Understanding First-Order Reaction Kinetics

A first-order reaction is a reaction whose rate depends linearly on the concentration of only one reactant. The rate of the reaction is proportional to the concentration of a single reactant.

For a first-order reaction, the integrated rate law can be expressed as:

\( \ln[A]_t = \ln[A]_0 - kt \)

or, rearranging to solve for the rate constant \(k\):

\( k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}\right) \)

Using base-10 logarithm, the equation becomes:

\( k = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]_t}\right) \)

Here:

• \( [A]_0 \) is the initial concentration of reactant A.
• \( [A]_t \) is the concentration of reactant A at time \(t\).
• \( k \) is the rate constant of the reaction.
• \( t \) is the time elapsed.

Calculating the Rate Constant from Given Data

We are given that the concentration of reactant A is reduced to 25% of its original value in 6.96 months. This means:

• \( t = 6.96 \) months
• \( [A]_t = 0.25 \times [A]_0 \)

Let's substitute these values into the integrated rate law equation:

\( k = \frac{2.303}{6.96 \text{ months}} \log\left(\frac{[A]_0}{0.25[A]_0}\right) \)

Simplify the concentration ratio:

\( k = \frac{2.303}{6.96} \log(4) \)

We know that \( \log(4) = \log(2^2) = 2 \times \log(2) \). We are given \( \log 2 = 0.3010 \).

So, \( \log(4) = 2 \times 0.3010 = 0.6020 \).

Substitute this value back into the equation for \(k\):

\( k = \frac{2.303}{6.96} \times 0.6020 \)

\( k = \frac{1.386}{6.96} \)

\( k \approx 0.1991 \text{ months}^{-1} \)

Relating Rate Constant to Half-Life

The half-life (\( t_{1/2} \)) of a reaction is the time required for the concentration of a reactant to be reduced to half of its initial value.

For a first-order reaction, the half-life is independent of the initial concentration and is related to the rate constant \(k\) by the formula:

\( t_{1/2} = \frac{\ln(2)}{k} \)

Since \( \ln(2) \approx 0.693 \), the formula is often written as:

\( t_{1/2} = \frac{0.693}{k} \)

Calculating the Half-Life of the Reaction

Now we can use the calculated value of \(k\) to find the half-life:

\( t_{1/2} = \frac{0.693}{0.1991 \text{ months}^{-1}} \)

\( t_{1/2} \approx 3.48 \text{ months} \)

Alternative Method Using Half-Life Concept

For a first-order reaction, the time taken for the concentration to reduce by a certain fraction can also be understood in terms of half-lives.

• After one half-life (\( t_{1/2} \)), the concentration is 50% of the original.
• After two half-lives (\( 2 \times t_{1/2} \)), the concentration is 50% of 50%, which is 25% of the original.
• After three half-lives (\( 3 \times t_{1/2} \)), the concentration is 50% of 25%, which is 12.5% of the original, and so on.

In this problem, the concentration is reduced to 25% of the original value, which corresponds to two half-lives.

So, the given time (6.96 months) is equal to two half-lives:

\( 2 \times t_{1/2} = 6.96 \text{ months} \)

Solving for \( t_{1/2} \):

\( t_{1/2} = \frac{6.96 \text{ months}}{2} \)

\( t_{1/2} = 3.48 \text{ months} \)

Both methods yield the same result.

Summary of Calculations

The half-life of the reaction is 3.48 months.

Revision Table: First-Order Kinetics Key Concepts

Additional Information: Reaction Order and Half-Life

The relationship between half-life and concentration (or lack thereof) is a key characteristic distinguishing reaction orders.

• Zero-Order Reaction: The half-life (\( t_{1/2} = \frac{[A]_0}{2k} \)) depends on the initial concentration \( [A]_0 \). As concentration decreases, the half-life decreases.
• First-Order Reaction: The half-life (\( t_{1/2} = \frac{0.693}{k} \)) is independent of the initial concentration \( [A]_0 \). The time taken for the concentration to reduce by half is constant throughout the reaction. This is why the shortcut method used above works for first-order reactions.
• Second-Order Reaction (Type 1: \( rate = k[A]^2 \)): The half-life (\( t_{1/2} = \frac{1}{k[A]_0} \)) depends on the initial concentration \( [A]_0 \). As concentration decreases, the half-life increases.

Understanding how half-life behaves with changing concentration can help identify the order of a reaction experimentally.