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 14.5% of its initial value?

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For a first-order reaction, the time $t$ required for a reactant to fall to a certain percentage of its initial concentration 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. The half-life ($t_{1/2}$) of a first-order reaction is given by $t_{1/2} = \frac{0.693}{k}$. Given that the half-life is 8.75 hours, we can find $k = \frac{0.693}{8.75}$ hours$^{-1}$. To find the time it takes for the concentration to fall to 14.5% of its initial value, we use the initial and final concentration ratio in the formula, where $\frac{[A]}{[A]_0} = 0.145$. Substituting the values and solving for $t$ gives a time significantly longer than the half-life, indicating that option D (24.4 hours) is the correct answer, as it represents a substantial decrease to 14.5% of the initial concentration, which is consistent with the nature of exponential decay in first-order reactions.

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