Question

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.

Answer 1

Answer : The partial pressure of
`SO_3` is, 67.009 atm

Solution : Given,

Partial pressure of
`SO_2` at equilibrium = 30.6 atm

Partial pressure of
`O_2` at equilibrium = 13.9 atm

Equilibrium constant =
`K_p=0.345`

The given balanced equilibrium reaction is,

`2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)`

The expression of
`K_p` will be,

`K_p=((p_(SO_3))^2)/((p_(SO_2))^2* (p_(O_2)))`

Now put all the values of partial pressure, we get

`0.345=((p_(SO_3))^2)/((30.6)^2* (13.9))`

`p_(SO_3)=67.009atm`

Therefore, the partial pressure of
`SO_3` is, 67.009 atm