Question

7) If the rate law of cyclopropane rearranging to propene is: Rate = 6.7 x 10s [C3H6], how many minutes will it take for the concentration of cyclopropane to drop by 12 %? a) 3165 c) 55.8 e) 1.41 b) 191 d) 3.18

Neat handwriting please. Does answering chemistry questions on here terrible handwriting. Also, you should get 3.18 minutes. Show me step by step how you get answer D.

Answer 1

we have:

[C3H6]o = 100 M [Let initial concentration be 100 M]

12 % of it has reacted.

So, remaining is 88 %.

[C3H6] = 88 % of 100 M

= 88*100/100 M

= 88 M

k = 6.7*10^-4 s-1

use integrated rate law for 1st order reaction

ln[C3H6] = ln[C3H6]o - k*t

ln(88) = ln(1*10^2) - 6.7*10^-4*t

4.477 = 4.605 - 6.7*10^-4*t

6.7*10^-4*t = 0.1278

t = 191 s

= 191/60 min

= 3.18 min

Answer: D