Consider the following reactions in which all | vedclass

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(C) The target reaction is: $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$.We are given:$(1)$ $2xy \rightleftharpoons x_2 +

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$1.0 	imes 10^{-5}$

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(C) The target reaction is: $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 ightleftharpoons xyz$.We are given:$(1)$ $2xy ightleftharpoons x_2 + y_2$ with $K_1 = 2.5 	imes 10^5$$(2)$ $xy + \frac{1}{2}z_2 ightleftharpoons xyz$ with $K_2 = 5 	imes 10^{-3}$To obtain the target reaction, we perform: (Reaction $2$) - $\frac{1}{2} 	imes$ (Reaction $1$).This corresponds to: $(xy + \frac{1}{2}z_2) - \frac{1}{2}(2xy) ightleftharpoons xyz - \frac{1}{2}(x_2 + y_2)$.Rearranging gives: $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 ightleftharpoons xyz$.The equilibrium constant $K_3$ is given by $K_3 = \frac{K_2}{(K_1)^{1/2}}$.$K_3 = \frac{5 	imes 10^{-3}}{\sqrt{2.5 	imes 10^5}} = \frac{5 	imes 10^{-3}}{\sqrt{25 	imes 10^4}} = \frac{5 	imes 10^{-3}}{500} = \frac{5 	imes 10^{-3}}{5 	imes 10^2} = 1.0 	imes 10^{-5}$.

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