At a certain temperature the equilibrium constant $K_{C}$ is 0.25 for the reaction
$A (g) + B (g) ⇌ C (g) + D (g)$
If we take 1 mole of each of four gases in a 10 litre container, what would be the equilibrium concentration of A(g)?

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TestHub · Accepted Answer

The give reaction will go in backward direction

$K_{c} = \frac{(A) \times (B)}{(C) \times (D)}$
$4 = \frac{(1 + α)^{2}}{(1 - α)^{2}}$
$2 = (\frac{1 + α}{1 - α})$
$2 - 2 α = 1 + α$
$α = \frac{1}{3} = 0.33$
$[A] = \frac{1 + α}{10}$
$= \frac{1 + 0.33}{10} = 0.133$ M

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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