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

Step 1: Write the general form of the rate law. For the reaction $2NO_2 + O_3 \rightarrow N_2O_5 + O_2$, the general rate law can be written as: $\text{rate} = k[NO_2]^x[O_3]^y$ 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 $\text{O}_3$.

Step 2: Determine the reaction order with respect to $\text{NO}_2$. The problem states that "Doubling $[NO_2]$ doubles the rate". If we double $[NO_2]$ while keeping $[O_3]$ constant, the rate becomes $2 \times \text{rate}$. $2 \times rate = k(2[NO_2])^x[O_3]^y$ Dividing this by the original rate law: $\frac{2 \times rate}{rate} = \frac{k(2[NO_2])^x[O_3]^y}{k[NO_2]^x[O_3]^y}$ $2 = (2)^x$ Therefore, $x = 1$. The reaction is first-order with respect to $\text{NO}_2$.

Step 3: Determine the reaction order with respect to $\text{O}_3$. The problem states that "doubling $[O_3]$ doubles the rate". If we double $[O_3]$ while keeping $[NO_2]$ constant, the rate becomes $2 \times \text{rate}$. $2 \times rate = k[NO_2]^x(2[O_3])^y$ Dividing this by the original rate law: $\frac{2 \times rate}{rate} = \frac{k[NO_2]^x(2[O_3])^y}{k[NO_2]^x[O_3]^y}$ $2 = (2)^y$ Therefore, $y = 1$. The reaction is first-order with respect to $\text{O}_3$.

Step 4: Write the complete rate law. Substitute the determined orders $x=1$ and $y=1$ into the general rate law: $\text{rate} = k[NO_2]^1[O_3]^1$ $\text{rate} = k[NO_2][O_3]$

Step 5: Compare the derived rate law with the given options. The derived rate law is $\text{rate} = k[NO_2][O_3]$. The visible options are: (a) $\text{rate} = k[NO_2]$ (b) $\text{rate} = k[NO_2]^2[O_3]$ 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:

$\boxed{rate = k[NO_2][O_3]}$ 3 done, 2 left today. You're making progress.