{"id":188015,"date":"2025-02-06T09:32:41","date_gmt":"2025-02-06T09:32:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188015"},"modified":"2025-02-06T09:32:43","modified_gmt":"2025-02-06T09:32:43","slug":"derive-integrated-rate-law-expressions-for-half-one-and-a-half-and-third-order-reactions-of-the-form-products-where-k-is-the-rate-constant","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/derive-integrated-rate-law-expressions-for-half-one-and-a-half-and-third-order-reactions-of-the-form-products-where-k-is-the-rate-constant\/","title":{"rendered":"Derive integrated rate law expressions for half-, one and a half-, and third-order reactions of the form products where k is the rate constant"},"content":{"rendered":"\n

Derive integrated rate law expressions for half-, one and a half-, and third-order reactions of the form products where k is the rate constant. Also, deduce the graphical linear relations, units of rate constant k, and expression for half-life, t1\/2 , for all cases. Assume that at t = 0 only A is present with an initial concentration of [A]0. Fill in the table below by following the example given for first-order reaction1. Show all your work. order Integrated Rate Law Graphical Linear Relation Unit of k Half-life 1 ln[A][A]0=\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u2122?? ln[A] vs t \u00c3\u00a2\u00e2\u20ac\u00a1\u00e2\u20ac\u2122 slope = \u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u2122k , y-intercept = ln[A]0 1? ?12\u00c3\u00a2\u00c2\ufffd\u00e2\u20ac\u017e=ln2? 1\/2 3\/2 3
Derive integrated rate law expressions for half-, one and a half-, and third-order reactions of the form A \u00c3\u201a\u00c2\u00ae products where k is the rate constant. Also, deduce the graphical linear relations, units of rate constant k, and expression for half-life, t1\/2 , for all cases. Assume that at t = 0 only A is present with an initial concentration of [A]0. Fill in the table below by following the example given for first-order reaction1.<\/p>\n\n\n\n

Show all your work.<\/p>\n\n\n\n

order Integrated Rate Law Graphical Linear Relation Unit of k Half-life<\/p>\n\n\n\n

1 ln[A][A]0=\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u2122?? ln[A] vs t \u00c3\u00a2\u00e2\u20ac\u00a1\u00e2\u20ac\u2122 slope = \u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u2122k , y-intercept = ln[A]0 1? ?12\u00c3\u00a2\u00c2\ufffd\u00e2\u20ac\u017e=ln2?<\/p>\n\n\n\n

1\/2<\/p>\n\n\n\n

3\/2<\/p>\n\n\n\n

3<\/p>\n\n\n\n

1Atkins\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2 Physical Chemistry, Atkins and de Paula, 10th Ed., 2014, OUP, pp. 827-828.<\/p>\n\n\n\n

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

To derive the integrated rate laws, graphical linear relations, units of the rate constant ( k ), and half-life expressions for reactions of orders \u00bd, 1\u00bd, and 3, we start with the general rate law for a reaction ( A \\rightarrow \\text{products} ):<\/p>\n\n\n\n

[ -\\frac{d[A]}{dt} = k[A]^n ]<\/p>\n\n\n\n

where ( n ) is the reaction order, ( [A] ) is the concentration of reactant ( A ) at time ( t ), and ( k ) is the rate constant.<\/p>\n\n\n\n

Half-Order (n = \u00bd):<\/strong><\/p>\n\n\n\n

    \n
  1. Integrated Rate Law:<\/strong> Separate variables and integrate: [ \\int [A]^{-1\/2} \\, d[A] = -k \\int dt ] This yields: [ 2[A]^{1\/2} = -kt + C ] Applying the initial condition ( [A] = [A]_0 ) at ( t = 0 ): [ 2[A]_0^{1\/2} = C ] Thus, the integrated rate law is: [ 2[A]^{1\/2} = -kt + 2[A]_0^{1\/2} ] Simplifying: [ [A]^{1\/2} = [A]_0^{1\/2} – \\frac{kt}{2} ]<\/li>\n\n\n\n
  2. Graphical Linear Relation:<\/strong> Plotting ( [A]^{1\/2} ) versus ( t ) yields a straight line with:<\/li>\n<\/ol>\n\n\n\n
      \n
    • Slope: ( -\\frac{k}{2} )<\/li>\n\n\n\n
    • Y-intercept: ( [A]_0^{1\/2} )<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Units of ( k ):<\/strong> From the rate law ( k[A]^{1\/2} ), the units of ( k ) are: [ \\text{(concentration\/time)} \\times (\\text{concentration})^{-1\/2} = \\text{concentration}^{1\/2} \\, \\text{time}^{-1} ]<\/li>\n\n\n\n
      2. Half-Life (( t_{1\/2} )):<\/strong> The half-life is the time at which ( [A] = \\frac{[A]_0}{2} ). Substituting into the integrated rate law: [ \\left( \\frac{[A]0}{2} \\right)^{1\/2} = [A]_0^{1\/2} – \\frac{kt<\/em>{1\/2}}{2} ] Solving for ( t_{1\/2} ): [ t_{1\/2} = \\frac{[A]_0^{1\/2}}{k} ]<\/li>\n<\/ol>\n\n\n\n

