Question

25% of first order reaction is completed in 30 minutes. Calculate the time taken in minutes for the reaction to go to 90% completion. 

Answer 1

Given: Initial concentration [A]₀ = 100

For 25% completion: 

x = 25

[A]₀ − x = 100 − 25 = 75

Time (t) = 30

For first-order:

k = `2.303/t log ([A]_0/([A]_0 - x))`

= `2.303/30 log (100/75)`

= `2.303/30 ln (4/3)`

= 0.7676 × 0.1249

k = 0.00959 min⁻¹

For 90% completion:

x = 90,

[A]₀ − x = 100 − 90 = 10

For first-order:

k = `2.303/t log ([A]_0/([A]_0 - x))`

⇒ t = `2.303/k log ([A]_0/([A]_0 - x))`

t = `2.303/0.00959 log (100/10)`

= `2.303/0.00959 xx 1`

= 240.14 min