Chemistry - Chemical Equilibrium Question with Solution | TestHub

Let the initial concentration be 'a'

$\begin{array}{cccccccc}& \text{A}& +& \text{B}& \to & \text{C}& +& \text{D}\\ \text{Initial}& \text{a}& & \text{a}& & 0& & 0\\ \text{at equilibrium}& \text{a}-\text{x}& & \text{a}-\text{x}& & x& & x\end{array}$

$\begin{array}{l}\text{given}\text{x}=2\left(\text{a}-\text{x}\right)=2\text{a}-2\text{x}\\ \text{3x}=\left(\text{2a}\right)\\ \text{x}=\frac{2}{3}\text{a}\end{array}$

at eq.$\begin{array}{ccccc}\Rightarrow & \frac{\text{a}}{3}& \frac{\text{a}}{3}& \frac{2\text{a}}{3}& \frac{2\text{a}}{3}\end{array}$

${\text{K}}_{\text{eq}}=\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}=\frac{\left(\frac{2\text{a}}{3}\right)\left(\frac{\text{2a}}{3}\right)}{\left(\frac{\text{a}}{3}\right)\left(\frac{\text{a}}{3}\right)}=4$