Q10 / 160 · TS EAMCET 2024 (09 May Shift 2) · Chemistry

At $\mathrm{T}(\mathrm{K}), \mathrm{K}_{\mathrm{c}}$ for the dissociation of $\mathrm{PCl}_5$ is $2 \times 10^{-2} \mathrm{~mol}$ $\mathrm{L}^{-1}$. The number of moles of $\mathrm{PCl}_5$ that must be taken in 1.0 L flask at the same temperature to get 0.2 mol of chlorine at equilibrium is

  • Correct answer
    A
    2.2
  • B
    1.8
  • C
    1.1
  • D
    4.4

Solution

Given $-K_c=2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$ $\begin{array}{lccc} & \mathrm{PCl}_5(\mathrm{g}) & \rightarrow & \mathrm{PCl}_3(\mathrm{g}) & + & \mathrm{Cl}_2(\mathrm{g}) \\ \text{moles at initial} & (\mathrm{x}) & & 0 & & 0 \\ \text{moles at equilibrium} & \mathrm{x}-0.2 & & 0.2 & & 0.2\end{array}$
Equilibrium constant $\mathrm{K}=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}$ $\begin{aligned} & 2 \times 10^{-2}=\frac{[0.2] \times[0.2]}{[x-0.2]} \\ & x=2.2 \text { moles } \end{aligned}$

Hence, the correct answer is option: 'A'.

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