1.

The rate of a first order reaction is 0.04mol L^(-1) sec^(-1) at 10 minute and 0.03mol L^(-1) at 20 minute after initiation. Find the half life of the reaction.

Answer»

Solution :RATE `=K[A]`
`0.04=k[A]_(10)` and `0.03=k[A]_(20)`
`([A]_(10))/([A]_(20))=(0.04)/(0.03)=(4)/(3)`
`t=(2.303)/(k)log.([A]_(10))/([A]_(20))` when `t=10^(1)`min.
`10=(2.303)/(k)log.(4)/(3)`
`k=(2.303)/(10)log.(4)/(3)=0.0288min^(-1)`
`t_((1)/(2))=(0.693)/(k)=(0.693)/(0.0288)=24.06min`.


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