Problem 62 - A Complete Resource Book in Chemistry for JEE Main

Question

For the first-order reaction half-life is \(14 \mathrm{~s}\). The time required for the initial concentration to reduce to \(1 / 8\) th of its value is (a) \(21 \mathrm{~s}\) (b) \(32 \mathrm{~s}\) (c) \(42 \mathrm{~s}\) (d) \(14^{2} \mathrm{~s}\)

Step-by-Step Solution

Answer

The time required is 42 seconds (option c).

Step 1: Understand the First-Order Reaction

In a first-order reaction, the rate at which the reaction proceeds is directly proportional to the concentration of the reactant. The half-life (
t_{1/2}
) for such reactions is constant and does not depend on the initial concentration.

Step 2: Calculate Half-Life Intervals

The half-life (
t_{1/2}
) is 14 s. It means that every 14 seconds, the concentration of the reactant is reduced to half. We need to determine how many half-lives are required to reduce the initial concentration to
1/8
of its initial value.

Step 3: Determine Number of Half-Lives Required

To reduce the concentration to
1/8
, we need to solve
(1/2)^n = 1/8
, where
n
is the number of half-lives. Solving,
n
is found to be 3, because
1/8 = (1/2)^3
.

Step 4: Calculate Total Time Required

Since 3 half-lives are required, the total time needed is
3 imes 14
seconds (as each half-life lasts for 14 seconds).

Step 5: Solve for Total Time

\[\text{Total time} = 3 \times 14 = 42 \text{ seconds}\]This means that 42 seconds are necessary to reduce the concentration to
1/8
of its initial value.

Key Concepts

Half-Life Calculation

Understanding the concept of half-life is crucial in chemistry, especially when dealing with first-order reactions. A half-life, denoted as \( t_{1/2} \), refers to the time it takes for the concentration of a reactant to reduce to half its original value. In a first-order reaction, this half-life is constant and does not depend on the initial concentration. This unique property allows us to predict how the concentration changes over time.
For example, if a reaction has a half-life of 14 seconds, it means that after every 14 seconds, the concentration reduces to half. The mathematical representation for determining how many half-lives are needed to achieve a certain concentration reduction is typically given by \((1/2)^n\), where \(n\) is the number of half-lives. Each cycle of reduction allows chemists to predict when a specific fraction of the reactant remains.

Concentration Reduction

Concentration reduction in a first-order reaction is a fundamental concept, especially when studying how reactions proceed over time. As the reaction progresses, the concentration of the reactant continuously decreases.
To find out how long it takes for the concentration to reduce to a specific fraction, such as \(1/8\), we use the idea of successive half-lives. For instance, when the concentration needs to go from its initial amount to \(1/8\) of its original value, you need to establish how many times it has to be halved to get to \(1/8\). The calculation is straightforward with the equation \((1/2)^n = 1/8 \), which helps determine that \(n = 3\). Hence, it takes three half-lives to reach this point.

Rate of Reaction

The rate of reaction is a central idea in kinetics, defining how fast or slow a reaction occurs. For first-order reactions, the rate is directly proportional to the concentration of the reactant remaining. This means that as the reactant is consumed, the reaction progressively slows down because there is less reactant available to convert into products.
Since first-order reactions have a constant half-life, the description of the rate connects deeply with the concept of half-life. As time passes and the concentration decreases, the rate decreases, but the time it takes for these reductions to reach any fraction of the initial concentration remains predictable and systematic. This proportional relationship is what makes first-order reactions particularly straightforward to model and study.