Answer
See explanation
Work Step by Step
(a) This reaction is the reverse of the known $\mathrm{H}_2 \mathrm{O}$ reaction. The equilibrium constant is then $1 / K_P$
(b) This reaction is the reverse of the known $\mathrm{H}_2 \mathrm{O}$ reaction at a different pressure. Since pressure has no effect on the equilibrium constant,
$$
1 / K_P
$$ (c) This reaction is the same as the known $\mathrm{H}_2 \mathrm{O}$ reaction multiplied by 3 . The quilibirium constant is then $$
K_P^3
$$ (d) This is the same as reaction (c) occurring at a different pressure. Since pressure has no effect on the equilibrium constant,$$
K_P^3
$$
