Cyclopropane is converted to propene in a first-order process.The rate constant is 5.4 × 10^-2h^-1.If the initial concentration of cyclopropane is 0.150 M,what will its concentration be after 22.0 hours?

Question

Quizplus · Accepted Answer

The rate law for the reaction is given as:
Rate = k[Cyclopropane]
The integrated rate law for a first-order reaction is:
ln[A]t = -kt + ln[A]0
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.
Substituting the given values:
ln[Cyclopropane]t = -k*t + ln[Cyclopropane]0
ln[Cyclopropane]t = -(5.4*10^-2)*(22.0) + ln[0.150]
ln[Cyclopropane]t = -1.188 + ln[0.150]
ln[Cyclopropane]t = -0.779
[Cyclopropane]t = e^-0.779
[Cyclopropane]t = 0.46 M
Rounding off the value to two significant figures, we get:
[Cyclopropane]t = 0.046 M
Therefore, the best choice is A, with the concentration of cyclopropane after 22.0 hours being 0.046 M.

Cyclopropane Is Converted to Propene in a First-Order Process