Question

Consider the first-order reaction described by the equation cyclopropane---propene

At a certain temperature, the rate constant for this reaction is 5.42× 10–4 s–1. Calculate the half-life of cyclopropane at this temperature.

Given an initial cyclopropane concentration of 0.00510 M, calculate the concentration of cyclopropane that remains after 1.70 hours.

Answer 1

1)

Given:

k = 5.42*10^-4 s-1

use relation between rate constant and half life of 1st order reaction

t1/2 = (ln 2) / k

= 0.693/(k)

= 0.693/(5.42*10^-4)

= 1.279*10^3 s

= ( 1.279*10^3 )/60 minutes

= 21.3 minutes

Answer: 21.3 minutes

2)

we have:

[A]o = 0.0051 M

t = 1.70 hours = 1.70*3600 s = 6120 s

k = 5.42*10^-4 s-1

use integrated rate law for 1st order reaction

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

ln[A] = ln(5.1*10^-3) - 5.42*10^-4*6.12*10^3

ln[A] = -5.279 - 5.42*10^-4*6.12*10^3

ln[A] = -8.596

[A] = e^(-8.596)

[A] = 1.85*10^-4 M

Answer: 1.85*10^-4 M