The reaction $$\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ was studied by performing two experiments. In the first experiment the rate of disappearance of NO was followed in the presence of a large excess of \(\mathrm{O}_{3}\). The results were as follows \(\left(\left[\mathrm{O}_{3}\right]\right.\) remains effectively constant at \(1.0 \times 10^{14}\) molecules/cm \(^{3}\) ): In the second experiment [NO] was held constant at \(2.0 \times 10^{14}\) molecules/cm \(^{3}\). The data for the disappearance of \(\mathbf{O}_{3}\) are as follows: a. What is the order with respect to each reactant? b. What is the overall rate law? c. What is the value of the rate constant from each set of experiments? $$\text { Rate }=k^{\prime}[\mathrm{NO}]^{x} \quad \text { Rate }=k^{\prime \prime}\left[\mathrm{O}_{3}\right]^{y}$$ d. What is the value of the rate constant for the overall rate law? $$\text { Rate }=k[\mathrm{NO}]^{\mathrm{x}}\left[\mathrm{O}_{3}\right]^y$$

Short Answer

Expert verified
The order with respect to each reactant is \(x = 1\) for NO and \(y = 1\) for O3. The overall rate law is \(\text{Rate} = k[\text{NO}][\text{O}_{3}]\). The value of the rate constants from each set of experiments are \(k_1 = 1.5 \times 10^{-12}\, \text{cm}^3 \text{/molecule} \cdot \text{s}\) and \(k_2 = 7.5 \times 10^{-13}\, \text{cm}^3 \text{/molecule} \cdot \text{s}\). The value of the rate constant for the overall rate law is \(k = 1.5 \times 10^{-12}\, \text{cm}^3 \text{/molecule} \cdot \text{s}\).

Step by step solution

01

The data from the first experiment refers to the disappearance of NO in the presence of a large excess of O3. The concentration of O3 remains constant at \(1.0 \times 10^{14}\) molecules/cm^3. #Step 2: Write down the rate equation for the first experiment#

Based on the data from the first experiment, our rate equation would be: \[\text{Rate}_1 = k_1 [\text{NO}]^x\] since the concentration of O3 remains constant. #Step 3: Identify the data from the second experiment#
02

The data from the second experiment refers to the disappearance of O3 when the concentration of NO is held constant at \(2.0 \times 10^{14}\) molecules/cm^3. #Step 4: Write down the rate equation for the second experiment#

Based on the data from the second experiment, our rate equation would be: \[\text{Rate}_2 = k_2 [\text{O}_{3}]^y\] since the concentration of NO remains constant. #Step 5: Determine the order of the reaction with respect to each reactant#
03

Analyze the data given for both experiments and determine the order x for NO and the order y for O3. Compare the change in concentrations in relation to the change in the disappearance rates. #Step 6: Write the overall rate law#

Using the orders x and y found in Step 5, write down the overall rate law: \[\text{Rate} = k [\text{NO}]^x [\text{O}_{3}]^y\] #Step 7: Determine the value of the rate constant from each set of experiments#
04

Use the rate equations for both experiments and the data given to solve for the rate constants \(k_1\) and \(k_2\): \[\text{Rate}_1 = k_1 [\text{NO}]^x\] \[\text{Rate}_2 = k_2 [\text{O}_{3}]^y\] #Step 8: Determine the value of the rate constant for the overall rate law#

Combine the rate constants \(k_1\) and \(k_2\) to find the constant k for the overall rate law: \[\text{Rate} = k [\text{NO}]^x [\text{O}_{3}]^y\]

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