For the given reaction: $\text{N}_2\text{O}_5(g) \rightarrow 2\text{NO}_2(g) + \frac{1}{2}\text{O}_2(g)$ The rate of the reaction is related to the rates of disappearance and formation of species as follows: $\text{Rate} = -\frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{1}{2} \frac{d[\text{NO}_2]}{dt} = \frac{1}{1/2} \frac{d[\text{O}_2]}{dt} = 2 \frac{d[\text{O}_2]}{dt}$ Given the rate of disappearance of $\text{N}_2\text{O}_5$: $-\frac{d[\text{N}_2\text{O}_5]}{dt} = 6.25 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}$

Rate of formation of $\text{NO}_2$: $\frac{d[\text{NO}_2]}{dt} = 2 \times \left( -\frac{d[\text{N}_2\text{O}_5]}{dt} \right) = 2 \times 6.25 \times 10^{-3} = 12.5 \times 10^{-3} = 1.25 \times 10^{-2} \text{ mol L}^{-1}\text{s}^{-1}$

Rate of formation of $\text{O}_2$: $\frac{d[\text{O}_2]}{dt} = \frac{1}{2} \times \left( -\frac{d[\text{N}_2\text{O}_5]}{dt} \right) = \frac{1}{2} \times 6.25 \times 10^{-3} = 3.125 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}$