Question

If the equilibrium constant for the reaction \(\mathrm{H}_{2}+\) \(\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}\) is \(K,\) the equilibrium constant for the reaction \(2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{H}_{2}+\mathrm{O}_{2}\) at the same temperature is \((a) 1 / K\) \((b) 1 /(2 K)\) \((c) 2 K\) \((d) K^{2}\) \((e) 1 / K^{2}\)

Short answer

Answer: (d) K^2

Step-by-step solution

Write down the given equilibrium constants

We are given the following equilibrium constants: Reaction 1: \(H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O\) with equilibrium constant \(K\). We need to find the equilibrium constant for this reaction: Reaction 2: \(2H_2O \rightleftharpoons 2H_2 + O_2\)

Multiply the first reaction to obtain the second reaction

In order to obtain the second reaction, we can multiply the first reaction by 2: \((H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O) \times 2 = (2H_2 + O_2 \rightleftharpoons 2H_2O)\)

Derive the new equilibrium constant

When we multiply a reaction by a factor (in this case 2), the new equilibrium constant is found by raising the original equilibrium constant to the power of that factor: New equilibrium constant for Reaction 2: \(K^2\)

Write down the final answer

Therefore, the equilibrium constant for the reaction \(2H_2O \rightleftharpoons 2H_2 + O_2\) at the same temperature as the first reaction is \(K^2\). The correct answer is \((d) K^{2}\).