Question

SO2 (0.522 mol.L-1) and O2 (0.633 mol.L-1) react and reach equilibrium. Calculate the equilibrium concentrations of the products and reactants given that KC = 5.66 x 10-10 for this reaction: 2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g).

Answer 1

Answer: Hence, the equilibrium concentration of the
`SO_2,O_2\text{ and }SO_3` are 0.521, 0.632 and
`10.88* 10^(-6)M`respectively.

Step-by-step explanation:

The balanced chemical reaction is:

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

At t = 0 0.522 0.633 0

At
`t=t_(eq)` 0.522-x 0.633-2x 2x

The expression for
`K_c` for the given reaction follows:

`K_c=([SO_2]^2)/([SO_2]* [O_2])`

We are given:

`K_c=5.66* 10^(-10)`

Putting values in above equation, we get:

`5.66* 10^(-10)=((2x)^2)/((0.522-x)* (0.633-2x)^2)`

`x=5.44* 10^(-6)M`

Thus equilibrium concentration of
`SO_2` is = (0.522-x )=
`(0.522-5.44* 10^(-6))=0.521`

equilibrium concentration of
`O_2` is = (0.633-2x )=
`(0.633-2* 5.44* 10^(-6))=0.632`

equilibrium concentration of
`SO_3` is = (2x )=
`(2* 5.44* 10^(-6))=10.88* 10^(-6)M`