Problem 47 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The acid catalysed reaction of acetic acid with ethanol: \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightarrow \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\) \(+\mathrm{H}_{2} \mathrm{O}\) follows the rate law: \(-\frac{\mathrm{d}\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}{\mathrm{d} t}\) \(=K\left[\mathrm{H}^{+}\right] \quad\left[\mathrm{CH}_{3} \mathrm{COOH}\right] \quad\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]\) \(=K^{\prime}\left[\mathrm{CH}_{3} \mathrm{COOH}\right]\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right] .\) When \(\left[\mathrm{CH}_{3} \mathrm{COOH}\right]_{0}=\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]_{0}=0.2 \mathrm{M}\) and \(\mathrm{pH}=3\), the half-life for the reaction is \(50 \mathrm{~min}\). The value of true rate constant, \(K\), of the reaction is (a) \(1.386 \times 10^{-2} \mathrm{~min}^{-1}\) (b) \(0.1 \mathrm{M}^{-1} \mathrm{~min}^{-1}\) (c) \(100 \mathrm{M}^{-2} \mathrm{~min}^{-1}\) (d) \(13.86 \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

The value of the true rate constant, \(K\), of the reaction is \(100 \mathrm{M}^{-2} \mathrm{~min}^{-1}\).

Step 1: Identify the relevant data from the question

The initial concentrations of acetic acid \(\mathrm{CH}_3\mathrm{COOH}\) and ethanol \(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\) are both 0.2 M, the pH of the solution is 3, and the half-life (\(t_{1/2}\)) of the reaction is 50 min. The given rate equation is \(\frac{-d[\mathrm{CH}_3\mathrm{COOH}]}{dt} = K'[\mathrm{CH}_3\mathrm{COOH}][\mathrm{C}_2\mathrm{H}_5\mathrm{OH}]\).

Step 2: Calculate the \(\mathrm{H}^+\) concentration using the pH

Using the formula \(\mathrm{pH} = -\log[\mathrm{H}^+]\), calculate the concentration of hydrogen ions: \[\mathrm{pH} = 3\] implies \[\log[\mathrm{H}^+] = -3\] which gives \[\mathrm{[H}^+] = 10^{-3} \mathrm{M}\].

Step 3: Connect the half-life to the rate constant for a second-order reaction

Since the reaction follows a second-order rate law given by \(K'\), we use the half-life formula for a second-order reaction, which is \(t_{1/2} = \frac{1}{K'[A]_0}\), where \(K'\) is the rate constant and \(\mathrm{[A]}_0\) is the initial concentration of the reactant.

Step 4: Calculate the pseudo rate constant \(K'\)

Substitute the given half-life and initial concentration into the half-life formula for a second-order reaction to solve for \(K'\): \[50 \mathrm{min} = \frac{1}{K' \times 0.2 \mathrm{M}}\] which simplifies to \[K' = \frac{1}{50 \mathrm{min} \times 0.2 \mathrm{M}} = 0.1 \mathrm{M^{-1}~min^{-1}}\].

Step 5: Calculate the true rate constant \(K\)

Substitute the values of \(K'\) and \(\mathrm{[H}^+]\) into the relation \(K' = K[\mathrm{H}^+]\) to find the true rate constant \(K\): \[0.1 \mathrm{M^{-1}~min^{-1}} = K \times 10^{-3} \mathrm{M}\] which leads to \[K = \frac{0.1 \mathrm{M^{-1}~min^{-1}}}{10^{-3} \mathrm{M}} = 100 \mathrm{M^{-2}~min^{-1}}\].

Key Concepts

Chemical Kinetics

In the study of chemical reactions, chemical kinetics is the branch of physical chemistry that is concerned with understanding the rates of chemical reactions. It involves the examination of how different experimental conditions can influence the speed of a chemical reaction and determines the sequence of steps that make up the reaction mechanism.

For instance, in the acid-catalyzed esterification of acetic acid with ethanol, the concentration of reactants and the presence of a catalyst (such as hydrogen ions) are key factors that affect the rate at which products are formed. By analyzing the relationship between the rate of reaction and the concentration of reactants, we apply the principles of chemical kinetics to deduce the rate law, which in this case is found to be second-order. The rate law mathematically expresses the connection between the rate of the reaction and the concentration of the reactants, providing insight into the mechanism of the reaction.

Second-order Reaction

A second-order reaction is identified by the dependence of the rate on the second power of concentration, typically involving either two first-order reactants or one second-order reactant. In the given problem, the esterification reaction rate depends on the product of the concentrations of acetic acid and ethanol, thus indicating that it is indeed a second-order reaction.

In terms of mathematical representation, a second-order rate law is typically written as rate = k[A][B] or rate = k[A]^2 for reactions involving two different reactants A and B, or a single reactant A, respectively. An important characteristic of second-order reactions is that the plot of the inverse of concentration versus time yields a straight line, confirming the second-order kinetics. The rate constant (k) has units of concentration^-1 time^-1, reflecting how the rate of reaction changes with a change in concentration of reactants.

Half-life of Reaction

The half-life of a reaction refers to the time it takes for a reactant's concentration to decrease by half its initial value. It is a crucial concept in both chemical kinetics and radioactive decay processes. For a second-order reaction, the half-life is inversely proportional to both the rate constant and the initial concentration of the reactant, expressed as t_{1/2} = 1 / (k[A]_0).

Unlike first-order reactions that have a constant half-life irrespective of concentration, the half-life for second-order reactions varies with the initial concentration of the reactants. The exercise provided explicitly deals with this concept by calculating the half-life given the rate constant and initial concentration. Knowing the half-life enables chemists to predict how long it will take for a reactant to reach a certain concentration, a particularly useful tool for controlling industrial chemical processes and managing the dosage of medications in pharmacokinetics.