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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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प्रश्न

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?

संख्यात्मक
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उत्तर १

Given: [A]0 = 0.8 M, [A]t = 0.2 M, t = 12 hours

To find: Half life of reaction (t1/2)

Formulae: 

  1. k = `2.303/"t" log_10  (["A"]_0)/(["A"]_"t")`
  2. `"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

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उत्तर २

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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