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 16.5% of its initial value?

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For a first-order reaction, the time $ t $ required for the concentration of a reactant to fall to a fraction $ f $ of its initial value is given by the formula $ t = \frac{\ln(1/f)}{k} $, where $ k $ is the rate constant. The half-life ($ t_{1/2} $) of a first-order reaction is related to the rate constant by $ t_{1/2} = \frac{\ln(2)}{k} $. Given that the half-life is 8.75 hours, we can find $ k $ and then calculate $ t $ for the concentration to fall to 16.5% (or 0.165) of its initial value.1. Calculate the rate constant $ k $ using the half-life formula:$$ k = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{8.75} $$2. Calculate the time $ t $ for the concentration to fall to 16.5% of its initial value:$$ t = \frac{\ln(1/0.165)}{k} = \frac{\ln(1/0.165)}{\ln(2)/8.75} $$This calculation will give $ t $ as approximately 22.7 hours, which means it takes 22.7 hours for the concentration to fall to 16.5% of its initial value.

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