Answer : The equilibrium constant [tex](K_p)[/tex] for the reaction is, 25.29
Solution : Given,
The given equilibrium reaction is,
[tex]Xe(g)+2F_2(g)\rightleftharpoons XeF_4(g)[/tex]
Initially 2.24 4.27 0
At equilibrium (2.24-x) (4.27-2x) x
The expression of [tex]K_p[/tex] will be,
[tex]K_p=\frac{(p_{XeF_4})}{(p_{Xe})(p_{F_2})^2}[/tex]
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 [tex]F_2[/tex] at equilibrium = 4.27 - 2x = 4.27 - 2(1.9) = 0.47 atm
The partial pressure of [tex]XeF_4[/tex] at equilibrium = x = 1.9 atm
Now put all the given values in this above expression of [tex]K_p[/tex], we get:
[tex]K_p=\frac{(1.9)}{(0.34)(0.47)^2}[/tex]
[tex]K_p=25.29[/tex]
Therefore, the equilibrium constant [tex](K_p)[/tex] for the reaction is, 25.29
