Question

The rate constant for a first-order reaction is \( 0.0693 \, \text{min}^{-1} \). What is the half-life of the reaction?

• A. \( 5 \, \text{min} \)
• B. \( 10 \, \text{min} \)
• C. \( 15 \, \text{min} \)
• D. \( 20 \, \text{min} \)

Hint:

For first-order reactions, the half-life is independent of the initial concentration and can be quickly calculated using \( t_{1/2} = \frac{\ln 2}{k} \). Memorize \( \ln 2 \approx 0.693 \) for efficiency.

Answer 1

The Correct Option is B

Step 1 — Recall the half-life formula for a first-order reaction

For a first-order reaction the half-life is independent of concentration and is given by the well-known formula:
\( \displaystyle t_{1/2} = \frac{\ln 2}{k} \).

Step 2 — Substitute values and compute

Use \(\ln 2 \approx 0.693147\) (often approximated as 0.693).
\[ t_{1/2} = \frac{0.693147\ldots}{0.0693\ \text{min}^{-1}} \approx \frac{0.693147}{0.0693}. \] Since \(0.0693\times 10 = 0.693\), the quotient is approximately \[ t_{1/2} \approx 10.00\ \text{min}. \] (Using the more precise ln2 gives \(t_{1/2}\approx 0.693147/0.0693 \approx 10.002\) min, which rounds to 10.0 min.)

Step 3 — Units and significant figures

• The rate constant k is given in min⁻¹, so the resulting time is in minutes.
• Given k = 0.0693 (4 significant digits) and ln2 ≈ 0.6931, the half-life computed to two or three significant figures is 10.0 min. For typical multiple-choice reporting, 10 min is the correct choice.

Quick verification

If \(t_{1/2}=10\) min, then \(k = \ln2 / t_{1/2} \approx 0.693147/10 = 0.0693147\ \text{min}^{-1}\), which matches the given k = 0.0693 min⁻¹ within the given precision.

Final answer (clear)

The half-life is \( \boxed{\,10\ \text{min}\,} \). (Select the option: 10 min.)

Note: This derivation follows directly from the integrated first-order rate law. The result is independent of initial concentration and depends only on k.