        One and a Half Order (n = 3\/2):<\/strong><\/p>\n\n\n\n

          \n
        1. Integrated Rate Law:<\/strong> Separate variables and integrate: [ \\int [A]^{-3\/2} \\, d[A] = -k \\int dt ] This yields: [ -2[A]^{-1\/2} = -kt + C ] Applying the initial condition ( [A] = [A]_0 ) at ( t = 0 ): [ -2[A]_0^{-1\/2} = C ] Thus, the integrated rate law is: [ -2[A]^{-1\/2} = -kt – 2[A]_0^{-1\/2} ] Simplifying: [ [A]^{-1\/2} = [A]_0^{-1\/2} + \\frac{kt}{2} ]<\/li>\n\n\n\n
        2. Graphical Linear Relation:<\/strong> Plotting ( [A]^{-1\/2} ) versus ( t ) yields a straight line with:<\/li>\n<\/ol>\n\n\n\n
            \n
          • Slope: ( \\frac{k}{2} )<\/li>\n\n\n\n
          • Y-intercept: ( [A]_0^{-1\/2} )<\/li>\n<\/ul>\n\n\n\n
              \n
            1. Units of ( k ):<\/strong> From the rate law ( k[A]^{3\/2} ), the units of ( k ) are: [ \\text{(concentration\/time)} \\times (\\text{concentration})^{-3\/2} = \\text{concentration}^{-1\/2} \\, \\text{time}^{-1} ]<\/li>\n\n\n\n
            2. Half-Life (( t_{1\/2} )):<\/strong> The half-life is the time at which ( [A] = \\frac{[A]_0}{2} ). Substituting into the integrated rate law: [ \\left( \\frac{[A]0}{2} \\right)^{-1\/2} = [A]_0^{-1\/2} + \\frac{kt<\/em>{1\/2}}{2} ] Solving for ( t_{1\/2} ): [ t_{1\/2} = \\frac{2(2^{1\/2} – 1)}{k[A]_0^{1\/2}} ]<\/li>\n<\/ol>\n\n\n\n

              Third Order (n = 3):<\/strong><\/p>\n\n\n\n

                \n
              1. Integrated Rate Law:<\/strong> Separate variables and integrate: [ \\int [A]^{-3} \\, d[A] = -k \\int dt ] This yields: [ \\frac{1}{2}[A]^{-2} = -kt + C ] Applying the initial condition ( [A] = [A]_0 ) at ( t = 0 ): [ \\frac{1}{2}[A]_0^{-2} = C ] Thus, the integrated rate law is: [ \\frac{1}{2}[A]^{-2} = -kt + \\frac{1}{2}[A]_0^{-2} ] Simplifying: [ \\frac{1}{[A]^2} = \\frac{1}{[A]_0^2} + 2kt ]<\/li>\n\n\n\n
              2. Graphical Linear Relation:<\/strong> Plotting ( \\frac{1}{[A]^2} ) versus ( t ) yields a straight line with:<\/li>\n<\/ol>\n\n\n\n
                  \n
                • Slope: ( 2k )<\/li>\n\n\n\n
                • Y-intercept: ( \\frac{1}{[A]_0^2} )<\/li>\n<\/ul>\n\n\n\n
                    \n
                  1. Units of ( k ):<\/strong> From the rate law ( k[A]^3 ), the units of ( k ) are: [ \\text{(concentration\/time)} \\times (\\text{concentration})^{-3} = \\text{concentration}^{-2} \\, \\text{time}^{-1} ]<\/li>\n\n\n\n
                  2. Half-Life (( t_{1\/2} )):<\/strong> The half-life is the time at which ( [A] = \\frac{[A]_0}{2} ). Substituting into the integrated rate law: [ \\frac{1}{\\left( \\frac{[A]0}{2} \\right)^2} = \\frac{1}{[A]_0^2} + 2kt<\/em>{1\/2} ] Solving for ( t_{1\/<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                    Derive integrated rate law expressions for half-, one and a half-, and third-order reactions of the form products where k is the rate constant. Also, deduce the graphical linear relations, units of rate constant k, and expression for half-life, t1\/2 , for all cases. Assume that at t = 0 only A is present with […]<\/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-188015","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188015","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=188015"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188015\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188015"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188015"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188015"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}