Question

The gas ${\text{A}}_2$ in the left flask allowed to react with gas ${\text{B}}_2$ present in right flask as


${\text{A}}_2\left(g\right)+{\text{B}}_2\left(g\right)\rightleftharpoons 2{\text{AB(g) ; K}}_C=4\text{at}27{}^{\circ }C$

What is the concentration of $AB$ when equilibrium is established?

• A. $1.33\text{M}$
• B. $2.66\text{M}$
• C. $0.66\text{M}$
• D. $0.33\text{M}$

Answer 1

The correct option is C $0.66\text{M}$

For the given reaction,
$\begin{array}{cccccc}& A_2& +& B_2& \rightleftharpoons & 2AB\\ t=0& 2& & 4& & 0\\ \text{at eqlibrium}& 2-x& & 4-x& & 2x\end{array}$

So equation for equilibrium constant gives,
$4=\frac{4x^2}{(2-x)\times (4-x)}$
$\Rightarrow x^2=8-6x+x^2$
$\Rightarrow x=\frac{8}{6}=1.33\text{mol}$
So at equilibrium,
number of moles of $AB=2x$
Total volume of system = $4\text{L}$
Therefore, concentration of $AB$ at equilibrium $=\left[\text{AB}\right]=\frac{2x}{4}=\frac{x}{2}=0.665\text{M}$