Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\begin{array}{ccccc}\text{A}& +& \text{B}& \rightleftharpoons & 2\text{C}\\ 1& & 1& & 0\\ 1-\text{x}& & 1-\text{x}& & 2\text{x}\end{array}$

$\therefore 81=\frac{4{\text{x}}^2}{{\left(1-\text{x}\right)}^2}$

or$\frac{2\text{x}}{\left(1-\text{x}\right)}=9$

$\therefore \text{x}=\frac{9}{11}$

Thus,$\begin{array}{ccccc}\text{A}& +& \text{B}& \rightleftharpoons & 2\text{C}\\ \frac{2\mathrm{}}{11}& & \frac{2\mathrm{}}{11}& & \frac{18}{11}\end{array}$

$($Total mole$=\frac{2\mathrm{}}{11}+\frac{2\mathrm{}}{11}+\frac{18}{11}=\frac{\mathrm{}22}{11})$

$\therefore$Mole fraction of$C=\frac{\frac{18}{11}}{\frac{22}{11}}=\frac{9}{11}$