Question

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

Answer 1

$A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right)K=50$

Above situation is initial picture and allowed to react

Total Volume $=1+2=3$ lit.

$A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right)K=50$

At $1$ $2$ $0$

$t=0$

At $1-\alpha$ $2(1-\alpha )$ $2\alpha$

$t=t_{eq}$

concentration $=\frac{mass}{volinlit}$

At

Eq'b $\frac{1-\alpha }{3}$ $\frac{2(1-\alpha )}{3}$ $2\alpha /3$

conc. of of of

$A_2$ $B_2$ $AB$

$K=\frac{{\left(2\alpha /3\right)}^2}{\frac{2(1-\alpha )}{3}\times \frac{(1-\alpha )}{3}}=50$

$=\frac{{2\alpha }^2}{{(1-\alpha )}^2}=50=\frac{\alpha }{1-\alpha }=5$

$\alpha =5/6$

Hence $2\alpha =5/3$ moles of $AB$ is formed.