{"id":185078,"date":"2025-01-22T05:45:30","date_gmt":"2025-01-22T05:45:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185078"},"modified":"2025-01-22T05:45:33","modified_gmt":"2025-01-22T05:45:33","slug":"in-the-discussion-of-the-lindemann-mechanism","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/22\/in-the-discussion-of-the-lindemann-mechanism\/","title":{"rendered":"In the discussion of the Lindemann mechanism"},"content":{"rendered":"\n

In the discussion of the Lindemann mechanism, it was assumed that the rate of activation by collision with another reactant molecule A was the same as collision with a nonreactant molecule M such as a buffer gas. What if the rates of activation for these two processes are different? In this case, the mechanism becomes<\/p>\n\n\n\n

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

a. Demonstrate that the rate law expression for this mechanism is<\/p>\n\n\n\n

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

b. Does this rate law reduce to the expected form when [M] = 0?<\/p>\n\n\n\n

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

To analyze the given Lindemann mechanism where the rates of activation for the reactant molecule AA and the buffer gas MM differ, we derive the rate law step-by-step.<\/p>\n\n\n\n

Mechanism:<\/h3>\n\n\n\n
    \n
  1. A+M\u21cck\u22121k1A\u2217+MA + M \\xrightleftharpoons[k_{-1}]{k_1} A^* + M (Activation of AA by MM)<\/li>\n\n\n\n
  2. A+A\u21cck\u22122k2A\u2217+AA + A \\xrightleftharpoons[k_{-2}]{k_2} A^* + A (Activation of AA by AA)<\/li>\n\n\n\n
  3. A\u2217\u2192k3productsA^* \\xrightarrow{k_3} \\text{products} (Deactivation and product formation)<\/li>\n<\/ol>\n\n\n\n

    Here, A\u2217A^* is the activated intermediate.<\/p>\n\n\n\n

    \n\n\n\n

    Derivation:<\/h3>\n\n\n\n

    The steady-state approximation is applied to A\u2217A^*, assuming its concentration does not change significantly during the reaction. d[A\u2217]dt=0=k1[A][M]+k2[A][A]\u2212k\u22121[A\u2217][M]\u2212k\u22122[A\u2217][A]\u2212k3[A\u2217]\\frac{d[A^*]}{dt} = 0 = k_1[A][M] + k_2[A][A] – k_{-1}[A^*][M] – k_{-2}[A^*][A] – k_3[A^*]<\/p>\n\n\n\n

    Rearranging for [A\u2217][A^*]: [A\u2217]=k1[A][M]+k2[A][A]k\u22121[M]+k\u22122[A]+k3[A^*] = \\frac{k_1[A][M] + k_2[A][A]}{k_{-1}[M] + k_{-2}[A] + k_3}<\/p>\n\n\n\n

    The rate of product formation (RR) is given by: R=k3[A\u2217]R = k_3[A^*]<\/p>\n\n\n\n

    Substituting [A\u2217][A^*]: R=k3(k1[A][M]+k2[A][A])k\u22121[M]+k\u22122[A]+k3R = \\frac{k_3 \\big( k_1[A][M] + k_2[A][A] \\big)}{k_{-1}[M] + k_{-2}[A] + k_3}<\/p>\n\n\n\n

    This is the rate law for the mechanism.<\/p>\n\n\n\n

    \n\n\n\n

    Special Case: [M]=0[M] = 0<\/h3>\n\n\n\n

    When [M]=0[M] = 0: R=k3\u22c5k2[A][A]k\u22122[A]+k3R = \\frac{k_3 \\cdot k_2[A][A]}{k_{-2}[A] + k_3}<\/p>\n\n\n\n

    Factoring AA out: R=k2k3[A]2k\u22122[A]+k3R = \\frac{k_2k_3[A]^2}{k_{-2}[A] + k_3}<\/p>\n\n\n\n

    This aligns with the expected Lindemann-Hinshelwood mechanism for unimolecular reactions in the absence of a buffer gas.<\/p>\n\n\n\n

    \n\n\n\n

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

    This derivation demonstrates that incorporating different rates for activation by MM and AA results in a more complex rate law, blending contributions from both reactant and buffer. When [M]=0[M] = 0, the system naturally reduces to a form dependent solely on AA, as expected for reactions without a buffer. This dual dependence allows for flexibility in modeling real systems where buffer gases influence activation differently.<\/p>\n","protected":false},"excerpt":{"rendered":"

    In the discussion of the Lindemann mechanism, it was assumed that the rate of activation by collision with another reactant molecule A was the same as collision with a nonreactant molecule M such as a buffer gas. What if the rates of activation for these two processes are different? In this case, the mechanism becomes […]<\/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-185078","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185078","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=185078"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185078\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185078"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185078"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185078"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}