A second-order reaction starts with an initial concentration of 0.100 mol/L of the reactant.If the rate constant is 1.4 × 10^-2 L/mol·s,what is the time required to decrease the initial concentration to 0.050 mol/L? 2A → B
Rate = k[A]²

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Quizplus · Accepted Answer

For a second-order reaction, the time $ t $ required for the concentration of a reactant to change from $ [A]_0 $ to $ [A] $ is given by the equation $ \frac{1}{[A]} - \frac{1}{[A]_0} = kt $. Substituting the given values, $ \frac{1}{0.050} - \frac{1}{0.100} = (1.4 \times 10^{-2})t $, solving for $ t $ gives approximately 710 seconds.

A Second-Order Reaction Starts with an Initial Concentration of 0