Question

The gas $A_2$ in the flask allowed to react with gas $B_2$ present in right flask as
$A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right)$ ; $K_c=4at27C$
What us the concentration of AB when equilibrium is established

Answer 1

$A_2+B_2\rightleftharpoons 2AB\left(g\right)\frac{1}{2}\frac{4}{3}0⟵\underset{–}{Att=0}(2-x)(\frac{4}{3}-x)2x⟵\underset{–}{Atequilibrium}$

Now, $K_c=\frac{(2x{)}^2}{(2-x)(\frac{4}{3}-x)}$

$\therefore 4=\frac{4x^2}{\frac{8}{3}-2x-\frac{4}{3}x+x^2}x=\frac{4}{5}$

$therefore$ Concentration of $AB$ at equilibrium $=2x2\times \frac{4}{5}=\frac{8}{5}molLi{t}^{-1}$