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 :
Key Idea \[\Delta {{H}_{reaction}}=\Sigma \] bond energy of reactant \[-\Sigma \] bond energy of product The formation of one mole of HCl can be represented as \[\frac{1}{2}{{H}_{2}}+\frac{1}{2}C{{l}_{2}}\to HCl\] \[\Delta \,{{H}_{reaction}}=\frac{1}{2}\Delta \,{{H}_{H-H}}+\frac{1}{2}\Delta {{H}_{Cl-Cl}}\] \[-\Delta {{H}_{H-Cl}}\] Here, \[\Delta H\] for \[HCl=-9\,k\,J\] \[\Delta {{H}_{H-H}}=430\,\,k\,J\,mo{{l}^{-1}}\] \[\Delta {{H}_{Cl-Cl}}=240\,\,k\,J\,mo{{l}^{-1}}\] therefore, \[-90=\frac{1}{2}\times 430+\frac{1}{2}\times 240-\Delta {{H}_{H-Cl}}\] \[-90=215+120-\Delta {{H}_{H-Cl}}\] \[=415\,kJ\,mo{{l}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
