Equilibrium constants for the following react | vedclass

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(D) The given reactions are:$(1) \ 2 \ H_2O_{(g)} ightleftharpoons 2 \ H_{2(g)} + O_{2(g)}, \ K_1 = 6.4 	imes 10^{-8}$$(2) \ 2 \ CO_{2(g)} ightle

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$5$

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(D) The given reactions are:$(1) \ 2 \ H_2O_{(g)} ightleftharpoons 2 \ H_{2(g)} + O_{2(g)}, \ K_1 = 6.4 	imes 10^{-8}$$(2) \ 2 \ CO_{2(g)} ightleftharpoons 2 \ CO_{(g)} + O_{2(g)}, \ K_2 = 1.6 	imes 10^{-6}$We need the equilibrium constant $(K)$ for the reaction: $H_{2(g)} + CO_{2(g)} ightleftharpoons CO_{(g)} + H_2O_{(g)}$Divide reaction $(1)$ by $2$ and reverse it: $H_{2(g)} + \frac{1}{2} O_{2(g)} ightleftharpoons H_2O_{(g)}$, $K_3 = \frac{1}{\sqrt{K_1}}$Divide reaction $(2)$ by $2$: $CO_{2(g)} ightleftharpoons CO_{(g)} + \frac{1}{2} O_{2(g)}$, $K_4 = \sqrt{K_2}$Adding these two reactions gives the target reaction, so $K = K_3 	imes K_4 = \frac{\sqrt{K_2}}{\sqrt{K_1}}$$K = \sqrt{\frac{1.6 	imes 10^{-6}}{6.4 	imes 10^{-8}}} = \sqrt{\frac{16 	imes 10^{-7}}{6.4 	imes 10^{-8}}} = \sqrt{\frac{160}{6.4}} = \sqrt{25} = 5$

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