After 55.0 min, 33.0% of a compound has decomposed. What is the half‑life of this reaction assuming first‑order kinetics?

Question

t1/2=

Answer 1

33% decomposed leaves 67% in 55 minutes

.67 = e^55r

take natural logs of both sides

ln.67 =55r

r = (ln.67)/55 = less than 1% per minute decay rate

.5 = e^[(ln.67)t/55]

take natural logs of both sides

ln.5=.(ln.67)t/55

use a calculator

t=-55ln.5/ln.67= 95.19 minutes = half life