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. $\frac{9}{11}$
• B. $\frac{1}{11}$
• C. $\frac{2}{11}$
• D. $\frac{7}{11}$

Answer 1

Answer: (A)

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

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

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