Question

a student compares the initial rate of run 1 (10 ml h₂o₂) to the initial rate of run 2 (5 ml h₂o₂), keeping the ki concentration and temperature constant. they observe that the rate of run 1 is approximately double the rate of run 2. what can be concluded about the order of the reaction with respect to h₂o₂? select one. the reaction is second - order in h₂o₂. the reaction is first - order in h₂o₂. the reaction order cannot be determined from this information. the reaction is zero - order in h₂o₂.

Explanation:

Step1: Recall rate law formula

The rate law is \( rate = k[H_2O_2]^n\), where \(n\) is the order of the reaction with respect to \(H_2O_2\).

Step2: Analyze the given data

When the concentration of \(H_2O_2\) doubles (\( [H_2O_2]_1 = 2[H_2O_2]_2\)), the rate doubles (\(rate_1 = 2rate_2\)).
Substitute into the rate law: \(2rate_2= k(2[H_2O_2]_2)^n\) and \(rate_2 = k[H_2O_2]_2^n\).
Divide the first equation by the second: \(\frac{2rate_2}{rate_2}=\frac{k(2[H_2O_2]_2)^n}{k[H_2O_2]_2^n}\), which simplifies to \(2 = 2^n\).

Step3: Solve for \(n\)

Since \(2 = 2^n\), then \(n = 1\).

Answer:

The reaction is first - order in \(H_2O_2\).