O KINETICS AND EQUILIBRIUM Writing the rate law implied by a simple mechanism Suppose the reaction between nitric oxide and bromine proceeds by the following mechanism: 1. NO(g) + Br2(g) ? NOBr2(g) 2. 2 NOBr2(g) + NO(g) ? 2 NOBr(g) Suppose also k1 > k2. That is, the first step is much faster than the second. Write the balanced chemical equation for the overall chemical reaction: Write the experimentally-observable rate law for the overall chemical reaction. rate = k Note: your answer should not contain the concentrations of any intermediates. Express the rate constant k for the overall chemical reaction in terms of k1, k2, and (if necessary) the rate constants k?1 and k?2 for the reverse of the two elementary reactions in the<\/p>\n\n\n\n
The Correct Answer and Explanatio<\/strong>n is:<\/mark><\/p>\n\n\n\n To solve this kinetics and equilibrium problem, we must identify the overall balanced chemical equation<\/strong>, the rate law<\/strong>, and then derive the rate constant expression<\/strong> eliminating intermediates.<\/p>\n\n\n\n Step 1 (Fast Equilibrium):<\/strong> Step 2 (Slow):<\/strong> To find the overall reaction, add both elementary steps and cancel any intermediates:<\/p>\n\n\n\n Add: Since Step 2 is the rate-determining step<\/strong>, the rate law is based on that step: rate=k2[NOBr2]2[NO]\\text{rate} = k_2[\\text{NOBr}_2]^2[\\text{NO}]<\/p>\n\n\n\n However, NOBr\u2082 is an intermediate<\/strong>, so we need to eliminate it.<\/p>\n\n\n\n From the fast equilibrium (Step 1), we assume: k1[NO][Br2]=k\u22121[NOBr2]k_1[\\text{NO}][\\text{Br}_2] = k_{-1}[\\text{NOBr}_2]<\/p>\n\n\n\n Solve for [NOBr\u2082]: [NOBr2]=k1k\u22121[NO][Br2][\\text{NOBr}_2] = \\frac{k_1}{k_{-1}}[\\text{NO}][\\text{Br}_2]<\/p>\n\n\n\n Substitute into the rate law: rate=k2(k1k\u22121[NO][Br2])2[NO]\\text{rate} = k_2 \\left( \\frac{k_1}{k_{-1}}[\\text{NO}][\\text{Br}_2] \\right)^2 [\\text{NO}]<\/p>\n\n\n\n Simplify: rate=k2(k12k\u221212)[NO]2[Br2]2[NO]\\text{rate} = k_2 \\left( \\frac{k_1^2}{k_{-1}^2} \\right)[\\text{NO}]^2[\\text{Br}_2]^2[\\text{NO}] rate=(k12k2k\u221212)[NO]3[Br2]2\\text{rate} = \\left( \\frac{k_1^2 k_2}{k_{-1}^2} \\right)[\\text{NO}]^3[\\text{Br}_2]^2<\/p>\n\n\n\n 3NO(g)+Br2(g)\u21923NOBr(g)\\boxed{3 \\text{NO}(g) + \\text{Br}_2(g) \\rightarrow 3 \\text{NOBr}(g)}<\/p>\n\n\n\n rate=k[NO]3[Br2]2\\boxed{\\text{rate} = k[\\text{NO}]^3[\\text{Br}_2]^2}<\/p>\n\n\n\n k=k12k2k\u221212\\boxed{k = \\frac{k_1^2 k_2}{k_{-1}^2}}<\/p>\n\n\n\n In reaction mechanisms, the rate-determining step (RDS)<\/strong> controls the observed kinetics. Here, we are given a two-step mechanism: the first step is a fast equilibrium, while the second is slow. When a fast step precedes the slow step, and the slow step involves an intermediate, we use the fast step to express the intermediate in terms of observable species.<\/p>\n\n\n\n Step 1 shows nitric oxide (NO) reacting with bromine (Br\u2082) to form an intermediate, NOBr\u2082. Because this step is fast and reversible, we assume it quickly reaches equilibrium. Step 2 is slower and thus dictates the overall rate. However, its rate law includes the concentration of the intermediate NOBr\u2082, which is not directly measurable. Therefore, we solve for [NOBr\u2082] using the equilibrium condition from Step 1.<\/p>\n\n\n\n By substituting [NOBr\u2082] in terms of [NO] and [Br\u2082], we eliminate the intermediate from the rate expression, resulting in a rate law expressed entirely in terms of reactants. The final rate law\u2014rate = k[NO]\u00b3[Br\u2082]\u00b2\u2014shows the reaction is third order in NO and second order in Br\u2082.<\/p>\n\n\n\n Lastly, to relate the overall rate constant to the elementary constants, we substitute the expressions, revealing how microscopic kinetics (k\u2081, k\u2082, k\u208b\u2081) combine to define the macroscopic rate constant k<\/strong>. This approach illustrates how mechanisms predict kinetics consistent with experiment while connecting observed behavior with molecular steps.<\/p>\n","protected":false},"excerpt":{"rendered":" O KINETICS AND EQUILIBRIUM Writing the rate law implied by a simple mechanism Suppose the reaction between nitric oxide and bromine proceeds by the following mechanism: 1. NO(g) + Br2(g) ? NOBr2(g) 2. 2 NOBr2(g) + NO(g) ? 2 NOBr(g) Suppose also k1 > k2. That is, the first step is much faster than the […]<\/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-213194","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213194","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=213194"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213194\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213194"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213194"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213194"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n\ud83d\udd39 Given Mechanism:<\/strong><\/h3>\n\n\n\n
NO(g) + Br\u2082(g) \u21cc NOBr\u2082(g)\u2003\u2003\u2003\u2003(k\u2081 forward, k\u208b\u2081 reverse)<\/p>\n\n\n\n
2 NOBr\u2082(g) + NO(g) \u2192 3 NOBr(g)\u2003\u2003\u2003(k\u2082)<\/p>\n\n\n\n
\n\n\n\n\ud83d\udd39 Step 1: Overall Balanced Equation<\/strong><\/h3>\n\n\n\n
Step 1:<\/strong> NO + Br\u2082 \u2192 NOBr\u2082
Step 2:<\/strong> 2 NOBr\u2082 + NO \u2192 3 NOBr<\/h2>\n\n\n\n
NO + Br\u2082 + 2 NOBr\u2082 + NO \u2192 NOBr\u2082 + 3 NOBr
Cancel NOBr\u2082 (intermediate):
Overall Reaction:<\/strong>
3 NO(g) + Br\u2082(g) \u2192 3 NOBr(g)<\/p>\n\n\n\n
\n\n\n\n\ud83d\udd39 Step 2: Rate Law<\/strong><\/h3>\n\n\n\n
\n\n\n\n\ud83d\udd39 Final Answers<\/strong><\/h3>\n\n\n\n
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\n\n\n\n\ud83d\udd0d Explanation (300+ words)<\/strong><\/h3>\n\n\n\n