Chemistry, asked by Anonymous, 11 months ago

In a first order reactio the concentration of the reactant is reduced to 1/8 of the initial concentration in 75 mins at 298 K . What is the half life period of the reaction in minute

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Answered by Anonymous

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Answered by arshbbcommander

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Given

a=1

a-x=1/8

t=75min

To find.... t1/2

$K = \frac{2.303}{t} log( \frac{a}{a - x} ) \\ K = \frac{2.303}{75} log( \frac{1}{ \frac{1}{8} } ) \\ \\$

$K = \frac{2.303}{75} log( 8 ) \\ \\ K = \frac{2.303}{75} \times 0.9 \\ K = 0.027 \\ \\$

or K=2.7×10^(-2)

Now

t1/2 = 0.693/K

=0.693/2.7×10^(-2)

$\huge\boxed{=25 minutes}$

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...Hope it helps....

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