Question

The equilibrium constant $K_c$ of the reaction, $A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right)$
at ${100}^{\circ }C$ is $64$. If a one litre flask containing one mole of $A_2$ is connected to a two litre flask containing one mole of $B_2,$ how many moles of $AB$ will be formed at $373K$?

• A. $1.5$
• B. $1.6$
• C. $2.6$
• D. $2.5$

Answer 1

The correct option is B $1.6$

$\begin{array}{cccccc}& A_2\left(g\right)& +& B_2\left(g\right)& \rightleftharpoons & 2AB\left(g\right)\\ \text{Initial no. of moles}& 1& & 1& & 0\\ \text{No. of moles at equilibrium}& (1-x)& & (1-x)& & 2x\\ \text{(Total volume = 3 litre)}& & & & & \\ \text{Active masses}& \frac{(1-x)}{3}& & \frac{(1-x)}{3}& & \frac{2x}{3}\end{array}$
Applying law of mass action,
$K_c=\frac{[AB{]}^2}{\left[A_2\right]\left[B_2\right]}=\frac{{\left(\frac{2x}{3}\right)}^2}{\left(\frac{1-x}{3}\right)\left(\frac{1-x}{3}\right)}=\frac{4x^2}{(1-x)(1-x)}$
But, $\frac{4x^2}{(1-x)(1-x)}=64\Rightarrow \frac{x^2}{(1-x{)}^2}=16\Rightarrow \frac{x}{1-x}=4\Rightarrow x=4-4x\Rightarrow 5x=4\Rightarrow x=0.8$

No. of moles of $AB=2x=(2\times 0.8)$
$=1.6$