Chemical Thermodynamics Question 346
Question: The standard enthalpy of formation of $ NH_3 $ is $ -46.0kJ/mol $ . If the enthalpy of formation of $ H_2 $ from its atoms 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 $
Show Answer
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,KJ/mol $
$ \Delta H _{f}(NH_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}}=-964kJ/mol $
