Chemical Thermodynamics Question 196
The standard enthalpy of formation of $ NH_3 $ is $ -46.0kJ/mol $ . If the bond dissociation energy of $ H_2 $ is $ +436kJ/mol $ and that of $ N_2 $ is $ +712kJ/mol $ , the average bond enthalpy of $ N-H $ bonding is:
Options:
A) $ -1102kJ/mol $
B) $ -964kJ/mol $
C) $ +352kJ/mol $
D) $ +1056kJ/mol $
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Answer:
Correct Answer: B
Solution:
- Given $ \frac{1}{2}N_2+\frac{3}{2}H_2\leftrightharpoons NH_3 $
$ \Delta H _{f}=-46.0kJ/mol $
$ H+H\leftrightharpoons H_2;\Delta H _{f}=+436kJ/mol $
$ N+N\leftrightharpoons N_2;\Delta H _{f}=-712,\text{kJ}/mol $
$ \Delta H {f}(NH_3)=\frac{1}{2}\Delta H{N_2}+\frac{3}{2}\Delta H_{H_2}-\Delta H_{N_2H_4}$
$ -46=\frac{1}{2}(-712)+\frac{3}{2}(-436)-\Delta {H _{N-F}} $ On calculation $ \Delta {H _{N-F}}=-964kJ/mol $
BETA
