{"id":47594,"date":"2025-07-02T13:16:07","date_gmt":"2025-07-02T13:16:07","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=47594"},"modified":"2025-07-02T13:16:09","modified_gmt":"2025-07-02T13:16:09","slug":"a-student-investigates-the-reaction-between-ags-and-hno3aq-represented-by-the-equation-above","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-student-investigates-the-reaction-between-ags-and-hno3aq-represented-by-the-equation-above\/","title":{"rendered":"A student investigates the reaction between Ag(s) and HNO3(aq) represented by the equation above."},"content":{"rendered":"\n<p>Ag(s) + 4 HNO3(aq) \u00e2\u2020\u2019 3 AgNO3(aq) + NO(g) + 2 H2O(l) A student investigates the reaction between Ag(s) and HNO3(aq) represented by the equation above. Predict the sign of the entropy change, \u00ce\u201dS, for the reaction. Justify your answer. b. Use the information in the table below to calculate the value \u00ce\u201dHrxn, the standard enthalpy change for the reaction, in kJ\/molrxn: Substance \u00ce\u201dHf\u00c2\u00b0 (kJ\/mol) Ag(s) 0 HNO3(aq) -207 AgNO3(aq) -101 NO(g) 90 H2O(l) -286 Based on your answers to parts (a) and (b), is the reaction more likely to be thermodynamically favorable at 25\u00c2\u00b0C or at 95\u00c2\u00b0C? Justify your answer. The student runs the reaction using a 3 to 4 mole ratio of Ag(s) to HNO3(aq). Suggest a method the student can use to isolate solid AgNO3 from the other products of the reaction.<\/p>\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><strong>a. Entropy Change (\u0394S):<\/strong><\/p>\n\n\n\n<p>To predict the sign of the entropy change (\u0394S) for the reaction, we need to consider the phases and the number of molecules involved in the reactants and products.<\/p>\n\n\n\n<p>The reaction is:<br><strong>Ag(s) + 4 HNO3(aq) \u2192 3 AgNO3(aq) + NO(g) + 2 H2O(l)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>On the left side (reactants), we have:\n<ul class=\"wp-block-list\">\n<li><strong>Ag(s)<\/strong> (solid): low entropy (solids generally have lower entropy than liquids or gases).<\/li>\n\n\n\n<li><strong>HNO3(aq)<\/strong> (aqueous): moderate entropy.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>On the right side (products), we have:\n<ul class=\"wp-block-list\">\n<li><strong>AgNO3(aq)<\/strong> (aqueous): moderate entropy.<\/li>\n\n\n\n<li><strong>NO(g)<\/strong> (gas): high entropy (gases have the highest entropy).<\/li>\n\n\n\n<li><strong>H2O(l)<\/strong> (liquid): lower entropy than gases but higher than solids.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>Entropy tends to increase when there is a phase change from solid to liquid or gas, or when the number of gas molecules increases.<\/strong> In this reaction:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The number of moles of gas increases from 0 (on the left) to 1 (on the right).<\/li>\n\n\n\n<li>The formation of NO(g), a gas, suggests an increase in disorder.<\/li>\n<\/ul>\n\n\n\n<p>Thus, <strong>\u0394S is positive<\/strong>, meaning there is an increase in entropy due to the production of a gas and the phase changes.<\/p>\n\n\n\n<p><strong>b. Standard Enthalpy Change (\u0394Hrxn):<\/strong><\/p>\n\n\n\n<p>We can calculate the standard enthalpy change (\u0394Hrxn) using the formula:\u0394Hrxn=\u2211\u0394Hf\u2218(products)\u2212\u2211\u0394Hf\u2218(reactants)\\Delta H_{\\text{rxn}} = \\sum \\Delta H_f^\\circ (\\text{products}) &#8211; \\sum \\Delta H_f^\\circ (\\text{reactants})\u0394Hrxn\u200b=\u2211\u0394Hf\u2218\u200b(products)\u2212\u2211\u0394Hf\u2218\u200b(reactants)<\/p>\n\n\n\n<p>From the table:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>\u0394Hf\u00b0 for Ag(s)<\/strong> = 0 kJ\/mol<\/li>\n\n\n\n<li><strong>\u0394Hf\u00b0 for HNO3(aq)<\/strong> = -207 kJ\/mol<\/li>\n\n\n\n<li><strong>\u0394Hf\u00b0 for AgNO3(aq)<\/strong> = -101 kJ\/mol<\/li>\n\n\n\n<li><strong>\u0394Hf\u00b0 for NO(g)<\/strong> = 90 kJ\/mol<\/li>\n\n\n\n<li><strong>\u0394Hf\u00b0 for H2O(l)<\/strong> = -286 kJ\/mol<\/li>\n<\/ul>\n\n\n\n<p>Now, calculate:\u0394Hrxn=[3\u00d7(\u2212101)+1\u00d790+2\u00d7(\u2212286)]\u2212[1\u00d70+4\u00d7(\u2212207)]\\Delta