{"id":219741,"date":"2025-05-26T18:18:17","date_gmt":"2025-05-26T18:18:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=219741"},"modified":"2025-05-26T18:18:19","modified_gmt":"2025-05-26T18:18:19","slug":"calculate-the-rate-constant-for-proton-transfer-to-nh3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/26\/calculate-the-rate-constant-for-proton-transfer-to-nh3\/","title":{"rendered":"Calculate the rate constant for proton transfer to NH3"},"content":{"rendered":"\n<p>a. The pKa of NH4+ is 9.25 at 25\u00b0C. The rate constant at 25\u00b0C for the reaction of NH4+ and OH\u2212 to form aqueous NH3 is 4.0 \u00d7 1010 dm3 mol\u22121 s\u22121. Calculate the rate constant for proton transfer to NH3. What relaxation time would be observed if a temperature jump were applied to a solution of 0.15 mol dm\u22123 NH3(aq) at 25\u00b0C?<\/p>\n\n\n\n<p>b. The equilibrium A 5B + C at 25\u00b0C is subjected to a temperature jump that slightly increases the concentrations of B and C. The measured relaxation time is 3.0 \u03bcs. The equilibrium constant for the system is 2.0 \u00d7 10\u221216 at 25\u00b0C, and the equilibrium concentrations of B and C at 25\u00b0C are both 2.0 \u00d7 10\u22124 mol dm\u22123. Calculate the rate constants for steps (1) and (2).<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s solve each part step-by-step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>a. Reaction of NH\u2084\u207a with OH\u207b and Relaxation Time<\/strong><\/h3>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>pKa of NH\u2084\u207a = 9.25 \u2192 Ka = 10^(-9.25) \u2248 5.62 \u00d7 10\u207b\u00b9\u2070<\/li>\n\n\n\n<li>Rate constant for forward reaction (NH\u2084\u207a + OH\u207b \u2192 NH\u2083 + H\u2082O):<br><strong>k\u2081 = 4.0 \u00d7 10\u00b9\u2070 dm\u00b3 mol\u207b\u00b9 s\u207b\u00b9<\/strong><\/li>\n\n\n\n<li>NH\u2084\u207a \u21cc NH\u2083 + H\u207a is the relevant acid-base equilibrium<\/li>\n\n\n\n<li>Equilibrium constant <strong>K = [NH\u2083][H\u207a]\/[NH\u2084\u207a] = Ka<\/strong><\/li>\n<\/ul>\n\n\n\n<p>So, we know:<br>K=k1k\u22121\u21d2k\u22121=k1KK = \\frac{k_1}{k_{-1}} \\Rightarrow k_{-1} = \\frac{k_1}{K}<\/p>\n\n\n\n<p>Substitute: k\u22121=4.0\u00d710105.62\u00d710\u221210\u22487.12\u00d71019&nbsp;s\u22121k_{-1} = \\frac{4.0 \\times 10^{10}}{5.62 \\times 10^{-10}} \u2248 7.12 \\times 10^{19} \\ \\text{s}^{-1}<\/p>\n\n\n\n<p><strong>Now, calculate the relaxation time (\u03c4)<\/strong> for a simple acid-base reaction: \u03c4=1k1[OH\u2212]+k\u22121\\tau = \\frac{1}{k_1[\\text{OH}^-] + k_{-1}}<\/p>\n\n\n\n<p>Assuming [OH\u207b] is from 0.15 mol dm\u207b\u00b3 NH\u2083:<br>NH\u2083&nbsp;+&nbsp;H\u2082O&nbsp;\u21cc&nbsp;NH\u2084\u207a&nbsp;+&nbsp;OH\u207b\u21d2Kb=KwKa=10\u2212145.62\u00d710\u221210\u22481.78\u00d710\u22125\\text{NH\u2083 + H\u2082O \u21cc NH\u2084\u207a + OH\u207b} \\Rightarrow K_b = \\frac{K_w}{K_a} = \\frac{10^{-14}}{5.62 \\times 10^{-10}} \\approx 1.78 \\times 10^{-5}<\/p>\n\n\n\n<p>For NH\u2083 (0.15 M): [OH\u2212]\u2248Kb\u22c5[NH\u2083]=1.78\u00d710\u22125\u22c50.15\u22481.63\u00d710\u22123&nbsp;mol&nbsp;dm\u22123[\\text{OH}^-] \u2248 \\sqrt{K_b \\cdot [\\text{NH\u2083}]} = \\sqrt{1.78 \\times 10^{-5} \\cdot 0.15} \u2248 1.63 \\times 10^{-3} \\ \\text{mol dm}^{-3} \u03c4=1(4.0\u00d71010)(1.63\u00d710\u22123)+7.12\u00d71019\u224817.12\u00d71019&nbsp;s\\tau = \\frac{1}{(4.0 \\times 10^{10})(1.63 \\times 10^{-3}) + 7.12 \\times 10^{19}} \u2248 \\frac{1}{7.12 \\times 10^{19}} \\ \\text{s} \u03c4\u22481.4\u00d710\u221220&nbsp;s\\tau \u2248 1.4 \\times 10^{-20} \\ \\text{s}<\/p>\n\n\n\n<p><strong>Answer (a):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>k\u208b\u2081 = 7.12 \u00d7 10\u00b9\u2079 s\u207b\u00b9<\/strong><\/li>\n\n\n\n<li><strong>\u03c4 \u2248 1.4 \u00d7 10\u207b\u00b2\u2070 s<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>b. Equilibrium A \u21cc 5B + C (relaxation time method)<\/strong><\/h3>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Relaxation time, <strong>\u03c4 = 3.0 \u03bcs = 3.0 \u00d7 10\u207b\u2076 s<\/strong><\/li>\n\n\n\n<li>Equilibrium constant, <strong>K = [B]^5[C]\/[A] = 2.0 \u00d7 10\u207b\u00b9\u2076<\/strong><\/li>\n\n\n\n<li>[B] = [C] = 2.0 \u00d7 10\u207b\u2074 mol dm\u207b\u00b3<\/li>\n<\/ul>\n\n\n\n<p>We