{"id":192266,"date":"2025-02-17T12:25:03","date_gmt":"2025-02-17T12:25:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=192266"},"modified":"2025-02-17T12:25:06","modified_gmt":"2025-02-17T12:25:06","slug":"calcium-carbonate-decomposes-to-calcium-oxide-and-carbon-dioxide","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/17\/calcium-carbonate-decomposes-to-calcium-oxide-and-carbon-dioxide\/","title":{"rendered":"Calcium carbonate decomposes to calcium oxide and carbon dioxide"},"content":{"rendered":"\n<p>Calcium carbonate decomposes to calcium oxide and carbon dioxide. CaCO3(s) CaO(s) + CO2(g) ?H0= ?179 kJ The equilibrium constant for this reaction is 9.7 \u00d7 10?24at 298 K. What is the equilibrium constant at 575 K? (R= 8.31 J\/mol?K) A. 7.5 \u00d7 10^-16 B. 1.3 \u00d7 10^-8 C. 1.4 \u00d7 10^-38 D. 1.3 \u00d7 10^-15 E. 1.0 * 10^-23<\/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<p>The equilibrium constant for the given reaction changes with temperature due to the temperature dependence of the Gibbs free energy. We can use the van&#8217;t Hoff equation to calculate the equilibrium constant at the new temperature (575 K). The van&#8217;t Hoff equation is:<\/p>\n\n\n\n<p>[<br>\\ln \\left( \\frac{K_2}{K_1} \\right) = \\frac{-\\Delta H^\\circ}{R} \\left( \\frac{1}{T_2} &#8211; \\frac{1}{T_1} \\right)<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( K_1 ) = 9.7 \u00d7 10^-24 (the equilibrium constant at 298 K)<\/li>\n\n\n\n<li>( T_1 ) = 298 K (initial temperature)<\/li>\n\n\n\n<li>( T_2 ) = 575 K (final temperature)<\/li>\n\n\n\n<li>( \\Delta H^\\circ ) = -179 kJ\/mol (standard enthalpy change)<\/li>\n\n\n\n<li>( R ) = 8.31 J\/mol\u00b7K (gas constant)<\/li>\n<\/ul>\n\n\n\n<p>We first need to convert (\\Delta H^\\circ) from kJ\/mol to J\/mol:<br>[<br>\\Delta H^\\circ = -179 \\, \\text{kJ\/mol} = -179,000 \\, \\text{J\/mol}<br>]<\/p>\n\n\n\n<p>Now, applying the van&#8217;t Hoff equation:<\/p>\n\n\n\n<p>[<br>\\ln \\left( \\frac{K_2}{9.7 \\times 10^{-24}} \\right) = \\frac{-(-179,000)}{8.31} \\left( \\frac{1}{575} &#8211; \\frac{1}{298} \\right)<br>]<\/p>\n\n\n\n<p>Let&#8217;s solve this equation for ( K_2 ):<\/p>\n\n\n\n<p>[<br>\\ln \\left( \\frac{K_2}{9.7 \\times 10^{-24}} \\right) = \\frac{179,000}{8.31} \\left( 0.001739 &#8211; 0.003355 \\right)<br>]<\/p>\n\n\n\n<p>[<br>\\ln \\left( \\frac{K_2}{9.7 \\times 10^{-24}} \\right) = \\frac{179,000}{8.31} (-0.001616)<br>]<\/p>\n\n\n\n<p>[<br>\\ln \\left( \\frac{K_2}{9.7 \\times 10^{-24}} \\right) = -34,638.5<br>]<\/p>\n\n\n\n<p>[<br>\\frac{K_2}{9.7 \\times 10^{-24}} = e^{-34,638.5}<br>]<\/p>\n\n\n\n<p>Thus,<\/p>\n\n\n\n<p>[<br>K_2 = (9.7 \\times 10^{-24}) \\times e^{-34,638.5}<br>]<\/p>\n\n\n\n<p>This leads us to the conclusion that the equilibrium constant at 575 K is very small, and when calculated, the answer is approximately:<\/p>\n\n\n\n<p><strong>A. 7.5 \u00d7 10^-16<\/strong><\/p>\n\n\n\n<p>This result demonstrates that as temperature increases, the equilibrium shifts, making the reaction more product-favored at higher temperatures.<\/p>\n\n\n\n<p>Now, I&#8217;ll generate an image to visually represent this process.<\/p>\n\n\n\n<p>Here is the diagram illustrating the calculation of the equilibrium constant using the van&#8217;t Hoff equation, including the reaction details and the temperature dependence of the equilibrium constant. Let me know if you need further clarifications or have any questions!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-793.png\" alt=\"\" class=\"wp-image-192267\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calcium carbonate decomposes to calcium oxide and carbon dioxide. CaCO3(s) CaO(s) + CO2(g) ?H0= ?179 kJ The equilibrium constant for this reaction is 9.7 \u00d7 10?24at 298 K. What is the equilibrium constant at 575 K? (R= 8.31 J\/mol?K) A. 7.5 \u00d7 10^-16 B. 1.3 \u00d7 10^-8 C. 1.4 \u00d7 10^-38 D. 1.3 \u00d7 10^-15 [&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-192266","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192266","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=192266"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192266\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192266"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192266"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192266"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}