Question

For the reactionSO2(g) + NO2(g) SO3(g) + NO(g), the equilibrium constant is 18.0 at 1,200ºC. If 2.0 moles of SO2 and 2.0 moles of NO2 are placed in a 20. L container, what concentration of SO3 will be present at equilibrium?

A) 0.081 mol/L
B) 0.019 mol/L
C) 0.11 mol/L
D) 1.00 mol/L
E) 18 mol/L

Answer 1

Answer : The correct option is, (A) 0.081 mol/L

Explanation : Given,

Equilibrium constant = 18.0

Initial concentration of
`SO_2` =
`\farc{Moles}{Volume}=(2.0)/(20)=0.1M`

Initial concentration of
`NO_2` =
`\farc{Moles}{Volume}=(2.0)/(20)=0.1M`

The balanced equilibrium reaction is,

`SO_2(g)+NO_2(g)\rightleftharpoons SO_3(g)+NO(g)`

Initial conc. 0.1 0.1 0 0

At eqm. (0.1-x) (0.1-x) x x

The expression of equilibrium constant for the reaction will be:

`K_c=([SO_3][NO])/([SO_2][NO_2])`

Now put all the values in this expression, we get :

`4.90=((x)* (x))/((0.1-x)* (0.1-x))`

By solving the term 'x' by quadratic equation, we get two value of 'x'.

`x=0.131M\text{ and }0.081M`

From this we conclude that, the value of x = 0.131 at equilibrium can not be more than the initial concentration. So, the value of 'x' which is equal to 0.131 M is not consider.

The concentration of
`SO_3` at equilibrium = x = 0.081 M

Therefore, the concentration of
`SO_3` at equilibrium is, 0.081 M