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: $$ \begin{array}{llll} \hline \text { Trial } & {\left[\mathrm{O}_{3}\right](\boldsymbol{M})} & {[\mathrm{H}](\boldsymbol{M})} & \text { Initial Rate }(\boldsymbol{M} / \mathrm{s}) \\ \hline 1 & 5.17 \times 10^{-33} & 3.22 \times 10^{-26} & 1.88 \times 10^{-14} \\\ 2 & 2.59 \times 10^{-33} & 3.25 \times 10^{-26} & 9.44 \times 10^{-15} \\ 3 & 5.19 \times 10^{-33} & 6.46 \times 10^{-26} & 3.77 \times 10^{-14} \\ \hline \end{array} $$ (a) Write the rate law for the reaction. (b) Calculate the rate constant.

Short answer

(a) The rate law for the reaction is \(rate=k[\text{O}_3][\text{H}]\). (b) The rate constant \(k \approx 1.15 \times 10^{42} \, \text{M}^{-1} \, \text{s}^{-1}\).

Step-by-step solution

Understand the Rate Law formula

The general rate law formula for a reaction is given by: \[rate = k[\text{A}]^m[\text{B}]^n\] Where k is the rate constant, A and B are the reactants, and m and n are the orders of reaction with respect to A and B respectively. In our case, the rate law would be in the form of: \[rate = k[\text{O}_3]^m[\text{H}]^n\]

Determine the Orders of Reaction

We need to determine m and n, the orders of reaction with respect to O3 and H. To do this, we'll utilize the provided data. First, compare trials 1 and 2, keeping in mind that only the initial concentration of O3 changes. Divide the rate in trial 2 by the rate in trial 1: \[\frac{9.44 \times 10^{-15}}{1.88 \times 10^{-14}} = \frac{[\text{O}_3]_2^m}{[\text{O}_3]_1^m}\] \[0.5 = \left(\frac{2.59 \times 10^{-31}}{5.17 \times 10^{-31}}\right)^m\] \[0.5 = (0.5)^m\] From this, it's clear that m = 1. Next, let's compare trials 1 and 3, where only the initial concentration of H changes. Divide the rate in trial 3 by the rate in trial 1: \[\frac{3.77 \times 10^{-14}}{1.88 \times 10^{-14}} = \frac{[\text{H}]_3^n}{[\text{H}]_1^n}\] \[2 = \left(\frac{6.46 \times 10^{-26}}{3.22 \times 10^{-26}}\right)^n\] \[2 = (2)^n\] From this, it's clear that n = 1.

Write the Rate Law

As the orders of reaction with respect to O3 (m) and H (n) are 1, the rate law for the reaction is: \[rate=k[\text{O}_3][\text{H}]\] So, the answer to part (a) is the rate law \(rate=k[\text{O}_3][\text{H}]\).

Calculate the Rate Constant (k)

Now, let's calculate the rate constant, k. We can use any trial data and the determined rate law to solve for k. We can use trial 1 for our calculations: \[1.88 \times 10^{-14} = k(5.17\times 10^{-31})(3.22\times 10^{-26})\] Now solve for k: \[k = \frac{1.88 \times 10^{-14}}{(5.17\times 10^{-31})(3.22\times 10^{-26})} \] \[k \approx 1.15 \times 10^{42} \, \text{M}^{-1} \, \text{s}^{-1}\] So, the answer to part (b) is the rate constant \(k \approx 1.15 \times 10^{42} \, \text{M}^{-1} \, \text{s}^{-1}\).

Key concepts

Chemical Kinetics

Chemical kinetics is the branch of chemistry that studies the rates of chemical processes and the factors influencing them. It helps scientists understand how different conditions, such as temperature, concentration of reactants, and catalysts, affect the speed at which reactions occur.

The speed or rate of a chemical reaction is quantified by the rate at which the reactants are consumed or the products are formed. A simple equation to represent this is:
\[ \text{Rate} = -\Delta[\text{Reactant}]/\Delta t \]\ where \(-\Delta[\text{Reactant}]\) symbolizes the change in the concentration of the reactant over the change in time (\(\Delta t\)). Note that it’s customary to express the rate of consumption of a reactant as a positive value, hence the negative sign in the formula.

Understanding chemical kinetics is crucial for predicting how long a reaction will take under certain conditions and for controlling industrial processes to optimize production rates.

Reaction Order

The reaction order is a crucial concept in the realm of chemical kinetics, referring to the power to which the concentration of a reactant is raised in the rate law. This reflects how the rate of the reaction is affected by the concentration of that reactant.

For a general two-reactant reaction, we can express the rate law as:
\[ \text{Rate} = k[\text{A}]^m[\text{B}]^n \]\ where \(k\) is the rate constant, \(m\) and \(n\) are the reaction orders with respect to reactants \(A\) and \(B\), respectively. If the rate is directly proportional to the concentration of one reactant (as in our example with O₃ and H), that reactant’s reaction order is one. Sometimes, the order can be zero (rate is not dependent on the concentration of the reactant), or two or more (rate is more than proportionally affected by changes in concentration).

In our exercise, by assessing the effect that varying concentrations of \(O_3\) and \(H\) has on the rate, we determined the reaction order for both to be 1, indicating a first-order dependency for each reactant.

Rate Constant

The rate constant is denoted by \(k\) in the rate law equation and is a measure of the intrinsic reactivity of the reaction. It's a constant of proportionality that links the concentration of the reactants to the rate of reaction. The value of \(k\) is determined experimentally and can vary significantly depending on factors such as temperature and pressure.

It's important to note that the rate constant is not affected by changes in reactant concentration; however, temperature does have a significant impact on its value. In the context of the textbook exercise, the rate constant is found by rearranging the rate law to solve for \(k\) using the given data. The calculated value provides us with an understanding of the reaction's speed under the specific conditions given in the problem.