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The transfer of an oxygen atom from \(\mathrm{NO}_{2}\) to CO has been studied at \(540 \mathrm{~K}\) : \(\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{NO}(\mathrm{g})\) These data were collected: \begin{tabular}{ccc} \hline Initial Rate & \multicolumn{2}{c} { Initial Concentration (mol/L) } \\ \cline { 2 - 3 } (mol \(\mathrm{L}^{-1} \mathrm{~h}^{-1}\) ) & {\(\left[\mathrm{NO}_{2}\right]\)} \\ \hline \(5.1 \times 10^{-4}\) & \(0.35 \times 10^{-4}\) & \(3.4 \times 10^{-8}\) \\ \(5.1 \times 10^{-4}\) & \(0.70 \times 10^{-4}\) & \(1.7 \times 10^{-8}\) \\ \(5.1 \times 10^{-4}\) & \(0.18 \times 10^{-4}\) & \(6.8 \times 10^{-8}\) \\ \(1.0 \times 10^{-3}\) & \(0.35 \times 10^{-4}\) & \(6.8 \times 10^{-8}\) \\ \(1.5 \times 10^{-3}\) & \(0.35 \times 10^{-4}\) & \(10.2 \times 10^{-8}\) \\ \hline \end{tabular} (a) Write the rate law. (b) Determine the reaction order with respect to each reactant (c) Calculate the rate constant and express it in appropriate units.

Step by step solution

01

Identify The Rate Law Form

The general form of the rate law for this reaction can be expressed as:\[\text{Rate} = k [\text{CO}]^m [\text{NO}_2]^n\]where \(k\) is the rate constant, \(m\) is the order of the reaction with respect to \(\text{CO}\), and \(n\) is the order of the reaction with respect to \(\text{NO}_2\).

02

Analyze Initial Rate Data

Inspect the provided experimental data to assess how changes in initial concentrations affect the rate.1. Compare Experiments 1 and 2: - Keeping \([\text{NO}_2]\) constant, doubling \([\text{CO}]\) doesn't change the rate. - This suggests that the reaction is zero-order with respect to \(\text{CO}\) (\(m = 0\)).2. Compare Experiment 1 to 4: - Increase \([\text{NO}_2]\) while \([\text{CO}]\) is constant, the rate doubles. - Indicates that the reaction is first-order with respect to \(\text{NO}_2\) (\(n = 1\)).

03

Write the Rate Law

Given the observations from step 2, we can deduce the following rate law:\[\text{Rate} = k [\text{NO}_2]^1\]This simplifies to:\[\text{Rate} = k [\text{NO}_2]\]

04

Calculate the Rate Constant k

Using the deduced rate law \(\text{Rate} = k [\text{NO}_2]\), and using data from an experiment where both the rate and \([\text{NO}_2]\) are known, compute \(k\):From Experiment 4, where Rate = \(1.0 \times 10^{-3}\) \(\text{mol L}^{-1} \text{ h}^{-1}\) and \([\text{NO}_2] = 6.8 \times 10^{-8}\) \(\text{mol L}^{-1}\), we have:\[k = \frac{\text{Rate}}{[\text{NO}_2]} = \frac{1.0 \times 10^{-3}}{6.8 \times 10^{-8}} \approx 1.47 \times 10^{4} \text{ L mol}^{-1} \text{ h}^{-1}\]

05

Summary of Findings

1. The rate law is \(\text{Rate} = k [\text{NO}_2]\).2. The reaction is zero-order with respect to \([\text{CO}]\) and first-order with respect to \([\text{NO}_2]\).3. The rate constant \(k\) is approximately \(1.47 \times 10^{4} \text{ L mol}^{-1} \text{ h}^{-1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Order

Understanding reaction order is crucial to comprehending how different reactants influence the rate of a chemical reaction. In this context, the order is an exponent that correlates changes in concentration to changes in rate.
The reaction order helps identify the dependency of the reaction rate on various reactants. In the given reaction of \[\text{CO(g)} + \text{NO}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + \text{NO}(\text{g})\]
we understand the influence by comparing different experimental results:

• If changing the concentration of a reactant does not change the reaction rate, the reaction is zero-order with respect to that reactant, as seen with \([\text{CO}])\). The order here is 0.
• When the rate is proportionate to the concentration of a reactant (i.e., doubling the concentration doubles the rate), it is first-order with respect to that reactant, as shown by \([\text{NO}_2])\). The order here is 1.

Thus, the overall reaction order is the sum of the orders with respect to all reactants.

Rate Constant

The rate constant, denoted as \(k\), is a crucial parameter in the rate law equation. It quantifies the speed of a reaction and depends on factors like temperature and nature of the reaction.
When we determined the rate law as\[\text{Rate} = k [\text{NO}_2]\],
we needed to calculate \(k\) using actual experimental data. With the rate and concentration known from Experiment 4, we use the formula:
\[k = \frac{\text{Rate}}{[\text{NO}_2]}\]

• For example, using the rate of \(1.0 \times 10^{-3}\) \(\text{mol L}^{-1} \text{ h}^{-1}\) and \([\text{NO}_2] = 6.8 \times 10^{-8}\) \(\text{mol L}^{-1}\), we calculate:
• \(k = \frac{1.0 \times 10^{-3}}{6.8 \times 10^{-8}} \approx 1.47 \times 10^{4} \text{ L mol}^{-1} \text{ h}^{-1}\)

The rate constant is specific for each reaction and facilitates the conversion of the rate law into a mathematical expression, indicating the rate at which a reaction occurs.

Chemical Kinetics

Chemical kinetics is the study of reaction rates and the steps involved in reactions. It delves into how various factors like concentration, temperature, and catalysts affect the speed of a reaction.
For students, this means understanding the nuances of reactions over time. In the given exercise, analyzing the reaction involving \[\text{CO(g)} + \text{NO}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + \text{NO}(\text{g})\]
requires looking at:

• How the concentration of \(\text{CO} \) and \(\text{NO}_2\) affects how quickly products form.
• Understanding the experimental data to derive a rate law.
• Seeing the rate constant used in practical calculations to predict reaction behavior.

Through these steps, chemical kinetics provides a toolkit for predicting how changing conditions affect reaction rates and mechanisms, which is vital for fields like pharmaceuticals, materials science, and environmental studies.