{"id":185401,"date":"2025-01-22T13:34:41","date_gmt":"2025-01-22T13:34:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185401"},"modified":"2025-01-22T13:34:43","modified_gmt":"2025-01-22T13:34:43","slug":"the-equilibrium-constant-for-the-equation-ch3coohaq-h2ol-h3o-aq-ch3coo-aq-is-ka-1-8-times-10-5-m-at-25-0-degree-c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/22\/the-equilibrium-constant-for-the-equation-ch3coohaq-h2ol-h3o-aq-ch3coo-aq-is-ka-1-8-times-10-5-m-at-25-0-degree-c\/","title":{"rendered":"The equilibrium constant for the equation CH3COOH(aq) + H2O(l) H3O + (aq) + CH3COO-(aq) is Ka = 1.8 Times 10-5 M at 25.0 degree C"},"content":{"rendered":"\n<p>The equilibrium constant for the equation CH3COOH(aq) + H2O(l) H3O + (aq) + CH3COO-(aq) is Ka = 1.8 Times 10-5 M at 25.0 degree C. Calculate the value of DeltaGdegreerXn at 25.0 degree C for acetic acid when [CH3COO-] = [H3O+] = [CH3COOH] = 1.00 M Will the acid spontaneously dissociate under these conditions? Calculate the value of DeltaGrxn at 25 degree C for acetic acid when [CH3COO-]=[H3OO + ] = 4.56 Times 10-5 M and [CH3COOH] = 1.140 M Will the acid spontaneously dissociate under these conditions?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Definitions and Formulae<\/h3>\n\n\n\n<p>The equilibrium constant ((K_a)) and Gibbs free energy change ((\\Delta G)) are related by the equation:<\/p>\n\n\n\n<p>[<br>\\Delta G^\\circ_{\\text{rxn}} = -RT \\ln K_a<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(R = 8.314 \\, \\text{J\/mol\u00b7K}), the gas constant.<\/li>\n\n\n\n<li>(T = 298.15 \\, \\text{K}), the temperature in Kelvin.<\/li>\n\n\n\n<li>(K_a = 1.8 \\times 10^{-5}), the equilibrium constant.<\/li>\n<\/ul>\n\n\n\n<p>Under non-standard conditions, the reaction Gibbs free energy ((\\Delta G_{\\text{rxn}})) is given by:<\/p>\n\n\n\n<p>[<br>\\Delta G_{\\text{rxn}} = \\Delta G^\\circ_{\\text{rxn}} + RT \\ln Q<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(Q = \\frac{[\\text{H}_3\\text{O}^+][\\text{CH}_3\\text{COO}^-]}{[\\text{CH}_3\\text{COOH}]}), the reaction quotient.<\/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\">Part 1: When ([\\text{CH}_3\\text{COOH}] = [\\text{H}_3\\text{O}^+] = [\\text{CH}_3\\text{COO}^-] = 1.00 \\, \\text{M})<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate (\\Delta G^\\circ_{\\text{rxn}}):<\/strong><br>[<br>\\Delta G^\\circ_{\\text{rxn}} = -RT \\ln K_a = -(8.314)(298.15) \\ln(1.8 \\times 10^{-5})<br>]<br>[<br>\\Delta G^\\circ_{\\text{rxn}} \\approx 26920 \\, \\text{J\/mol} = 26.92 \\, \\text{kJ\/mol}<br>]<\/li>\n\n\n\n<li><strong>Calculate (Q):<\/strong><br>[<br>Q = \\frac{(1.00)(1.00)}{1.00} = 1.00<br>]<\/li>\n\n\n\n<li><strong>Calculate (\\Delta G_{\\text{rxn}}):<\/strong><br>[<br>\\Delta G_{\\text{rxn}} = \\Delta G^\\circ_{\\text{rxn}} + RT \\ln Q = 26.92 + 0 = 26.92 \\, \\text{kJ\/mol}<br>]<\/li>\n<\/ol>\n\n\n\n<p>Since (\\Delta G_{\\text{rxn}} &gt; 0), the dissociation is non-spontaneous under these conditions.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Part 2: When ([\\text{CH}_3\\text{COO}^-] = [\\text{H}_3\\text{O}^+] = 4.56 \\times 10^{-5} \\, \\text{M}) and ([\\text{CH}_3\\text{COOH}] = 1.140 \\, \\text{M})<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate (Q):<\/strong><br>[<br>Q = \\frac{(4.56 \\times 10^{-5})(4.56 \\times 10^{-5})}{1.140}<br>]<br>[<br>Q \\approx 1.82 \\times 10^{-9}<br>]<\/li>\n\n\n\n<li><strong>Calculate (\\Delta G_{\\text{rxn}}):<\/strong><br>[<br>\\Delta G_{\\text{rxn}} = \\Delta G^\\circ_{\\text{rxn}} + RT \\ln Q<br>]<br>[<br>\\Delta G_{\\text{rxn}} = 26.92 + (8.314)(298.15) \\ln(1.82 \\times 10^{-9})<br>]<br>[<br>\\Delta G_{\\text{rxn}} \\approx 26.92 &#8211; 50.87 \\approx -23.95 \\, \\text{kJ\/mol}<br>]<\/li>\n<\/ol>\n\n\n\n<p>Since (\\Delta G_{\\text{rxn}} &lt; 0), the dissociation is spontaneous under these conditions.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>When ([\\text{CH}_3\\text{COOH}] = [\\text{H}_3\\text{O}^+] = [\\text{CH}_3\\text{COO}^-] = 1.00 \\, \\text{M}), dissociation is <strong>non-spontaneous<\/strong>.<\/li>\n\n\n\n<li>When ([\\text{CH}_3\\text{COO}^-] = [\\text{H}_3\\text{O}^+] = 4.56 \\times 10^{-5} \\, \\text{M}) and ([\\text{CH}_3\\text{COOH}] = 1.140 \\, \\text{M}), dissociation is <strong>spontaneous<\/strong>.<\/li>\n<\/ol>\n\n\n\n<p>This demonstrates the importance of the reaction quotient ((Q)) in determining spontaneity, as it compares actual concentrations to equilibrium concentrations.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The equilibrium constant for the equation CH3COOH(aq) + H2O(l) H3O + (aq) + CH3COO-(aq) is Ka = 1.8 Times 10-5 M at 25.0 degree C. Calculate the value of DeltaGdegreerXn at 25.0 degree C for acetic acid when [CH3COO-] = [H3O+] = [CH3COOH] = 1.00 M Will the acid spontaneously dissociate under these conditions? Calculate [&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-185401","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185401","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=185401"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185401\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185401"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185401"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185401"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}