Question

the half life of a first order reaction is $7.80 \times 10^{2}$ sec. calculate the rate constant, k.
$0.217 \text{ sec}^{-1}$
$0.313 \text{ sec}^{-1}$
$8.88 \times 10^{-4} \text{ sec}^{-1}$
$0.150 \text{ sec}^{-1}$

Explanation:

Step1: Recall first-order half-life formula

For a first-order reaction, the half-life $t_{1/2}$ is related to the rate constant $k$ by:
$$t_{1/2} = \frac{\ln 2}{k}$$
Rearranged to solve for $k$:
$$k = \frac{\ln 2}{t_{1/2}}$$

Step2: Substitute given values

Given $t_{1/2} = 7.80 \times 10^2$ sec, and $\ln 2 \approx 0.6931$:
$$k = \frac{0.6931}{7.80 \times 10^2}$$

Step3: Calculate the value

$$k = \frac{0.6931}{780} \approx 8.88 \times 10^{-4} \text{ sec}^{-1}$$

Answer:

8.88 x 10⁻⁴ sec⁻¹