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Question
What is half life of first order reaction if time required to decrease concentration of reactants from 0.8 M to 0.2 M is 12 hours?
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Solution 1
Given: [A]0 = 0.8 M, [A]t = 0.2 M, t = 12 hours
To find: Half life of reaction (t1/2)
Formulae:
- k = `2.303/"t" log_10 (["A"]_0)/(["A"]_"t")`
- `"t"_(1/2) = 0.693/"k"`
Calculation: Substituting given value in
k = `2.303/"t" log_10 (["A"]_0)/(["A"]_"t")`
k = `2.303/(12 "hr") log_10 0.8/0.2`
= `2.303/(12 "hr") log_10 (4)`
= `2.303/(12 "hr") xx 0.6020`
= Antilog10 (log10 2.303 + log10 0.6020 − log10 12)
= Antilog10 (0.3623 + `bar(1).7796` − 1.0792 )
= Antilog10 `(bar(1).0627 )`
= 0.1115 hr–1
`"t"_(1/2) = 0.693/"k" = 0.693/(0.1155 "hr"^-1)` = 6 hr
Solution 2
Concentration is reduced to 25%. It means it takes two half-lives to decrease the concentration of reactant from 0.8 M to 0.2 M in first-order reaction. Hence, half-life of the reaction is 12/2 = 6 hours.
The half life of reaction is 6 hours.
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