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Chemical Kinetics — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryChemical Kinetics3 Marks
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, $t_1 = 2t_2$
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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