• # question_answer 26) Given: ${{N}_{2}}(g)+3{{H}_{2}}(g)\to 2N{{H}_{3}}(g);{{\Delta }_{r}}{{H}^{{}^\circ }}=-92.4\,kJmo{{l}^{-1}}$ What is the standard enthalpy of formation of gas?

The given reaction is: ${{N}_{2}}(g)+3{{H}_{2}}(g)\to 2N{{H}_{3}}(g);{{\Delta }_{r}}{{H}^{{}^\circ }}=-92.4\,kJmo{{l}^{-1}}$ ${{\Delta }_{r}}H_{{{N}_{2}}(g)}^{{}^\circ }=0;\,\,\,\,\,{{\Delta }_{f}}H_{{{H}_{2}}(g)}^{{}^\circ }=0;\,\,\,{{\Delta }_{f}}H_{N{{H}_{3}}(g)}^{{}^\circ }=x$ ${{\Delta }_{r}}H=\Sigma \,\,{{\Delta }_{f}}H_{(products)}^{{}^\circ }-\Sigma {{\Delta }_{f}}H_{(reactants)}^{{}^\circ }$ $=2{{\Delta }_{f}}H_{(N{{H}_{3}})}^{{}^\circ }-[{{\Delta }_{f}}H_{{{N}_{2}}(g)}^{{}^\circ }+3{{\Delta }_{f}}H_{{{H}_{2}}(g)}^{{}^\circ }]$ $-92.4=2x-[0+3\times 0]$ $x=-\frac{1}{2}\times 92.4$ $=-46.2kJ\,mo{{l}^{-1}}$ $i.e.,\,{{\Delta }_{f}}H_{N{{H}_{3}}(g)}^{{}^\circ }=-46.2kJ\,mo{{l}^{-1}}$