Consider the following reactions in which all the reactants and...

The given reactions are:
$2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5 \times 10^5$
$xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5 \times 10^{-3}$

We need to find the equilibrium constant $K_3$ for the reaction:
$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$

Reversing the first reaction and multiplying it by $\dfrac{1}{2}$, we get:
$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 \rightleftharpoons xy$
The equilibrium constant for this modified reaction is:
$K' = \left(\dfrac{1}{K_1}\right)^{\dfrac{1}{2}} = \dfrac{1}{\sqrt{K_1}}$

Adding this modified reaction to the second reaction:
$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 \rightleftharpoons xy \quad (K')$
$xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad (K_2)$
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$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$

The equilibrium constant $K_3$ for the overall reaction is the product of the equilibrium constants of the added reactions:
$K_3 = K' \times K_2 = \dfrac{K_2}{\sqrt{K_1}}$

Substituting the given values:
$K_3 = \dfrac{5 \times 10^{-3}}{\sqrt{2.5 \times 10^5}}$

$K_3 = \dfrac{5 \times 10^{-3}}{\sqrt{25 \times 10^4}}$

$K_3 = \dfrac{5 \times 10^{-3}}{5 \times 10^2}$

$K_3 = 1.0 \times 10^{-5}$

Answer: $1.0\times 10^{-5}$