Question

When the concentration of 'A' is 0.1 M, it decomposes to give ' \(\mathrm{X}\) ' by a firstorder process with a rate constant of \(6.93 \times 10^{-2} \mathrm{~min}^{-1}\). The reactant 'A', in the presence of catalyst, gives ' \(\mathrm{Y}\) ' by a secondorder mechanism with the rate constant of \(0.2 \mathrm{~min}^{-1} \mathrm{M}^{-1} .\) In order to make half-life of both the processes, same, one should start the second-order reaction with an initial concentration of 'A' equal to (a) \(0.01 \mathrm{M}\) (b) \(2.0 \mathrm{M}\) (c) \(1.0 \mathrm{M}\) (d) \(0.5 \mathrm{M}\)

Short answer

0.5 M

Step-by-step solution

Determine the half-life for the first-order reaction

For a first-order reaction, the half-life is given by the formula \( t_{1/2} = \frac{0.693}{k} \), where \(k\) is the rate constant. Here, the rate constant \(k\) for the first-order reaction is \(6.93 \times 10^{-2} \mathrm{min}^{-1}\). The half-life \(t_{1/2}\) can be calculated by substituting the value of \(k\): \[\begin{equation} t_{1/2} = \frac{0.693}{6.93 \times 10^{-2}} \end{equation}\]

Calculate the half-life for the first-order reaction

Using the given formula, \(t_{1/2} = \frac{0.693}{6.93 \times 10^{-2} \mathrm{min}^{-1}} = 10 \mathrm{min}\). This is the half-life for the first-order reaction where 'A' decomposes to give 'X'.

Determine half-life for the second-order reaction

For a second-order reaction, the half-life is given by the formula \( t_{1/2} = \frac{1}{k[A]_0} \), where \(k\) is the rate constant and \( [A]_0 \) is the initial concentration. Here, the rate constant \(k\) for the second-order reaction is \(0.2 \mathrm{min}^{-1}\mathrm{M}^{-1}\) and the initial concentration \( [A]_0 \) is what we need to find so that the half-life is equal to that of the first-order reaction, which is 10 min.

Set up and solve the equation for the initial concentration 'A'

Equating the half-lives of both reactions, we have \[\begin{equation} \frac{1}{0.2[A]_0} = 10 \end{equation}\] Solving for \( [A]_0 \) gives us \[\begin{equation} [A]_0 = \frac{1}{0.2 \times 10} = 0.5 \mathrm{M} \end{equation}\] Hence, one should start the second-order reaction with an initial concentration of 0.5 M for 'A' to make the half-lives of both processes the same.

Key concepts

First-Order Reaction

A first-order reaction is one in which the rate of reaction is directly proportional to the concentration of one of the reactants. In simpler terms, as the reactant's concentration goes down during the reaction, so does the rate at which the reaction occurs. This type of reaction has a constant half-life, meaning the time it takes for half of the reactant to be used up is consistent, regardless of how much of the reactant you started with.

If we have a reaction where compound 'A' decays to product 'X' with the rate constant \( k \), we can represent this process mathematically as: \[ -\frac{d[A]}{dt} = k[A] \]
Here, \( [A] \) stands for the concentration of 'A' at any given time, and \( \frac{d[A]}{dt} \) represents the rate of the reaction. For such a reaction, the half-life (\( t_{1/2} \) ) can be calculated using the equation: \[ t_{1/2} = \frac{0.693}{k} \]
By utilizing this relationship, teachers often help students understand the concept of half-life and how it is unaffected by initial concentration in first-order kinetics.

Second-Order Reaction

In contrast to a first-order reaction, a second-order reaction has a rate that is proportional to the square of the concentration of the reactant, or to the product of the concentrations of two different reactants. This implies that the rate at which the reaction proceeds speeds up or slows down more dramatically based on changes in concentrations than it would for a first-order reaction.

When looking at a reaction where 'A' converts to product 'Y' via a second-order mechanism with rate constant \( k \) and initial concentration \( [A]_0 \), the rate law is expressed as: \[ -\frac{d[A]}{dt} = k[A]^2 \]
Here too, \( [A] \) is the concentration of 'A' and \( \frac{d[A]}{dt} \) is the rate of the reaction. As for the half-life (\( t_{1/2} \) ) for a second-order reaction, the formula differs from that of a first-order reaction: \[ t_{1/2} = \frac{1}{k[A]_0} \]
This formula shows that for second-order reactions, the half-life is inversely proportional to the initial concentration. This intricate relationship is crucial for students to understand, as it demonstrates how the half-life varies according to how much of the reactant is present at the start.

Reaction Half-Life

The half-life of a reaction is a concept used to describe the time required for half of a reactant to be converted into products under certain conditions. It's a commonly used metric because it gives a clear and easily understood way of representing the speed of a reaction. The concept is not limited to nuclear reactions, as sometimes believed, but is also used across chemical kinetics.

In general, the half-life for various order reactions is found using different formulas. For a first-order reaction, the half-life formula as mentioned earlier is: \[ t_{1/2} = \frac{0.693}{k} \]
Meanwhile, for a second-order reaction, the half-life is dependent on the initial concentration as well as the rate constant: \[ t_{1/2} = \frac{1}{k[A]_0} \]
Through exercises that involve half-life calculations, students learn the practical implications of half-life and are able to predict how long a given concentration of reactant will last under different reaction orders, which is vital for fields such as pharmaceuticals, where drug concentration over time is of utmost importance.