Problem 24 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Hydrolysis of ethyl acetate is catalysed by \(\mathrm{HCl}\). The rate is proportional to the concentration of both the ester and HCl. The rate constant is \(0.1 \mathrm{M}^{-1} \mathrm{~h}^{-1}\). What is the half-life, if the initial concentrations are \(0.02 \mathrm{M}\) for the ester and \(0.01 \mathrm{M}\) for the catalysing acid? (a) \(347 \mathrm{~h}\) (b) \(519 \mathrm{~h}\) (c) \(836 \mathrm{~h}\) (d) \(693 \mathrm{~h}\)

Step-by-Step Solution

Answer

693 hours

Step 1: Understand the Reaction and Rate Law

Recognize that hydrolysis of ethyl acetate is a reaction that follows a rate law where the rate is proportional to the concentration of the ester and HCl, so the rate law can be written as rate = k[ester][HCl], where k is the rate constant.

Step 2: Write the Differential Rate Equation

Set up the differential rate equation based on the rate law. The rate of change in ethyl acetate concentration with respect to time is given by \(\frac{d[ester]}{dt} = -k[ester][HCl]\), where the negative sign indicates that the concentration of ester decreases over time.

Step 3: Integrate the Rate Equation

Since the rate of reaction depends on the concentration of both the ester and HCl which are both reactants, we can assume a pseudo first order reaction where [HCl] remains fairly constant due to its excess amount. The integrated form will be \(\ln([ester]_0/[ester]) = k[HCl]t\), where \(\ln\) is the natural logarithm, \(t\) is time, \(k\) is the rate constant, [ester] is the variable concentration of ester and \([ester]_0\) is the initial concentration of ester.

Step 4: Apply Half-Life Concept

For the half-life \(t_{1/2}\), the concentration of the ester [ester] falls to half its initial value \(\frac{[ester]_0}{2}\). Substitute \(\frac{[ester]_0}{2}\) for [ester] and solve for \(t_{1/2}\).

Step 5: Calculate the Half-Life

Substitute the values into the integrated rate equation and solve for \(t_{1/2}\): \(\ln(2) = k[HCl]t_{1/2}\). Plugging in the values, \(\ln(2) = 0.1 \mathrm{M}^{-1} \mathrm{~h}^{-1} \times 0.01 \mathrm{M} \times t_{1/2}\), we get \(t_{1/2} = \frac{\ln(2)}{0.1 \times 0.01}\).

Key Concepts

Understanding the Half-Life of a Reaction

Chemical reactions proceed at varying rates, and one of the pivotal factors in understanding this is the 'half-life' of a reaction. The half-life, denoted as \( t_{1/2} \), is the time required for the concentration of a reactant to decrease by half its initial amount. In kinetics, it provides a clear gauge of how quickly a reaction progresses. For instance, in the example given with the hydrolysis of ethyl acetate, the half-life can reveal how swiftly the concentration of the ester diminishes.

Understanding half-life is crucial because it's a constant value for a given reaction at a constant temperature, irrespective of the initial concentration, for first order reactions. However, as we've seen in the provided exercise solution, reactions with more than one reactant, like the hydrolysis of ethyl acetate in the presence of HCl, can exhibit complexities. In situations like these, it's useful to apply the concept of a 'pseudo first order reaction' to simplify our calculations of half-life, which arise from the assumption that the concentration of one reactant remains relatively unchanged.

Decoding the Rate Law

The 'rate law' of a chemical reaction expresses the relationship between the rate of the reaction and the concentrations of the reactants. It's typically written in the form \( \text{rate} = k[A]^{m}[B]^{n} \), where \( k \) is the rate constant, \( [A] \) and \( [B] \) are the molar concentrations of the reactants, and \( m \) and \( n \) are the reaction orders with respect to each reactant. These reaction orders are determined experimentally and are not necessarily related to the stoichiometry of the reaction.

In our textbook exercise, the hydrolysis reaction rate is dependent on both the ethyl acetate and HCl concentration, leading to a bimolecular rate law. Here, understanding the rate law is imperative for it allows prediction of how changes in concentration affect the reaction rate, and consequently, considering its direct impact on components such as the half-life of a reaction.

Simplifying with Pseudo First Order Reaction

A 'pseudo first order reaction' is a neat trick used in kinetics when a reaction involves more than one reactant, but one reactant's concentration is so high compared to the others that it remains nearly constant throughout the reaction. Hence, it simplifies the mathematics by not including this reactant in the rate law equation.

In our example, the hydrolysis of ethyl acetate is catalyzed by hydrochloric acid (HCl), and we treat it as a pseudo first order reaction because the acid is in excess. This makes the concentration of HCl effectively constant, thereby allowing us to simplify the differential rate equation that usually involves the concentration of both reactants. Due to this simplification, we can calculate the half-life of the reaction easily, as shown in the step-by-step solution. Remember that the pseudo first order approach is an approximation and is most effective when one reactant's concentration is significantly greater than the other's.