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 $SO_2$ = $\farc{Moles}{Volume}=\frac{2.0}{20}=0.1M$

Initial concentration of $NO_2$ = $\farc{Moles}{Volume}=\frac{2.0}{20}=0.1M$

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:

$K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}$

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

$4.90=\frac{(x)\times (x)}{(0.1-x)\times (0.1-x)}$

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