Answer
The equilibrium constant value for this reaction is equal to 136.
Work Step by Step
1. Write the equilibrium constant expression:
$K_{eq} = \frac{[CH_3OH]}{[CO][H_2]^2}$
2. Use the given values to find the $K_{eq}$ value.
$K_{eq} = \frac{0.185}{(0.105)(0.114)^2} = 136$
