{"id":257958,"date":"2025-07-17T18:58:00","date_gmt":"2025-07-17T18:58:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257958"},"modified":"2025-07-17T18:58:06","modified_gmt":"2025-07-17T18:58:06","slug":"use-the-thermochemical-equations-shown-below-to-determine-the-enthalpy-of-the-final-reaction","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/use-the-thermochemical-equations-shown-below-to-determine-the-enthalpy-of-the-final-reaction\/","title":{"rendered":"Use the thermochemical equations shown below to determine the enthalpy of the final reaction"},"content":{"rendered":"\n

Use the thermochemical equations shown below to determine the enthalpy of the final reaction: (1) CO2 (g) ? C (s) + O2 (g) ?? = 551 kJ (2) H2(g) + 1\/2 O2 (g) ? H2O (l) ?? = -400 kJ (3) CH4 (g) + 2O2 (g) ? CO2 (g) + 2H2O (l) ?? = -1247 kJ CH4 (g) ? C (s) + 2H2 (g) ?? =? 2) The heat released on combustion of benzoic acid (C6H5COOH) is 26.38 kJ\/g. When 1.000 g of benzoic acid was combusted in a bomb calorimeter, the temperature of the calorimeter rose from 24.32 \u00b0C to 27.05 \u00b0C. What is the heat capacity, C, of the calorimeter (in kJ\/\u00b0C)? NOTE: Marks are deducted if units and significant figures are ignored.<\/p>\n\n\n\n

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

Problem 1: Enthalpy of the Final Reaction<\/h3>\n\n\n\n

We are given the following thermochemical equations:<\/p>\n\n\n\n

    \n
  1. CO2(g)\u2192C(s)+O2(g)\u0394H1=551\u2009kJCO_2 (g) \\rightarrow C (s) + O_2 (g) \\quad \\Delta H_1 = 551 \\, \\text{kJ}CO2\u200b(g)\u2192C(s)+O2\u200b(g)\u0394H1\u200b=551kJ<\/li>\n\n\n\n
  2. H2(g)+12O2(g)\u2192H2O(l)\u0394H2=\u2212400\u2009kJH_2 (g) + \\frac{1}{2} O_2 (g) \\rightarrow H_2O (l) \\quad \\Delta H_2 = -400 \\, \\text{kJ}H2\u200b(g)+21\u200bO2\u200b(g)\u2192H2\u200bO(l)\u0394H2\u200b=\u2212400kJ<\/li>\n\n\n\n
  3. CH4(g)+2O2(g)\u2192CO2(g)+2H2O(l)\u0394H3=\u22121247\u2009kJCH_4 (g) + 2O_2 (g) \\rightarrow CO_2 (g) + 2H_2O (l) \\quad \\Delta H_3 = -1247 \\, \\text{kJ}CH4\u200b(g)+2O2\u200b(g)\u2192CO2\u200b(g)+2H2\u200bO(l)\u0394H3\u200b=\u22121247kJ<\/li>\n<\/ol>\n\n\n\n

    We need to determine the enthalpy of the reaction:
    CH4(g)\u2192C(s)+2H2(g)CH_4 (g) \\rightarrow C (s) + 2H_2 (g)CH4\u200b(g)\u2192C(s)+2H2\u200b(g).<\/p>\n\n\n\n

    Solution Strategy:<\/h4>\n\n\n\n

    We can manipulate the given reactions to derive the desired reaction. Here\u2019s the breakdown of steps:<\/p>\n\n\n\n

      \n
    1. Start with the combustion of methane (reaction 3) and reverse it so that methane (CH4) is on the left: CO2(g)+2H2O(l)\u2192CH4(g)+2O2(g)CO_2 (g) + 2H_2O (l) \\rightarrow CH_4 (g) + 2O_2 (g)CO2\u200b(g)+2H2\u200bO(l)\u2192CH4\u200b(g)+2O2\u200b(g) The enthalpy change for this reversed reaction is +1247\u2009kJ+1247 \\, \\text{kJ}+1247kJ.<\/li>\n\n\n\n
    2. Add reaction (1) to remove CO2(g)CO_2 (g)CO2\u200b(g) from the equation: CO2(g)\u2192C(s)+O2(g)\u0394H1=551\u2009kJCO_2 (g) \\rightarrow C (s) + O_2 (g) \\quad \\Delta H_1 = 551 \\, \\text{kJ}CO2\u200b(g)\u2192C(s)+O2\u200b(g)\u0394H1\u200b=551kJ This results in the cancellation of CO2(g)CO_2 (g)CO2\u200b(g).<\/li>\n\n\n\n
    3. Add reaction (2) to remove H2O(l)H_2O (l)H2\u200bO(l) and form H2(g)H_2 (g)H2\u200b(g): H2(g)+12O2(g)\u2192H2O(l)\u0394H2=\u2212400\u2009kJH_2 (g) + \\frac{1}{2} O_2 (g) \\rightarrow H_2O (l) \\quad \\Delta H_2 = -400 \\, \\text{kJ}H2\u200b(g)+21\u200bO2\u200b(g)\u2192H2\u200bO(l)\u0394H2\u200b=\u2212400kJ<\/li>\n<\/ol>\n\n\n\n

