What is the average rate of this reaction?

Question

You are studying the decomposition of sulfur dioxide gas. You start the reaction by putting 52.710g of sulfur dioxide gas in a 6.70L vessel. After 4.96 minutes, there is 19.453g of sulfur dioxide left. What is the average rate of this reaction? ( units of "M/min")

Answer 1

Decomposition of sulfur dioxide:

SO₂ ==> S + O₂

The question is asking for rate of reaction in units of M/min which is moles/L/min (M = moles/liter)

So, let us first find the initial molarity of the SO2:

52.710 g SO2 x 1 mol SO2 / 64.064 g = 0.82277 mols

0.82277 mols / 6.70 L = 0.1228 M initial concentration

Final concentration of SO2 after 4.96 min is:

19.453 g SO2 x 1 mol SO2 / 64.064 g = 0.30365 mols

0.30365 mols / 6.70 L = 0.04532 M final concentration

The change between initial and final concentration is

0.82277 M - 0.04532 M = 0.77745 M

The change in time is 4.96 min

The average rate of reaction is 0.77745 M / 4.96 min = 0.157 M/min (3 sig.figs.)