H2(g)+I2(g)↔2HI(g)
initial concentration
H2=1mMI2=2mM
HI=0
at equilibrium let's assume
H2=(1−x)mMI2=(2−x)mMHI=2x
we are given equilibrium concentration of HI is 1.87mM
2x=1.87x=0.935
at equilibrium,
H2 =1−0.935=0.065mM
I2=2−0.935=1.065mM
HI=2×0.935=1.87mM
Kc of the reaction =H2I2[HI]2=1.065×0.0651.87×1.87=50.5
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