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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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M manish

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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