In a first-order reaction, the concentration of the reactant is reduced to $12.5$ % in one hour. Calculate its half-life.

3

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20

min

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30

min

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15

min

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Solution

The correct option is B $20$ min
For a first order reaction;

$k = \frac{2.303}{t} l o g \frac{[A]_{o}}{[A]_{t}}$

$k = \frac{2.303}{60} l o g \frac{100}{12.5} = \frac{2.303}{60} l o g 8$

$t_{1 / 2} = \frac{2.303}{k} l o g 2 = \frac{60}{3} = 20 m i n$

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In a first-order reaction, the concentration of the reactant is reduced to 12.5% in one hour. Calculate its half-life.

The correct option is B 20 minFor a first order reaction;k=2.303tlog[A]o[A]tk=2.30360log10012.5=2.30360log 8t1/2=2.303klog 2=603=20 min

In a first-order reaction, the concentration of the reactant is reduced to $12.5$ % in one hour. Calculate its half-life.

The correct option is B $20$ min
For a first order reaction;

$k = \frac{2.303}{t} l o g \frac{[A]_{o}}{[A]_{t}}$

$k = \frac{2.303}{60} l o g \frac{100}{12.5} = \frac{2.303}{60} l o g 8$

$t_{1 / 2} = \frac{2.303}{k} l o g 2 = \frac{60}{3} = 20 m i n$