Question

The rate of the reaction $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q) &+\mathrm{H}_{2} \mathrm{O}(l) \\ \longrightarrow & \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \end{aligned} $$ shows first-order characteristics-that is, rate \(=\) \(k\left[\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\right]\) - even though this is a second- order reaction (first order in \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\) and first order in \(\mathrm{H}_{2} \mathrm{O}\) ). Explain.

Short answer

Even though this reaction is technically second order, it shows first order characteristics because the water, being the solvent, has its concentration held constant. Hence, its effect on reaction rate can be incorporated into the rate constant.

Step-by-step solution

Understanding Reaction Order

An order of reaction refers to how the rate of reaction responds to the concentration of one or more reactants. It's the power to which its concentration term in the rate equation is raised. A first order reaction implies that the rate of reaction is directly proportional to the concentration of one reactant. A second order implies the rate is proportional to the square of the concentration of one reactant or to the concentrations of two different reactants.

Analyze the Rate Expression

Looking at the given rate of reaction which is \( k\left[\mathrm{CH}_{3}\mathrm{COOC}_{2}\mathrm{H}_{5}\right] \), it shows the reaction is first order with respect to the ester reactant. However, it's said that the reaction is second order which implies that the rate of reaction is influenced by the concentration of another reactant, which is water in this case.

Explain The Contradiction In Reaction Order

In the given reaction, water is a reactant. However, because water is a solvent and its concentration remains effectively constant during the reaction, it can be incorporated into the rate constant to give an observed rate constant. Hence, despite the reaction being technically second order (first order to the ester, first order to water), it's experimental rate law shows first-order characteristics, where the rate of the reaction depends on the concentration of the ester, because the concentration of water is held constant.

Key concepts

Rate Law

The rate law of a chemical reaction is a mathematical expression that links the reaction rate to the concentration of its reactants.
It provides vital insight into how variations in reactant concentrations can influence the speed of a reaction, and is typically determined experimentally.

In the rate law, the rate of a reaction is often expressed as: \[ \text{Rate} = k[A]^m[B]^n \]where:

• \( k \) is the rate constant, a unique value for each reaction at a given temperature.
• \( [A] \) and \( [B] \) represent the concentrations of reactants, usually in moles per liter.
• \( m \) and \( n \) are the reaction orders with respect to each reactant, which dictate how changes in concentration affect the reaction rate.

These orders do not always match the stoichiometric coefficients of the chemical equation, but they do outline how sensitive the rate is to concentration changes. This is why understanding the rate law is pivotal in reaction kinetics, allowing chemists to make predictions about how a reaction will behave under different conditions.

First Order Reaction

A first-order reaction is a type of chemical reaction where the rate is directly proportional to the concentration of one reactant.
This characteristic means that if the concentration of the reactant is doubled, the reaction rate also doubles.

The general form of the rate law for a first-order reaction is: \[ \text{Rate} = k[A] \]where:

• \( k \) is the rate constant, which remains constant throughout the reaction.
• \( [A] \) is the concentration of the reactant.

This type of reaction exhibits exponential decay, meaning that the concentration of the reactant decreases exponentially over time.
The equation that describes this behavior is given by: \[ [A] = [A]_0 e^{-kt}\]where \([A]_0\) is the initial concentration of the reactant, and \(t\) is time.

Understanding first-order kinetics is crucial because many processes, such as radioactive decay and the hydrolysis of esters in aqueous solutions, follow this order. It also simplifies calculations because the half-life of a first-order reaction is independent of the initial concentration.

Second Order Reaction

Second-order reactions introduce additional complexity compared to first-order reactions.
In these reactions, the rate depends on the square of the concentration of one reactant or the product of the concentrations of two different reactants.

The rate law for a typical second-order reaction is written as: \[ \text{Rate} = k[A]^2 \text{ or } \text{Rate} = k[A][B] \]where:

• \( k \) is again the rate constant.
• \( [A] \) and \( [B] \) signify the reactant concentrations.

If the reaction depends on the concentration of two different reactants, doubling one concentration can quadruple the reaction rate.
Second-order reactions don't follow exponential decay; instead, they conform to a more complex mathematical model.
The integrated rate law in its simplest form (for one reactant) is:\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]
This equation highlights how the reaction slows down more dramatically over time compared to first-order reactions.
Second-order kinetics are crucial to understand, as they can often be observed in many bimolecular reactions, such as those involving enzyme kinetics or formation of dimers in chemistry.