{"id":192441,"date":"2025-02-17T20:34:59","date_gmt":"2025-02-17T20:34:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=192441"},"modified":"2025-02-17T20:35:03","modified_gmt":"2025-02-17T20:35:03","slug":"consider-the-equilibrium-system-described-by-the-chemical-reaction-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/17\/consider-the-equilibrium-system-described-by-the-chemical-reaction-below\/","title":{"rendered":"Consider the equilibrium system described by the chemical reaction below"},"content":{"rendered":"\n

Consider the equilibrium system described by the chemical reaction below. At equilibrium, a 2.0 L reaction vessel contained a mixture of 1.2 mol Fe, 1.0 \u00d7 10-\u00b3 mol O2, and 2.0 mol of Fe2O, at 2000.0 \u00b0C. What are the values of Kc and Kp for this reaction? 4Fe(s) + 302(g) = 2Fe2O2(s) NEXT Based on the given data, set up the expression for Kc. Each reaction participant must be represented by one tile. Do not combine terms. [1.2] Once the expression is constructed, solve for Kc. Complete Parts 1-2 before submitting your answer. 2[2.0] 2[1.2] [1.0 \u00d7 10\u00b3] [0.50] 8.0 \u00d7 10\u00b0 ?? 1 = [2.0] [5.0 x 10\u00b2] 1.0 \u00d7 10? [1.2]\u00b9 [0.50]* 1.3 x 10-1\u00b0 [1.0 \u00d7 10.\u00b31\u00b3 [5.0 \u00d7 10.\u00b91\u00b3 2 = [2.0]\u00b2 [1.2]\u00b2 4[1.2] 4[0.50] > RESET 3[1.0 \u00d7 10\u00b3] 3[5.0 x 10+]<\/p>\n\n\n\n

The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Step 1: Write the Expression for KcK_c<\/strong><\/h3>\n\n\n\n

The equilibrium constant KcK_c is expressed as: Kc=[Products]coefficients[Reactants]coefficientsK_c = \\frac{[\\text{Products}]^{\\text{coefficients}}}{[\\text{Reactants}]^{\\text{coefficients}}}<\/p>\n\n\n\n

For the given reaction: 4Fe(s)+3O2(g)\u21cc2Fe2O2(s)4Fe(s) + 3O_2(g) \\rightleftharpoons 2Fe_2O_2(s)<\/p>\n\n\n\n

Since solids (FeFe and Fe2O2Fe_2O_2) are not included in the equilibrium expression: Kc=[Fe2O2]2[O2]3K_c = \\frac{[Fe_2O_2]^2}{[O_2]^3}<\/p>\n\n\n\n

Using the given equilibrium concentrations:<\/p>\n\n\n\n

    \n
  • [Fe2O2]=2.0\u00a0mol2.0\u00a0L=1.0\u00a0M[Fe_2O_2] = \\frac{2.0 \\text{ mol}}{2.0 \\text{ L}} = 1.0 \\text{ M}<\/li>\n\n\n\n
  • [O2]=1.0\u00d710\u22123\u00a0mol2.0\u00a0L=5.0\u00d710\u22124\u00a0M[O_2] = \\frac{1.0 \\times 10^{-3} \\text{ mol}}{2.0 \\text{ L}} = 5.0 \\times 10^{-4} \\text{ M}<\/li>\n<\/ul>\n\n\n\n

    Substituting these values: Kc=(1.0)2(5.0\u00d710\u22124)3K_c = \\frac{(1.0)^2}{(5.0 \\times 10^{-4})^3} Kc=1.0(1.25\u00d710\u221210)K_c = \\frac{1.0}{(1.25 \\times 10^{-10})} Kc=8.0\u00d7109K_c = 8.0 \\times 10^9<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Calculate KpK_p<\/strong><\/h3>\n\n\n\n

    The relationship between KpK_p and KcK_c is: Kp=Kc(RT)\u0394nK_p = K_c(RT)^{\\Delta n}<\/p>\n\n\n\n

    where:<\/p>\n\n\n\n

      \n
    • R=0.0821R = 0.0821 L\u00b7atm\/(mol\u00b7K) (Ideal gas constant)<\/li>\n\n\n\n
    • T=2000.0T = 2000.0 K (given temperature)<\/li>\n\n\n\n
    • \u0394n=\\Delta n = (moles of gas products \u2013 moles of gas reactants)\n
        \n
      • Products: 00 (since Fe\u2082O\u2082 is solid)<\/li>\n\n\n\n
      • Reactants: 33 (from O\u2082 gas)<\/li>\n\n\n\n
      • \u0394n=0\u22123=\u22123\\Delta n = 0 – 3 = -3<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

        Kp=(8.0\u00d7109)\u00d7(0.0821\u00d72000)\u22123K_p = (8.0 \\times 10^9) \\times (0.0821 \\times 2000)^{-3} Kp=(8.0\u00d7109)\u00d7(164.2)\u22123K_p = (8.0 \\times 10^9) \\times (164.2)^{-3} Kp=(8.0\u00d7109)\u00d7(2.24\u00d710\u22126)K_p = (8.0 \\times 10^9) \\times (2.24 \\times 10^{-6}) Kp=1.79\u00d7104K_p = 1.79 \\times 10^4<\/p>\n\n\n\n

        \n\n\n\n

        Final Answer:<\/strong><\/h3>\n\n\n\n
          \n
        • Kc=8.0\u00d7109K_c = 8.0 \\times 10^9<\/li>\n\n\n\n
        • Kp=1.79\u00d7104K_p = 1.79 \\times 10^4<\/li>\n<\/ul>\n\n\n\n

          This indicates that the equilibrium heavily favors the formation of Fe2O2Fe_2O_2 over oxygen gas.<\/p>\n\n\n\n

          Now, generating an image illustrating the chemical equilibrium:<\/p>\n\n\n\n

          Here is a scientific illustration of the chemical equilibrium reaction at 2000\u00b0C, showing the balance between solid iron (Fe), gaseous oxygen (O\u2082), and iron oxide (Fe\u2082O\u2082) in a reaction vessel. Let me know if you need any modifications!<\/p>\n\n\n\n

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

          Consider the equilibrium system described by the chemical reaction below. At equilibrium, a 2.0 L reaction vessel contained a mixture of 1.2 mol Fe, 1.0 \u00d7 10-\u00b3 mol O2, and 2.0 mol of Fe2O, at 2000.0 \u00b0C. What are the values of Kc and Kp for this reaction? 4Fe(s) + 302(g) = 2Fe2O2(s) NEXT Based […]<\/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-192441","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192441","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=192441"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192441\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192441"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192441"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192441"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}