Question

After 58.0 min, 24.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?

Answer 1

we have:

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

[A] = 76 M (24.0 % has decomposed. So, remaining is 76 %)

t = 58.0 min

use integrated rate law for 1st order reaction

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

ln(76) = ln(100) - k*58

4.3307 = 4.6052 - k*58

k*58 = 0.2744

k = 4.732*10^-3 min-1

we have:

k = 4.732*10^-3 min-1

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

t1/2 = (ln 2) / k

= 0.693/(k)

= 0.693/(4.732*10^-3)

= 1.465*10^2 min

Answer: 146 min