A) -1102 kJ/mol
B) - 964 kJ/mol
C) + 352 kJ/mol
D) + 1056 kJ/mol
Correct Answer: B
Solution :
Given\[\frac{1}{2}{{N}_{2}}+\frac{3}{2}{{H}_{2}}N{{H}_{3}};\] \[\Delta {{H}_{f}}=-46.0kJ\text{/}mol\] \[H+H{{H}_{2}};\Delta {{H}_{f}}=-436kJ/mol\] \[N+N{{N}_{2}};\Delta {{H}_{f}}=-712kJ/mol\] \[\Delta {{H}_{f}}\left( N{{H}_{3}} \right)=\frac{1}{2}\Delta {{H}_{N-N}}+\frac{3}{2}\Delta {{H}_{H-H}}-\Delta {{H}_{N-F}}\] \[-46=\frac{1}{2}\left( -712 \right)+\frac{3}{2}\left( -436 \right)-\Delta {{H}_{N-F}}\] On calculation\[\Delta {{H}_{N-F}}=-964kJ/mol\]You need to login to perform this action.
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