Question

For the second-order reaction: \(2 \mathrm{~A} \rightarrow \mathrm{B}\), time taken for the \([\mathrm{A}]\) to fall to one-fourth value is how many times the time it takes for \([\mathrm{A}]\) to fall to half of its initial value? (a) two (b) three (c) four (d) seven

Short answer

The time taken for the concentration to fall to one-fourth is three times the time it takes to fall to half, so the correct answer is (b) three.

Step-by-step solution

Write down the second-order rate equation

For a second-order reaction of the form \(2A \rightarrow B\), the rate equation can be written as \(\frac{1}{[A]} - \frac{1}{[A]_0} = kt\), where \(k\) is the rate constant, \([A]\) is the concentration of A at time \(t\), and \([A]_0\) is the initial concentration of A.

Time to reduce the concentration to half

Let the time taken for \([A]\) to fall to half of its initial value be \(t_{1/2}\). Substituting \([A] = \frac{1}{2}[A]_0\) into the rate equation, we get \(\frac{2}{[A]_0} - \frac{1}{[A]_0} = kt_{1/2}\). Simplifying, we find that \(kt_{1/2} = \frac{1}{[A]_0}\).

Time to reduce the concentration to one-fourth

Let the time taken for \([A]\) to fall to one-fourth of its initial value be \(t_{1/4}\). Substituting \([A] = \frac{1}{4}[A]_0\) into the rate equation, we get \(\frac{4}{[A]_0} - \frac{1}{[A]_0} = kt_{1/4}\). Simplifying, we find that \(kt_{1/4} = \frac{3}{[A]_0}\).

Compare \(t_{1/4}\) with \(t_{1/2}\)

Divide the equation \(kt_{1/4} = \frac{3}{[A]_0}\) by \(kt_{1/2} = \frac{1}{[A]_0}\) to find the ratio \(\frac{t_{1/4}}{t_{1/2}} = \frac{\frac{3}{[A]_0}}{\frac{1}{[A]_0}} = 3\). Therefore, the time taken for the concentration to fall to one-fourth is three times the time it takes to fall to half.

Key concepts

Chemical Kinetics

Chemical kinetics is a branch of physical chemistry that studies the rates at which chemical reactions occur and determines the factors influencing them. Understanding kinetics is crucial for controlling how fast reactions proceed, which is essential in industries like pharmaceuticals and environmental engineering.

For instance, in the context of the second-order reaction presented in the exercise, kinetics not only tell us how fast the reactant concentrations decline, but also how these changes depend on concentration and time. The kinetic analysis of reactions, especially those of higher order like the second-order reaction in our example, requires a thorough grasp of the mathematical relationship between reaction rate and concentration.

Reaction Rate Equation

The reaction rate equation, or rate law, expresses the relationship between the rate of a chemical reaction and the concentration of the reactants. For a second-order reaction such as the one given by the equation 2A → B, the rate is proportional to the square of the concentration of A.

Expressed mathematically, the second-order rate law is written as rate = k[A]², where k is the rate constant specific to the reaction at a given temperature. By integrating this rate law over time, we can find the relationship between the initial concentration of the reactant and its concentration at any later time, which is crucial for determining how long a certain process will take.

Half-life of Reaction

The half-life of a reaction refers to the time it takes for the concentration of a reactant to decrease to half of its initial value. It's a widely used metric in chemical kinetics because it provides a straightforward way of comparing the rates of different reactions and also has implications in areas such as pharmacology and radioactivity.

For first-order reactions, the half-life remains constant throughout the reaction. However, for second-order reactions, the half-life depends on the initial concentration of the reactants, as seen in the step-by-step solution provided. By understanding the concept of half-life, chemists can predict how quickly a reactant will be consumed or a product formed, which is essential for the design and operation of chemical reactors.