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## Answers

t3/4 = ?

k = 2.303log(Co/C) / t

k = 0.693/(t1/2)

0.693/32 = 2.303log{Co/(3Co/4)} / (t3/4)

0.693/32 = 2.303log(4/3) / (t3/4)

0.693/32 = 2.303(log4-log3) / (t3/4)

0.693/32 = 2.303(0.6020-0.4771) / (t3/4)

0.693/32 = 0.2876 / (t3/4)

t3/4 = 0.2876*32/0.693

t3/4 = 13.28 min.

t 1/2 (or half life) implies time in which half the reactant is consumed.

t 3/4 implies time in which 3/4 of the reactant is consumed.

After 1 half life, 1/2 of the reactant will be consumed.

After 2 half lives, 1/2 of 1/2 i.e. 3/4 of the reactant will be consumed.

so, time for t 3/4 is 2 x 32 = 64 minutes.