Chemical Thermodynamics Question 384
Question: The standard Gibb’s free energy change, $ \Delta G{}^\circ $ is related to equilibrium constant, $ K_{P} $ as
Options:
A) $ K_{P}=-RT $ In $ \Delta {{G}^{{}^\circ }} $
B) $ K_{P}={{[ \frac{e}{RT} ]}^{\Delta {{G}^{{}^\circ }}}} $
C) $ K_{P}=-\frac{\Delta G}{RT} $
D) $ K_{P}={{e}^{-\Delta {{G}^{{}^\circ }}/RT}} $
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Answer:
Correct Answer: D
Solution:
- $ \Delta G=-RT $ ln $ K_{P} $ or $ K_{P}={{e}^{-\Delta G/RT}} $
