{"id":221802,"date":"2025-05-30T14:28:37","date_gmt":"2025-05-30T14:28:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221802"},"modified":"2025-05-30T14:28:39","modified_gmt":"2025-05-30T14:28:39","slug":"draw-enantiomers-for-the-following-compounds-using","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/30\/draw-enantiomers-for-the-following-compounds-using\/","title":{"rendered":"Draw enantiomers for the following compounds using"},"content":{"rendered":"\n<p>Draw enantiomers for the following compounds using<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Perspective Formulas<\/li>\n\n\n\n<li>Fischer Projections<\/li>\n<\/ul>\n\n\n\n<p>The reaction shown here has a K<sub>p<\/sub>=4.5*10<sup>2<\/sup>&nbsp;at 825K. Find Kc for the reaction at this temperature:<\/p>\n\n\n\n<p>CH<sub>4<\/sub>(g) + CO<sub>2<\/sub>(g) &lt;&#8212;&gt; 2CO(g) + 2H<sub>2<\/sub>(g)<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part 1: <strong>Drawing Enantiomers<\/strong><\/h3>\n\n\n\n<p>Let\u2019s assume the compound of interest is <strong>lactic acid<\/strong> (CH\u2083CH(OH)COOH) \u2014 a common example with one chiral center. If your compound is different, please specify.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">a) <strong>Perspective Formulas (3D Wedge-Dash)<\/strong><\/h4>\n\n\n\n<p>Lactic acid has one chiral center at the second carbon (C2).<\/p>\n\n\n\n<p><strong>(R)-Lactic Acid:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>         OH\n          |\nH3C \u2014 C \u2014 COOH\n          |\n         H\n<\/code><\/pre>\n\n\n\n<p><em>Perspective (Wedge-Dash):<\/em><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>       OH\n        |\nH3C \u2014 C \u2014 COOH\n     \/   \\\n   H     (wedge) OH\n          (dash) H\n<\/code><\/pre>\n\n\n\n<p><strong>(S)-Lactic Acid:<\/strong><br><em>Swap two groups to get the enantiomer:<\/em><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>       OH\n        |\nH3C \u2014 C \u2014 COOH\n     \/   \\\n(wedge) H   (dash) OH\n<\/code><\/pre>\n\n\n\n<h4 class=\"wp-block-heading\">b) <strong>Fischer Projections<\/strong><\/h4>\n\n\n\n<p><strong>(R)-Lactic Acid:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>      COOH\n       |\nH \u2014 C \u2014 OH\n       |\n      CH3\n<\/code><\/pre>\n\n\n\n<p><strong>(S)-Lactic Acid:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>      COOH\n       |\nOH \u2014 C \u2014 H\n       |\n      CH3\n<\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Part 2: <strong>Finding KcK_c from KpK_p<\/strong><\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Kp=4.5\u00d7102K_p = 4.5 \\times 10^2<\/li>\n\n\n\n<li>Temperature, T=825T = 825 K<\/li>\n\n\n\n<li>Reaction: CH4(g)+CO2(g)\u21cc2CO(g)+2H2(g)\\text{CH}_4(g) + \\text{CO}_2(g) \\rightleftharpoons 2\\text{CO}(g) + 2\\text{H}_2(g)<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1: Use the equation to relate KpK_p and KcK_c:<\/strong> Kp=Kc(RT)\u0394nK_p = K_c(RT)^{\\Delta n}<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>R=0.0821\u2009L\\cdotpatm\/mol\\cdotpKR = 0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K}<\/li>\n\n\n\n<li>\u0394n=moles\u00a0of\u00a0gas\u00a0products\u2212moles\u00a0of\u00a0gas\u00a0reactants=(2+2)\u2212(1+1)=2\\Delta n = \\text{moles of gas products} &#8211; \\text{moles of gas reactants} = (2+2) &#8211; (1+1) = 2<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Solve for KcK_c:<\/strong> Kc=Kp(RT)\u0394n=4.5\u00d7102(0.0821\u00d7825)2K_c = \\frac{K_p}{(RT)^{\\Delta n}} = \\frac{4.5 \\times 10^2}{(0.0821 \\times 825)^2}<\/p>\n\n\n\n<p>Calculate: RT=0.0821\u00d7825=67.7325RT = 0.0821 \\times 825 = 67.7325 (RT)2=(67.7325)2\u22484586.73(RT)^2 = (67.7325)^2 \\approx 4586.73 Kc=4504586.73\u22480.0981K_c = \\frac{450}{4586.73} \\approx 0.0981<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>Kc\u22480.0981\\boxed{K_c \\approx 0.0981}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation <\/h3>\n\n\n\n<p>In gas-phase equilibrium reactions, the equilibrium constants KpK_p and KcK_c are related by the formula: Kp=Kc(RT)\u0394nK_p = K_c(RT)^{\\Delta n}<\/p>\n\n\n\n<p>where RR is the gas constant, TT is the absolute temperature in Kelvin, and \u0394n\\Delta n is the change in the number of moles of gas between products and reactants. This relationship accounts for how pressure and concentration differ in behavior for gases.<\/p>\n\n\n\n<p>In this case, the reaction involves 2 moles of CO and 2 moles of H\u2082 (total 4) on the product side, and 1 mole of CH\u2084 and 1 mole of CO\u2082 (total 2) on the reactant side. So \u0394n=4\u22122=2\\Delta n = 4 &#8211; 2 = 2.<\/p>\n\n\n\n<p>We are given Kp=4.5\u00d7102K_p = 4.5 \\times 10^2 at 825 K. Using the ideal gas constant R=0.0821\u2009L\\cdotpatm\/mol\\cdotpKR = 0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K}, we first compute RTRT, square it (since \u0394n=2\\Delta n = 2), and then solve for KcK_c by dividing KpK_p by this quantity.<\/p>\n\n\n\n<p>The calculation gives: Kc=450(0.0821\u00d7825)2=4504586.73\u22480.0981K_c = \\frac{450}{(0.0821 \\times 825)^2} = \\frac{450}{4586.73} \\approx 0.0981<\/p>\n\n\n\n<p>This result means that even though the partial pressures strongly favor the products (large KpK_p), the actual concentration-based constant KcK_c is lower due to the pressure-concentration conversion that accounts for the increase in gas moles (which dilutes concentration but increases pressure).<\/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-banner4-135.jpeg\" alt=\"\" class=\"wp-image-221803\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw enantiomers for the following compounds using The reaction shown here has a Kp=4.5*102&nbsp;at 825K. Find Kc for the reaction at this temperature: CH4(g) + CO2(g) &lt;&#8212;&gt; 2CO(g) + 2H2(g) The Correct Answer and Explanation is: Part 1: Drawing Enantiomers Let\u2019s assume the compound of interest is lactic acid (CH\u2083CH(OH)COOH) \u2014 a common example with [&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-221802","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221802","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=221802"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221802\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221802"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221802"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221802"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}