Consider the following reactions in which all the reactants and products are present in gaseous state
$$2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5 \times 10^5$$
$$xy + \frac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5 \times 10^{-3}$$
The value of $$K_3$$ for the equilibrium $$\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$$ is :

Target Reaction:

$$\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz \quad \left(K_3\right)$$

1. Modify Reaction 1:

   We need $$\frac{1}{2}x_2$$ and $$\frac{1}{2}y_2$$ as reactants.
   Therefore, we reverse Reaction 1 and multiply the entire equation by $$\frac{1}{2}$$:

   • Reversing Reaction 1 changes the equilibrium constant to its reciprocal:
     $$x_2 + y_2 \rightleftharpoons 2xy \quad \implies \quad K_1' = \frac{1}{K_1}$$
   • Multiplying the reaction by a factor of $$\frac{1}{2}$$ raises the equilibrium constant to the power of $$\frac{1}{2}$$:
     $$\frac{1}{2}x_2 + \frac{1}{2}y_2 \rightleftharpoons xy \quad \implies \quad K_1'' = \left(\frac{1}{K_1}\right)^{1/2} = \frac{1}{\sqrt{K_1}}$$
2. Combine with Reaction 2:

   Now, add this modified reaction to Reaction 2:

   $$\left(\frac{1}{2}x_2 + \frac{1}{2}y_2 \right) + \left(xy + \frac{1}{2}z_2\right) \rightleftharpoons xy + xyz$$

   Canceling the common species $$xy$$ from both sides gives our target reaction:

   $$\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$$
3. Calculate $$K_3$$:

   When two reactions are added, their equilibrium constants are multiplied:

   $$K_3 = K_1'' \times K_2 = \frac{1}{\sqrt{K_1}} \times K_2 = \frac{K_2}{\sqrt{K_1}}$$
4. Substitute the given values:
   $$K_3 = \frac{5 \times 10^{-3}}{\sqrt{2.5 \times 10^5}}$$
   $$K_3 = \frac{5 \times 10^{-3}}{\sqrt{250000}} = \frac{5 \times 10^{-3}}{500}$$
   $$K_3 = \frac{5 \times 10^{-3}}{5 \times 10^2} = 1.0 \times 10^{-5}$$

Conclusion:

The value of $$K_3$$ is $$1.0 \times 10^{-5}$$