• # question_answer Calculate the heat change in the reaction $4\,N{{H}_{3}}(g)+3{{O}_{2}}(g)\to 2{{N}_{2}}(g)+6{{H}_{2}}O(\ell )$ at 298 K, given that the heats of formation at 298 K  for $N{{H}_{3}}(g)$ and ${{H}_{2}}O(\ell )$ are - 46.0 and - 28.60 kJ $mo{{l}^{-1}}$ respectively. A) - 1861 kJ B) - 1361 kJ C) - 1261 kJ D) - 1532 kJ

 [d] $\Delta {{H}^{o}}$ for the reaction $4N{{H}_{3}}\left( g \right)+3{{O}_{2}}\left( g \right)\to 6{{H}_{2}}O\left( l \right)+2{{N}_{2}}\left( g \right)$ $\Delta {{H}^{o}}=\Delta H_{f}^{o}\left( \text{products} \right)-\Delta H_{f}^{o}\left( \text{reactants} \right)$ $=\left\{ 6H_{f}^{o}\left[ {{H}_{2}}O\left( l \right) \right]+\Delta {{H}^{\circ }}_{f}\left[ {{N}_{2}}\left( g \right) \right] \right\}$ $-\left\{ 4\Delta H\left[ N{{H}_{3}}\left( g \right) \right]+3\Delta {{H}_{f}}\left[ {{O}_{2}}\left( g \right) \right] \right\}$
 $\Delta H_{f}^{o}\left[ {{H}_{2}}O\left( l \right) \right]=-286.0\,\,kJ\,\,mo{{l}^{-1}}$ $\Delta H_{f}^{o}\left[ N{{H}_{2}}\left( g \right) \right]=0$ and $\Delta H_{f}^{o}\left[ {{N}_{2}}\left( g \right) \right]$ = 0 (by convention) $\Delta {{H}^{o}}=\left\{ 6\left( -286 \right)+2\left( 0 \right) \right\}-\left\{ 4\left( -46.0 \right)+3\left( 0 \right) \right\}$ $=-1716+184=-1532\,kJ$