A) - 1861 kJ
B) - 1361 kJ
C) - 1261 kJ
D) - 1532 kJ
Correct Answer: D
Solution :
[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) |
\[\Delta {{H}^{o}}=\left\{ 6\left( -286 \right)+2\left( 0 \right) \right\}-\left\{ 4\left( -46.0 \right)+3\left( 0 \right) \right\}\] |
\[=-1716+184=-1532\,kJ\] |
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