Question

The following set of data was obtained by the method of initial rates for the reaction: \(\left(\mathrm{H}_{3} \mathrm{C}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \rightarrow\left(\mathrm{H}_{3} \mathrm{C}\right)_{3} \mathrm{COH}+\mathrm{Br}\) What is the order of reaction with respect to ion, \(\mathrm{OH}^{-2}\) $$ \begin{array}{lcl} \hline\left[\left(\mathrm{H}_{3} \mathrm{C}\right)_{3} \mathrm{CBr}\right], \mathrm{M} & {\left[\mathrm{OH}^{-}\right], \mathrm{M}} & \begin{array}{l} \text { Initial rate, } \\ \mathrm{M} / \mathrm{s} \end{array} \\ \hline 0.25 & 0.25 & 1.1 \times 10^{-4} \\ 0.50 & 0.25 & 2.2 \times 10^{-4} \\ 0.50 & 0.50 & 2.2 \times 10^{-4} \\ \hline \end{array} $$ a. First b. Second c. Third d. Zero

Short answer

The reaction is zero order with respect to \(\mathrm{OH}^-\). (d) Zero

Step-by-step solution

Write the Rate Law Expression

For the given reaction \((\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr} + \mathrm{OH}^- \rightarrow (\mathrm{H}_3 \mathrm{C})_3 \mathrm{COH} + \mathrm{Br}^-\), the general rate law expression is \(\text{Rate} = k[(\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}]^m[\mathrm{OH}^-]^n\), where \(k\) is the rate constant, and \(m\) and \(n\) are the orders of the reaction with respect to each reactant. We need to find \(n\), the order with respect to \(\mathrm{OH}^-\).

Compare Experiments 1 and 2 to Determine m

**Experiment 1:** - \([(\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}] = 0.25\, \text{M}, [\mathrm{OH}^-] = 0.25\, \text{M}\) - Rate = \(1.1 \times 10^{-4}\, \text{M/s}\) **Experiment 2:** - \([(\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}] = 0.50\, \text{M}, [\mathrm{OH}^-] = 0.25\, \text{M}\)- Rate = \(2.2 \times 10^{-4}\, \text{M/s}\) The concentration of \(\mathrm{OH}^-\) stays constant while the concentration of \((\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}\) doubles, and the initial rate also doubles, implying \((\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}\) is first order \((m = 1)\).

Compare Experiments 2 and 3 to Determine n

**Experiment 2:** - \([(\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}] = 0.50\, \text{M}, [\mathrm{OH}^-] = 0.25\, \text{M}\) - Rate = \(2.2 \times 10^{-4}\, \text{M/s}\) **Experiment 3:** - \([(\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}] = 0.50\, \text{M}, [\mathrm{OH}^-] = 0.50\, \text{M}\)- Rate = \(2.2 \times 10^{-4}\, \text{M/s}\) The concentration of \(\mathrm{OH}^-\) doubles but the initial rate remains the same, indicating that the order with respect to \(\mathrm{OH}^-\) is zero \((n = 0)\).

Conclusion

Thus, the overall reaction is first order with respect to \((\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}\) and zero order with respect to \(\mathrm{OH}^-\). Therefore, the correct answer is (d) Zero for \(\mathrm{OH}^-\) ion.

Key concepts

Rate Law

Rate law is a mathematical equation that relates the rate of a chemical reaction to the concentration of the reactants. For a reaction like \[(\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr} + \mathrm{OH}^- \rightarrow (\mathrm{H}_3 \mathrm{C})_3 \mathrm{COH} + \mathrm{Br}^- \], the rate law expression is given by \[\text{Rate} = k[(\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}]^m[\mathrm{OH}^-]^n \]. Here, \( k \) is the rate constant, and \( m \) and \( n \) are the reaction orders with respect to each reactant. Understanding these orders is crucial as they indicate how changes in concentrations affect the rate of the reaction.

• If the concentration of a reactant is doubled and the reaction rate remains the same, the order with respect to that reactant is zero.
• If the reaction rate doubles when the concentration of a reactant doubles, the order is one.

The rate law provides insightful information about the reaction mechanism and the role each reactant plays. It isn't necessarily related to the coefficients of the balanced equation, rather it's determined experimentally.

Initial Rates Method

The initial rates method is an experimental approach to determine the reaction order of a reactant. By measuring the initial rate of the reaction at different concentrations of reactants, one can deduce the orders by comparing these initial rates. This method involves:

• Conducting a series of experiments where the concentration of one reactant is changed while others are kept constant.
• Calculating the initial rate of reaction for each experiment to observe how changes in concentration affect the rate.

In the provided exercise, this method is used to determine the order of the reaction for the reactants. For instance, by comparing two experiments where the concentration of \((\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}\) doubled while keeping \(\mathrm{OH}^-\) constant, the increase in rate indicated that the reaction order with respect to \((\mathrm{H}_3 \mathrm{C})_3 \mathrm{CBr}\) is one. Similarly, if doubling the concentration of \(\mathrm{OH}^-\) does not change the initial rate, the reaction is zero order with respect to \(\mathrm{OH}^-\). This method is handy for deducing complex reactions where multiple reactants are involved.

Chemical Kinetics

Chemical kinetics is the branch of chemistry that studies the speed or rate at which chemical reactions occur. It also explores how different conditions affect the reaction rate and sheds light on the mechanism of the reaction, providing a deeper understanding of the process. Studying chemical kinetics involves:

• Determining rate laws through experiments, which gives insight into the role each reactant plays.
• Using methods like the initial rates method to establish how changes in concentration impact the reaction rate.
• Understanding factors such as temperature, presence of a catalyst, and pressure that affect how fast reactions proceed.

Chemical kinetics helps chemists not only determine how to control reaction rates but also provides crucial applications in numerous fields such as pharmacy, where it is essential to know how fast a drug or medicine will act once administered. By understanding the principles of chemical kinetics, one can predict and influence the necessary conditions to achieve the desired reaction speed.