Question

$A+B\rightleftharpoons C+D$.If initially the concentration of A and B both equal but at equilibrium,concentration of D will be twice of that of A,then what will be the equilibrium constant of the reaction?

• A. $\frac{4}{9}$
• B. $\frac{9}{4}$
• C. $\frac{1}{9}$
• D. $4$

Answer 1

The correct option is D $4$

Option $D$ is the correct answer which is $4$

Consider the reaction is $A+BC\rightleftharpoons D$

initially $\begin{array}{c}\begin{array}{cc}a& a\end{array}\\ a-x\end{array}$ $\begin{array}{c}\begin{array}{cc}0& 0\end{array}\\ \begin{array}{cc}x& x\end{array}\end{array}$

we have,

$\begin{array}{c}\left[D\right]=2\left[A\right]\\ x=2\left(a-x\right)\\ \Rightarrow x=2a-2x\\ x=\frac{2a}{3}\end{array}$

$\begin{array}{c}{k}_{eq}=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}\\ =\frac{\left(\frac{2a}{3}\right)\left(\frac{2a}{3}\right)}{\left(a-\frac{2a}{3}\right)\left(a-\frac{2a}{3}\right)}=\frac{{\left(\frac{2a}{3}\right)}^2}{{\left(\frac{a}{3}\right)}^2}=2^2\\ =4\end{array}$