If the half-life of a particular radioactive substance (which follows first-order kinetics) is exact — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

$t_{1/2} = 36 \text{ mins}$. Let's calculate the number of half-lives in 2 hours ($120 \text{ mins}$): $n = 120 / 36 = 10 / 3 = 3.333$. Alternatively, use the rate law: $k = \ln 2 / 36 \text{ min}^{-1}$. Using the integrated rate law: $\log([A]_0/[A]_t) = \frac{kt}{2.303} = \frac{0.693}{36} \times \frac{120}{2.303} = \frac{0.301}{36} \times 120 = \frac{36.12}{36} \approx 1.00$. Therefore, $\log(10/A_t) = 1 \Rightarrow 10/A_t = 10 \Rightarrow A_t = 1 \text{ g}$.