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From the following data for the liequied phase reactino `Ato`B. determine the order of reaction and calculate its rate constant :
image

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(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.

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