A) \[180\,kJ\,mo{{l}^{-1}}\]
B) \[360\,kJ\,mo{{l}^{-1}}\]
C) \[213\,kJ\,mo{{l}^{-1}}\]
D) \[425\,kJ\,mo{{l}^{-1}}\]
Correct Answer: D
Solution :
\[\frac{1}{2}{{H}_{2}}+\frac{1}{2}C{{l}_{2}}\to HCl,\,\,\,\Delta H=-90\,KJ\] \[\therefore \Delta H=\frac{1}{2}{{E}_{H-H}}+\frac{1}{2}{{E}_{Cl-Cl}}\] or \[-90=\frac{1}{2}\times 430+\frac{1}{2}\times 240-{{E}_{HCl}}\] \[\therefore {{E}_{H-Cl}}=425\,kJ\,mo{{l}^{-1}}\].
You need to login to perform this action.
You will be redirected in
3 sec
