The rate constant for the first-order decomposition of C₄H₈ at 500^$\circ$C is 9.2 * 10^-3 s^-1. How long will it take for 10.0% of a 0.100 M sample of C₄H₈ to decompose at 500^$\circ$C

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The integrated rate law for a first-order reaction is:ln[A]t/[A]0 = -ktwhere [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, and k is the rate constant.We are given that the rate constant for the first-order decomposition of C4H8 at 500°C is 9.2 * 10^-3 s^-1. We are also given that the initial concentration of C4H8 is 0.100 M. We want to find the time it will take for 10.0% of the C4H8 to decompose, which means that [A]t = 0.100 M * 0.100 = 0.0100 M.Substituting these values into the integrated rate law, we get:ln(0.0100 M/0.100 M) = -(9.2 * 10^-3 s^-1) * tSolving for t, we get:t = 11 sTherefore, the correct answer is A) 11 s.

The Rate Constant for the First-Order Decomposition of C4H8 at 500

The Rate Constant for the First-Order Decomposition of C₄H₈ at 500