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}}=(0.800mol)/(1.50L)=1.2M`

and,

`\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=(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=([SO_2]^2[O_2])/([SO_3]^2)`

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

`K_c=((2x)^2* (x))/((1.2-2x)^2)`

`K_c=((2* 0.1)^2* (0.1))/((1.2-2* 0.1)^2)`

`K_c=0.004`

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