# For a first order reaction,

Question:

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

Solution:

For a first order reaction, the time required for 99% completion is

$t_{1}=\frac{2.303}{k} \log \frac{100}{100-99}$

$=\frac{2.303}{k} \log 100$

$=2 \times \frac{2.303}{k}$

For a first order reaction, the time required for 90% completion is

$t_{2}=\frac{2.303}{k} \log \frac{100}{100-90}$

$=\frac{2.303}{k} \log 10$

$=\frac{2.303}{k}$

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.

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