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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial pressure of SO2 is 30.6 atm and that of O2 is 13.9 atm. The partial pressure of SO3 is __________ atm.

Respuesta :

Answer : The partial pressure of [tex]SO_3[/tex] is, 67.009 atm

Solution :  Given,

Partial pressure of [tex]SO_2[/tex] at equilibrium = 30.6 atm

Partial pressure of [tex]O_2[/tex] at equilibrium = 13.9 atm

Equilibrium constant = [tex]K_p=0.345[/tex]

The given balanced equilibrium reaction is,

[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

The expression of [tex]K_p[/tex] will be,

[tex]K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}[/tex]

Now put all the values of partial pressure, we get

[tex]0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}[/tex]

[tex]p_{SO_3}=67.009atm[/tex]

Therefore, the partial pressure of [tex]SO_3[/tex] is, 67.009 atm

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