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Sum

**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?

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#### Solution

**Given: **

[A]_{0 }= 20 mmol dm^{-3 }, [A]_{t} = 8 mmol dm^{-3 }, t = 38 min

**To find:**

Half life of reaction t_{1/2}

**Formulae:**

**i. **`"k" = 2.303/"t" "log"_10 ["A"]_0/["A"]_"t"`

**ii. **t_{1/2 }= `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}

t_{1/2 }= `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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