Final answer:
The concentration of cyclopropane after 8.8 minutes can be found using the first-order rate equation, converting time to seconds, and solving for concentration. To determine the time it takes for the concentration to decrease from 0.25 M to 0.15 M, one must solve for time using the given concentrations and rate constant.
Step-by-step explanation:
The conversion of cyclopropane to propene in the gas phase is a first-order reaction. For a first-order reaction, the concentration of the reactant decreases exponentially with time according to the equation ln[A] = -kt + ln[A0], where [A] is the concentration at time t, k is the rate constant, and [A0] is the initial concentration.
a) To find the concentration of cyclopropane after 8.8 minutes, we first convert minutes to seconds because the rate constant is given in seconds. 8.8 minutes = 528 seconds. Plugging in the values into the first-order rate equation: ln[A] = -(6.7 x 10^-4 s^-1)(528 s) + ln(0.25 M). Solving for [A] will give the concentration of cyclopropane after 528 seconds.
b) To find out how long it takes for the concentration to decrease from 0.25 M to 0.15 M, we set up the first-order rate equation with these concentrations and solve for t: ln[0.15 M] = -(6.7 x 10^-4 s^-1)t + ln(0.25 M). Solving for t will give the time in seconds, which we can then convert to minutes.