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) \(14^{2} \mathrm{~s}\)

Short answer

The time required is 42 seconds, option (c).

Step-by-step solution

Identify the Relationship Between Half-life and Time

First, note that for a first-order reaction, the time for the initial concentration to reduce by half is known as the half-life, denoted by \(t_{1/2}\). For this problem, \(t_{1/2} = 14 \ \mathrm{s}.\)

Find the Relationship for Reduction to 1/8th

The concentration reducing to \(\frac{1}{8}\) of its initial value can be seen as an occurrence of three half-lives because \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\).

Calculate the Time for Three Half-lives

Since three half-lives are required to reduce the concentration to \(\frac{1}{8}\) of its original value, multiply the half-life by 3: \[ \text{Time required} = 3 \times t_{1/2} = 3 \times 14 \ \mathrm{s} = 42 \ \mathrm{s}. \]

Select the Correct Answer

From the calculated result, the time required is \(42 \ \mathrm{s}\), which corresponds to option (c) \(42 \ \mathrm{s}\).

Key concepts

Understanding Half-life in First-order Reactions

In the world of chemical reactions, the concept of half-life is particularly important for understanding how a substance transforms over time. For a first-order reaction, the half-life is the time it takes for the concentration of a reactant to decrease by half. This period, denoted as \( t_{1/2} \), remains constant throughout the reaction. This means regardless of how much substance you start with, it will always take the same amount of time to reach half its initial concentration. The half-life for first-order reactions can be derived from the equation: \[ t_{1/2} = \frac{0.693}{k} \] where \( k \) is the rate constant. Since the half-life tells us how quickly a substance reacts over a given time, it allows for straightforward predictions about the progress of a chemical reaction.

• If you start with a certain mass of a reactant, after one half-life, only half of that mass will remain.
• After two half-lives, the amount will be halved again, leaving one-quarter of the original amount.

This process continues until either all reactants are exhausted, or external conditions alter the reaction.

Exploring Concentration Reduction

When analyzing first-order reactions, concentration reduction plays a critical role. It involves determining how much of a reactant remains after a certain period. The initial concentration reduces exponentially in these reactions. For instance, reducing a substance to \( \frac{1}{8} \)th of its initial concentration in a first-order reaction means observing the span of three half-lives. Here's why:

• In the first half-life, the concentration reduces from its initial value to half.
• In the second half-life, this remaining concentration halves again, totalling to \( \frac{1}{4} \)th of the initial concentration.
• Finally, in the third half-life, the concentration halves yet again, resulting in \( \frac{1}{8} \)th of the initial concentration.

This illustrates the exponential nature of concentration reduction within first-order reactions, making it predictable, as each half-life signifies a consistent division.

The Process of Kinetics Calculation

Kinetics calculation provides a methodical approach to understanding the rate at which reactions proceed. For a first-order reaction, it's particularly tailored in the half-life and exponential decay concept, needing only the rate constant to describe how quickly a reaction occurs. The kinetics calculation of the time needed for reducing the concentration by a factor uses the derived formula: \[ t = \frac{\ln(N_0/N)}{k} \] where \( N_0 \) is the initial concentration, \( N \) is the concentration after time \( t \), and \( k \) is the rate constant. From this equation:

• To calculate time for concentration reduction to \( \frac{1}{8} \)th of its initial, it involves solving \( t = 3 \times t_{1/2} \).
• Here it's simplified as multiplying the half-life by the number of periods needed (in this case, three), which gives \( 42 \ \textrm{s} \).

Through a blend of logical deduction and arithmetical operations, kinetic calculations provide a reliable framework for predicting how quickly reactions manage the transformation of substances over time.