The rate constant for a particular second-order reaction is 0.47M^-1s^-1. If the initial concentration of reactant is 0.25 mol/L, it takes __________ s for the concentration to decrease to 0.080 mol/L.

Question

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Explanation:The integrated rate law for a second-order reaction is:1/[A]t - 1/[A]0 = ktwhere [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, k is the rate constant, and t is time.Substituting the given values into the integrated rate law, we get:1/0.080 - 1/0.25 = 0.47tt = 18 s

The Rate Constant for a Particular Second-Order Reaction Is 0