Question

For a first order reaction, the time taken to reduce initial concentration by a factor of 1/4 is 10 min. The time required to reduce initial concentration by a factor of 1/6 will be

Answer 1

log 4/3 =0.13
log 6/5=0.08
use formula : kt = 2.303log(C/C at time t)

Answer 2

Answer: $12.80 min^{-1}$

Explanation: Rate law expression for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken for decay process

a = initial amount of the reactant

a - x = amount left after decay process

Given $t_\frac{1}{4}$ = 10 min   

a-x =$\frac{1}{4}\times a$

Thus $t_\frac{1}{4}=\frac{2.303}{k}\log\frac{a}{a/4}$

Thus $10 min=\frac{2.303}{k}\log\frac{a}{a/4}$

$k=0.14min^{-1}$

Thus $t_\frac{1}{6}=\frac{2.303}{0.14}\log\frac{a}{a/6}$

$t_\frac{1}{6}=\frac{2.303}{0.14}\log{6}$

$t_\frac{1}{6}=12.80min$