If we will consider this following equations:
ΔG = -RTlnK eq1
ΔG = ΔH - TΔS eq2
we can rewrite eq1 to solve for K
K = e^(-ΔG/RT) eq3
In this form, we can clearly see the relationship of ΔG and K.
An exothermic reaction has -ΔH thus the value of ΔG would be negative (assume that it is enthalpy driven).
Let me show you a hypothetical example.
if ΔH = -4550kJ and ΔS = .10kJ/K at 298 K
ΔG = -4550 - (298 x 0.1)
ΔG = -4579.8 kJ
K = e^(-ΔG/RT)
K = e^(-(-4579.8 kJ/8.31447*298))
K = 6.34
if ΔH value is changed to -4000 kJ (assume ΔS would be the same) 298 K.
ΔG = -4000 - (298 x 0.1)
ΔG = -4029.8 kJ
K = e^(-ΔG/RT)
K = e^(-(-4029.8/8.31447*298))
K = 5.08
we can summarize it as (enthalpy driven):
ΔH -4550 ----> -4000
ΔG -4579.8 ----> -4029.8
K 6.34 ----> 5.08
Conclusion:
As the heat is decreased in an exothermic reaction, the value of the delta H becomes more positive (-4550 to -4000), the value of K decrease.
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