Problem 87 - Chemistry The Central Science

Question

The following data were collected for the destruction of \(\mathrm{O}_{3}\) by \(\mathrm{H}\left(\mathrm{O}_{3}+\mathrm{H} \longrightarrow \mathrm{O}_{2}+\mathrm{OH}\right)\) at very low concentrations: $$\begin{array}{lll}{\text { Trial }} & {\left[0_{3}\right](M)} & {[\mathrm{H}](M)} & {\text { Initial Rate }(M / s)} \\ \hline 1 & {5.17 \times 10^{-33}} & {3.22 \times 10^{-26}} & {1.88 \times 10^{-14}} \\ {2} & {2.59 \times 10^{-33}} & {3.25 \times 10^{-26}} & {9.44 \times 10^{-15}} \\\ {3} & {5.19 \times 10^{-33}} & {6.46 \times 10^{-26}} & {3.77 \times 10^{-14}}\end{array}$$ \begin{equation}\begin{array}{l}{\text { (a) Write the rate law for the reaction. }} \\ {\text { (b) Calculate the rate constant. }}\end{array}\end{equation}

Step-by-Step Solution

Answer

The rate law for the ozone destruction reaction is given by \(rate = k[\mathrm{O}_{3}][\mathrm{H}]\), where the rate constant (k) is approximately \(1.177 \times 10^{18}\ \mathrm{M^{-1} s^{-1}}\).

Step 1: Determine the order of the reaction regarding O3 and H concentrations

From the table, we see that in Trial 1 and Trial 3, [H] is doubled while [O3] remains almost the same. This change causes the initial rate to double as well. Therefore, the reaction order with respect to H is 1. Now, let's compare Trial 1 and Trial 2. In these trials, [O3] is halved while [H] remains almost the same. This change also causes the initial rate to halve. Therefore, the reaction order with respect to O3 is also 1.

Step 2: Write the rate law for the reaction

Now, based on the order determined in Step 1, we can write the rate law for the reaction as follows: \(rate = k[\mathrm{O}_{3}]^1[\mathrm{H}]^1\) The rate law can be simplified to: \(rate = k[\mathrm{O}_{3}][\mathrm{H}]\)

Step 3: Calculate the rate constant

Now, we need to calculate the rate constant (k) using any of the trials given. Let's use Trial 1: \(rate_{1} = k[\mathrm{O}_{3_{1}}][\mathrm{H}_{1}]\) \(k = \frac{rate_{1}}{[\mathrm{O}_{3_{1}}][\mathrm{H}_{1}]}\) Substitute the values from Trial 1: \(k = \frac{1.88 \times 10^{-14}\ \mathrm{M/s}}{(5.17 \times 10^{-33}\ \mathrm{M})(3.22 \times 10^{-26}\ \mathrm{M})}\) Now, calculate the value of k: \(k \approx 1.177 \times 10^{18}\ \mathrm{M^{-1} s^{-1}}\)

Step 4: Result

The rate constant for this reaction is approximately \(1.177 \times 10^{18}\ \mathrm{M^{-1} s^{-1}}\), and the rate law is given by: \(rate = k[\mathrm{O}_{3}][\mathrm{H}]\), where \(k\approx 1.177 \times 10^{18}\ \mathrm{M^{-1} s^{-1}}\).

Key Concepts

Understanding Rate Law

The rate law is a mathematical expression that relates the rate of a chemical reaction to the concentration of its reactants. It provides insight into the kinetics of a reaction and is crucial in predicting how a reaction's rate will change with varying concentrations of reactants. In the context of our problem, the rate law is established by observing how changes in reactant concentrations affect the rate of reaction. More specifically, we determined that the reaction is first order with respect to both \(\mathrm{O}_{3}\) and \(\mathrm{H}\), since doubling the concentration of each reactant, individually, resulted in a doubling of the initial reaction rate.

In mathematical terms, this relationship is summarized as \(rate = k[\mathrm{O}_{3}][\mathrm{H}]\), where \(k\) is the rate constant, and the reactant concentrations are raised to powers corresponding to their reaction orders—in this case, first order, denoted by the exponent of 1. It's essential for students to recognize that while the rate laws can often be deduced from experimental data, they cannot be reliably predicted solely from the stoichiometry of the reaction equation.

Grasping Reaction Order

Reaction order describes the dependence of the reaction rate on the concentration of each reactant. Each reactant in a chemical reaction can have its own reaction order, and the overall reaction order is the sum of the orders for all reactants. As discerned from the problem's data, both \(\mathrm{O}_{3}\) and \(\mathrm{H}\) have a reaction order of 1, which makes the overall reaction order 2, since 1+1=2.

It's fundamental to know that the reaction order can be zero, one, two, or even fractional, and can only be determined empirically. A reaction order of zero implies that the rate is not dependent on the concentration of that reactant, one means the rate is directly proportional to its concentration, and two indicates the rate is proportional to the square of its concentration. Understanding this concept is critical for predicting how a reaction will behave under different concentrations, a skill that is essential in various industrial and research applications.

Deciphering Rate Constant

The rate constant, represented by \(k\), is a proportionality constant in the rate law equation that is specific to a particular reaction at a given temperature. While the rate law expression tells us how rate depends on concentration, the rate constant tells us how fast the reaction proceeds under those conditions. The units of the rate constant vary depending on the overall order of reaction; in our case, with an overall reaction order of 2, the unit is \(\mathrm{M^{-1} s^{-1}}\).

In our exercise, the calculated rate constant is approximately \(1.177 \times 10^{18}\ \mathrm{M^{-1} s^{-1}}\), which is incredibly high, suggesting a very rapid reaction under the given conditions. It is critical to understand that the rate constant is affected by temperature, with higher temperatures generally increasing the rate constant, thus accelerating the reaction. Learning how to calculate and interpret the rate constant empowers students to quantify reaction kinetics and make predictions regarding reaction rates.