Question

The decomposition of dinitrogentetroxide to nitrogen dioxide N_{2}2​O_{4}4​(g) \rightarrow→ 2 NO_{2}2​(g) is a first order reaction with a half-life of 1.3 x 10^{-5}−5 seconds. How many seconds are required for 26% of the original sample to remain?

Answer 1

Given:

Half life = 1.3*10^-5 s

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

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(1.3*10^-5)

= 5.331*10^4 s-1

we have:

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

[A] = 26 (26 % of 100 remains)

k = 5.331*10^4 s-1

use integrated rate law for 1st order reaction

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

ln(26) = ln(1*10^2) - 5.331*10^4*t

3.258 = 4.605 - 5.331*10^4*t

5.331*10^4*t = 1.347

t = 2.527*10^-5 s

Answer: 2.5*10^-5 s