{"id":44172,"date":"2025-06-30T13:44:21","date_gmt":"2025-06-30T13:44:21","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=44172"},"modified":"2025-06-30T13:44:23","modified_gmt":"2025-06-30T13:44:23","slug":"the-standard-molar-enthalpy-of-formation-of-nhg-is-45-9-kj-mol","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-standard-molar-enthalpy-of-formation-of-nhg-is-45-9-kj-mol\/","title":{"rendered":"The standard molar enthalpy of formation of NH(g) is -45.9 kJ\/mol."},"content":{"rendered":"\n<p>The standard molar enthalpy of formation of NH<br>(g) is -45.9 kJ\/mol. What is the enthalpy change if 6.31 g N<br>(s) and 1.96 g H<br>(g) react to produce NH<br>(g)?<\/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>To determine the enthalpy change for the given reaction, we first need to write the reaction for the formation of ammonia (NH\u2083) from nitrogen (N\u2082) and hydrogen (H\u2082) gases: N2(g)+3H2(g)\u21922NH3(g)N_2(g) + 3H_2(g) \\rightarrow 2NH_3(g)N2\u200b(g)+3H2\u200b(g)\u21922NH3\u200b(g)<\/p>\n\n\n\n<p>The reaction involves the formation of ammonia (NH\u2083) from nitrogen (N\u2082) and hydrogen (H\u2082). You are given the standard molar enthalpy of formation of ammonia (NH\u2083(g)) as -45.9 kJ\/mol.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Determine moles of nitrogen and hydrogen<\/h3>\n\n\n\n<p>You are given the following masses:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>6.31 g of nitrogen (N\u2082)<\/li>\n\n\n\n<li>1.96 g of hydrogen (H\u2082)<\/li>\n<\/ul>\n\n\n\n<p>First, we calculate the moles of nitrogen and hydrogen:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of nitrogen (N\u2082) = 28.0 g\/mol<\/li>\n\n\n\n<li>Molar mass of hydrogen (H\u2082) = 2.0 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>For nitrogen: moles&nbsp;of&nbsp;N2=6.31\u2009g28.0\u2009g\/mol=0.225\u2009mol\\text{moles of N}_2 = \\frac{6.31 \\, \\text{g}}{28.0 \\, \\text{g\/mol}} = 0.225 \\, \\text{mol}moles&nbsp;of&nbsp;N2\u200b=28.0g\/mol6.31g\u200b=0.225mol<\/p>\n\n\n\n<p>For hydrogen: moles&nbsp;of&nbsp;H2=1.96\u2009g2.0\u2009g\/mol=0.98\u2009mol\\text{moles of H}_2 = \\frac{1.96 \\, \\text{g}}{2.0 \\, \\text{g\/mol}} = 0.98 \\, \\text{mol}moles&nbsp;of&nbsp;H2\u200b=2.0g\/mol1.96g\u200b=0.98mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine limiting reactant<\/h3>\n\n\n\n<p>From the balanced equation, 1 mole of N\u2082 reacts with 3 moles of H\u2082 to form 2 moles of NH\u2083. We now compare the mole ratio:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For 0.225 moles of N\u2082, the required moles of H\u2082 would be:<\/li>\n<\/ul>\n\n\n\n<p>0.225\u2009mol&nbsp;N2\u00d73=0.675\u2009mol&nbsp;H20.225 \\, \\text{mol N}_2 \\times 3 = 0.675 \\, \\text{mol H}_20.225mol&nbsp;N2\u200b\u00d73=0.675mol&nbsp;H2\u200b<\/p>\n\n\n\n<p>Since we have 0.98 moles of H\u2082 (which is more than 0.675 moles), nitrogen (N\u2082) is the limiting reactant.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the enthalpy change<\/h3>\n\n\n\n<p>Since 1 mole of N\u2082 produces 2 moles of NH\u2083, 0.225 moles of N\u2082 would produce: 0.225\u2009mol&nbsp;N2\u00d72=0.45\u2009mol&nbsp;NH30.225 \\, \\text{mol N}_2 \\times 2 = 0.45 \\, \\text{mol NH}_30.225mol&nbsp;N2\u200b\u00d72=0.45mol&nbsp;NH3\u200b<\/p>\n\n\n\n<p>The standard enthalpy of formation for NH\u2083 is -45.9 kJ\/mol. The total enthalpy change is: Enthalpy&nbsp;change=0.45\u2009mol\u00d7(\u221245.9\u2009kJ\/mol)=\u221220.66\u2009kJ\\text{Enthalpy change} = 0.45 \\, \\text{mol} \\times (-45.9 \\, \\text{kJ\/mol}) = -20.66 \\, \\text{kJ}Enthalpy&nbsp;change=0.45mol\u00d7(\u221245.9kJ\/mol)=\u221220.66kJ<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The enthalpy change for the reaction is <strong>-20.66 kJ<\/strong>.<\/p>\n\n\n\n<p>This negative value indicates that the reaction is exothermic, meaning that heat is released when ammonia is formed from nitrogen and hydrogen.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1470.jpeg\" alt=\"\" class=\"wp-image-44173\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1470.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1470-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1470-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The standard molar enthalpy of formation of NH(g) is -45.9 kJ\/mol. What is the enthalpy change if 6.31 g N(s) and 1.96 g H(g) react to produce NH(g)? The Correct Answer and Explanation is: To determine the enthalpy change for the given reaction, we first need to write the reaction for the formation of ammonia [&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-44172","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44172","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=44172"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44172\/revisions"}],"predecessor-version":[{"id":44175,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44172\/revisions\/44175"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=44172"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=44172"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=44172"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}