{"id":196460,"date":"2025-03-05T04:35:03","date_gmt":"2025-03-05T04:35:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196460"},"modified":"2025-03-05T04:35:06","modified_gmt":"2025-03-05T04:35:06","slug":"calculate-the-standard-reaction-entropy-at-298-k","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/05\/calculate-the-standard-reaction-entropy-at-298-k\/","title":{"rendered":"Calculate the standard reaction entropy at 298 K"},"content":{"rendered":"\n<p>Calculate the standard reaction entropy at 298 K of<\/p>\n\n\n\n<p>(1) 2 CH3CHO(g) + O2(g)\u21922 CH3COOH(l)<\/p>\n\n\n\n<p>(2) 2 AgCl(s) + Br2(l)\u21922 AgBr(s) + Cl2(g)<\/p>\n\n\n\n<p>(3) Hg(l) + Cl2(g)\u2192HgCl2(s)<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To calculate the standard reaction entropy (\u0394S\u00b0) at 298 K for each of the given reactions, we use the formula:<\/p>\n\n\n\n<p>\u0394S\u00b0 = \u03a3S\u00b0(products) \u2212 \u03a3S\u00b0(reactants)<\/p>\n\n\n\n<p>where S\u00b0 represents the standard molar entropy of each substance involved. Standard entropy values are typically expressed in J\/mol\u00b7K.<\/p>\n\n\n\n<p><strong>Reaction (1):<\/strong><\/p>\n\n\n\n<p>2 CH\u2083CHO(g) + O\u2082(g) \u2192 2 CH\u2083COOH(l)\ue206<\/p>\n\n\n\n<p>Using standard entropy values at 298 K:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>S\u00b0(CH\u2083CHO(g)) = 250.3 J\/mol\u00b7K\ue206<\/li>\n\n\n\n<li>S\u00b0(O\u2082(g)) = 205.0 J\/mol\u00b7K\ue206<\/li>\n\n\n\n<li>S\u00b0(CH\u2083COOH(l)) = 159.8 J\/mol\u00b7K\ue206<\/li>\n<\/ul>\n\n\n\n<p>Calculating \u0394S\u00b0:<\/p>\n\n\n\n<p>\u0394S\u00b0 = [2 \u00d7 S\u00b0(CH\u2083COOH(l))] \u2212 [2 \u00d7 S\u00b0(CH\u2083CHO(g)) + S\u00b0(O\u2082(g))]<br>\u0394S\u00b0 = [2 \u00d7 159.8] \u2212 [2 \u00d7 250.3 + 205.0]<br>\u0394S\u00b0 = 319.6 \u2212 [500.6 + 205.0]<br>\u0394S\u00b0 = 319.6 \u2212 705.6<br>\u0394S\u00b0 = \u2212386.0 J\/mol\u00b7K<\/p>\n\n\n\n<p><strong>Reaction (2):<\/strong><\/p>\n\n\n\n<p>2 AgCl(s) + Br\u2082(l) \u2192 2 AgBr(s) + Cl\u2082(g)\ue206<\/p>\n\n\n\n<p>Standard entropy values:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>S\u00b0(AgCl(s)) = 96.2 J\/mol\u00b7K\ue206<\/li>\n\n\n\n<li>S\u00b0(Br\u2082(l)) = 152.2 J\/mol\u00b7K\ue206<\/li>\n\n\n\n<li>S\u00b0(AgBr(s)) = 107.1 J\/mol\u00b7K\ue206<\/li>\n\n\n\n<li>S\u00b0(Cl\u2082(g)) = 223.0 J\/mol\u00b7K\ue206<\/li>\n<\/ul>\n\n\n\n<p>Calculating \u0394S\u00b0:<\/p>\n\n\n\n<p>\u0394S\u00b0 = [2 \u00d7 S\u00b0(AgBr(s)) + S\u00b0(Cl\u2082(g))] \u2212 [2 \u00d7 S\u00b0(AgCl(s)) + S\u00b0(Br\u2082(l))]<br>\u0394S\u00b0 = [2 \u00d7 107.1 + 223.0] \u2212 [2 \u00d7 96.2 + 152.2]<br>\u0394S\u00b0 = [214.2 + 223.0] \u2212 [192.4 + 152.2]<br>\u0394S\u00b0 = 437.2 \u2212 344.6<br>\u0394S\u00b0 = 92.6 J\/mol\u00b7K<\/p>\n\n\n\n<p><strong>Reaction (3):<\/strong><\/p>\n\n\n\n<p>Hg(l) + Cl\u2082(g) \u2192 HgCl\u2082(s)\ue206<\/p>\n\n\n\n<p>Standard entropy values:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>S\u00b0(Hg(l)) = 76.0 J\/mol\u00b7K\ue206<\/li>\n\n\n\n<li>S\u00b0(Cl\u2082(g)) = 223.0 J\/mol\u00b7K\ue206<\/li>\n\n\n\n<li>S\u00b0(HgCl\u2082(s)) = 144.3 J\/mol\u00b7K\ue206<\/li>\n<\/ul>\n\n\n\n<p>Calculating \u0394S\u00b0:<\/p>\n\n\n\n<p>\u0394S\u00b0 = S\u00b0(HgCl\u2082(s)) \u2212 [S\u00b0(Hg(l)) + S\u00b0(Cl\u2082(g))]<br>\u0394S\u00b0 = 144.3 \u2212 [76.0 + 223.0]<br>\u0394S\u00b0 = 144.3 \u2212 299.0<br>\u0394S\u00b0 = \u2212154.7 J\/mol\u00b7K<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>The standard reaction entropy (\u0394S\u00b0) measures the change in entropy, or disorder, between reactants and products under standard conditions (298 K and 1 atm). A negative \u0394S\u00b0 indicates a decrease in disorder, often associated with the formation of more ordered products from less ordered reactants. Conversely, a positive \u0394S\u00b0 suggests an increase in disorder.\ue206<\/p>\n\n\n\n<p>In reactions (1) and (3), the negative \u0394S\u00b0 values imply that the products are more ordered than the reactants, leading to a decrease in entropy. In reaction (2), the positive \u0394S\u00b0 indicates that the products are less ordered, resulting in an increase in entropy.\ue206<\/p>\n\n\n\n<p><strong>Note on Image:<\/strong><\/p>\n\n\n\n<p>Providing an image of the entropy values or reaction mechanisms would require visual content, which is not possible in this text-based format. However, you can refer to standard chemistry textbooks or reputable online databases for visual representations and more detailed explanations.\ue206<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the standard reaction entropy at 298 K of (1) 2 CH3CHO(g) + O2(g)\u21922 CH3COOH(l) (2) 2 AgCl(s) + Br2(l)\u21922 AgBr(s) + Cl2(g) (3) Hg(l) + Cl2(g)\u2192HgCl2(s) The Correct Answer and Explanation is : To calculate the standard reaction entropy (\u0394S\u00b0) at 298 K for each of the given reactions, we use the formula: \u0394S\u00b0 [&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-196460","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196460","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=196460"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196460\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196460"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196460"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196460"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}