A) \[11.64\times {{10}^{30}}\]
B) \[1.64\times {{10}^{37}}\]
C) \[0.1064\times {{10}^{35}}\]
D) \[21.64\times {{10}^{39}}\]
Correct Answer: B
Solution :
\[\Delta G{}^\circ =-RT\] In\[{{K}_{c}}\] \[-212300J=-(\text{ }8.314\text{ }J{{K}^{-1}}mo{{l}^{-1}})(298K)\]In\[{{K}_{c}}\] ln \[{{K}_{c}}=\frac{212300}{8.314\times 293}=85.69\] \[{{K}_{c}}=1.64\times {{10}^{37}}\]
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