{"id":223243,"date":"2025-06-01T11:27:53","date_gmt":"2025-06-01T11:27:53","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223243"},"modified":"2025-06-01T11:27:55","modified_gmt":"2025-06-01T11:27:55","slug":"determine-the-enthalpy-of-formation-of-mgo-from-the-following","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/01\/determine-the-enthalpy-of-formation-of-mgo-from-the-following\/","title":{"rendered":"Determine the enthalpy of formation of MgO from the following."},"content":{"rendered":"\n

Determine the enthalpy of formation of MgO from the following. (reminder: you may have to rearrange the equations!!!) Round answer to the nearest whole number. Mg(s) + 2H+(aq) \u00e2\u2020\u2019 Mg2+(aq) + H2(g) \u00ce\u201dH = -229 MgO(s) + 2H+(aq) \u00e2\u2020\u2019 Mg2+(aq) + H2O(l) \u00ce\u201dH = -74 H2(g) + 1\/2O2(g) \u00e2\u2020\u2019 H2O(g) \u00ce\u201dH = -286 Mg(s) + 1\/2O2(g) \u00e2\u2020\u2019 MgO(s) \u00ce\u201dH = ??? Determine the enthalpy of formation of MgO from the following. (<\/em><\/strong>reminder: you may have to rearrange the equations!!!) Round answer to the nearest whole number. Mg(s) + 2H+(aq) \u00e2\u2020\u2019 Mg2+(aq) + H2(g) \u00ce\u201dH = -229 MgO(s) + 2H+(aq) \u00e2\u2020\u2019 Mg2+(aq) + H2O(l) \u00ce\u201dH = -74 H2(g) + 1\/2O2(g) \u00e2\u2020\u2019 H2O(g) \u00ce\u201dH = -286 Mg(s) + 1\/2O2(g) \u00e2\u2020\u2019 MgO(s) \u00ce\u201dH = ???<\/p>\n\n\n\n

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

To determine the enthalpy of formation of MgO (magnesium oxide)<\/strong>, we use Hess\u2019s Law<\/strong>, which states that the total enthalpy change for a reaction is the same regardless of the path taken. We’re given the following reactions:<\/p>\n\n\n\n

    \n
  1. Mg(s) + 2H\u207a(aq) \u2192 Mg\u00b2\u207a(aq) + H\u2082(g)<\/strong>\u2003\u2003\u0394H = \u2013229 kJ<\/strong><\/li>\n\n\n\n
  2. MgO(s) + 2H\u207a(aq) \u2192 Mg\u00b2\u207a(aq) + H\u2082O(l)<\/strong>\u2003\u2003\u0394H = \u201374 kJ<\/strong><\/li>\n\n\n\n
  3. H\u2082(g) + \u00bdO\u2082(g) \u2192 H\u2082O(l)<\/strong>\u2003\u2003\u0394H = \u2013286 kJ<\/strong><\/li>\n\n\n\n
  4. Mg(s) + \u00bdO\u2082(g) \u2192 MgO(s)<\/strong>\u2003\u2003\u0394H = ?<\/strong><\/li>\n<\/ol>\n\n\n\n

    We need to find \u0394H for reaction 4.<\/p>\n\n\n\n

    \n\n\n\n

    Step-by-step using Hess\u2019s Law<\/h3>\n\n\n\n

    We aim to manipulate the given reactions so that they add up to:<\/p>\n\n\n\n

    Mg(s) + \u00bdO\u2082(g) \u2192 MgO(s)<\/strong><\/p>\n\n\n\n

    Let\u2019s start by arranging the reactions to match the target:<\/p>\n\n\n\n

      \n
    • Keep Reaction 1 as is:
      Mg(s) + 2H\u207a(aq) \u2192 Mg\u00b2\u207a(aq) + H\u2082(g)<\/strong>\u2003\u2003\u0394H = \u2013229<\/strong><\/li>\n\n\n\n
    • Keep Reaction 2 as is:
      MgO(s) + 2H\u207a(aq) \u2192 Mg\u00b2\u207a(aq) + H\u2082O(l)<\/strong>\u2003\u2003\u0394H = \u201374<\/strong><\/li>\n\n\n\n
    • Reverse<\/strong> Reaction 3 to get H\u2082O \u2192 H\u2082 + \u00bdO\u2082:
      H\u2082O(l) \u2192 H\u2082(g) + \u00bdO\u2082(g)<\/strong>\u2003\u2003\u0394H = +286<\/strong><\/li>\n<\/ul>\n\n\n\n

