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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? - Chemistry

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प्रश्न

In a first-order reaction, the concentration of the reactant decreases from 20 mmol dm3 to 8 mmol dm−3 in 38 minutes. What is the half-life of reaction? 

संख्यात्मक
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उत्तर

Given: 

[A]= 20 m mol dm3,

[A]t = 8 m mol dm3,

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 min1

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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अध्याय 6: Chemical Kinetics - Exercises [पृष्ठ १३७]

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बालभारती Chemistry [English] Standard 12 Maharashtra State Board
अध्याय 6 Chemical Kinetics
Exercises | Q 4. i. | पृष्ठ १३७

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