Problem 87 - Chemistry The Central Science In Si Units

Question

The following data were collected for the decomposition of \(\mathrm{O}_{3} \mathrm{by}\left(\mathrm{O}_{3}+\mathrm{H} \longrightarrow \mathrm{O}_{2}+\mathrm{OH}\right)\) at very low concentrations: $$\begin{array}{llll} \hline \text { Trial } & {\left[\mathrm{O}_{3}\right](M)} & {[\mathrm{H}](M)} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 3.25 \times 10^{-33} & 2.25 \times 10^{-26} & 8.10 \times 10^{-15} \\\ 2 & 6.50 \times 10^{-33} & 4.50 \times 10^{-26} & 3.25 \times 10^{-14} \\ 3 & 6.48 \times 10^{-33} & 2.23 \times 10^{-26} & 1.62 \times 10^{-14} \\ \hline \end{array}$$ (a) Write the rate law for the reaction. (b) Calculate the rate constant.

Step-by-Step Solution

Answer

Rate law: \( \text{Rate} = k [\mathrm{O}_3][\mathrm{H}] \), \( k \approx 1.11 \times 10^{44} \text{ M}^{-1}\text{s}^{-1} \).

Step 1: Determine the Reaction Order with Respect to Each Component

To find the reaction order, compare the trials where only one concentration changes.\- **Comparing Trials 1 and 2:**\ - The concentration of \( [\mathrm{O}_3]\) is doubled from \(3.25 \times 10^{-33}\) M to \(6.50 \times 10^{-33}\) M, and \([ \mathrm{H}]\) is also doubled, resulting in a quadruple increase in the rate from \(8.10 \times 10^{-15}\) M/s to \(3.25 \times 10^{-14}\) M/s.\ - This suggests that the reaction is first-order with respect to \([ \mathrm{O}_3]\) and first-order with respect to \([ \mathrm{H}]\) since doubling both reactants quadruples the reaction rate.\- From these observations, the overall order of the reaction is \(1 + 1 = 2\).

Step 2: Write the Rate Law

Since we determined the reaction is first-order with respect to both \([ \mathrm{O}_3]\) and \([ \mathrm{H}]\), the rate law expression is:\\[ \text{Rate} = k [\mathrm{O}_3][\mathrm{H}] \]

Step 3: Calculate the Rate Constant k

Using data from any of the trials, substitute the known values into the rate law to solve for the rate constant \(k\).\- **Using Trial 1:**\ - \(\text{Rate} = 8.10 \times 10^{-15}\) M/s, \([ \mathrm{O}_3] = 3.25 \times 10^{-33}\) M, \([ \mathrm{H}] = 2.25 \times 10^{-26}\) M\ - Substitute into the rate law: \[8.10 \times 10^{-15} = k (3.25 \times 10^{-33})(2.25 \times 10^{-26})\]- Solving for \(k\) gives: \[k = \frac{8.10 \times 10^{-15}}{(3.25 \times 10^{-33})(2.25 \times 10^{-26})} \approx 1.11 \times 10^{44} \text{ M}^{-1}\text{s}^{-1}\]

Key Concepts

Reaction Order

The concept of reaction order is fundamental in understanding how the concentration of reactants affects the rate of a chemical reaction. Reaction order tells us the power to which the concentration of a reactant is raised in the rate law equation. It essentially shows how changes in the reactant concentration can affect the reaction rate.
In the given exercise, by comparing different trials, the orders with respect to each reactant — \(\mathrm{O}_3\) and \[\mathrm{H}\] — are determined. For instance, when the concentration of both reactants is doubled in the given data, the reaction rate quadruples. This indicates that the reaction order is 1 for each reactant. So, the overall reaction order is the sum of individual orders, which in this case is 1 (for \(\mathrm{O}_3\)) plus 1 (for \(\mathrm{H}\)), giving an overall order of 2.
Determining reaction order is crucial because it helps chemists understand the relationship between reactant concentration and the reaction speed. It also assists in predicting the outcome and behavior of reactions under different concentrations.

Rate Constant

The rate constant, denoted as \(k\), is a proportionality factor in the rate law equation. It provides insight into the speed or efficiency of a reaction under specific conditions. The units of \(k\) vary depending on the overall order of the reaction. For example, a second-order reaction like ours typically has units of \(\text{M}^{-1}\text{s}^{-1}\).
To calculate the rate constant, the rate law expression \[\text{Rate} = k [\mathrm{O}_3][\mathrm{H}]\] is used. By substituting the known values from one of the experimental trials, \(k\) can be independently calculated. Using data from Trial 1: \[\text{Rate} = 8.10 \times 10^{-15}\, \text{M/s},\] \[\mathrm{O}_3 = 3.25 \times 10^{-33}\, \text{M}\], and \[\mathrm{H} = 2.25 \times 10^{-26}\, \text{M}\], applying these to the rate equation allows us to solve for \(k\).
This calculation underscores the utility of the rate constant, which remains unchanged under constant conditions and provides significant predictive power over the reaction kinetics.

Chemical Kinetics

Chemical kinetics is the branch of chemistry that focuses on understanding the rate at which chemical reactions occur. It involves studying the rate laws, reaction orders, and rate constants to analyze how the concentration of reactants and reaction conditions affect the speed of a chemical reaction.
The study of chemical kinetics helps in elucidating reaction mechanisms by connecting experimental data to theoretical models. By understanding the kinetics, chemists can propose mechanisms that justify the rate laws observed.
The practical applications of chemical kinetics are vast, including:

• Designing and optimizing industrial chemical processes for maximum efficiency.
• Predicting and controlling reaction rates in pharmaceuticals.
• Understanding environmental processes, like the decomposition of pollutants.

Chemical kinetics, therefore, is not just about solving equations; it provides valuable insights into both theoretical and industrial applications in chemistry.