Question

The reversible reaction: 2SO2(g) O2(g) darrow-tn.gif 2SO3(g) has come to equilibrium in a vessel of specific volume at a given temperature. Before the reaction began, the concentrations of the reactants were 0.060 mol/L of SO2 and 0.050 mol/L of O2. After equilibrium is reached, the concentration of SO3 is 0.040 mol/L. What is the equilibrium concentration of O2

Answer 1

`[O_2]_(eq)=0.030M`

Step-by-step explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the mathematical expression for the concentration of oxygen at equilibrium, given the initial one and the change due to the reaction extent:

`[O_2]_(eq)=0.050M-x`

Whereas
`x` can be found considering the equilibrium of SO3:

`[SO_3]_(eq)=2x=0.040M`

Which means:

`x=(0.040M)/(2) =0.020M`

Thus, the equilibrium concentration of oxygen gas turns out:

`[O_2]_(eq)=0.050M-0.020M=0.030M`

Regards!