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.