Calculate the half-life of a first-order reaction with a rate constant of 4.5 × 10⁻³ sec⁻¹.

Question

ScanSolve AI Tutor · Accepted Answer

To determine the half-life of a first-order reaction, use the formula:

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

where $t_{1/2}$ is the half-life and $k$ is the rate constant.

Identify the given rate constant.
The rate constant $k = 4.5 \times 10^{-3} \text{ sec}^{-1}$ .
Substitute the value of $k$ into the half-life formula.
$t_{1/2} = \frac{\ln(2)}{4.5 \times 10^{-3} \text{ sec}^{-1}}$
Calculate the value.
We know that $\ln(2) \approx 0.693$ .
$t_{1/2} = \frac{0.693}{4.5 \times 10^{-3} \text{ sec}^{-1}}$ $t_{1/2} = \frac{0.693}{0.0045} \text{ s}$ $t_{1/2} \approx 154 \text{ s}$
Comparing this result with the given options:
(A) 1540 s
(B) 15.4 s
(C) 1.54 s
(D) 154 s

The calculated half-life matches option (D).

The final answer is $\boxed{\text{D) 154 s}}$ .

Calculate the half-life of a first-order reaction with a rate constant of 4.5 x 10⁻³ sec⁻¹.