{"id":231656,"date":"2025-06-11T09:36:07","date_gmt":"2025-06-11T09:36:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231656"},"modified":"2025-06-11T09:36:09","modified_gmt":"2025-06-11T09:36:09","slug":"the-integrated-rate-law-for-a-third-order-reaction-is-frac1a_t2-frac1a_02-2kt-write-an-expression-for-the-half-life-of-a-reactant-in-a-third-order-reaction","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/the-integrated-rate-law-for-a-third-order-reaction-is-frac1a_t2-frac1a_02-2kt-write-an-expression-for-the-half-life-of-a-reactant-in-a-third-order-reaction\/","title":{"rendered":"The integrated rate law for a third order reaction is \\frac{1}{[A]_t^2} = \\frac{1}{[A]_0^2} + 2kt Write an expression for the half life of a reactant in a third order reaction."},"content":{"rendered":"\n

The integrated rate law for a third order reaction is \\frac{1}{[A]_t^2} = \\frac{1}{[A]_0^2} + 2kt Write an expression for the half life of a reactant in a third order reaction.<\/p>\n\n\n\n

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

To find the half-life expression<\/strong> for a third-order reaction<\/strong>, we start from the integrated rate law:1[A]t2=1[A]02+2kt\\frac{1}{[A]_t^2} = \\frac{1}{[A]_0^2} + 2kt[A]t2\u200b1\u200b=[A]02\u200b1\u200b+2kt<\/p>\n\n\n\n

At half-life t1\/2t_{1\/2}t1\/2\u200b, the concentration [A]t=[A]02[A]_t = \\frac{[A]_0}{2} [A]t\u200b=2[A]0\u200b\u200b. Plug this into the equation:1([A]02)2=1[A]02+2kt1\/2\\frac{1}{\\left( \\frac{[A]_0}{2} \\right)^2} = \\frac{1}{[A]_0^2} + 2k t_{1\/2}(2[A]0\u200b\u200b)21\u200b=[A]02\u200b1\u200b+2kt1\/2\u200b<\/p>\n\n\n\n

Simplify the left-hand side:1([A]024)=4[A]02\\frac{1}{\\left( \\frac{[A]_0^2}{4} \\right)} = \\frac{4}{[A]_0^2}(4[A]02\u200b\u200b)1\u200b=[A]02\u200b4\u200b<\/p>\n\n\n\n

Now the equation becomes:4[A]02=1[A]02+2kt1\/2\\frac{4}{[A]_0^2} = \\frac{1}{[A]_0^2} + 2k t_{1\/2}[A]02\u200b4\u200b=[A]02\u200b1\u200b+2kt1\/2\u200b<\/p>\n\n\n\n

Subtract 1[A]02\\frac{1}{[A]_0^2}[A]02\u200b1\u200b from both sides:3[A]02=2kt1\/2\\frac{3}{[A]_0^2} = 2k t_{1\/2}[A]02\u200b3\u200b=2kt1\/2\u200b<\/p>\n\n\n\n

Now solve for t1\/2t_{1\/2}t1\/2\u200b:t1\/2=32k[A]02t_{1\/2} = \\frac{3}{2k [A]_0^2}t1\/2\u200b=2k[A]02\u200b3\u200b<\/p>\n\n\n\n

\n\n\n\n

\u2705 Final Answer:<\/h3>\n\n\n\n

t1\/2=32k[A]02\\boxed{t_{1\/2} = \\frac{3}{2k [A]_0^2}}t1\/2\u200b=2k[A]02\u200b3\u200b\u200b<\/p>\n\n\n\n

This is the half-life expression for a third-order reaction.<\/p>\n\n\n\n

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

The integrated rate law for a third order reaction is \\frac{1}{[A]_t^2} = \\frac{1}{[A]_0^2} + 2kt Write an expression for the half life of a reactant in a third order reaction. To find the half-life expression for a third-order reaction, we start from the integrated rate law:1[A]t2=1[A]02+2kt\\frac{1}{[A]_t^2} = \\frac{1}{[A]_0^2} + 2kt[A]t2\u200b1\u200b=[A]02\u200b1\u200b+2kt At half-life t1\/2t_{1\/2}t1\/2\u200b, the concentration […]<\/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-231656","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231656","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=231656"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231656\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231656"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231656"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231656"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}