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 $16.$ Initially equal moles of $A_2\&B_2$ are taken in $2L$ container. Then find mole % of $A_2$ in equilibrium mixture.

Answer 1

The given reaction is :-

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

Initial moles : $M$ $M$ $0$ (let)

At eqm : $M-m$ $M-m$ $2m$ (let)

Now, At equation, $\left[A_2\right]=\frac{M-m}{2};\left[B_2\right]=\frac{M-m}{2};\left[AB\right]=\frac{2m}{2}$

Now, $K_C=\frac{{\left[AB\right]}^2}{\left[A_2\right]\left[B_2\right]}=\frac{m^2}{\frac{(M-m)}{2}.\frac{(M-m)}{2}}$

$\Rightarrow 16=\frac{{4m}^2}{{(M-m)}^2}$

$\Rightarrow \frac{2m}{M-m}=4or\frac{2m}{M-m}=-4$

$\Rightarrow m=2M-2morm=-2M+2m$

$\Rightarrow 2M=3mor\Rightarrow -m=-2m$

$\Rightarrow \frac{m}{M}=\frac{2}{3}=0.66or\Rightarrow \frac{m}{M}=0.5$

At equation, Total no. of moles $=M-m+M-m+2m=2M$

moles of $A_2$ at equation $=M-m$

mole % of $A_2$ at equation $=\frac{M-m}{2M}\times 100$% $=(\frac{1}{2}-\frac{m}{2M})\times 100$%

$=(\frac{1}{2}-\frac{1}{4})\times 100$% $=25$%

or $(\frac{1}{2}-\frac{1}{3})\times 100=\frac{1}{6}\times 100$% $=17$%