{"id":232198,"date":"2025-06-11T19:20:28","date_gmt":"2025-06-11T19:20:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=232198"},"modified":"2025-06-11T19:20:30","modified_gmt":"2025-06-11T19:20:30","slug":"standard-enthalpy-of-formation-of-nh3-g-is-46-11-kj-mol-1-at-298-k","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/standard-enthalpy-of-formation-of-nh3-g-is-46-11-kj-mol-1-at-298-k\/","title":{"rendered":"Standard enthalpy of formation of NH3 (g) is -46.11 kJ mol-1 at 298 K."},"content":{"rendered":"\n

Standard enthalpy of formation of NH3 (g) is -46.11 kJ mol-1 at 298 K. Given the heat capacity data below, calculate the standard enthalpy of formation at 1200 K. CP,m (H2 (g))\/ J mol-1 K-1 = 29.1 – (0.84 x 10-3 K-1)T CP,m (N2 (g))\/ J mol-1 K-1 = 26.98 + (5.9 x 10-3 K-1)T CP,m (NH3 (g))\/ J mol-1 K-1 = 25.9 + (33 x 10-3 K-1)T (-53.1 kJ\/mol)<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To calculate the standard enthalpy of formation of ammonia (NH\u2083(g)) at 1200 K<\/strong>, we use the temperature correction<\/strong> method based on heat capacity data. The relationship used is:\u0394Hf\u2218(1200 K)=\u0394Hf\u2218(298 K)+\u0394H\u2218(1200 K)\u2212\u0394H\u2218(298 K)\\Delta H_f^\\circ(1200\\ \\text{K}) = \\Delta H_f^\\circ(298\\ \\text{K}) + \\Delta H^\\circ(1200\\ \\text{K}) – \\Delta H^\\circ(298\\ \\text{K})\u0394Hf\u2218\u200b(1200 K)=\u0394Hf\u2218\u200b(298 K)+\u0394H\u2218(1200 K)\u2212\u0394H\u2218(298 K)<\/p>\n\n\n\n

Where the change in enthalpy is calculated from:\u0394H=\u222b2981200\u2211\u03bdCP,m\u2218(products) dT\u2212\u222b2981200\u2211\u03bdCP,m\u2218(reactants) dT\\Delta H = \\int_{298}^{1200} \\sum \\nu C_{P,m}^\\circ(\\text{products})\\ dT – \\int_{298}^{1200} \\sum \\nu C_{P,m}^\\circ(\\text{reactants})\\ dT\u0394H=\u222b2981200\u200b\u2211\u03bdCP,m\u2218\u200b(products) dT\u2212\u222b2981200\u200b\u2211\u03bdCP,m\u2218\u200b(reactants) dT<\/p>\n\n\n\n

\n\n\n\n

Step 1: Write the formation reaction<\/h3>\n\n\n\n

12N2(g)+32H2(g)\u2192NH3(g)\\frac{1}{2}N_2(g) + \\frac{3}{2}H_2(g) \\rightarrow NH_3(g)21\u200bN2\u200b(g)+23\u200bH2\u200b(g)\u2192NH3\u200b(g)<\/p>\n\n\n\n

\n\n\n\n

Step 2: Heat capacity equations (in J\/mol\u00b7K)<\/h3>\n\n\n\n
    \n
  • CP,m(H2)=29.1\u22120.00084TC_{P,m}(H_2) = 29.1 – 0.00084TCP,m\u200b(H2\u200b)=29.1\u22120.00084T<\/li>\n\n\n\n
  • CP,m(N2)=26.98+0.0059TC_{P,m}(N_2) = 26.98 + 0.0059TCP,m\u200b(N2\u200b)=26.98+0.0059T<\/li>\n\n\n\n
  • CP,m(NH3)=25.9+0.033TC_{P,m}(NH_3) = 25.9 + 0.033TCP,m\u200b(NH3\u200b)=25.9+0.033T<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step 3: Compute enthalpy change from 298 K to 1200 K<\/h3>\n\n\n\n

