Cyclopropane rearranges to form propene in a reaction that follows first order kinetics. At 800 K, the specific rate constant for this reaction is 2.74 x 10 - 3 s - 1. Suppose we start with a cyclopropane concentration of 0.290 M. How long will it take for 99.0 % of the cyclopropane to disappear according to this reaction

1866.4 secs

Explanation:

The integrated rate law for first order kinetics is given by;

[A] = [Ao] ^-kt

Where;

[A] = concentration of cyclopropane at time = t

[Ao] = initial concentration of cyclopropane

k = rate constant

t = time taken

From the question, we are told that 99% of cyclopropane disappeared hence;

[A] = 0.290 - (0.99 * 0.290)

[A] = 0.290 - 0.2871

[A] = 0.0029 M

Hence

0.0029 = 0.290 e^-2.74 x 10^ - 3 * t

0.0029/0.290 = e^-2.74 x 10^ - 3 * t

0.01 = e^-2.74 x 10^ - 3 * t

ln(0.01) = -2.74 x 10^ - 3 * t

-4.61 = -2.74 x 10^ - 3 * t

t= -4.61/-2.74 x 10^ - 3

t = 1866.4 secs

the answer is barium and copper bromide