For 2NO2 + O3 -> N2O5 + O2: Doubling [NO2] doubles the rate, doubling [O3] doubles the rate. What is the rate law?

For the reaction $2\text{NO}_2 + \text{O}_3 \rightarrow \text{N}_2\text{O}_5 + \text{O}_2$, the general rate law can be written as:

$$\text{rate} = k[\text{NO}_2]^x[\text{O}_3]^y$$

where $k$ is the rate constant, $x$ is the order of reaction with respect to $\text{NO}_2$, and $y$ is the order of reaction with respect to $\text{O}_3$.

The problem states that "Doubling $[\text{NO}_2]$ doubles the rate".
If we double $[\text{NO}_2]$ while keeping $[\text{O}_3]$ constant, the rate becomes $2 \times \text{rate}$.

$$2 \times \text{rate} = k(2[\text{NO}_2])^x[\text{O}_3]^y$$

Dividing this by the original rate law:

$$\frac{2 \times \text{rate}}{\text{rate}} = \frac{k(2[\text{NO}_2])^x[\text{O}_3]^y}{k[\text{NO}_2]^x[\text{O}_3]^y}$$ $$2 = (2)^x$$

Therefore, $x = 1$. The reaction is first-order with respect to $\text{NO}_2$.

The problem states that "doubling $[\text{O}_3]$ doubles the rate".
If we double $[\text{O}_3]$ while keeping $[\text{NO}_2]$ constant, the rate becomes $2 \times \text{rate}$.

$$2 \times \text{rate} = k[\text{NO}_2]^x(2[\text{O}_3])^y$$

Dividing this by the original rate law:

$$\frac{2 \times \text{rate}}{\text{rate}} = \frac{k[\text{NO}_2]^x(2[\text{O}_3])^y}{k[\text{NO}_2]^x[\text{O}_3]^y}$$ $$2 = (2)^y$$

Therefore, $y = 1$. The reaction is first-order with respect to $\text{O}_3$.

Substitute the determined orders $x=1$ and $y=1$ into the general rate law:

$$\text{rate} = k[\text{NO}_2]^1[\text{O}_3]^1$$ $$\text{rate} = k[\text{NO}_2][\text{O}_3]$$

The derived rate law is $\text{rate} = k[\text{NO}_2][\text{O}_3]$.
The visible options are:

(a) $\text{rate} = k[\text{NO}_2]$

(b) $\text{rate} = k[\text{NO}_2]^2[\text{O}_3]$
Neither of the visible options matches the derived rate law. Assuming there are other options not shown in the image, the correct rate law based on the experimental data is:

$$\boxed{\text{rate} = k[\text{NO}_2][\text{O}_3]}$$