Question

If the rate constant for a first order reaction is k, the time (t) required for the completion of $99\%$ of the reaction is given by:

• A. $t=\frac{4.606}{k}$
• B. $t=\frac{2.303}{k}$
• C. $t=\frac{6.909}{k}$
• D. $t=\frac{0.693}{k}$

Answer 1

The correct option is A $t=\frac{4.606}{k}$

For 1st order reaction,
$t=\frac{2.303}{k}log\frac{a}{a-x}=\frac{2.303}{k}log\frac{100}{100-99}\frac{2.303}{k}log{10}^2\times log10=\frac{4.606}{k}$