A) \[100\text{ }kJ\,mo{{l}^{-1}}\]
B) \[150\text{ }kJ\,mo{{l}^{-1}}\]
C) \[163\text{ }kJ\,mo{{l}^{-1}}\]
D) \[2303\text{ }kJ\,mo{{l}^{-1}}\]
Correct Answer: C
Solution :
\[{{\Delta }_{r}}G{}^\circ =2.303\,\,\log {{K}_{p}}(R=8.314\,J{{K}^{-1}}mo{{l}^{-1}})\] \[{{\Delta }_{r}}G{}^\circ =-\,2.303\times 8.314\,J{{K}^{-1}}mo{{l}^{-1}}\times 298\,K\] \[\times \log 2.47\times {{10}^{-29}}\] \[=163000\text{ }J\,mo{{l}^{-1}}=163\text{ }kJ\,mo{{l}^{-1}}\]
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