A) \[{{K}_{P}}=-RT\] In \[\Delta {{G}^{o}}\]
B) \[{{K}_{P}}={{\left[ \frac{e}{RT} \right]}^{\Delta {{G}^{o}}}}\]
C) \[{{K}_{P}}=-\frac{\Delta G}{RT}\]
D) \[{{K}_{P}}={{e}^{-\Delta {{G}^{o}}/RT}}\]
Correct Answer: D
Solution :
Standard free energy, \[\Delta {{G}^{o}}\] and equilibrium constant, \[{{K}_{p}}\] are related as \[\Delta {{G}^{o}}=-RT\] In \[{{K}_{P}}\] In \[{{K}_{P}}=-\frac{\Delta {{G}^{o}}}{RT}\] \[{{K}_{P}}={{e}^{-\Delta {{G}^{o}}/RT}}\]
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