Chemical Kinetics - Decomposes in a First-order Reaction

Question

At 1000 degrees Celsius cyclobutene C₄H₈ decomposes in a first order reaction, with the very high-rate constant of 87 s^-1 to two molecules of ethylene C₂H₄.

a) the initial C₄H₈ concentration if 2.00M. What is the concentration after 0.010 s?

b) How long will it take for 70.0% of the C₄H₈ to decompose?

Answer 1

Recall that for a first order reaction, t_1/2 = ln 2 / k = 0.693 / k. Since we know the value of k, we can calculate t_1/2 and then use that value to answer both parts of the question.

t_1/2 = 0.693 / 87 s^-1 = 7.97x10^-3 sec

(a) Also, for a 1st order reaction, the fraction remaining (FR) = 0.5ⁿ where n = number of half lives that have elapsed. In this problem for part (a), that would be 0.010 s x 1 half life / 7.97x10^-3 sec = 1.25 half lives

FR = 0.5^1.25 = 0.420 .. this is the fraction remaining. Since we began with 2.00 M, the fraction remaining will be 2.00 M x 0.420 = 0.84 M = concentration after 0.01 sec

(b) If 70% decomposes, the fraction remaining (FR) will be 30%. So we now determine how many half lives must elapse so that FR is 0.3.

FR = 0.5ⁿ

0.30 = 0.5ⁿ

take log of both sides to get log 0.30 = n log 0.5

-0.523 = n x -0.301

n = 1.74 half lives

1.74 half lives x 7.97x10^-3 sec / half life = 0.0139 sec = time for 70% to decompose

Of course, you could use the more conventional (but more math intensive) equations for the integrated rate law for a 1st order reaction: ln[A] = -kt + ln[A_o]. For part (a) you would solve for [A] when t = 0.01 s, [A_o] = 2.00 M and k = 87 s^-1. For part (b), you would solve for t when [A] = 0.6 M (30% of 2.00 M), and [A_o] and k are as in part (a). You should get the same answers as above.