Question

The equilibrium constant KP for the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g) is 1.05 atm at 250ºC. The reaction starts with a mixture of PCl5, PCl3, and Cl2 at pressures 0.177 atm, 0.223 atm, and 0.111 atm, respectively, at 250ºC. When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.

Answer 1

Step-by-step explanation:

It is known that relation between equilibrium pressure and equilibrium partial pressure is as follows.

`K_(p) = \frac{p_{PCl_(3)} * p_{Cl_(2)}}{p_{PCl_(5)}}`

Initial pressures are given for
`PCl_5`,
`PCl_3`, and
`Cl_2` at pressures as 0.177 atm, 0.223 atm, and 0.111 atm, respectively.

Substituting these values into the above formula as follows.

`K_(p) = \frac{p_{PCl_(3)} * p_{Cl_(2)}}{p_{PCl_(5)}}`

=
`(0.223 atm * 0.111 atm)/(0.177 atm)`

= 0.139 atm

or, = 0.140 (approx)

So,
`K_(p)` > 0.14 atm. As calculated value is less than the given value of
`K_(p)`. Therefore, reaction will proceed in the forward reaction.

Therefore, partial pressure of
`PCl_3` and
`Cl_2` needs to increase and
`PCl_5` needs to decrease.