Problem 62 - Chemistry The Science in Context

Question

The following kinetic data were obtained at \(298 \mathrm{K}\) for the reaction: $$\begin{array}{cccc} \hline \text { Experiment } & \left[\mathrm{ClO}_{2}\right]_{0}(\mathrm{M}) & \left[\mathrm{OH}^{-}\right]_{0}(\mathrm{M}) & \begin{array}{c} \text { Initial Rate } \\ (\mathrm{M} / \mathrm{s}) \end{array} \\ \hline 1 & 0.060 & 0.030 & 0.0248 \\ \hline 2 & 0.020 & 0.030 & 0.00827 \\ \hline 3 & 0.020 & 0.090 & 0.0247 \\ \hline \end{array}$$ Determine the rate law and the rate constant for this reaction at \(298 \mathrm{K}.\) The following kinetic data were collected at \(298 \mathrm{K}\) for the reaction of ozone with nitrite ion, producing nitrate and oxygen: $$ \mathrm{NO}_{2}^{-}(a q)+\mathrm{O}_{3}(g) \rightarrow \mathrm{NO}_{3}^{-}(a q)+\mathrm{O}_{2}(g) $$ $$\begin{array}{cccc} \hline \text { Experiment } & \left[\mathrm{NO}_{2}\right]_{0}(\mathrm{M}) & \left[\mathrm{O}_{3}\right]_{0}(\mathrm{M}) & \begin{array}{c} \text { Initial Rate } \\ (M / \mathrm{s}) \end{array} \\ \hline 1 & 0.0100 & 0.0050 & 25 \\ \hline 2 & 0.0150 & 0.0050 & 37.5 \\ \hline 3 & 0.0200 & 0.0050 & 50.0 \\ \hline 4 & 0.0200 & 0.0200 & 200.0 \\ \hline \end{array}$$ Determine the rate law for the reaction and the value of the rate constant.

Step-by-Step Solution

Answer

Based on the provided information and calculations, the rate laws and rate constants for the two reactions are as follows: Reaction 1: Rate = (13.78 M⁻¹s⁻¹)[ClO₂][OH⁻] Reaction 2: Rate = (500 M⁻¹s⁻¹)[NO₂⁻][O₃]

Step 1: Compare Experiments

Focus on the experiments where only one reactant concentration changes, and the other remains constant. Comparing Experiment 1 and 2, we observe that when the initial concentration of \(\mathrm{ClO_2}\) is tripled, the initial rate triples. Therefore, the reaction is first-order with respect to \(\mathrm{ClO_2}\). Comparing Experiment 2 and 3, we notice that when the initial concentration of \(\mathrm{OH^-}\) is tripled, the rate triples. Therefore, the reaction is first-order with respect to \(\mathrm{OH^-}\) as well.

Step 2: Determine the Rate Law

Based on the observations from Step 1, the rate law for this reaction is given by: Rate \(=k\left[\mathrm{ClO_2}\right]\left[\mathrm{OH^-}\right]\) Now we need to determine the rate constant, k, for the reaction at 298 K.

Step 3: Calculate the Rate Constant

Use the initial rate and concentrations for any of the experiments to calculate the rate constant. We will use the data from Experiment 1. Rate \(= k\left[\mathrm{ClO_2}\right]\left[\mathrm{OH^-}\right]\) \(0.0248 \mathrm{M/s} = k(0.060 \mathrm{M})(0.030 \mathrm{M})\) Now, solve for k: \(k = \frac{0.0248}{(0.060)(0.030)} = 13.78 \mathrm{M^{-1}s^{-1}}\) As a result, the rate law for the reaction is: Rate \(= (13.78 \mathrm{M^{-1}s^{-1}})\left[\mathrm{ClO_2}\right]\left[\mathrm{OH^-}\right]\) Reaction 2: \(\mathrm{NO_2^-(aq) + O_3(g)} \rightarrow \mathrm{NO_3^-(aq) + O_2(g)}\) Determine the rate law for this reaction.

Step 1: Comparing Experiments

Again, focus on experiments where only one reactant concentration changes, and the other remains constant. Comparing Experiments 1, 2, and 3, we observe that as the initial concentration of \(\mathrm{NO_2^-}\) increases, the initial rate increases proportionally. Therefore, the reaction is first-order with respect to \(\mathrm{NO_2^-}\). Comparing Experiment 3 and 4, we notice that when the initial concentration of \(\mathrm{O_3}\) is quadrupled, the rate quadruples. Therefore, the reaction is first-order with respect to \(\mathrm{O_3}\) as well.

