Substance | \[{{N}_{2}}\] | \[{{H}_{2}}\] | \[N{{H}_{3}}\], |
\[{{C}_{p}}\] | 3.5R | 3.5R | 4R |
A) \[+44.42\]
B) \[-44.42\]
C) \[+88.85~\]
D) \[-88.85\]
Correct Answer: B
Solution :
Using Kirchhoff?s equations, \[\Delta {{H}_{2}}(1000K)\,=\Delta {{H}_{1}}(300K)\,+\Delta {{C}_{p}}(1000-300)\] \[\Delta {{H}_{2}}(1000K)\,=-123.77\,KJ\,mo{{l}^{-1}}\] \[\Delta {{H}_{1}}(300K)=?\] \[\Delta {{C}_{p}}=2{{C}_{p}}(N{{H}_{3}})-[{{C}_{p}}({{N}_{2}})\,+3{{C}_{p}}({{H}_{2}})]\] \[=-6R\times 8.314\,\times {{10}^{-3}}\,KJ\,mo{{l}^{-1}}\,{{K}^{-1}}\] \[\therefore \,\,-123.\,77=\Delta {{H}_{1}}(300K)\,-6\times 8.314\,\times {{10}^{-3}}\times 700\]\[\therefore \,\,\,\Delta {{H}_{1}}(300K)\,=-88.85\,KJ\] for two moles of \[N{{H}_{3}}\] \[\therefore \,\,\,N{{H}_{3}}\,\Delta {{H}_{1}}(300K)=-\,88.85\,KJ\] \[\therefore \,\,\,\,\Delta {{H}_{1}}=-44.42KJ\,mo{{l}^{-1}}\]You need to login to perform this action.
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