Question

In a first-order reaction, 10% of the reactant is consumed in 25 minutes. Calculate:

1. The half-life period of the reaction.
2. The time required for completing 87.5% of the reaction.

Answer 1

Given: t = 25 min

a = 100

a − x = 90

k = ?

1. For a first-order reaction,

k = `2.303/t log  a/(a - x)`

= `2.303/25 log  100/90`

= `2.303/25 [log 10 - log 9]`

k = `2.303/25 [1.000 - 0.9542]`

= `2.303/25 xx 0.0458`

k = 0.0042 min⁻¹

t_1/2 = `0.693/k`

= `0.693/0.0042`

= 165 min

2. t_87.5% = `2.303/k log  100/(100 - 87.5)`

= `2.303/0.0042 log  100/12.5`

= `2.303/0.0042 xx log 8`

t_87.5% = 495 min