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 : The equilibrium constant for the reaction is, 25.29

Solution : Given,

The given equilibrium reaction is,

Initially                2.24      4.27         0

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

The expression of will be,

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 at equilibrium = 4.27 - 2x = 4.27 - 2(1.9) = 0.47 atm

The partial pressure of at equilibrium = x = 1.9 atm

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

Therefore, the equilibrium constant for the reaction is, 25.29

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