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

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=(4.606/k) (B) t=(2.303/k) (C) t=(0.693/k) (D) t=(6.909/k).

Tardigrade Admin · Accepted Answer

t =(2.303/ k ) log (100/1) =(2.303/ k ) × 2 t =(4.606/ k )

Q. 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

$t = \frac{4.606}{k}$

$t = \frac{2.303}{k}$

$t = \frac{0.693}{k}$

$t = \frac{6.909}{k}$ .

$t = \frac{2.303}{k} lo g \frac{100}{1}$
$= \frac{2.303}{k} \times 2$
$t = \frac{4.606}{k}$

Q. 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

t=k4.606​

t=k2.303​

t=k0.693​

t=k6.909​.