(i) We require to calculate the rate constant at different time intervals.

`(a) k_(1)=(2*303)/(t)"log"_(10)([A_(0)])/([A_(t)])`

`[A_(0)]=0.624`

`[A_(t)]=0.446`

`t=600s`

`k_(1)=(2.303)/(600)"log" ([0.624])/([0.446])`

`:. k_(1)=3.838xx10^(-3)"log"[1.399]`

`:. k_(1)=3.838xx10^(-3)xx0.1458`

`:. k_(1)=5.595xx10^(-3)s^(-1)`

(b) `k_(2)=(2.303)/(t)"log"([A_(0)])/([A_(t)])`

`[A_(0)]=0624`

`[A_(t)]=0.318`

`t=1200s`

`k_(2)=(2.303)/(1200)"log"([0.624])/(([0.318])`

`:. k_(2)=1.919xx10^(-3)log [1.962]`

`:. k_(2)=1.919xx10^(-3) xx0.2927`

`:.k_(2)=5.616xx10^(-4)s^(-1)`

(c) `K_(3)=(2.303)/(t)"log"_(10)([A_(0)])/([A_(t)])`

`[A_(0)]=0.624`

`[A_(t)]=0.226`

`t=1800 s`

`k_(3)=(2.303)/(1200)"log"([0.624])/([0.226])`

`:.k_(3)=1.279xx10^(-3)log[2.7615]`

`:. k_(3)1.279xx10^(-3)xx0.4411`

`:. k_(3)=5.641xx10^(-4)s^(-1)`

All the k values calculated at different time intervals are the same. This implies that the reaction obeys the intgrated rate equation of first order reaction. Hence. the reaction is a first order reaction.