Problem 30 - Chemistry: The Central Science

Question

The reaction \(2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\) \(\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) was studied with the following results: \begin{tabular}{llll} \hline Experiment & {\(\left[\mathrm{ClO}_{2}\right](M)\)} & {\(\left[\mathrm{OH}^{-}\right](M)\)} & Rate \((M / \mathrm{s})\) \\ \hline 1 & \(0.060\) & \(0.030\) & \(0.0248\) \\ 2 & \(0.020\) & \(0.030\) & \(0.00276\) \\ 3 & \(0.020\) & \(0.090\) & \(0.00828\) \\ \hline \end{tabular} (a) Determine the rate law for the reaction. (b) Calculate the rate constant. (c) Calculate the rate when \(\left[\mathrm{ClO}_{2}\right]=\) \(0.100 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]=0.050 \mathrm{M}\)

Step-by-Step Solution

Answer

The rate law for the reaction is Rate = k [ClO_2]^2 [OH^-], with a rate constant (k) of approximately 2.76 M⁻²s⁻¹. When \([\text{ClO}_2] = 0.100\text{ M}\) and \([\text{OH}^-] = 0.050\text{ M}\), the reaction rate is about 0.069 M/s.

Step 1: Determine the order of the reaction with respect to ClO_2 and OH^-

First, let's compare experiments 1 and 2. In these experiments, the concentration of OH^- is the same, while the concentration of ClO_2 changes by a factor of 3 (0.060 M to 0.020 M). We can see that when the concentration of ClO_2 is reduced, the rate drops from 0.0248 M/s to 0.00276 M/s, which is approximately reduced by a factor of 9. So, with respect to ClO_2, the reaction is second order. To find the order with respect to OH^-, let's compare experiments 2 and 3. Now, let's compare Experiments 2 and 3. In these experiments, the concentration of ClO_2 is the same, while the concentration of OH^- changes by a factor of 3 (0.030 M to 0.090 M). When the concentration of OH^- is increased by this factor, the rate rises from 0.00276 M/s to 0.00828 M/s, which is approximately an increase by a factor of 3. So, with respect to OH^-, the reaction is first-order.

Step 2: Write down the rate law for the reaction

Now that we know the order with respect to each reactant, we can write down the rate law for the reaction. The rate of the reaction is given by the following expression: Rate = k [ClO_2]^2 [OH^-] Where k is the rate constant for the reaction.

Step 3: Calculate the rate constant

Let's calculate the rate constant (k) by using the first experiment. Based on the experimental data: Rate = 0.0248 M/s [ClO_2] = 0.060 M [OH^-] = 0.030 M Substitute these values into the rate law we found in Step 2: 0.0248 = k (0.060)^2 (0.030) Now, solve for the rate constant k: k = 0.0248 / ((0.060)^2 * (0.030)) k ≈ 2.76 M⁻²s⁻¹

Step 4: Calculate the reaction rate when given concentrations of ClO_2 and OH^-

Finally, we will calculate the reaction rate when [ClO_2] = 0.100 M and [OH^-] = 0.050 M. Substitute these values, along with the calculated rate constant (k ≈ 2.76 M⁻²s⁻¹), into the rate law: Rate = 2.76 (0.100)^2 (0.050) Rate ≈ 0.069 M/s So, when [ClO_2] = 0.100 M and [OH^-] = 0.050 M, the reaction rate is about 0.069 M/s.

Key Concepts

Reaction Order

Understanding the reaction order is crucial in chemical kinetics, as it helps us know how the concentration of reactants affects the rate of a reaction. In this case, we are examining the reaction \(2 \text{ClO}_2(aq) + 2 \text{OH}^-(aq) \rightarrow \text{ClO}_3^-(aq) + \text{ClO}_2^-(aq) + \text{H}_2\text{O}(l)\). For such reactions, determining the reaction order with respect to each reactant is essential. To find out, we compare experimental data where only one reactant concentration changes, while the other remains constant. In comparing experiments 1 and 2: The concentration of \(\text{ClO}_2\) decreases by a factor of 3, but the rate drops by a factor of 9. This indicates that the reaction is

• Second order with respect to \(\text{ClO}_2\)

Similarly, by comparing experiments 2 and 3 where the concentration of \(\text{OH}^-\) changes by a factor of 3, the rate also changes by the same factor. This means:

• First order with respect to \(\text{OH}^-\)

Thus, the overall order of the reaction is based on the sum of these individual orders.

Rate Law

The rate law is a mathematical expression that relates the reaction rate to the concentrations of reactants, each raised to a power corresponding to their respective reaction orders.For the given reaction, the rate law is determined once the reaction orders have been established: \[\text{Rate} = k [\text{ClO}_2]^2 [\text{OH}^-]^1\]Here,

• \(k\) is the rate constant, a unique value for a given reaction at a specified temperature.
• \([\text{ClO}_2]^2\) denotes that the reaction is second order in \(\text{ClO}_2\).
• \([\text{OH}^-]^1\) denotes first order in \(\text{OH}^-\).

The rate law is fundamental because it allows us to predict how changes in concentration affect reaction rates. It provides insights into the stoichiometric relationship between reactants and the kinetics of the chemical process.

Rate Constant Calculation

The rate constant, \(k\), is a crucial parameter in the rate law that is determined through experimental data and calculations. It provides insights into the speed of a reaction under certain conditions.To find \(k\), we use experimental results, inserting values into the rate law. Let's take Experiment 1 as an example:

• Rate = 0.0248 M/s
• \([\text{ClO}_2] = 0.060\) M
• \([\text{OH}^-] = 0.030\) M

Inserting these into the rate law, \[0.0248 = k (0.060)^2 (0.030)\]Solving for \(k\), we find: \[k = \frac{0.0248}{(0.060)^2 \times 0.030} \approx 2.76 \, \text{M}^{-2}\text{s}^{-1}\]This constant is crucial for calculating the reaction rate at different concentrations and understanding the reaction's nature and speed more deeply.

Reaction Rate

The reaction rate describes how quickly a reaction proceeds over time. It depends on the concentration of reactants, which influences how frequently molecules collide to react.To determine the reaction rate with given concentrations, we use the rate law together with the rate constant. For instance, let's calculate the rate when \([\text{ClO}_2] = 0.100\) M and \([\text{OH}^-] = 0.050\) M:Using the rate law: \[\text{Rate} = k [\text{ClO}_2]^2 [\text{OH}^-]\]Where \(k \approx 2.76 \, \text{M}^{-2}\text{s}^{-1}\), we substitute the values:\[\text{Rate} = 2.76 \times (0.100)^2 \times 0.050\]Calculating this, we find:\[\text{Rate} \approx 0.069 \, \text{M/s}\]This rate indicates how fast the reaction progresses at these specific concentrations. Understanding reaction rates helps in controlling reaction conditions effectively in both laboratory and industrial settings.