The partial pressures of CH_3OH,CO and H_2 in | vedclass

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(D) The given reaction is $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$.For this reaction,the equilibrium constant $K_{P1}$ is given by:$K_{P1} = \fr

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$5 	imes 10^{-3} \ atm^2$

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(D) The given reaction is $CO(g) + 2H_2(g) ightleftharpoons CH_3OH(g)$.For this reaction,the equilibrium constant $K_{P1}$ is given by:$K_{P1} = \frac{P_{CH_3OH}}{P_{CO} \cdot P_{H_2}^2} = \frac{2.0}{1.0 	imes (0.1)^2} = \frac{2.0}{0.01} = 200 \ atm^{-2}$.The reaction for the decomposition of $CH_3OH$ is the reverse of the given reaction:$CH_3OH(g) ightleftharpoons CO(g) + 2H_2(g)$.The equilibrium constant $K_P$ for this reverse reaction is the reciprocal of $K_{P1}$:$K_P = \frac{1}{K_{P1}} = \frac{1}{200} = 0.005 = 5 	imes 10^{-3} \ atm^2$.

The partial pressures of $CH_3OH$,$CO$ and $H_2$ in the equilibrium mixture for the reaction $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$ at $427 \ ^oC$ are $2.0 \ atm$,$1.0 \ atm$ and $0.1 \ atm$ respectively. The value of $K_P$ for the decomposition of $CH_3OH$ to $CO$ and $H_2$ is:

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