Question

The decomposition of chloroform at 504°C is first order with a half-life of 1570 seconds. What percent of any initial amount of chloroform remains after 1.25 hours?

Answer 1

Given:

Half life = 1570 s

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

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(1570)

= 4.414*10^-4 s-1

we have:

[A]o = 100 (Let initial concentration be 100)

t = 1.25 h = 1.25*3600 s = 4500 s

k = 4.414*10^-4 s-1

use integrated rate law for 1st order reaction

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

ln[A] = ln(100) - 4.414*10^-4*4.5*10^3

ln[A] = 4.605 - 4.414*10^-4*4.5*10^3

ln[A] = 2.619

[A] = e^(2.619)

[A] = 13.7 M

So, 13.7 M of 100 M remains which is 13.7 %

Answer: 13.7 %