The rate constant for the reaction 3A → 4B is 6.00 × 10^-3 L mol^-1min^-1. How long will it take the concentration of A to drop from 0.75 M to 0.25 M?

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We will use the integrated rate law for a first-order reaction, which is: ln([A]t/[A]0) = -kt where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is the time. We rearrange this equation to solve for t: t = (ln([A]0/[A]t))/k Substituting the given values: t = (ln(0.75/0.25))/(6.00 × 10^-3 L mol^-1min^-1) t = 440 min Therefore, it will take 440 min for the concentration of A to drop from 0.75 M to 0.25 M.

The Rate Constant for the Reaction 3A → 4B Is