The first-order reaction,SO₂Cl₂ → SO₂ + Cl₂,has a half-life of 8.75 hours at 593 K.How long will it take for the concentration of SO₂Cl₂ to fall to 12.5% of its initial value?

Question

Quizplus · Accepted Answer

For a first-order reaction, the time it takes for the concentration to fall to a certain percentage of its initial value can be calculated using the formula $t = \frac{\ln(\frac{[A]_0}{[A]})}{k}$, where $[A]_0$ is the initial concentration, $[A]$ is the final concentration, and $k$ is the rate constant. Falling to 12.5% of its initial value means the concentration falls to $\frac{1}{8}$ of its initial value, which corresponds to 3 half-lives ($2^3 = 8$). Since the half-life is 8.75 hours, the total time is $3 \times 8.75 = 26.25$ hours, which is approximately 26.2 hours.

The First-Order Reaction,SO2Cl2 → SO2 + Cl2,has a Half-Life of 8.75

The First-Order Reaction,SO₂Cl₂ → SO₂ + Cl₂,has a Half-Life of 8.75