H_{\\text{rxn}} = [3 \\times (-101) + 1 \\times 90 + 2 \\times (-286)] &#8211; [1 \\times 0 + 4 \\times (-207)]\u0394Hrxn\u200b=[3\u00d7(\u2212101)+1\u00d790+2\u00d7(\u2212286)]\u2212[1\u00d70+4\u00d7(\u2212207)]\u0394Hrxn=[\u2212303+90\u2212572]\u2212[0\u2212828]\\Delta H_{\\text{rxn}} = [-303 + 90 &#8211; 572] &#8211; [0 &#8211; 828]\u0394Hrxn\u200b=[\u2212303+90\u2212572]\u2212[0\u2212828]\u0394Hrxn=\u2212785\u2212(\u2212828)=\u2212785+828=43\u2009kJ\/mol\\Delta H_{\\text{rxn}} = -785 &#8211; (-828) = -785 + 828 = 43 \\, \\text{kJ\/mol}\u0394Hrxn\u200b=\u2212785\u2212(\u2212828)=\u2212785+828=43kJ\/mol<\/p>\n\n\n\n<p>Therefore, the standard enthalpy change (\u0394Hrxn) for the reaction is <strong>+43 kJ\/mol<\/strong>.<\/p>\n\n\n\n<p><strong>c. Thermodynamic Favorability at 25\u00b0C vs 95\u00b0C:<\/strong><\/p>\n\n\n\n<p>To determine whether the reaction is more likely to be thermodynamically favorable at 25\u00b0C or 95\u00b0C, we need to consider the relationship between enthalpy (\u0394H) and entropy (\u0394S) changes. The Gibbs free energy (\u0394G) determines whether a reaction is thermodynamically favorable:\u0394G=\u0394H\u2212T\u0394S\\Delta G = \\Delta H &#8211; T\\Delta S\u0394G=\u0394H\u2212T\u0394S<\/p>\n\n\n\n<p>At 25\u00b0C (298 K):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394H = +43 kJ\/mol (positive, endothermic)<\/li>\n\n\n\n<li>\u0394S is positive, so at a lower temperature, the <strong>T\u0394S<\/strong> term is smaller.<\/li>\n\n\n\n<li>Therefore, <strong>\u0394G may be positive<\/strong>, indicating that the reaction is less likely to be spontaneous at lower temperatures.<\/li>\n<\/ul>\n\n\n\n<p>At 95\u00b0C (368 K):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The temperature is higher, and <strong>T\u0394S increases<\/strong>, which makes the <strong>\u0394G more negative<\/strong> (because T\u0394S becomes larger than \u0394H).<\/li>\n\n\n\n<li>Hence, the reaction is more likely to be thermodynamically favorable at <strong>95\u00b0C<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>d. Method to Isolate Solid AgNO3:<\/strong><\/p>\n\n\n\n<p>To isolate solid AgNO3 from the other products, the student can:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Evaporate the water<\/strong>: Heat the reaction mixture gently to remove the water (H2O) and obtain AgNO3 as a solid since it is highly soluble in water and will remain in the aqueous solution.<\/li>\n\n\n\n<li><strong>Filter the NO(g)<\/strong>: Since NO is a gas, it will escape from the solution, leaving behind the aqueous AgNO3.<\/li>\n\n\n\n<li><strong>Crystallize AgNO3<\/strong>: Once the solution is concentrated, the student can cool it to allow the AgNO3 to crystallize out of the solution, then filter it off to isolate the solid AgNO3.<\/li>\n<\/ol>\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-banner5-257.jpeg\" alt=\"\" class=\"wp-image-47596\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-257.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-257-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-257-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-257-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Ag(s) + 4 HNO3(aq) \u00e2\u2020\u2019 3 AgNO3(aq) + NO(g) + 2 H2O(l) A student investigates the reaction between Ag(s) and HNO3(aq) represented by the equation above. Predict the sign of the entropy change, \u00ce\u201dS, for the reaction. Justify your answer. b. Use the information in the table below to calculate the value \u00ce\u201dHrxn, the standard [&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-47594","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47594","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=47594"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47594\/revisions"}],"predecessor-version":[{"id":47600,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47594\/revisions\/47600"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=47594"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=47594"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=47594"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}