| Substance | \[{{N}_{2}}\] | \[{{H}_{2}}\] | \[N{{H}_{3}}\] |
| P/R | 3.5 | 3.5 | 4 |
A) \[-44.42\,\,kJ\,\,mo{{l}^{-1}}\]
B) \[-88.85\,\,kJ\,\,mo{{l}^{-1}}\]
C) \[+44.42\,\,kJ\,\,mo{{l}^{-1}}\]
D) \[+88.85\,\,kJ\,\,mo{{l}^{-1}}\]
Correct Answer: A
Solution :
From Kirchhoffs equation \[\Delta {{H}_{2}}\,(1000\,K)=\Delta {{H}_{1}}\,(300\,K)\] \[+\Delta {{C}_{p}}\,(1000-300)\] Here, \[\Delta {{H}_{2}}(1000\,K)=-123.77\,kJ\,mo{{l}^{-1}}\] \[\Delta {{H}_{1}}(300\,\,K)=?\] \[\Delta {{C}_{p}}=2{{C}_{p}}\,(N{{H}_{3}})-[{{C}_{p}}(N)+3{{C}_{p}}\,\,({{H}_{2}})]\] = - 6 R \[=-6\times 8.314\times {{10}^{-3}}kJ\] \[\therefore \] \[-123.77=\Delta {{H}_{1}}\,(300\,\,K)\] \[-6\times 8.314\times {{10}^{-3}}\times 700\] or \[\Delta {{H}_{1}}\,(300\,K)=-88.85\,kJ\] For two moles of \[N{{H}_{3}}\] \[\therefore \] \[\Delta {{H}_{f}}\,(N{{H}_{3}})=\frac{\Delta {{H}_{1}}(300\,K)}{2}\] \[=-\frac{88.85}{2}\] \[=-44.42\,\,kJ\,\,mo{{l}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
