Q12 / 160 · TS EAMCET 2018 (07 May Shift 1) · Chemistry
$\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{PCl}_2$
If the equilibrium constant $\left(K_C\right)$ for the above reaction at $500 \mathrm{~K}$ is 1.79 and the equilibrium concentration of $\mathrm{PCl}_5$ and $\mathrm{PCl}_3$ are $1.41 \mathrm{M}$ and $1.59 \mathrm{M}$, respectively, then the concentration of $\mathrm{Cl}_2$ is approximately.
- A$1.26 \mathrm{M}$
- B$3.59 \mathrm{M}$
- C$0.59 \mathrm{M}$
- Correct answerD$1.59 \mathrm{M}$
Solution
Equilibrium constant $\left(K_C\right)=\frac{\left[\mathrm{PCI}_3\right]\left[\mathrm{CI}_2\right]}{\left[\mathrm{PCI}_5\right]}$
$\begin{aligned}
1.79 & =\frac{1.59 \times\left[\mathrm{CI}_2\right]}{1.41} \\
{\left[\mathrm{CI}_2\right] } & =\frac{1.79 \times 1.41}{1.59}=1.587 \approx 1.59 \mathrm{~m}
\end{aligned}$
Hence, the correct answer is option: 'D'.