Question

At a certain temperature, 3.67 mol SO 2 and 1.83 mol O 2 are placed in a container. 2 SO 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 SO 3 ( g ) At equilibrium, there is 1.92 mol SO 3 present. Determine the number of moles of SO 2 and O 2 that are present when the reaction is at equilibrium.

Answer 1

Answer : The number of moles of $SO_2$ and $O_2$ at equilibrium is, 1.75 mol and 0.87 mol respectively.

Explanation :

The given chemical reaction is:

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

Initial mol.       3.67         1.83          0

At eqm.        (3.67-2x)   (1.83-x)       2x

As we are given:

Number of moles of $SO_3$ at equilibrium = 1.92

That means,

2x = 1.92

x = 0.96 mol

Number of moles of $SO_2$ at equilibrium = (3.67-2x) = [3.67-2(0.96)] = 1.75 mol

Number of moles of $O_2$ at equilibrium = (1.83-x) = (1.83-0.96) = 0.87 mol

Thus, the number of moles of $SO_2$ and $O_2$ at equilibrium is, 1.75 mol and 0.87 mol respectively.