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28/01/2022 5:43 pm
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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.
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28/01/2022 5:47 pm
For a first order reaction, the time required for 99% completion is
t1 = \(\frac{2.303}{k}\)log \(\frac{100}{100 - 99}\)
= \(\frac{2.303}{k}\)log 100
= 2 x \(\frac{2.303}{k}\)
For a first order reaction, the time required for 90% completion is
t2 = \(\frac{2.303}{k}\)log \(\frac{100}{100 - 90}\)
= \(\frac{2.303}{k}\)log 10
= \(\frac{2.303}{k}\)
Therefore, t1 = 2t2
Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.