Problem 34 - Chemistry The Central Science In Si Units

Question

The reaction \(2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\) \(\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(I)\) was studied with the following results: \begin{tabular}{lccc} \hline Experiment & {\(\left[\mathrm{ClO}_{2}\right](M)\)} & {\(\left[0 \mathrm{H}^{-}\right](M)\)} & Initial Rate \((\mathrm{M} / \mathrm{s})\) \\ \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 with proper units. (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

(a) Rate law: Rate = k[ClO₂]²[OH⁻]. (b) k ≈ 22.96 M⁻²s⁻¹. (c) Rate ≈ 0.01148 M/s.

Step 1: Determine the Order of Reaction with Respect to Each Reactant

Start by using the rate data from the experiments to determine the order with respect to each reactant. Assume the rate law is of the form \[\text{Rate} = k[\text{ClO}_2]^m [\text{OH}^-]^n\]where \(m\) and \(n\) are the orders of the reaction with respect to \(\text{ClO}_2\) and \(\text{OH}^-\) respectively.1. **For ClO₂:** Compare experiments 1 and 2. Here, \([\text{OH}^-]\) is constant, so:\[\frac{\text{Rate in Exp 1}}{\text{Rate in Exp 2}} = \left(\frac{[\text{ClO}_2]_{1}}{[\text{ClO}_2]_{2}}\right)^m = \left(\frac{0.060}{0.020}\right)^m = \frac{0.0248}{0.00276}\]\[3^m = 9 \implies m = 2\]2. **For OH⁻:** Compare experiments 2 and 3. Here, \([\text{ClO}_2]\) is constant, so:\[\frac{\text{Rate in Exp 3}}{\text{Rate in Exp 2}} = \left(\frac{[\text{OH}^-]_{3}}{[\text{OH}^-]_{2}}\right)^n = \left(\frac{0.090}{0.030}\right)^n = \frac{0.00828}{0.00276}\]\[3^n = 3 \implies n = 1\]

Step 2: Write the Rate Law

Based on the order determinations from step 1:The order with respect to \(\text{ClO}_2\) is 2 and with respect to \(\text{OH}^-\) is 1. Therefore, the rate law is:\[\text{Rate} = k [\text{ClO}_2]^2 [\text{OH}^-]^1\]

Step 3: Determine the Rate Constant (k)

Choose one of the experiments to find the rate constant, \(k\). We use experiment 1:\[0.0248 = k (0.060)^2 (0.030)\]\[k = \frac{0.0248}{0.060^2 \times 0.030} = \frac{0.0248}{0.00108} = 22.96\]The rate constant, \(k\), is approximately 22.96 M\(^{-2}\)s\(^{-1}\).

Step 4: Calculate the Rate for Given Concentrations

Use the rate law to find the rate when \([\text{ClO}_2] = 0.100 \text{ M}\) and \([\text{OH}^-] = 0.050 \text{ M}\) using the rate law and the constant determined earlier:\[\text{Rate} = 22.96 (0.100)^2 (0.050) = 22.96 \times 0.010 \times 0.050 = 0.01148 \]The calculated rate is 0.01148 M/s.

Key Concepts

Reaction Order

In chemical kinetics, understanding the order of a reaction is crucial for predicting how the concentration of reactants affects the rate of a reaction. The reaction order is determined experimentally and indicates how the rate is affected by the concentration of one or more reactants. This is represented in the rate law equation. For a reaction of the form \(aA + bB \rightarrow \, ext{products} \,\), the rate law can be expressed as: \[\text{Rate} = k[A]^m[B]^n\] Here, \([A]\) and \([B]\) are the concentrations of reactants, \(k\) is the rate constant, and \(m\) and \(n\) are the reaction orders with respect to each reactant.

• If \(m = 0\), the reaction is zero-order with respect to \(A\), implying the rate is independent of \([A]\).
• If \(m = 1\), the reaction is first-order, meaning the rate is directly proportional to the concentration of \(A\).
• If \(m = 2\), the reaction is second-order with respect to \(A\), showing that the rate is proportional to the square of the concentration of \(A\).

In the example provided, the reaction order with respect to \(\text{ClO}_2\) is 2 and with respect to \(\text{OH}^-\) is 1. This means the rate of reaction quadruples if \([\text{ClO}_2]\) is doubled and triples if \([\text{OH}^-]\) is tripled.

Rate Constant

The rate constant, represented by \(k\), is a vital part of the rate law equation. It ties together how fast a reaction proceeds at a certain concentration of reactants. The rate constant is specific to a particular reaction at a particular temperature and does not alter with changes in concentration. Calculating the rate constant requires experimental data and using the rearranged form of the rate law: \[k = \frac{\text{Rate}}{[\text{ClO}_2]^2 [\text{OH}^-]^1}\]

Units of the Rate Constant

The units of the rate constant are derived from the rate law. In this second-order reaction (total order \(m+n=3\)), the units for \(k\) are \(\text{M}^{-2}\text{s}^{-1}\). In practical applications, knowing \(k\) allows chemists to predict how fast a reaction will occur under different conditions and is essential for designing chemical processes efficiently. In the given example, the rate constant was calculated to be approximately 22.96 \(\text{M}^{-2}\text{s}^{-1}\), reflecting how swiftly the reaction moves forward under the specified conditions.

Initial Rate

The initial rate of a reaction refers to the change in concentration of a reactant or product per unit time at the very start of the reaction. It is often measured by observing the concentration change immediately after the reactants are mixed. Initial rates are particularly useful as they provide a snapshot of the reaction kinetics before any significant changes in concentration occur that could affect the outcome. By studying the initial rate across different experiments, one can deduce the reaction order and subsequently the rate law for the reaction, as showcased in the exercise. For example, in the provided data, experiment 1 had an initial rate of 0.0248 M/s, while experiment 2, with a lower concentration of \([\text{ClO}_2]\), had a reduced rate of 0.00276 M/s. These values help compare how changes in reactant concentrations influence the reaction's speed.

Concentration

Concentration is a measure of how much of a given substance is present in a mixture or solution. It is usually expressed in terms of molarity \((\text{M})\), which is moles per liter of solution. In the context of chemical kinetics, concentration plays a key role in determining the rate at which reactions occur. According to the law of mass action, the rate of a chemical reaction is directly proportional to the product of the reactant concentrations, each raised to some power (the reaction order). In the exercise, the concentrations of \(\text{ClO}_2\) and \(\text{OH}^-\) were manipulated during experiments to observe changes in the initial rate. By comparing these changes, one can ascertain the reaction orders and establish the rate law. This demonstrates that by altering concentrations, one observes how rate changes, illustrating real-time chemical dynamics.