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