{"id":221914,"date":"2025-05-30T17:34:56","date_gmt":"2025-05-30T17:34:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221914"},"modified":"2025-05-30T17:34:58","modified_gmt":"2025-05-30T17:34:58","slug":"for-the-decomposition-of-barium-carbonate-consider-the-following-thermodynamic-data","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/30\/for-the-decomposition-of-barium-carbonate-consider-the-following-thermodynamic-data\/","title":{"rendered":"For the decomposition of barium carbonate, consider the following thermodynamic data"},"content":{"rendered":"\n

Use four significant figures please.<\/p>\n\n\n\n

For the decomposition of barium carbonate, consider the following thermodynamic data: Calculate the temperature in kelvins above which this reaction is spontaneous. Calculate the equilibrium constant for the following reaction at room temperature, 25 degree C: BaCO3(s) rightarrow BaO(s) + CO2(g) Express your answer numerically to three significant figures. Barium oxide, BaO, is manufactured by decomposition of barium carbonate, BaCO3, in a furnace: BaCO3(s) rightarrow BaO(s) + CO2(g) The heat produced by a furnace maintained at a temperature of 600 K can decompose barium carbonate to barium oxide. The spontaneity of a reaction can be determined from the free energy change for the reaction, Delta G degree degree.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To determine when the decomposition of barium carbonate is spontaneous<\/strong>, we use the Gibbs free energy equation<\/strong>: \u0394G\u2218=\u0394H\u2218\u2212T\u0394S\u2218\\Delta G^\\circ = \\Delta H^\\circ – T\\Delta S^\\circ<\/p>\n\n\n\n

A reaction becomes spontaneous<\/strong> when: \u0394G\u2218<0\\Delta G^\\circ < 0<\/p>\n\n\n\n

So we solve for the temperature TT<\/strong> at which: \u0394G\u2218=0\u21d2T=\u0394H\u2218\u0394S\u2218\\Delta G^\\circ = 0 \\Rightarrow T = \\frac{\\Delta H^\\circ}{\\Delta S^\\circ}<\/p>\n\n\n\n

Step 1: Convert units<\/h3>\n\n\n\n
    \n
  • \u0394H\u2218=243.5\u2009kJ\/mol=243500\u2009J\/mol\\Delta H^\\circ = 243.5 \\, \\text{kJ\/mol} = 243500 \\, \\text{J\/mol}<\/li>\n\n\n\n
  • \u0394S\u2218=172.0\u2009J\/mol\\cdotpK\\Delta S^\\circ = 172.0 \\, \\text{J\/mol\u00b7K}<\/li>\n<\/ul>\n\n\n\n

    Step 2: Calculate the temperature<\/h3>\n\n\n\n

    T=243500\u2009J\/mol172.0\u2009J\/mol\\cdotpK=1415.7\u2009KT = \\frac{243500 \\, \\text{J\/mol}}{172.0 \\, \\text{J\/mol\u00b7K}} = 1415.7 \\, \\text{K}<\/p>\n\n\n\n

    \u2705 Answer<\/strong>:<\/h3>\n\n\n\n

    1415 K<\/strong> (to four significant figures: 1415.698 K<\/strong>)<\/p>\n\n\n\n

    \n\n\n\n

    Part 2: Calculate the equilibrium constant KK<\/strong> at 25\u00b0C (298 K)<\/h3>\n\n\n\n

    We use: \u0394G\u2218=\u2212RTln\u2061K\u21d2ln\u2061K=\u2212\u0394G\u2218RT\\Delta G^\\circ = -RT \\ln K \\Rightarrow \\ln K = -\\frac{\\Delta G^\\circ}{RT}<\/p>\n\n\n\n

    We first compute \u0394G\u2218\\Delta G^\\circ at 298 K: \u0394G\u2218=\u0394H\u2218\u2212T\u0394S\u2218=243500\u2212(298)(172.0)=243500\u221251256=192244\u2009J\/mol\\Delta G^\\circ = \\Delta H^\\circ – T\\Delta S^\\circ = 243500 – (298)(172.0) = 243500 – 51256 = 192244 \\, \\text{J\/mol}<\/p>\n\n\n\n

    Now, using R=8.314\u2009J\/mol\\cdotpKR = 8.314 \\, \\text{J\/mol\u00b7K}: ln\u2061K=\u2212192244(8.314)(298)=\u221277.46\u21d2K=e\u221277.46\u22481.00\u00d710\u221234\\ln K = -\\frac{192244}{(8.314)(298)} = -77.46 \\Rightarrow K = e^{-77.46} \\approx 1.00 \\times 10^{-34}<\/p>\n\n\n\n

    \u2705 Equilibrium Constant at 25\u00b0C<\/strong>:<\/h3>\n\n\n\n

    K=1.00\u00d710\u221234K = 1.00 \\times 10^{-34}<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Explanation<\/h3>\n\n\n\n

    The spontaneity of a chemical reaction depends on the sign of the Gibbs free energy change, \u0394G\u2218\\Delta G^\\circ. For the decomposition of barium carbonate (BaCO3(s)\u2192BaO(s)+CO2(g)\\text{BaCO}_3(s) \\rightarrow \\text{BaO}(s) + \\text{CO}_2(g)), we are provided with the enthalpy change (\u0394H\u2218=243.5\u2009kJ\/mol\\Delta H^\\circ = 243.5 \\, \\text{kJ\/mol}) and the entropy change (\u0394S\u2218=172.0\u2009J\/mol\\cdotpK\\Delta S^\\circ = 172.0 \\, \\text{J\/mol\u00b7K}).<\/p>\n\n\n\n

    Spontaneity is achieved when \u0394G\u2218<0\\Delta G^\\circ < 0. Setting \u0394G\u2218=0\\Delta G^\\circ = 0 allows us to calculate the threshold temperature<\/strong> above which the reaction becomes spontaneous. Converting enthalpy to joules and dividing by the entropy gives T=243500172.0=1415.698\u2009KT = \\frac{243500}{172.0} = 1415.698 \\, \\text{K}. Thus, the decomposition is spontaneous at temperatures above approximately 1415.7 K<\/strong>.<\/p>\n\n\n\n

    To further analyze the reaction\u2019s behavior under standard conditions (room temperature, 25\u00b0C or 298 K), we use the Gibbs free energy change to calculate the equilibrium constant KK. A positive \u0394G\u2218\\Delta G^\\circ at this temperature indicates the reaction is non-spontaneous<\/strong> under standard conditions. The computed equilibrium constant, K=1.00\u00d710\u221234K = 1.00 \\times 10^{-34}, is extremely small<\/strong>, confirming that at room temperature, barium carbonate hardly decomposes.<\/p>\n\n\n\n

    This helps explain why a furnace (at around 600 K) is used industrially to drive this decomposition. Although 600 K is below the threshold for spontaneity, external energy helps shift the reaction toward products. The calculations illustrate the strong temperature dependence of reaction spontaneity and equilibrium in thermodynamically controlled processes<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    .<\/p>\n","protected":false},"excerpt":{"rendered":"

    Use four significant figures please. For the decomposition of barium carbonate, consider the following thermodynamic data: Calculate the temperature in kelvins above which this reaction is spontaneous. Calculate the equilibrium constant for the following reaction at room temperature, 25 degree C: BaCO3(s) rightarrow BaO(s) + CO2(g) Express your answer numerically to three significant figures. Barium […]<\/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-221914","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221914","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=221914"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221914\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221914"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221914"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221914"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}