Question

Given the following concentrations and rates, which one corresponds to the rate law for the reaction? \[ [A] = 0.10 \quad [D] = 0.2 \quad \text{Rate} = 0.2 \] \[ [A] = 0.2 \quad [D] = 0.2 \quad \text{Rate} = 0.4 \] \[ [A] = 0.10 \quad [D] = 0.1 \quad \text{Rate} = 0.05 \]

• A. Rate law is \( \text{Rate} = k[A][D] \)
• B. Rate law is \( \text{Rate} = k[A]^2[D] \)
• C. Rate law is \( \text{Rate} = k[A][D]^2 \)
• D. Rate law is \( \text{Rate} = k[A]^2[D]^2 \)

Hint:

When determining the rate law from experimental data, look for changes in concentration and how the rate changes to identify the order of reaction with respect to each reactant.

Answer 1

The Correct Option is A

We are given a table with concentrations of reactants and corresponding reaction rates. To determine the rate law, we need to analyze the data by comparing how the rate changes with the concentrations of \( A \) and \( D \). ### Step 1: Compare Trials 1 and 2 In trial 1, \( [A] = 0.10 \) and \( [D] = 0.2 \) with a rate of 0.2. In trial 2, \( [A] = 0.2 \) and \( [D] = 0.2 \) with a rate of 0.4. Here, \( [D] \) is constant, so we can compare the rates based on \( [A] \) alone. When \( [A] \) doubles from 0.10 to 0.20, the rate also doubles from 0.2 to 0.4. This indicates that the rate is directly proportional to \( [A] \), suggesting a first-order dependence on \( [A] \). ### Step 2: Compare Trials 1 and 3 In trial 1, \( [A] = 0.10 \) and \( [D] = 0.2 \) with a rate of 0.2. In trial 3, \( [A] = 0.10 \) and \( [D] = 0.1 \) with a rate of 0.05. Here, \( [A] \) is constant, so we compare the rates based on \( [D] \). When \( [D] \) is halved from 0.2 to 0.1, the rate is also halved from 0.2 to 0.05. This indicates that the rate is directly proportional to \( [D] \), suggesting a first-order dependence on \( [D] \). ### Step 3: Conclusion Thus, the rate law is: \[ \text{Rate} = k[A][D] \] \[ \boxed{(A) \, \text{Rate law is } \text{Rate} = k[A][D]} \]