Question

Write expressions for \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) b. \(\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)\) c. \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) d. \(4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)\)

Short answer

The expressions for \(K_p\) for the given reactions are as follows: a. \(K_{p} = \frac{1}{[\mathrm{O}_{2}]^{\frac{3}{2}}}\) b. \(K_{p} = \frac{1}{[\mathrm{CO}_{2}]}\) c. \(K_{p} = \frac{[\mathrm{CO}][\mathrm{H}_{2}]}{[\mathrm{H}_{2}\mathrm{O}]}\) d. \(K_{p} = \frac{[\mathrm{O}_{2}]^3}{[\mathrm{H}_{2}\mathrm{O}]^2}\)

Step-by-step solution

Reaction a

\(2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) For reaction a, we have the following equilibrium: \(2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) Since the Kp expression includes only the partial pressures of gases, we will not include Fe(s) and Fe₂O₃(s) in the expression. Thus, the expression for Kp for reaction a is: \(K_{p} = \frac{1}{[\mathrm{O}_{2}]^{\frac{3}{2}}}\)

Reaction b

\(\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)\) For reaction b, we have the following equilibrium: \(\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)\) Since the Kp expression includes only the partial pressures of gases, we will not include MgO(s) and MgCO₃(s) in the expression. Thus, the expression for Kp for reaction b is: \(K_{p} = \frac{1}{[\mathrm{CO}_{2}]}\)

Reaction c

\(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) For reaction c, we have the following equilibrium: \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) Since the Kp expression includes only the partial pressures of gases, we will not include C(s) in the expression. Thus, the expression for Kp for reaction c is: \(K_{p} = \frac{[\mathrm{CO}][\mathrm{H}_{2}]}{[\mathrm{H}_{2}\mathrm{O}]}\)

Reaction d

\(4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)\) For reaction d, we have the following equilibrium: \(4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)\) Since the Kp expression includes only the partial pressures of gases, we will not include KO₂(s) and KOH(s) in the expression. Thus, the expression for Kp for reaction d is: \(K_{p} = \frac{[\mathrm{O}_{2}]^3}{[\mathrm{H}_{2}\mathrm{O}]^2}\)