# 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? - Chemistry

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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?

#### Solution

Given:

[A]= 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.

Concept: Integrated Rate Law
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#### APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 6 Chemical Kinetics
Exercises | Q 4.1 | Page 137