Question

At a certain temperature, 0.660 mol SO 3 is placed in a 4.00 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equilibrium, 0.110 mol O 2 is present. Calculate K c .

Answer 1

`Kc=6.875x10^(-3)`

Step-by-step explanation:

Hello,

In this case, for the given chemical reaction at equilibrium:

`2 SO_3 ( g ) \rightleftharpoons 2 SO_ 2 ( g ) + O_ 2 ( g )`

The initial concentration of sulfur trioxide is:

`[SO_3]_0=(0.660mol)/(4.00L)=0.165M`

Hence, the law of mass action to compute Kc results:

`Kc=([SO_2]^2[O_2])/([SO_3]^2)`

In such a way, in terms of the change
`x` due to the reaction extent, by using the ICE method, it is modified as:

`Kc=((2x)^2*x)/((0.165-2x)^2)`

In that case, as at equilibrium 0.11 moles of oxygen are present,
`x` equals:

`x=[O_2]=(0.110mol)/(4.00L)=0.0275M`

Therefore, the equilibrium constant finally turns out:

`Kc=((2*0.0275)^2*0.0275)/((0.165-2*0.0275)^2) \\\\Kc=6.875x10^(-3)`

Best regards.