Problem 27 - Chemistry: The Central Science

Question

Consider the following reaction: $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M}\), the reaction rate at \(298 \mathrm{~K}\) is \(0.0432 \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of \(\mathrm{OH}^{-}\) were tripled?

Step-by-Step Solution

Answer

(a) The value of the rate constant is \(17.28 \, \mathrm{M^{-1} \, s^{-1}}\). (b) The units of the rate constant are \(\mathrm{M^{-1} \, s^{-1}}\). (c) If the concentration of \(\mathrm{OH}^{-}\) is tripled, the new rate would be \(0.1296 \, \mathrm{M} / \, \mathrm{s}\).

Step 1: Find the rate law expression

The rate law expression is given by the rate of the reaction being dependent on the initial concentrations of the reactants. The rates are first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^-\), so we can write the rate law as: \[Rate = k \times [\mathrm{CH}_{3} \mathrm{Br}] \times [\mathrm{OH}^-]\]

Step 2: Find the rate constant (k)

Given that the initial concentration of \(\mathrm{CH}_{3} \mathrm{Br}\) is \(5.0 \times 10^{-3} \mathrm{M}\), the concentration of \(\mathrm{OH}^-\) is \(0.050 \mathrm{M}\), and the rate of the reaction at \(298 \mathrm{K}\) is \(0.0432 \mathrm{M} \, / \, \mathrm{s}\). We can plug these values into the rate law expression: \[0.0432 = k \times (5.0 \times 10^{-3}) \times (0.050)\] Now, we can solve for k: \(k = \frac{0.0432}{(5.0 \times 10^{-3}) \times (0.050)}\) \(k = 17.28\, \mathrm{M^{-1}}\, \mathrm{s^{-1}}\) The value of the rate constant k is \(17.28 \, \mathrm{M^{-1} \, s^{-1}}\). (a) The value of the rate constant is \(17.28 \, \mathrm{M^{-1} \, s^{-1}}\).

Step 3: Determine the units of the rate constant

We have already found the units of the rate constant in step 2. (b) The units of the rate constant are \(\mathrm{M^{-1} \, s^{-1}}\).

Step 4: Analyze the effect of tripling the concentration of \(\mathrm{OH}^{-}\) on the rate

We are asked to find the new rate if the concentration of \(\mathrm{OH}^{-}\) is tripled. We can use the rate law expression with the new concentration value: \[New \, Rate = k \times [\mathrm{CH}_{3} \mathrm{Br}] \times [3 \times \mathrm{OH}^-]\] Using the rate constant value from step 2, we get: \[New \, Rate = 17.28 \times (5.0 \times 10^{-3}) \times (3 \times 0.050)\] \[New \, Rate = 0.1296 \, \mathrm{M} / \, \mathrm{s}\] (c) If the concentration of \(\mathrm{OH}^{-}\) is tripled, the new rate would be \(0.1296 \, \mathrm{M} / \, \mathrm{s}\).

Key Concepts

Reaction Rate

Understanding how fast a chemical reaction proceeds is a fundamental aspect of chemistry called the reaction rate. It's defined as the change in concentration of a reactant or product per unit time. For instance, in the given exercise where \( \text{CH}_3\text{Br} \) reacts with \( \text{OH}^- \), the reaction rate reflects how quickly \( \text{CH}_3\text{Br} \) and \( \text{OH}^- \) are being converted to products. A higher reaction rate means a faster reaction and vice versa. Measuring reaction rates is crucial for industries where controlling the speed of a reaction can be just as important as the outcome itself––such as in pharmaceuticals where reaction speed can affect the yield and purity of a drug.

Several factors can affect the reaction rate, including the concentrations of reactants, temperature, catalysts, and surface area. In our case, we're looking at the impact of concentrations where, given the first-order dependence on both \( \text{CH}_3\text{Br} \) and \( \text{OH}^- \), doubling one reactant's concentration would double the rate, while doubling both would quadruple the rate. This proportional relationship is what makes studying kinetics so valuable in predicting reaction behavior under different conditions.

Rate Law

The rate law for a chemical reaction is an equation that links the reaction rate with the concentrations of reactants. It is determined experimentally and reflects the actual relationship between concentrations and the rate. The general form of a rate law is \( \text{Rate} = k \times [\text{A}]^{m} \times [\text{B}]^{n} \) where \( [\text{A}] \) and \( [\text{B}] \) are the concentrations of the reactants, \( k \) is the rate constant, and \( m \) and \( n \) are the reaction orders for A and B respectively.

The given exercise provided that the reaction is first order in both \( \text{CH}_3\text{Br} \) and \( \text{OH}^- \). This information is imperative because it tells us that the rate is directly proportional to the concentration of each reactant—one of the key principles in chemical kinetics. If either concentration is increased, the reaction rate increases accordingly. The rate law can also reveal the mechanism of a reaction, which describes the sequence of events at the molecular level.

Rate Constant

The rate constant symbolized as \( k \) in our rate law, is a proportionality coefficient that is specific to a particular reaction at a given temperature. It is independent of reactant concentrations but can vary with temperature or if a catalyst is present. The value of \( k \) provides insight into the intrinsic speed of the reaction––a larger \( k \) means a faster reaction under the same conditions.

In this textbook exercise, the rate constant was calculated to be \( 17.28 \, \mathrm{M^{-1}} \, \mathrm{s^{-1}} \), indicating the speed at which the compounds react when the chemicals are at the concentrations given. The units of the rate constant can also provide clues to the reaction order as they must balance out the units of concentration in the rate expression to give the units of rate (M/s). If \( k \) had different units, it would suggest a different overall reaction order. Remember that \( k \) is a crucial part of predicting how a reaction will respond to changes, such as increasing the concentration of \( \text{OH}^- \) which, as per our exercise, would increase the rate of the reaction.