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.
✓
Answer
For a first order reaction, the time required for 99% completion is
$\text{t}_1=\frac{2.303}{\text{k}}\text{log}\frac{100}{100-99}$
$=\frac{2.303}{\text{k}}\text{log}{100}$
$=2\times\frac{2.303}{\text{k}}$
For a first order reaction, the time required for 90% completion is
$\text{t}_2=\frac{2.303}{\text{k}}\text{log}\frac{100}{100-99}$
$=\frac{2.303}{\text{k}}\text{log}{10}$
$=\frac{2.303}{\text{k}}$
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.
$\text{t}_1=\frac{2.303}{\text{k}}\text{log}\frac{100}{100-99}$
$=\frac{2.303}{\text{k}}\text{log}{100}$
$=2\times\frac{2.303}{\text{k}}$
For a first order reaction, the time required for 90% completion is
$\text{t}_2=\frac{2.303}{\text{k}}\text{log}\frac{100}{100-99}$
$=\frac{2.303}{\text{k}}\text{log}{10}$
$=\frac{2.303}{\text{k}}$
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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