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The equilibrium constant of the reaction $H_{2} (g) + {CO}_{2} (g) ⇌ H_{2} O (g) + CO (g)$ is $1.60.$ If $0.4 mol$ of each of $H_{2}$ and ${CO}_{2}$ are enclosed in a 1.0 litre vessel, the equilibrium concentration of $H_{2} O (g)$ will be

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0.26 M

0.24 M

0.22 M

0.20 M

answer is C.

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Detailed Solution

$H_{2} (g) + {CO}_{2} (g) ⇌ H_{2} O (g) + CO (g)$

$0.4 mol - x 0.4 mol - x x x$

$\frac{x^{2}}{(0.4 mol - x)^{2}} = 1.6$ or $x = 0.22 m o l$

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