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# For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

4.18   For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

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case 1-

for 99% complition,
$t^{1} = \frac{2.303}{k}\log\frac{100}{100-99}$
$= \frac{2.303}{k}\log100$
$=2\times (\frac{2.303}{k})$

CASE- II
for 90% complition,

$t^{2}=\frac{2.303}{k}\log\frac{100}{100-90}$
$=\frac{2.303}{k}\log10$
$=(\frac{t^{1}}{2})$

$t^{1}=2t^{2}$
Hence proved.

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