Question

One mole of a compound reacts with one mole of a compound CD according to the equation

$AB\left(g\right)+CD\left(g\right)\rightleftharpoons AD\left(g\right)+CB\left(g\right)$

When equilibrium had been established it was found that $\frac{3}{4}$ mole each of reactants AB and CD had converted to AD and CB. There is no change in volume. The equilibrium constant for the reaction is :

• A. $\frac{6}{15}$
• B. $\frac{1}{9}$
• C. $\frac{19}{5}$
• D. 9

Answer 1

The correct option is D 9

$\begin{array}{c}\text{equilibrium constant =?}\\ \begin{array}{cc}\text{equilibrium constant}k=\frac{\left[AD\right]\left[CB\right]}{\left[AB\right]\left[CD\right]}=\frac{Rf}{Rb}& \\ K& =\frac{\frac{3}{4}\times \frac{3}{4}}{\frac{1}{4}\times \frac{1}{4}}\\ K=9& \end{array}\end{array}$