Determine the half-life of a first order reaction with constant 4.5 x 10$^{-3}$ sec$^{-1}$.

Analysis

To solve this problem, we need to understand the relationship between the rate constant ($k$) and the half-life ($t_{1/2}$) for a first-order reaction.

Concept

For any first-order reaction, the half-life is the time required for the concentration of a reactant to decrease to half of its initial value. The formula connecting the rate constant ($k$) and the half-life ($t_{1/2}$) is:

Since $\ln(2) \approx 0.693$, the formula simplifies to:

Solution

Step 1: Identify the given values.

• Rate constant ($k$) = $4.5 \times 10^{-3} \text{ s}^{-1}$

Step 2: Substitute the values into the formula.

Step 3: Perform the calculation.

• First, handle the power of 10: $\frac{1}{10^{-3}} = 10^3 = 1000$
• Now, calculate the division: $\frac{0.693}{4.5} \times 1000$
• $0.693 \div 4.5 = 0.154$
• $0.154 \times 1000 = 154 \text{ s}$

The half-life of the reaction is 154 seconds.

Conclusion

Comparing our result with the provided options, we can see that it matches option A.

Exam Tip

When dealing with first-order kinetics in JAMB, always remember the constant 0.693. If you ever forget the formula, just remember that the half-life is inversely proportional to the rate constant. If $k$ is a very small number (like $10^{-3}$), your answer for $t_{1/2}$ should be a relatively large number. This helps you quickly eliminate impossible options!