    Integrate CP(T)C_P(T)CP\u200b(T) for each species:<\/p>\n\n\n\n

    For NH\u2083:\u0394HNH3=\u222b2981200(25.9+0.033T)\u2009dT=25.9(T)+0.033T22\u21d2[25.9T+0.0165T2]2981200\\Delta H_{NH_3} = \\int_{298}^{1200}(25.9 + 0.033T)\\,dT = 25.9(T) + 0.033\\frac{T^2}{2} \\Rightarrow [25.9T + 0.0165T^2]_{298}^{1200}\u0394HNH3\u200b\u200b=\u222b2981200\u200b(25.9+0.033T)dT=25.9(T)+0.0332T2\u200b\u21d2[25.9T+0.0165T2]2981200\u200b=(25.9\u00d71200+0.0165\u00d712002)\u2212(25.9\u00d7298+0.0165\u00d72982)=48828+23760\u22127718.2\u22121465.4=63304.4J\/mol= (25.9\u00d71200 + 0.0165\u00d71200^2) – (25.9\u00d7298 + 0.0165\u00d7298^2) = 48828 + 23760 – 7718.2 – 1465.4 = 63304.4 J\/mol=(25.9\u00d71200+0.0165\u00d712002)\u2212(25.9\u00d7298+0.0165\u00d72982)=48828+23760\u22127718.2\u22121465.4=63304.4J\/mol<\/p>\n\n\n\n

    For 12N2\\frac{1}{2}N_221\u200bN2\u200b:\u0394H=0.5\u00d7[26.98T+0.00295T2]2981200=0.5\u00d7[(26.98\u00d71200+0.00295\u00d712002)\u2212(26.98\u00d7298+0.00295\u00d72982)]=0.5\u00d7(32376+4248\u22128036.04\u2212261.2)=0.5\u00d7(36624\u22128297.24)=14163.4J\/mol\\Delta H = 0.5 \\times \\left[26.98T + 0.00295T^2\\right]_{298}^{1200} = 0.5 \\times [(26.98\u00d71200 + 0.00295\u00d71200^2) – (26.98\u00d7298 + 0.00295\u00d7298^2)] = 0.5 \\times (32376 + 4248 – 8036.04 – 261.2) = 0.5 \\times (36624 – 8297.24) = 14163.4 J\/mol\u0394H=0.5\u00d7[26.98T+0.00295T2]2981200\u200b=0.5\u00d7[(26.98\u00d71200+0.00295\u00d712002)\u2212(26.98\u00d7298+0.00295\u00d72982)]=0.5\u00d7(32376+4248\u22128036.04\u2212261.2)=0.5\u00d7(36624\u22128297.24)=14163.4J\/mol<\/p>\n\n\n\n

    For 32H2\\frac{3}{2}H_223\u200bH2\u200b:\u0394H=1.5\u00d7[29.1T\u22120.00042T2]2981200=1.5\u00d7[(29.1\u00d71200\u22120.00042\u00d712002)\u2212(29.1\u00d7298\u22120.00042\u00d72982)]=1.5\u00d7(34920\u2212604.8\u22128671.8+37.3)=1.5\u00d725680.7=38521.0J\/mol\\Delta H = 1.5 \\times [29.1T – 0.00042T^2]_{298}^{1200} = 1.5 \u00d7 [(29.1\u00d71200 – 0.00042\u00d71200^2) – (29.1\u00d7298 – 0.00042\u00d7298^2)] = 1.5 \u00d7 (34920 – 604.8 – 8671.8 + 37.3) = 1.5 \u00d7 25680.7 = 38521.0 J\/mol\u0394H=1.5\u00d7[29.1T\u22120.00042T2]2981200\u200b=1.5\u00d7[(29.1\u00d71200\u22120.00042\u00d712002)\u2212(29.1\u00d7298\u22120.00042\u00d72982)]=1.5\u00d7(34920\u2212604.8\u22128671.8+37.3)=1.5\u00d725680.7=38521.0J\/mol<\/p>\n\n\n\n

