Question

The equilibrium constant of the reaction, $A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right)$ at ${100}^o$ C is 50. If a one-litre flask containing one mole of $A_2$ is connected to a two-litre flask containing two moles of $B_2$ how many motes of $AB$ will be formed at 373 K?

Answer 1

$A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right)$

At $t=0:-$ $\frac{1}{3}$ $\frac{2}{3}$ $0$

At equilibrium :- $(\frac{1}{3}-x)$ $(\frac{2}{3}-x)$ $2x$

Now, $K_C=\frac{4x^2}{(\frac{1}{3}-x)(\frac{2}{3}-x)}=50$

$\therefore \frac{4x^2}{\frac{2}{9}-\frac{x}{3}-\frac{2x}{3}+x^2}=50$

$46x^2-50x+\frac{100}{9}=0$

$x=0.311$

Therefore, moles of $AB$ formed $=2x\times 3=2\times 0.311\times 3=1.866$ $moles.$