A first order reaction is 40% complete in $20 min$ . Using $lo g 2 = 0.3010$ and $lo g 3 = 0.4771$ , calculate the time (in minutes, nearest integer) for the reaction to be 80% complete.

Numerical answer type, answer in min, tolerance 0.5.

Answer: 63 min

Remaining fractions are 0.60 and 0.20, so

t = (2.303/ k) lo g (1/0.20)

with

k = (2.303/20) lo g (1/0.60)

. Then

t /20 = lo g 5/ lo g (5/3)

with

lo g 5 = 1 - 0.3010 = 0.6990

and

lo g (5/3) = 0.6990 - 0.4771 = 0.2219

. So

t = 20 \times 0.6990/0.2219 = 63.00 min