    \n\n\n\n

    Step 4: Total \u0394H correction<\/h3>\n\n\n\n

    \u0394H=HNH3\u2212HN2+H2=63304.4\u2212(14163.4+38521.0)=10620J\/mol=10.62kJ\/mol\\Delta H = H_{NH_3} – H_{N_2+H_2} = 63304.4 – (14163.4 + 38521.0) = 10620 J\/mol = 10.62 kJ\/mol\u0394H=HNH3\u200b\u200b\u2212HN2\u200b+H2\u200b\u200b=63304.4\u2212(14163.4+38521.0)=10620J\/mol=10.62kJ\/mol<\/p>\n\n\n\n

    \n\n\n\n

    Step 5: Apply correction<\/h3>\n\n\n\n

    \u0394Hf\u2218(1200 K)=\u221246.11 kJ\/mol+10.62 kJ\/mol=\u221235.49 kJ\/mol\\Delta H_f^\\circ(1200\\ K) = -46.11\\ \\text{kJ\/mol} + 10.62\\ \\text{kJ\/mol} = \\boxed{-35.49\\ \\text{kJ\/mol}}\u0394Hf\u2218\u200b(1200 K)=\u221246.11 kJ\/mol+10.62 kJ\/mol=\u221235.49 kJ\/mol\u200b<\/p>\n\n\n\n

    \n\n\n\n

    Explanation<\/h3>\n\n\n\n

    The standard enthalpy of formation, \u0394Hf\u2218\\Delta H_f^\\circ\u0394Hf\u2218\u200b, of a compound is the enthalpy change when one mole of the substance forms from its elements in their standard states. For NH\u2083(g), this value is -46.11 kJ\/mol at 298 K. However, enthalpy is temperature-dependent because molecular heat capacities vary with temperature.<\/p>\n\n\n\n

    To determine the enthalpy of formation at 1200 K, we account for the temperature-dependent change in enthalpy using heat capacity expressions. The reaction considered is:12N2(g)+32H2(g)\u2192NH3(g)\\frac{1}{2}N_2(g) + \\frac{3}{2}H_2(g) \\rightarrow NH_3(g)21\u200bN2\u200b(g)+23\u200bH2\u200b(g)\u2192NH3\u200b(g)<\/p>\n\n\n\n

    For each species, we integrate their molar heat capacity (CP,mC_{P,m}CP,m\u200b) from 298 K to 1200 K to obtain the change in enthalpy with temperature. The integral of a linear function CP=a+bTC_P = a + bTCP\u200b=a+bT over temperature gives:\u0394H=a(T2\u2212T1)+b2(T22\u2212T12)\\Delta H = a(T_2 – T_1) + \\frac{b}{2}(T_2^2 – T_1^2)\u0394H=a(T2\u200b\u2212T1\u200b)+2b\u200b(T22\u200b\u2212T12\u200b)<\/p>\n\n\n\n

    By applying this to each component in the reaction and calculating the net enthalpy difference between products and reactants, we obtain the total enthalpy correction. This correction (10.62 kJ\/mol) is added to the standard enthalpy of formation at 298 K to yield the value at 1200 K.<\/p>\n\n\n\n

    Thus, the standard enthalpy of formation of NH\u2083(g) at 1200 K is -35.49 kJ\/mol<\/strong>, reflecting the increased thermal energy at elevated temperatures and reduced thermodynamic driving force for ammonia formation.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Standard enthalpy of formation of NH3 (g) is -46.11 kJ mol-1 at 298 K. Given the heat capacity data below, calculate the standard enthalpy of formation at 1200 K. CP,m (H2 (g))\/ J mol-1 K-1 = 29.1 – (0.84 x 10-3 K-1)T CP,m (N2 (g))\/ J mol-1 K-1 = 26.98 + (5.9 x 10-3 K-1)T […]<\/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-232198","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232198","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=232198"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232198\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=232198"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=232198"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=232198"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}