Question

At a certain temperature the equilibrium constant $K_c$ is 0.25 for the reaction $A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons C_2\left(g\right)+D_2\left(g\right)$
If we take 1 mole of each of the four gases in a 10 litre container, what would be equilibrium concentration of $A_2\left(g\right).$

• A. 0.0
• B. 0.13
• C. 0.25
• D. 1

Answer 1

The correct option is B 0.13

Reaction Quotient, $Q=\frac{1\times 1}{1\times 1}\Rightarrow 1$
$\because Q>K_c$ so reaction will proceed in backward direction
$\begin{array}{ccccccc}& A_2\left(g\right)& +& B_2\left(g\right)& \rightleftharpoons & C_2\left(g\right)& +D_2\left(g\right)\\ \text{concentration at equilibrium}& \frac{1.0+x}{10}& & \frac{1.0+x}{10}& & \frac{1.0-x}{10}& \frac{1.0-x}{10}\end{array}$
$0.25=\frac{{\left(\frac{1-x}{10}\right)}^2}{{\left(\frac{1+x}{10}\right)}^2}$
$\Rightarrow 0.5=\frac{1-x}{1+x}$
$\Rightarrow 0.5+0.5x=1-x$
$\Rightarrow 1.5x=0.5$
$\Rightarrow x=0.333$
$\left[A_2\right(g\left)\right]=\frac{1+x}{10}$
$\left[A_2\right(g\left)\right]=\frac{1.333}{10}=0.13$