A) \[20.16\]
B) \[2.303\]
C) \[2.016\]
D) \[13.83\]
Correct Answer: A
Solution :
\[\Delta {{G}^{o}}=-115\times {{10}^{3}}J,\] \[T=298K,\,\,R=8.314J{{K}^{-1}}mo{{l}^{-1}}\] \[-\Delta {{G}^{o}}=2.303RT\,\,{{\log }_{10}}\,{{K}_{P}}\] \[-(-115\times {{10}^{3}})=2.303\times 8.314\times 298{{\log }_{10}}{{K}_{p}}\] \[{{\log }_{10}}\,\,{{K}_{P}}=\frac{115000}{2.303\times 8.314\times 298}\] \[{{\log }_{10}}\,\,{{K}_{P}}=20.16\]
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