Question

Given the following reactions and their equilibrium constants, $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) & \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) & & K=1.6 \\ \mathrm{FeO}(s)+\mathrm{CO}(g) & \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) & & K=0.67 \end{aligned} $$ calculate \(K\) for the reaction $$ \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{H}_{2}(g) $$

Short answer

Based on the step-by-step solution, the equilibrium constant (K) for the reaction: Fe(s) + H₂O(g) ⇄ FeO(s) + H₂(g) is 2.39.

Step-by-step solution

Write the given reactions and equilibrium constants

We are given the following reactions and their equilibrium constants: Reaction 1: \(\mathrm{H}_{2}\mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \quad K_1=1.6\) Reaction 2: \(\mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \quad K_2=0.67\) We want to find the equilibrium constant \(K\) for the reaction: Reaction 3: \(\mathrm{Fe}(s)+\mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{H}_{2}(g)\)

Manipulate the given reactions to match the desired reaction

Observe that if we reverse Reaction 2 and then add Reaction 1 to the reversed Reaction 2, we obtain the desired Reaction 3. Let's perform this operation: Reverse Reaction 2: \(\mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{CO}(g)\) Add Reversing Reaction 2 and Reaction 1: $ \left( \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{CO}(g) \right)+ \left( \mathrm{H}_{2}\mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \right) \newline \Longrightarrow \mathrm{Fe}(s)+\mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons \mathrm{FeO}(s)+\mathrm{H}_{2}(g)$ The resulting operation happens to be our desired Reaction 3.

Calculate the equilibrium constant for the desired reaction

Since we reversed Reaction 2 and added it to Reaction 1, we need to first find the equilibrium constant for the reversal of Reaction 2 (by taking the reciprocal), then multiply the equilibrium constants of both modified reactions to find the equilibrium constant of the desired reaction: Reversal of Reaction 2: \(K'_{2} = \frac{1}{K_2} = \frac{1}{0.67} = 1.49\) Calculate \(K\) for Reaction 3: \(K = K_1 \cdot K'_{2} = 1.6 \cdot 1.49 = 2.39\) So, the equilibrium constant \(K\) for the desired reaction is \(2.39\).