Calculate Equilibrium Constant for Combined Reaction

To find the equilibrium constant $K_3$ for the target reaction, we manipulate the given equilibrium reactions and their corresponding constants.

Given Reactions:

1. $2 \text{xy} \rightleftharpoons \text{x}_2 + \text{y}_2 \quad \text{with } K_1 = 2.5 \times 10^5$
2. $\text{xy} + \frac{1}{2} \text{z}_2 \rightleftharpoons \text{xyz} \quad \text{with } K_2 = 5 \times 10^{-3}$

Target Reaction: $\frac{1}{2} \text{x}_2 + \frac{1}{2} \text{y}_2 + \frac{1}{2} \text{z}_2 \rightleftharpoons \text{xyz} \quad \text{with } K_3 = ?$

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Step-by-Step Derivation:

1. Modify Reaction (1): Reverse reaction (1) and multiply all the stoichiometric coefficients by $\frac{1}{2}$: $\frac{1}{2} \text{x}_2 + \frac{1}{2} \text{y}_2 \rightleftharpoons \text{xy}$

   The equilibrium constant for this modified reaction, $K_1'$, is given by: $K_1' = \left(\frac{1}{K_1}\right)^{\frac{1}{2}} = \frac{1}{\sqrt{2.5 \times 10^5}} = \frac{1}{\sqrt{25 \times 10^4}} = \frac{1}{5 \times 10^2} = 2 \times 10^{-3}$

2. Combine with Reaction (2): Adding the modified reaction (1) to reaction (2): $\left(\frac{1}{2} \text{x}_2 + \frac{1}{2} \text{y}_2 \right) + \left(\text{xy} + \frac{1}{2} \text{z}_2 \right) \rightleftharpoons \text{xy} + \text{xyz}$

   Canceling the common species $\text{xy}$ from both sides yields the target reaction: $\frac{1}{2} \text{x}_2 + \frac{1}{2} \text{y}_2 + \frac{1}{2} \text{z}_2 \rightleftharpoons \text{xyz}$

3. Calculate $K_3$: Since the target reaction is the sum of these two individual steps, its equilibrium constant $K_3$ is the product of their respective equilibrium constants: $K_3 = K_1' \times K_2$ $K_3 = (2 \times 10^{-3}) \times (5 \times 10^{-3}) = 10 \times 10^{-6} = 1.0 \times 10^{-5}$

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Correct Answer: Option C ($1.0 \times 10^{-5}$)