Question:
For a reaction $\mathrm{X}+\mathrm{Y} \rightleftharpoons 2 \mathrm{Z}, 1.0 \mathrm{~mol}$ of $\mathrm{X}, 1.5$ mol of $\mathrm{Y}$ and $0.5 \mathrm{~mol}$ of $\mathrm{Z}$ were taken in a 1 $\mathrm{L}$ vessel and allowed to react. At equilibrium, the concentration of $\mathrm{Z}$ was $1.0 \mathrm{~mol} \mathrm{~L}^{-1}$. The equilibrium constant of the reaction is
$\frac{x}{15}$. The value of $x$ is
Solution:
$\begin{array}{llll} & \mathrm{X} & +\mathrm{Y} & =2 \mathrm{Z} \\ \mathrm{t}=0 & 1 & & 1.5 & 0.5 \\ \text { At eq. } & 0.75 & 1.25 & 1\end{array}$
$\mathrm{K}_{\text {eq. }}=\frac{1^{2}}{\frac{3}{4} \times \frac{5}{4}}=\frac{16}{15}$