Question

At a definite temperature, the equilibrium constant for a reaction, $\text{A} + \text{B} \rightleftharpoons 2 \text{C}$ , was found to be 81. Starting with 1 mole A and 1 mole B, the mole fraction of C at equilibrium is :

• A. $\frac{9}{1 1}$
• B. $\frac{1}{1 1}$
• C. $\frac{2}{1 1}$
• D. $\frac{7}{1 1}$

Answer 1

Answer: (A)

$\begin{array}{cccc} A & + & B & \rightleftharpoons & 2 C \\ 1 & & 1 & 0 \\ 1- & x & 1- x & 2 x \\ \therefore & 81=\frac{4 x ^{2}}{(1- x )^{2}} & & \\ \text { or } & \frac{2 x }{(1- x )}=9 & \\ \therefore \quad x & =\frac{9}{11} \\ & A + B \rightleftharpoons 2 C \end{array}$

Thus, $\begin{array}{ccc}\frac{2}{11} & \frac{2}{11} & \frac{18}{11} \\ \text { (Total mole } & \left.=\frac{2}{11}+\frac{2}{11}+\frac{18}{11}=\frac{22}{11}\right)\end{array}$

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