Answer 1

Given:

[A]₀ = 100%, [A]_t = 100 - 30 = 70%, t = 40 min

To find:

Half life of reaction (t_1/2)

Formula: 

`"k" = 2.303/"t" "log"_10 ["A"]_0/["A"]_"t"`

Calculation:

Substitution of these in above

`"k" = 2.303/"t" "log"_10 100/70`

= `2.303/"40 min" "log"_10 (1.429)`

= `2.303/"40 min" xx 0.155`

= 0.008924 min^-1

t_1/2 = `0.693/"k" = 0.693/(0.008924  "min"^-1) = 77.66  "min"`

The half life of reaction is 77.66 min.