      By adding all the reactions, we can now derive the following: CH4(g)\u2192C(s)+2H2(g)\u0394H=1247+551\u2212400=1398\u2009kJCH_4 (g) \\rightarrow C (s) + 2H_2 (g) \\quad \\Delta H = 1247 + 551 – 400 = 1398 \\, \\text{kJ}CH4\u200b(g)\u2192C(s)+2H2\u200b(g)\u0394H=1247+551\u2212400=1398kJ<\/p>\n\n\n\n

      Thus, the enthalpy of the reaction CH4(g)\u2192C(s)+2H2(g)CH_4 (g) \\rightarrow C (s) + 2H_2 (g)CH4\u200b(g)\u2192C(s)+2H2\u200b(g) is \u0394H=1398\u2009kJ\\Delta H = 1398 \\, \\text{kJ}\u0394H=1398kJ.<\/p>\n\n\n\n

      \n\n\n\n

      Problem 2: Heat Capacity of the Calorimeter<\/h3>\n\n\n\n

      We are told the heat released by the combustion of benzoic acid is 26.38 kJ\/g. When 1.000 g of benzoic acid is combusted, the temperature of the calorimeter rises from 24.32 \u00b0C to 27.05 \u00b0C.<\/p>\n\n\n\n

      Solution Strategy:<\/h4>\n\n\n\n
        \n
      1. Determine the heat released during combustion<\/strong>:
        The heat released by the combustion of 1.000 g of benzoic acid is: Q=26.38\u2009kJ\/g\u00d71.000\u2009g=26.38\u2009kJQ = 26.38 \\, \\text{kJ\/g} \\times 1.000 \\, \\text{g} = 26.38 \\, \\text{kJ}Q=26.38kJ\/g\u00d71.000g=26.38kJ<\/li>\n\n\n\n
      2. Calculate the temperature change<\/strong>:
        The temperature change of the calorimeter is: \u0394T=27.05\u2009\u00b0C\u221224.32\u2009\u00b0C=2.73\u2009\u00b0C\\Delta T = 27.05 \\, \\text{\u00b0C} – 24.32 \\, \\text{\u00b0C} = 2.73 \\, \\text{\u00b0C}\u0394T=27.05\u00b0C\u221224.32\u00b0C=2.73\u00b0C<\/li>\n\n\n\n
      3. Use the heat capacity formula<\/strong>:
        The formula for the heat absorbed by the calorimeter is: Q=C\u00d7\u0394TQ = C \\times \\Delta TQ=C\u00d7\u0394T Where:
        • Q=26.38\u2009kJQ = 26.38 \\, \\text{kJ}Q=26.38kJ<\/li>
        • \u0394T=2.73\u2009\u00b0C\\Delta T = 2.73 \\, \\text{\u00b0C}\u0394T=2.73\u00b0C<\/li>
        • CCC is the heat capacity of the calorimeter.<\/li><\/ul>Rearranging the formula to solve for CCC: C=Q\u0394T=26.38\u2009kJ2.73\u2009\u00b0C=9.66\u2009kJ\/\u00b0CC = \\frac{Q}{\\Delta T} = \\frac{26.38 \\, \\text{kJ}}{2.73 \\, \\text{\u00b0C}} = 9.66 \\, \\text{kJ\/\u00b0C}C=\u0394TQ\u200b=2.73\u00b0C26.38kJ\u200b=9.66kJ\/\u00b0C<\/li>\n<\/ol>\n\n\n\n

          Thus, the heat capacity of the calorimeter is C=9.66\u2009kJ\/\u00b0CC = 9.66 \\, \\text{kJ\/\u00b0C}C=9.66kJ\/\u00b0C.<\/p>\n\n\n\n

          \n\n\n\n

          Final Answers:<\/h3>\n\n\n\n
            \n
          1. The enthalpy of the final reaction CH4(g)\u2192C(s)+2H2(g)CH_4 (g) \\rightarrow C (s) + 2H_2 (g)CH4\u200b(g)\u2192C(s)+2H2\u200b(g) is 1398 kJ<\/strong>.<\/li>\n\n\n\n
          2. The heat capacity of the calorimeter is 9.66 kJ\/\u00b0C<\/strong>.<\/li>\n<\/ol>\n\n\n\n

            Both results are rounded to the correct number of significant figures based on the given data.<\/p>\n\n\n\n

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

            Use the thermochemical equations shown below to determine the enthalpy of the final reaction: (1) CO2 (g) ? C (s) + O2 (g) ?? = 551 kJ (2) H2(g) + 1\/2 O2 (g) ? H2O (l) ?? = -400 kJ (3) CH4 (g) + 2O2 (g) ? CO2 (g) + 2H2O (l) ?? = -1247 […]<\/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-257958","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257958","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=257958"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257958\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257958"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257958"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257958"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}