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Question
Solve
In a first-order reaction, the concentration of the reactant decreases from 20 mmol dm-3 to 8 mmol dm-3 in 38 minutes. What is the half-life of reaction?
Solution
Given:
[A]0 = 20 mmol dm-3 , [A]t = 8 mmol dm-3 , t = 38 min
To find:
Half life of reaction t1/2
Formulae:
i. `"k" = 2.303/"t" "log"_10 ["A"]_0/["A"]_"t"`
ii. t1/2 = `0.693/"k"`
Calculation:
Substituting given value in
`"k" = 2.303/"t" "log"_10 ["A"]_0/["A"]_"t"`
`"k" = 2.303/(38 "min") "log"_10 20/8`
= `2.303/"38 min" "log"_10 (2.5)`
= `2.303/"38 min" xx 0.3979 = 0.0241` min-1
t1/2 = `0.693/"k" = 0.693/0.0241 = 28.7` min
The half life of reaction is 28.7 min.
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