Question

The equilibrium constant (K p) for the interconversion of PCl 5 and PCl 3 is 0.0121:
PCl5 (g) → PCl3 (g) + Cl2 (g)
A vessel is charged with PCl 5 giving an initial pressure of 0.123 atm and yields PCl 3 and Cl 2. At equilibrium, the partial pressure of PCl 3 is ________ atm.
A) 0.0782.
B) 0.0455.
C) 0.0908.
D) 0.0330.
E) 0.123.

Answer 1

Answer: At equilibrium, the partial pressure of $PCl_{3}$ is 0.0330 atm.

Explanation:

The partial pressure of $PCl_{3}$ is equal to the partial pressure of $Cl_{2}$. Hence, let us assume that x quantity of $PCl_{5}$ is decomposed and gives x quantity of $PCl_{3}$ and x quantity of $Cl_{2}$.

Therefore, at equilibrium the species along with their partial pressures are as follows.

                         $PCl_{5}(g) \rightarrow PCl_{3}(g) + Cl_{2}(g)$

At equilibrium:  0.123-x          x              x

Now, expression for $K_{p}$ of this reaction is as follows.

$K_{p} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}\\0.0121 = \frac{x \times x}{(0.123 - x)}\\x = 0.0330$

Thus, we can conclude that at equilibrium, the partial pressure of $PCl_{3}$ is 0.0330 atm.