Question

20% of a first-order reaction is completed in five minutes. How much time will the 60% reaction take to complete? Calculate the half-life period (t_1/2) for the above reaction.

Answer 1

In 5 minutes, 20% reaction is complete

[A]₀ = 100 (initial concentration)

[A] = 100 - 20 = 80 (Final concentration)

K = `2.303/"t" log  (["A"]_0)/(["A"])`

K = `2.303/5  log  100/80`

K = 0.0446 min^-1

Now, the reaction is 60% complete

[A]₀ = 100, [A] = 100 - 60 = 40

t = `2.303/"K" log  (["A"]_0)/("A")`

t = `2.303/0.0446  log  100/40`

t = 20.54 min

Half life period (t_1/2) = `0.693/"K"`

where K = 0.0446 min^-1

`"t"_(1//2) = 0.693/0.0446`

`"t"_(1//2) = 15.538` min