Calculate the rate of disappearance of N₂O₅ given 2.24 L of O₂ is produced in 30 minutes at STP.

The decomposition of dinitrogen pentoxide (N₂O₅) produces nitrogen dioxide (NO₂) and oxygen (O₂).

$$2 \text{N}_2\text{O}_5(g) \rightarrow 4 \text{NO}_2(g) + \text{O}_2(g)$$

At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 litres.
Given volume of O₂ = 2.24 litres.

$$\text{Moles of O}_2 = \frac{\text{Volume of O}_2}{\text{Molar volume at STP}}$$ $$\text{Moles of O}_2 = \frac{2.24 \text{ L}}{22.4 \text{ L/mol}}$$ $$\text{Moles of O}_2 = 0.1 \text{ mol}$$

The time taken is 30 minutes.

$$\text{Rate of O}_2 \text{ production} = \frac{\text{Moles of O}_2}{\text{Time}}$$ $$\text{Rate of O}_2 \text{ production} = \frac{0.1 \text{ mol}}{30 \text{ min}}$$ $$\text{Rate of O}_2 \text{ production} = 0.00333 \text{ mol/min}$$

From the balanced equation, 2 moles of N₂O₅ decompose for every 1 mole of O₂ produced.

$$\text{Rate of disappearance of N}_2\text{O}_5 = 2 \times \text{Rate of O}_2 \text{ production}$$ $$\text{Rate of disappearance of N}_2\text{O}_5 = 2 \times 0.00333 \text{ mol/min}$$ $$\text{Rate of disappearance of N}_2\text{O}_5 = 0.00666 \text{ mol/min}$$

The rate of disappearance of N₂O₅ is:

$$\text{0.00666 mol/min}$$