use the relation: \u03c4=1kf+kr\\tau = \\frac{1}{k_f + k_r}<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Forward: A \u2192 5B + C \u2192 rate = <strong>k_f [A]<\/strong><\/li>\n\n\n\n<li>Reverse: 5B + C \u2192 A \u2192 rate = <strong>k_r [B]^5[C]<\/strong><\/li>\n<\/ul>\n\n\n\n<p>From K: K=kfkr=2.0\u00d710\u221216\u21d2kf=K\u22c5krK = \\frac{k_f}{k_r} = 2.0 \\times 10^{-16} \\Rightarrow k_f = K \\cdot k_r<\/p>\n\n\n\n<p>Also: \u03c4=1kf+kr=1Kkr+kr=1kr(1+K)\\tau = \\frac{1}{k_f + k_r} = \\frac{1}{K k_r + k_r} = \\frac{1}{k_r(1 + K)} kr=1\u03c4(1+K)\u22481(3.0\u00d710\u22126)(1+2.0\u00d710\u221216)\u22483.33\u00d7105&nbsp;s\u22121k_r = \\frac{1}{\\tau (1 + K)} \u2248 \\frac{1}{(3.0 \\times 10^{-6})(1 + 2.0 \\times 10^{-16})} \u2248 3.33 \\times 10^{5} \\ \\text{s}^{-1} kf=K\u22c5kr=(2.0\u00d710\u221216)(3.33\u00d7105)\u22486.67\u00d710\u221211&nbsp;s\u22121k_f = K \\cdot k_r = (2.0 \\times 10^{-16})(3.33 \\times 10^5) \u2248 6.67 \\times 10^{-11} \\ \\text{s}^{-1}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answers:<\/strong><\/h3>\n\n\n\n<p><strong>(a)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Proton transfer rate constant (k\u208b\u2081):<\/strong> 7.12 \u00d7 10\u00b9\u2079 s\u207b\u00b9<\/li>\n\n\n\n<li><strong>Relaxation time (\u03c4):<\/strong> 1.4 \u00d7 10\u207b\u00b2\u2070 s<\/li>\n<\/ul>\n\n\n\n<p><strong>(b)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>k_f = 6.67 \u00d7 10\u207b\u00b9\u00b9 s\u207b\u00b9<\/strong><\/li>\n\n\n\n<li><strong>k_r = 3.33 \u00d7 10\u2075 s\u207b\u00b9<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation <\/strong><\/h3>\n\n\n\n<p>In acid-base equilibrium systems, understanding forward and reverse rate constants provides insights into dynamic behavior. For part (a), the ammonium ion (NH\u2084\u207a) reacts rapidly with hydroxide (OH\u207b) to form ammonia (NH\u2083). The forward rate constant, k\u2081, is known. Using the acid dissociation constant (Ka), the reverse rate constant (k\u208b\u2081) is found via the relation K=k1\/k\u22121K = k\u2081\/k\u208b\u2081. A large k\u208b\u2081 (~10\u00b9\u2079 s\u207b\u00b9) indicates proton transfer is extremely fast, approaching diffusion control. The relaxation time \u03c4, representing how quickly equilibrium re-establishes after a perturbation (like a temperature jump), is dominated by this large k\u208b\u2081 value, making \u03c4 extraordinarily short (\u223c10\u207b\u00b2\u2070 s), implying virtually instantaneous relaxation.<\/p>\n\n\n\n<p>In part (b), we use the temperature-jump relaxation method to analyze a more complex equilibrium: A \u21cc 5B + C. Given the relaxation time (\u03c4 = 3 \u03bcs), equilibrium constant, and concentrations, we derive rate constants for both directions. The reverse rate constant (k_r) dominates due to the extremely small equilibrium constant (K = 2.0 \u00d7 10\u207b\u00b9\u2076), reflecting a reaction that heavily favors reactants. The forward rate constant (k_f) is extremely small, consistent with the low equilibrium product concentrations. The combination of a fast reverse step and slow forward step produces the observed relaxation behavior. This method illustrates how kinetic data can reveal the detailed mechanism and speed of chemical equilibria, beyond what equilibrium constants alone can show.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner8-29.jpeg\" alt=\"\" class=\"wp-image-219742\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>a. The pKa of NH4+ is 9.25 at 25\u00b0C. The rate constant at 25\u00b0C for the reaction of NH4+ and OH\u2212 to form aqueous NH3 is 4.0 \u00d7 1010 dm3 mol\u22121 s\u22121. Calculate the rate constant for proton transfer to NH3. What relaxation time would be observed if a temperature jump were applied to a [&hellip;]<\/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-219741","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219741","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=219741"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219741\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219741"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219741"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219741"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}