy change $\\Delta_r S^\\circ$:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">From:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\Delta_r G^\\circ = \\Delta_r H^\\circ &#8211; T \\Delta_r S^\\circ<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\Delta_r S^\\circ = \\frac{\\Delta_r H^\\circ &#8211; \\Delta_r G^\\circ}{T} = \\frac{160096 &#8211; (-66586)}{700.15} = \\frac{226682}{700.15} \\approx \\boxed{323.8 \\, \\text{J mol}^{-1} \\text{K}^{-1}}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>$K_p = 92416 \\, \\text{kPa}^2$<\/li>\n\n\n\n<li>$\\Delta_r G^\\circ = -66.6 \\, \\text{kJ\/mol}$<\/li>\n\n\n\n<li>$\\Delta_r H^\\circ = 160.1 \\, \\text{kJ\/mol}$<\/li>\n\n\n\n<li>$\\Delta_r S^\\circ = 323.8 \\, \\text{J\/mol\u00b7K}$<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-109.png\" alt=\"\" class=\"wp-image-217985\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Solid ammonium chloride dissociates according to NH4Cl(s) &#8211; NH3(g) + HCl(g). The dissociation vapour pressure of NH4Cl at 427 \u00b0C is 608 kPa but at 459 \u00b0C it has risen to 1115 kPa. Calculate (i) the equilibrium constant, (ii) the standard reaction Gibbs energy, (iii) the standard enthalpy, (iv) the standard entropy of dissociation, all [&hellip;]<\/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-217984","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217984","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=217984"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217984\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217984"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217984"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217984"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}