Question

The half-life of a first order reaction is 1386 s. The rate constant \(k\) is:

• A. \(5 \times 10^{-4}\ \text{s}^{-1}\)
• B. \(2 \times 10^{-4}\ \text{s}^{-1}\)
• C. \(1 \times 10^{-3}\ \text{s}^{-1}\)
• D. \(4 \times 10^{-3}\ \text{s}^{-1}\)

Hint:

For first order reactions: \[ t_{1/2} \propto \frac{1}{k} \] Half-life is independent of initial concentration.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a first-order reaction, the rate depends only on the concentration of one reactant raised to the power of one.
A unique property of first-order kinetics is that the half-life (\(t_{1/2}\))—the time it takes for the concentration to drop to half its initial value—is constant and independent of the starting concentration.
The relationship between the rate constant (\(k\)) and half-life (\(t_{1/2}\)) is derived from the integrated rate law.
Key Formula or Approach:
The formula for the half-life of a first-order reaction is:
\[ t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k} \]
To find the rate constant \(k\), we rearrange the formula:
\[ k = \frac{0.693}{t_{1/2}} \]
Step 2: Detailed Explanation:
Given data:
Half-life (\(t_{1/2}\)) = \(1386\) s
Substituting the value into the rearranged formula:
\[ k = \frac{0.693}{1386} \]
To simplify the division without a calculator, look for mathematical patterns. Notice that \(693 \times 2 = 1386\).
Therefore, we can rewrite \(0.693\) as \(693 \times 10^{-3}\).
\[ k = \frac{693 \times 10^{-3}}{1386} \]
\[ k = \frac{1}{2} \times 10^{-3} \]
\[ k = 0.5 \times 10^{-3} \]
\[ k = 5 \times 10^{-4} s^{-1} \]
The units for a first-order rate constant are always \(time^{-1}\), in this case, \(s^{-1}\).
Step 3: Final Answer:
The rate constant \(k\) is \(5 \times 10^{-4} s^{-1}\).
This matches Option (A).