Question

How much $PC{l}_5$ must be added to a one-litre vessel at ${250}^{o}C$ in order to obtain a concentration of 0.1 moles of $C{l}_2$?

$K_c$ for $PC{l}_5⟺PC{l}_3+C{l}_2$ is 0.0414 mol/litre.

• A. 3.415 mole
• B. 34.15 mole
• C. 0.03415 mole
• D. 0.3415 mole

Answer 1

Answer: (D)

The equilibrium reaction is $PC{l}_5\left(g\right)\iff PC{l}_3\left(g\right)+C{l}_2\left(g\right)$.

Let a be the initial number of moles of $PC{l}_5$.

Following table lists quantities of various species.

$PC{l}_5⟶$ $PC{l}_3$ + $C{l}_2$

Initial number of moles $a$ 0 0

Moles at equilibrium $a-0.1$ $0.1$ $0.1$

$\because$ The equilibrium number of moles of chlorine is 0.1.

Hence, the equilibrium concentration of $PC{l}_5,PC{l}_3$ and $C{l}_2$ are $a-0.1,0.1$ and $0.1$ respectively.

The expression for the equilibrium constant is $K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}=\frac{0.1\times 0.1}{a-0.1}=0.0414$.

Hence, $a=0.3415$ moles.

Hence, option D is correct.