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