For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is
(Given : ln 10 = 2.303 and log 2 = 0.3010)
Solution
$\mathrm{A} \rightarrow$ Products<br/><br/>
For a first order reaction,<br/><br/>
$\mathrm{t}_{1 / 2}=\frac{\ln 2}{\mathrm{k}}=\frac{0.693}{\mathrm{k}}$<br/><br/>
Time for $90 \%$ conversion,<br/><br/>
$t_{90 \%}=\frac{1}{k} \ln \frac{100}{10}=\frac{\ln 10}{k}=\frac{2.303}{k}$<br/><br/>
$t_{90\%}=\frac{2.303}{0.693} t_{1 / 2}=3.32 t_{1 / 2}$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate Laws and Order
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