Question

A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.

(log 2 = 0.3010)

Answer 1

For the given first−order reaction, the rate constant for 50% completion is given by

`k=2.303/t`

Here,

t=Time taken for 50% completion=30 min

[R]_o=Initial concentration of the reactant

[R]=Final concentration of the reactant

Let [R]0 be 100 and due to 50% completion of reaction, [R] will be 100−50, i.e. 50.

Putting values in (i), we get

`k=2.303/30`

  `=2.303/30`

For the same reaction, the time required for 90% completion of the reaction can be computed using the expression

`k=2.303/t`

Here,

[R]=Final concentration of reactant=100−90=10

(ii) Rate = `(Δ x)/(Δ t)`

= `2.303/k log10 = 2.303/k`

By putting the value of k here,we get

t = `(2.303 xx 1800)/(2.303 xx 0.3010)`

= `5.98 xx 10^3 "sec"`