A) \[{{10}^{8}}\]
B) \[{{10}^{2222}}\]
C) \[{{10}^{33.33}}\]
D) \[None\]
Correct Answer: C
Solution :
[c] \[{{10}^{3333}}\] \[\Delta \,{{G}^{{}^\circ }}=46.05Kcal/mole\] \[=46.06\times 1000\times 4.184J\,mo{{l}^{-1}}\] \[A{{G}^{{}^\circ }}=-\,RT\ell n\,{{K}_{p}}\] \[=-2.303\text{ }RT\text{ }log\,{{K}_{p}}\] \[46.06\times 1000\times 4.184=\] \[-\,2.303\times 8.314\times 300\text{ }log\] \[log\text{ }Kp=-33.55\] \[or\text{ }{{K}_{p}}={{10}^{-33-55}}\]
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