The first-order decomposition of N₂O at 1000 K has a rate constant of 0.76 s^-1. If the initial concentration of N₂O is 10.9 mol L^-1, what is the concentration of N₂O after 9.6 s?

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The integrated rate law for a first-order reaction is:ln[A]t = -kt + ln[A]0where [A]t is the concentration of the reactant at time t, k is the rate constant, and [A]0 is the initial concentration of the reactant.Substituting the given values into the integrated rate law, we get:ln[N2O]t = -(0.76 s-1)(9.6 s) + ln(10.9 mol L-1)ln[N2O]t = -7.344 + 2.415ln[N2O]t = -4.929[N2O]t = e-4.929 = 7.4 × 10-3 mol L-1Therefore, the concentration of N2O after 9.6 s is 7.4 × 10-3 mol L-1.

The First-Order Decomposition of N2O at 1000 K Has a Rate

The First-Order Decomposition of N₂O at 1000 K Has a Rate