{"id":46536,"date":"2025-07-02T07:09:48","date_gmt":"2025-07-02T07:09:48","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=46536"},"modified":"2025-07-02T07:09:49","modified_gmt":"2025-07-02T07:09:49","slug":"the-molar-heat-of-vaporization-of-water-is-42-kj-mol","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-molar-heat-of-vaporization-of-water-is-42-kj-mol\/","title":{"rendered":"The molar heat of vaporization of water is 42 kJ\/mol"},"content":{"rendered":"\n<ul class=\"wp-block-list\">\n<li>The molar heat of vaporization of water is 42 kJ\/mol. How much energy is released by the condensation of 3.0 g of water? (1 mark) a. 0.88 kJ b. 7.0 kJ c. 126 kJ d. 130 kJ 6- An equilibrium that strongly favors products has (1 mark) a. a value of Q >> 1 b. a value of K >> 1 c. a value of Q &lt;&lt; 1 d. K = Q 7- Consider the reaction CH (g) + 2O (g) &lt;=> CO<br>(g) + 2H<br>O(g), find the numbers instead of the letters a to g: (each 1 mark) CH<br>O<br>CO<br>H<br>O I 0.5 0.5 C a b +0.10 c E d e f g a= b= c= d= e= f= g<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"700\" height=\"636\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-89.png\" alt=\"\" class=\"wp-image-46540\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-89.png 700w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-89-300x273.png 300w\" sizes=\"auto, (max-width: 700px) 100vw, 700px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here are the correct answers for the questions presented in the image.<\/p>\n\n\n\n<p><strong>5.<\/strong>&nbsp;b. 7.0 kJ<br><strong>6.<\/strong>&nbsp;b. a value of K &gt;&gt; 1<br><strong>7.<\/strong><br>a = -0.10<br>b = -0.20<br>c = +0.20<br>d = 0.40<br>e = 0.30<br>f = 0.10<br>g = 0.20<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p><strong>Question 5:<\/strong>&nbsp;This question involves a thermochemical calculation. The molar heat of vaporization is the energy required to turn one mole of a liquid into a gas. Condensation is the opposite process (gas to liquid), so it releases the same amount of energy. The molar heat of condensation is therefore 42 kJ\/mol. To find the total energy released for 3.0 g of water, we first need to convert the mass of water into moles. The molar mass of water (H\u2082O) is approximately 18.02&nbsp;g\/mol&nbsp;.<\/p>\n\n\n\n<p>Moles of water = Mass \/ Molar Mass = 3.0 g \/ 18.02&nbsp;g\/mol&nbsp;\u2248 0.1665 mol.<\/p>\n\n\n\n<p>Next, we multiply the number of moles by the molar heat of condensation to find the total energy released:<\/p>\n\n\n\n<p>Energy released = Moles \u00d7 Molar Heat of Condensation = 0.1665 mol \u00d7 42 kJ\/mol \u2248 7.0 kJ. Thus, the correct answer is 7.0 kJ.<\/p>\n\n\n\n<p><strong>Question 6:<\/strong>&nbsp;This question tests the understanding of the equilibrium constant, K. The equilibrium constant is a ratio of the concentration of products to reactants at equilibrium. A large value for K (K &gt;&gt; 1) signifies that at equilibrium, the concentration of products is significantly greater than the concentration of reactants. This indicates that the forward reaction is highly favored, and the equilibrium &#8220;lies to the right,&#8221; strongly favoring the formation of products. In contrast, the reaction quotient, Q, describes the ratio at any given moment, not necessarily at equilibrium, and indicates the direction a reaction will shift to reach equilibrium.<\/p>\n\n\n\n<p><strong>Question 7:<\/strong>&nbsp;This problem requires completing an ICE (Initial, Change, Equilibrium) table for the reaction CH\u2084(g) + 2O\u2082(g) \u21cc CO\u2082(g) + 2H\u2082O(g). The changes in the amounts of reactants and products are dictated by the stoichiometric coefficients in the balanced equation. We are given that the change (C) for CO\u2082 is +0.10.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Finding a, b, c (Changes):<\/strong>\u00a0Based on the 1:2:1:2 molar ratio:\n<ul class=\"wp-block-list\">\n<li><strong>a (CH\u2084):<\/strong>\u00a0Reactant, ratio 1:1 with CO\u2082. Change = -(1\/1) * 0.10 = -0.10.<\/li>\n\n\n\n<li><strong>b (O\u2082):<\/strong>\u00a0Reactant, ratio 2:1 with CO\u2082. Change = -(2\/1) * 0.10 = -0.20.<\/li>\n\n\n\n<li><strong>c (H\u2082O):<\/strong>\u00a0Product, ratio 2:1 with CO\u2082. Change = +(2\/1) * 0.10 = +0.20.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Finding d, e, f, g (Equilibrium):<\/strong>\u00a0The equilibrium amount (E) is the initial amount (I) plus the change (C).\n<ul class=\"wp-block-list\">\n<li><strong>d (CH\u2084):<\/strong>\u00a0E = 0.5 + (-0.10) = 0.40.<\/li>\n\n\n\n<li><strong>e (O\u2082):<\/strong>\u00a0E = 0.5 + (-0.20) = 0.30.<\/li>\n\n\n\n<li><strong>f (CO\u2082):<\/strong>\u00a0Assuming initial amount is 0, E = 0 + 0.10 = 0.10.<\/li>\n\n\n\n<li><strong>g (H\u2082O):<\/strong>\u00a0Assuming initial amount is 0, E = 0 + 0.20 = 0.20.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-162.jpeg\" alt=\"\" class=\"wp-image-46545\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-162.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-162-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-162-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-162-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The Correct Answer and Explanation is: Here are the correct answers for the questions presented in the image. 5.&nbsp;b. 7.0 kJ6.&nbsp;b. a value of K &gt;&gt; 17.a = -0.10b = -0.20c = +0.20d = 0.40e = 0.30f = 0.10g = 0.20 Explanation Question 5:&nbsp;This question involves a thermochemical calculation. The molar heat of vaporization is [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-46536","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46536","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=46536"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46536\/revisions"}],"predecessor-version":[{"id":46549,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46536\/revisions\/46549"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=46536"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=46536"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=46536"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}