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

Answer 1

The correct option is C $0.133M$

$Q_c=\frac{1\times 1}{1\times 1}=1$
$\because Q_c>K_c$
So reaction will proceed in backward direction.
$$\begin{array}{ccccc}& A_2\left(g\right)+& B_2\left(g\right)\rightleftharpoons & C_2\left(g\right)+& D_2\left(g\right)\\ \text{Conc. eqm}& \frac{1+x}{10}& \frac{1+x}{10}& \frac{1-x}{10}& \frac{1-x}{10}\end{array}$$
$0.25=\frac{{\left(\frac{1-x}{10}\right)}^2}{{\left(\frac{1+x}{10}\right)}^2}\Rightarrow x=0.333$
$\left[A_2\right(g\left)\right]=\frac{1+x}{10}=\frac{1.333}{10}=0.133$