The half-life of a first order reaction is $25 min$ . Using $lo g 2 = 0.3010$ , calculate the time (in minutes, two decimals) needed for 90% of the reactant to be consumed.

Numerical answer type, answer in min, tolerance 0.05.

Answer: 83.06 min

For first order

t = (2.303/ k) lo g ([A]_{0} / [A])

with

k = 2.303 \times 0.3010/ t_{1/2}

. For 90% conversion

lo g (10) = 1

, so

t = t_{1/2} /0.3010 = 25/0.3010 = 83.06 min