Question

At a certain temperature, 0.800 mol SO 3 is placed in a 1.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equilibrium, 0.150 mol O 2 is present. Calculate K c .

Answer 1

Answer : The value of equilibrium constant (K) is, 0.004

Explanation :

First we have to calculate the concentration of $SO_3\text{ and }O_2$

$\text{Concentration of }SO_3=\frac{\text{Moles of }SO_3}{\text{Volume of solution}}=\frac{0.800mol}{1.50L}=1.2M$

and,

$\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{0.150mol}{1.50L}=0.1M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                       $2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)$

Initial conc.       1.2             0               0

At eqm.          (1.2-2x)        2x               x

As we are given:

Concentration of $O_2$ at equilibrium = x = 0.1 M

The expression for equilibrium constant is:

$K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}$

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

$K_c=\frac{(2x)^2\times (x)}{(1.2-2x)^2}$

$K_c=\frac{(2\times 0.1)^2\times (0.1)}{(1.2-2\times 0.1)^2}$

$K_c=0.004$

Thus, the value of equilibrium constant (K) is, 0.004