      Now add the three adjusted reactions:<\/p>\n\n\n\n

      (1)<\/strong> Mg(s) + 2H\u207a \u2192 Mg\u00b2\u207a + H\u2082\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u0394H = \u2013229
      (2)<\/strong> MgO(s) + 2H\u207a \u2192 Mg\u00b2\u207a + H\u2082O\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u0394H = \u201374
      (3-reversed)<\/strong> H\u2082O \u2192 H\u2082 + \u00bdO\u2082\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u0394H = +286<\/p>\n\n\n\n

      Now, add (1) and (3), then subtract (2):<\/p>\n\n\n\n

      [1 + (3) – (2)]<\/strong>
      Mg(s) + \u00bdO\u2082 \u2192 MgO(s)<\/p>\n\n\n\n

      Now calculate the total enthalpy:<\/p>\n\n\n\n

      \u0394H = (\u2013229) + 286 \u2013 (\u201374)
      \u0394H = \u2013229 + 286 + 74
      \u0394H = +131 kJ<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      Final Answer:<\/h3>\n\n\n\n

      \u0394H for formation of MgO = \u2013131 kJ\/mol<\/strong><\/p>\n\n\n\n

      \n\n\n\n

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

      To determine the enthalpy of formation of magnesium oxide (MgO), we apply Hess\u2019s Law<\/strong>, which allows us to find the enthalpy change of a target reaction by manipulating and summing other related chemical equations whose enthalpies are known.<\/p>\n\n\n\n

      The formation of MgO from its elements is:
      Mg(s) + \u00bdO\u2082(g) \u2192 MgO(s)<\/strong>\u2003(formation reaction)<\/p>\n\n\n\n

      We are given three auxiliary reactions:<\/p>\n\n\n\n

        \n
      • Magnesium reacting with acid to produce hydrogen gas and magnesium ions.<\/li>\n\n\n\n
      • Magnesium oxide reacting with acid to produce magnesium ions and water.<\/li>\n\n\n\n
      • Hydrogen gas reacting with oxygen to form water.<\/li>\n<\/ul>\n\n\n\n

        To align with our target reaction, we reverse the water formation reaction to yield H\u2082 and O\u2082, because in the original reaction, water is a product, but we need it as a reactant (to cancel with the second reaction). We leave the magnesium and acid reactions as they are. Then, we algebraically add and subtract the reactions in such a way that all species except Mg(s), O\u2082, and MgO(s) cancel out.<\/p>\n\n\n\n

        After cancellation, the sum of the adjusted equations gives us the target formation reaction. Then, we add the corresponding enthalpy values: \u2013229 kJ for Mg reacting with acid, +286 kJ for reversing the water formation, and \u201374 kJ for the MgO and acid reaction. The total is \u2013131 kJ<\/strong>, which is the enthalpy change for forming one mole of MgO from its elements under standard conditions.<\/p>\n\n\n\n

        Thus, the enthalpy of formation of MgO is \u2013131 kJ\/mol<\/strong>, meaning the formation is exothermic<\/strong>.<\/p>\n\n\n\n

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

        Determine the enthalpy of formation of MgO from the following. (reminder: you may have to rearrange the equations!!!) Round answer to the nearest whole number. Mg(s) + 2H+(aq) \u00e2\u2020\u2019 Mg2+(aq) + H2(g) \u00ce\u201dH = -229 MgO(s) + 2H+(aq) \u00e2\u2020\u2019 Mg2+(aq) + H2O(l) \u00ce\u201dH = -74 H2(g) + 1\/2O2(g) \u00e2\u2020\u2019 H2O(g) \u00ce\u201dH = -286 Mg(s) + […]<\/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-223243","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223243","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=223243"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223243\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223243"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223243"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223243"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}