Question

Dinitrogen pentoxide, N2O5, decomposes by first-order kinetics with a rate constant of 3.7 × 10–5 s–1 at 298 K.

a. What is the half-life, in hours, of N2O5 at 298 K?

b. If [N2O5]0 = 0.0648 mol·L–1, what will be the concentration of N2O5 after 2.1 hours?

c. How much time, in minutes, will elapse before the N2O5 concentration decreases from 0.0648 mol·L–1 to 0.0246 mol·L–1?

Answer 1

a)

Given:

k = 3.7*10^-5 s-1

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

t1/2 = (ln 2) / k

= 0.693/(k)

= 0.693/(3.7*10^-5)

= 1.873*10^4 s

= (1.873*10^4)/3600 hr

= 5.20 hours

Answer: 5.20 hours

b)

we have:

[A]o = 0.0648 M

t = 2.1 hr = 2.1*3600 s = 7560 s

k = 3.7*10^-5 s-1

use integrated rate law for 1st order reaction

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

ln[A] = ln(6.48*10^-2) - 3.7*10^-5*7.56*10^3

ln[A] = -2.736 - 3.7*10^-5*7.56*10^3

ln[A] = -3.016

[A] = e^(-3.016)

[A] = 4.899*10^-2 M

Answer: 4.9*10^-2 M

c)

we have:

[A]o = 6.48*10^-2 M

[A] = 2.46*10^-2 M

k = 3.7*10^-5 s-1

use integrated rate law for 1st order reaction

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

ln(2.46*10^-2) = ln(6.48*10^-2) - 3.7*10^-5*t

-3.705 = -2.736 - 3.7*10^-5*t

3.7*10^-5*t = 0.9686

t = 2.618*10^4 s

t = (2.618*10^4)/60 min

= 436 min

Answer: 436 min