Question

Studying decomposition of sulfur dioxide. You put 53.811g of sulfur dioxide in 6.00L vessel. After 2.17 minutes, there is a 19.73g of sulfur dioxide left, what is the average rate of the reaction? in M/min

Answer 1

The average rate of the reaction is the rate the change channge of concentration = ∆ Concentration / ∆ Time

Initial Mass of SO₂ = 53.811g

Final Mass of SO₂ left after reaction = 19.73 g

Therefore, Change in mass of SO₂ reacted , ∆m= 53.811g - 19.73g = 34.081g

We know, Molar mass of SO₂ ,M= 64 g.mol^-1

So, Change in number of moles  ∆n= ∆m /M = 34.081g / 64 g.mol^-1 = 0.5325 moles

Given : Volume, V = 6.00 L

Thus ,Change in concentration, ∆Concentration = ∆n / V = 0.5325 moles / 6.00 L = 0.08875 moles/L = 0.08875 M

Given : Iniitial time , t₁ = 0 min , Final time, t₂ = 2.17 min ; Change in time, ∆ t = t₂ - t₁ = 2.17 min - 0 min = 2.17 min

So, Average Rate of reaction =   ∆ Concentration / ∆ Time = 0.08875 M / 2.17 min = 0.0409 M/min