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: (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 \(\left[\mathrm{OH}^{-}\right]=0.050 \mathrm{M}\).

Short answer

The rate law for the reaction is Rate = k[ClO₂]^m[OH⁻]^n. Using the given experimental data, we can find the values of m and n. Then, calculate the rate constant (k) using one of the trials, and finally, compute the rate when [ClO₂] = 0.100 M and [OH⁻] = 0.050 M by plugging the values into the rate law equation.

Step-by-step solution

(a) Determine the rate law for the reaction

To determine the rate law for the reaction, we need to use experimental data. Here, let's assume that you have been given the results from several trials in which the concentrations of the reactants are varied, and the initial rates are measured. For example, three trials with initial concentrations and rates may be given as follows: Trial 1 : [ClO₂]₀ = 0.100 M, [OH⁻]₀ = 0.100 M, Initial rate = x M/s Trial 2 : [ClO₂]₀ = 0.200 M, [OH⁻]₀ = 0.100 M, Initial rate = y M/s Trial 3 : [ClO₂]₀ = 0.100 M, [OH⁻]₀ = 0.200 M, Initial rate = z M/s Comparing trial 1 with trial 2 doubles the concentration of ClO₂ but keeps the concentration of OH⁻ the same. In contrast, comparing trial 1 with trial 3, the concentration of ClO₂ is the same, but the concentration of OH⁻ is doubled. With the help of this experimental data, we can determine the order of the reaction with respect to each reactant. Assuming the rate law is in the form: Rate = k[ClO₂]^m[OH⁻]^n, we will find the values of m and n.

(b) Calculate the rate constant

Once we've determined the order of the reaction with respect to each reactant, we can then plug in the values from any one of the trials to find the rate constant k. For example, using trial 1, we have: x M/s = k(0.100 M)^m(0.100 M)^n Solve for k.

(c) Calculate the rate when [ClO₂] = 0.100 M and [OH⁻] = 0.050 M

Now that we have the rate law and the rate constant, we can calculate the initial rate of the reaction when [ClO₂] = 0.100 M and [OH⁻] = 0.050 M. Plug in these values and the rate constant k into the rate law equation: Rate = k(0.100 M)^m(0.050 M)^n Compute the rate.

Key concepts

Rate Law

Understanding the rate law of a chemical reaction is pivotal when studying its kinetics. The rate law is an equation that relates the rate of a reaction to the concentration of the reactants and the reaction rate constant. It is often expressed in the form:

\( \text{Rate} = k[\text{A}]^m[\text{B}]^n \)

Here, \( [\text{A}] \) and \( [\text{B}] \) represent the molar concentrations of reactants A and B, while \( m \) and \( n \) are the reaction orders with respect to A and B, respectively. The term \( k \) denotes the reaction rate constant, which is unique for each chemical reaction at a given temperature.

In the exercise provided, one needs to analyze experimental data to derive the rate law. By comparing rates from different trials with varying concentrations of reactants, you can deduce the order of the reaction with respect to each reactant. This is achieved by observing the effect of changing reactant concentrations on the rate of the reaction. A systematic approach involving keeping one reactant concentration constant while varying the other enables the determination of the individual effect of each reactant on the reaction rate.

Once these orders are found, they can be plugged back into the rate law formula to establish a complete rate equation for the reaction, which then serves as a tool for predicting reaction rates under different reactant concentrations. It's important to treat this step with care, as it establishes the foundation for further calculations in the problem.

Reaction Rate Constant

The reaction rate constant, denoted as \( k \), is a proportionality factor that relates the reaction rate to the concentrations of reactants raised to their orders in the rate law. It is a crucial parameter that varies with temperature and the presence of catalysts but remains constant for a given reaction under fixed conditions. The units of the rate constant change depending on the overall order of the reaction, which can lead to initial confusion but is vital for dimensional consistency in rate equations.

To calculate the rate constant for the given exercise, one must first discern the rate law by determining the orders of the reactants. Using data from a selected trial—such as Trial 1, which states the initial concentration and rate—you can then solve for \( k \) while holding the reaction orders constant. For example:

\( x \text{ M/s} = k(0.100 \text{ M})^m(0.100 \text{ M})^n \)

Solving this equation provides the value of \( k \), which then can be used to predict the rate of the reaction under different conditions. The calculated reaction rate constant is fundamentally important, as it encapsulates the intrinsic rate at which the reactants transform into products, and touches on the effect of external factors such as temperature or catalysts.

Reaction Order

The reaction order indicates the dependence of the reaction rate on the concentration of each reactant. It is typically determined through experimentation and is part of the rate law expression. The overall reaction order is the sum of the individual orders with respect to each reactant.

For instance, in the exercise provided, the reaction order with respect to \( \text{ClO}_2 \) and \( \text{OH}^- \) would be determined by examining how the rate changes when their concentrations vary. If the rate changes proportionally with the concentration of a reactant, it is first-order with respect to that reactant. If the rate changes by the square of the concentration change, it is second-order, and so on.

In mathematical terms, for a rate law expressed as \( \text{Rate} = k[\text{A}]^m[\text{B}]^n \), if doubling the concentration of A leads to a doubling of the rate while B's concentration remains constant, the reaction is first-order with respect to A (i.e., \( m=1 \)). If the rate e.g. quadruples instead, the reaction is second-order with respect to A (\( m=2 \)).

Determining the correct reaction orders is vital as it not only helps in deriving the rate law but also assists in understanding the mechanism of the reaction. A clear understanding of the relationship between the concentration of reactants and the rate of the reaction is essential for the study of reaction kinetics and to predict the outcomes of the reaction under various scenarios.