Question

0.60 atm of SO3 and 0.30 atm of SO2are placed in a container and the system is allowed to reach equilibrium. Calculate the pressure of O2(g) at equilibrium.

Answer 1

Complete Question

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Answer:

The pressure is
`[O_2] = 4.8 *10^(-5) \ atm`

Step-by-step explanation:

From the question we are told that

The pressure of
`SO_3` is
`[SO_3 ] = 0.63 \ atm`

The pressure of
`SO_2` is
`[SO_ 2] = 0.30 \ atm`

The equilibrium constant is
`K_p = 1.2 *10^(-5)`

The reaction is

`2SO_3 _((g))` ⇔
`2SO_2_((g)) + O_2 _((g))`

Generally the equilibrium constant is mathematically represented as

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

=>
`[O_2] = (k_p * [SO_3] ^2 )/([SO_2]^2)`

substituting values

`[O_2] = (1.2 *10^(-5) * 0.60 ^2 )/(0.30^2)`

`[O_2] = 4.8 *10^(-5) \ atm`