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.

Step-by-step 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) = ([PCl_(3)][Cl_(2)])/([PCl_(5)])\\0.0121 = (x * x)/((0.123 - x))\\x = 0.0330`

Thus, we can conclude that at equilibrium, the partial pressure of
`PCl_(3)` is 0.0330 atm.