Question

Consider the reaction \(\mathrm{A} \rightleftarrows \mathrm{B}\), which is first order in each direction with rate constants \(k\) and \(k^{\prime}\). Initially, only A is present. Show that the concentrations approach their equilibrium values at a rate that depends on \(k\) and \(k^{\prime}\).

Short answer

The concentrations of A and B approach equilibrium as an exponential function of time, with the rate of approach determined by the rate constants k and k'.

Step-by-step solution

Define the rate of reaction for forward and reverse reactions

The rate of the forward reaction is given by Rate_forward = k[A], where [A] is the concentration of A. The rate of the reverse reaction is given by Rate_reverse = k'[B], where [B] is the concentration of B.

Write the differential equations for the reactions

Since the reaction is first-order in both directions, we can write the differential equations for the concentrations of A and B as: \( \frac{d[A]}{dt} = -k[A] + k'[B] \) and \( \frac{d[B]}{dt} = k[A] - k'[B] \).

Define the equilibrium condition

At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. Therefore, \( k[A]_{eq} = k'[B]_{eq} \), where \( [A]_{eq} \) and \( [B]_{eq} \) are the equilibrium concentrations of A and B, respectively.

Relate the rate of approach to equilibrium with rate constants

Since initially only A is present, as the reaction proceeds, [A] decreases while [B] increases, but the sum of the concentrations remains constant (assuming the volume does not change). Thus at any time t, \( [A] + [B] = [A]_0 \), where \( [A]_0 \) is the initial concentration of A. Using the equilibrium condition from Step 3, we can express \( [B] \) in terms of \( [A] \) and substitute it into the differential equations to get an equation that shows how the concentration of A changes with time and depends on k and k'.

Solve the differential equation

By solving the differential equation from Step 4, \( \frac{d[A]}{dt} = -k[A] + k'([A]_0 - [A]) \), we can find an expression for \( [A] \) as a function of time that shows how it approaches equilibrium exponentially with a rate that depends on both \( k \) and \( k' \). This can be solved by separation of variables or other appropriate techniques for solving first-order linear differential equations.

Key concepts

Rate Constants

The concept of a rate constant is fundamental in the study of chemical kinetics, which deals with the speeds, or rates, of chemical reactions. A rate constant, denoted by the symbol k, relates the reaction rate to the concentrations of reactants in a rate equation, encapsulating the effect of various factors including temperature and catalyst presence on the rate of a reaction.

For the straightforward reaction \(\mathrm{A} \rightleftarrows \mathrm{B}\) involving the conversion of species A to species B (forward reaction) and vice versa (reverse reaction), we have two rate constants, k for the forward reaction and k' for the backward reaction. These constants help us understand how quickly a system moves towards equilibrium, which is a state where the rates of the forward and backward reactions are the same. Without knowledge of the rate constants, it would be impossible to predict how long it will take for a given system to reach equilibrium or how changing conditions could affect the speed of that approach.

First-Order Reaction

In the context of chemical kinetics, a first-order reaction is one where the rate is directly proportional to the concentration of one reactant. Mathematically, for a reaction \( A \rightarrow products \), the rate can be expressed as \(\text{Rate} = k[A]\), where k is the rate constant and [A] is the concentration of reactant A.

This type of reaction is characterized by its simplicity in mathematical treatment. A key feature of first-order reactions is their exponential decay of reactant concentration over time, which can be described using a simple differential equation. For the reaction \(\mathrm{A} \rightleftarrows \mathrm{B}\), both the forward and reverse reactions are first-order, meaning each rate depends on the concentration of its respective reactant. This fact allows us to predict how the concentrations of A and B will change over time and is especially useful for calculating the time it will take for the concentrations to get close to their equilibrium values.

Equilibrium Concentration

The equilibrium concentration refers to the amounts of reactants and products in a reaction mixture when the rates of the forward and reverse reactions are equal, resulting in no net change in concentration over time. At equilibrium, the concentration of reactants and products remains constant, though not necessarily equal, and is determined by the rate constants of the forward and reverse reactions.

In the system \(\mathrm{A} \rightleftarrows \mathrm{B}\), where only A is initially present, the equilibrium concentrations \( [A]_{eq} \) and \( [B]_{eq} \) satisfy the relationship \( k[A]_{eq} = k'[B]_{eq} \). By considering the total concentration of A initially present, denoted \( [A]_0 \), we can define a relation such that \( [A] + [B] = [A]_0 \) at any point before reaching equilibrium. The journey to equilibrium can thus be mapped by tracking the changing concentrations of A and B over time using the integrated rate equation, and the equilibrium concentration is a crucial component for quantifying the position of equilibrium in a chemical system.