Question

Half-life \( (t_{1/2}) \) of a first order reaction is 1386 s. The value of rate constant is :

• A. \( 0 \cdot 5 \times 10^4 \text{ s}^{-1} \)
• B. \( 5 \cdot 0 \times 10^{-4} \text{ s}^{-1} \)
• C. \( 0 \cdot 5 \times 10^{-5} \text{ s}^{-1} \)
• D. \( 0 \cdot 5 \times 10^{-2} \text{ s}^{-1} \)

Hint:

Always remember the constant 0.693 for first-order half-life calculations.
Ensure that units of time in the rate constant are inverse to the units of half-life.
Simplifying numbers into powers of 10 makes the division easier without a calculator.

Answer 1

The Correct Option is B

Concept:

• For a first-order reaction, the rate constant \( (k) \) is independent of the initial concentration of the reactant.

• The relationship between the half-life \( (t_{1/2}) \) and the rate constant \( (k) \) is given by a specific mathematical expression derived from the integrated rate law.
Step 1: Recall the formula for the rate constant of a first-order reaction
The relationship is:
\[ k = \frac{0.693}{t_{1/2}} \]

Step 2: Substitute the given values into the formula
Given \( t_{1/2} = 1386 \text{ s} \).
\[ k = \frac{0.693}{1386} \]

Step 3: Perform the calculation
\[ \begin{aligned} k &= \frac{693 \times 10^{-3}}{1386} \\ k &= \frac{1}{2} \times 10^{-3} \\ k &= 0.5 \times 10^{-3} \\ k &= 5.0 \times 10^{-4} \text{ s}^{-1} \end{aligned} \]
The final answer is (B).