Question

Consider the following reaction:Xe(g) 2 F2(g) ⇌XeF4(g)A reaction mixture initially contains 2.24 atm Xe and 4.27 atm F2. If the equilibrium pressure of Xe is 0.34 atm, find the equilibrium constant (Kp) for the reaction

Answer 1

Answer : The equilibrium constant
`(K_p)` for the reaction is, 25.29

Solution : Given,

The given equilibrium reaction is,

`Xe(g)+2F_2(g)\rightleftharpoons XeF_4(g)`

Initially 2.24 4.27 0

At equilibrium (2.24-x) (4.27-2x) x

The expression of
`K_p` will be,

`K_p=((p_(XeF_4)))/((p_(Xe))(p_(F_2))^2)`

As we are given that,

The partial pressure of Xe at equilibrium = 0.34 atm

That means,

2.24 - x = 0.34

x = 1.9 atm

The partial pressure of
`F_2` at equilibrium = 4.27 - 2x = 4.27 - 2(1.9) = 0.47 atm

The partial pressure of
`XeF_4` at equilibrium = x = 1.9 atm

Now put all the given values in this above expression of
`K_p`, we get:

`K_p=((1.9))/((0.34)(0.47)^2)`

`K_p=25.29`

Therefore, the equilibrium constant
`(K_p)` for the reaction is, 25.29