Question

At 250°C the equilibrium constant (Kp) for the reaction below is 1.81. PCl5(g) PCl3(g) + Cl2(g) Sufficient PCl5 is put into a vessel to give an initial pressure of 2.75 atm at 250°C. What will be the final pressure at this temperature after the system has reached equilibrium?

Answer 1

the partial pressure of the
`PCl_5` at equilibrium is 1.25 atm

Step-by-step explanation:

Initial pressure of the
`PCl_5 =2.75 atm`

The value of the equilibrium constant =
`K_p=1.81`

`PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)`

Initially

2.75 atm 0 0

At equilibrium

(2.74- x) x x

The expression for
`K_p` for the given reaction

`K_p=(p_(PCl_3)* p_(Cl_2))/(p_(PCl_5))`

we get:

`1.81=(x* x)/(2.75-x)x=1.50`

Pressure of the
`PCl_5` at equilibrium :

`= (2.75 - p) atm \\= 2.75 - 1.50 atm \\= 1.25 atm`

Hence, the partial pressure of the
`PCl_5` at equilibrium is 1.25 atm