From equations 1 and 2,CO_2 rightleftharpoons | vedclass

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(D) The given reactions are:$(i) \, CO_2 ightleftharpoons CO + \frac{1}{2} O_2, \, K_{C_1} = 9.1 	imes 10^{-12}$$(ii) \, H_2O ightleftharpoons H_

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$1.28$

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(D) The given reactions are:$(i) \, CO_2 ightleftharpoons CO + \frac{1}{2} O_2, \, K_{C_1} = 9.1 	imes 10^{-12}$$(ii) \, H_2O ightleftharpoons H_2 + \frac{1}{2} O_2, \, K_{C_2} = 7.1 	imes 10^{-12}$We want to find the equilibrium constant $K$ for the reaction:$CO_2 + H_2 ightleftharpoons CO + H_2O \quad (iii)$To obtain reaction $(iii)$,we reverse reaction $(ii)$ and add it to reaction $(i)$:Reverse of $(ii): \, H_2 + \frac{1}{2} O_2 ightleftharpoons H_2O, \, K_{C_3} = \frac{1}{K_{C_2}} = \frac{1}{7.1 	imes 10^{-12}}$Adding $(i)$ and reversed $(ii)$:$CO_2 + H_2 + \frac{1}{2} O_2 ightleftharpoons CO + \frac{1}{2} O_2 + H_2O$$CO_2 + H_2 ightleftharpoons CO + H_2O$The equilibrium constant $K$ is given by $K = K_{C_1} 	imes \frac{1}{K_{C_2}} = \frac{9.1 	imes 10^{-12}}{7.1 	imes 10^{-12}} = 1.28$.

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