Question

At a certain temperature the equilibrium constant $K_{C}$ is $0.25$ for the reaction

$A\left(\right.g\left.\right)+B\left(\right.g\left.\right)\rightleftharpoons C\left(\right.g\left.\right)+D\left(\right.g\left.\right)$

If we take $1$ mole of each of four gases in a $10$ litre container, what would be the equilibrium concentration of $A(g)$ ?

• A. 0.331 M
• B. 0.033 M
• C. 0.133 M
• D. 1.33 M

Answer 1

Answer: (C)

The give reaction will go in backward direction



$K_{c}=\frac{\left(\right. A \left.\right) \times \left(\right. B \left.\right)}{\left(\right. C \left.\right) \times \left(\right. D \left.\right)}$

$4=\frac{(1+\alpha)^{2}}{(1-\alpha)^{2}}$

$2=\left(\frac{1 + \alpha }{1 - \alpha }\right)$

$2-2\alpha =1+\alpha$

$\alpha =\frac{1}{3}=0.33$

$\left[\right.A\left]\right.=\frac{1 + \alpha }{10}$

$=\frac{1 + 0.33}{10}=0.133$ M