The decomposition of n-butyl ether at 504°C is first order with half-life of 1670 sec. What fraction of initial amount of n-butyl ether remain after 4710 second?

Step 1: Calculate the rate constant (k) for the first-order reaction. For a first-order reaction, the half-life (t_1/2) is related to the rate constant (k) by the formula: t_1/2 = ((2))/(k) Rearranging to solve for k: k = ((2))/(t_1/2) Substitute the given half-life t_1/2 = 1670 s: k = (0.6931)/(1670 s) k ≈ 0.0004150 s^-1 Step 2: Calculate the fraction of diethyl ether remaining after 4710 s. The integrated rate law for a first-order reaction is: (([A]_t)/([A]_0)) = -kt Where [A]_t is the concentration at time t, and [A]_0 is the initial concentration. To find the fraction remaining, ([A]_t)/([A]_0), we exponentiate both sides: ([A]_t)/([A]_0) = e^-kt Substitute the calculated k and the given time t = 4710 s: ([A]_t)/([A]_0) = e^-(0.0004150 s^-1 × 4710 s) ([A]_t)/([A]_0) = e^-1.9545 ([A]_t)/([A]_0) ≈ 0.1416 The fraction of diethyl ether remaining after 4710 s is approximately 0.1416. Send me the next one 📸