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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)
Solution
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"`
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