The reaction has an equilibrium constant of Kp=2.26x10^4 at 298 K.

Question

CO(g) + 2H_2(g) <-> CH_3OH(g)

Calculate Kp for the reactions and predict whether reactants or products will be favored at equilibrium.
A. CH_3OH(g) <-> CO(g) + 2H_2(g)
B. 1/2CO(g) + H_2(g) <-> 1/2CH_3OH(g)

Answer 1

A. The reaction for A is just the reverse of the original; therefore, the new Kp(that is K'p) for the new rxn will be 1/Kp

B. B is just 1/2 the original; therefore, the new Kp (that's K"p) will be sqrt(Kp)

For any reaction of A ==> B, then
Kp = pB/pA. So when Kp is a large number that means pB is large and pA is small so B (products) are favored.
When Kp is a small number that mans pB is small and pA is large so A (the reactants) are favored.