Step 2: Determine the Rate Law

Based on our observations from Step 1, the rate law for this reaction is given by: Rate \(= k\left[\mathrm{NO_2^-}\right]\left[\mathrm{O_3}\right]\) Now we need to determine the rate constant, k, for the reaction.

Step 3: Calculating the Rate Constant

Use the initial rate and concentrations for any of the experiments to calculate the rate constant. We will use the data from Experiment 1. Rate \(= k\left[\mathrm{NO_2^-}\right]\left[\mathrm{O_3}\right]\) \(25 \mathrm{M/s} = k(0.0100 \mathrm{M})(0.0050 \mathrm{M})\) Now, solve for k: \(k = \frac{25}{(0.0100)(0.0050)} = 500 \mathrm{M^{-1}s^{-1}}\) Thus, the rate law for the reaction is: Rate \(= (500 \mathrm{M^{-1}s^{-1}})\left[\mathrm{NO_2^-}\right]\left[\mathrm{O_3}\right]\)

Key Concepts

Rate Law Determination

In chemical kinetics, understanding how to determine a rate law is a key skill. A rate law shows how the rate of a reaction depends on the concentration of the reactants. To determine the rate law, we look closely at experimental data, comparing how changes in the concentration of reactants affect the reaction rate.

Here's a typical process:

• Choose experiments where only one reactant's concentration changes while the others stay the same.
• Observe how changing this concentration affects the rate.
• If the rate changes proportionally with the concentration, the reaction is first-order regarding that reactant.

In our example, comparing Experiments 1 and 2, the concentration of \(\mathrm{ClO_2}\) is tripled, and the rate triples, indicating that the reaction is first-order in \(\mathrm{ClO_2}\). Meanwhile, comparing Experiments 2 and 3, we see that the rate changes proportionally with \(\mathrm{OH^-}\) concentration, making it first-order in \(\mathrm{OH^-}\) as well. Thus, the rate law for this reaction is:\[ \text{Rate} = k[\mathrm{ClO_2}][\mathrm{OH^-}]\]Understanding this approach will prepare you to identify rate laws for various reactions effectively.

Reaction Order

Reaction order in the context of chemical kinetics determines how the rate is affected by the concentration of the reactants. Each reactant's concentration is raised to some power in the rate law equation, which tells you the order of that reactant. The sum of these powers gives us the overall reaction order.

For example, in the reaction involving \(\mathrm{ClO_2}\) and \(\mathrm{OH^-}\), each is first-order as described earlier. We express this mathematically as:\[\text{Rate} = k[\mathrm{ClO_2}]^1[\mathrm{OH^-}]^1\]Consequently, the overall reaction order is 1 (from \(\mathrm{ClO_2}\)) + 1 (from \(\mathrm{OH^-}\)) = 2.

Knowing the reaction order helps in predicting how changes in concentrations affect the reaction rate and is essential in understanding complex reactions. In chemical reactions with higher orders (>2), rates can indicate very sensitive changes to concentration, which could suggest different mechanisms at play.

Rate Constant Calculation

Once the rate law is determined, the next step is calculating the rate constant \(k\). This constant is specific to each reaction at a given temperature, reflecting how efficiently a reaction proceeds. With the rate law equation, we use experimental data to solve for \(k\). Here's a simple method:

• Identify the rate of reaction and the concentrations of reactants from the experiment.
• Plug these values into the rate law equation.
• Solve for \(k\) using algebraic manipulation.

For the reaction involving \(\mathrm{ClO_2}\) and \(\mathrm{OH^-}\), the equation from Experiment 1 is:\[0.0248 \mathrm{M/s} = k(0.060 \mathrm{M})(0.030 \mathrm{M})\]Solving this gives \(k = 13.78 \mathrm{M^{-1}s^{-1}}\). Similarly, for the \(\mathrm{NO_2^-}\) and \(\mathrm{O_3}\) reaction, a similar calculation yields \(k = 500 \mathrm{M^{-1}s^{-1}}\).

The value of \(k\) influences the speed of the reaction and provides insight into the reaction mechanism and temperature effects.