Question

A first order reaction is 50% complete in 25min. Calculate time for 80% completion reaction

Answer 1

Answer:

let the initial concentration[a] be 100

then decomposed concentration = 50

final concentration[A] =50

K= 2.303/t log [a]/[A]

K=2.303/25 log 100/50

k= 2.303/25 log2

k=2.303/25 0.3010......................1

For 80% completion

again let {a]=100

decomposed concentration= 80

final concentration[A]=20

k=2.303/t log 100/20

from eq 1

2.303/25 0.3010 = 2.303/t log5

0.3010/25 = 0.6990/t

t= 0.6990 * 25 /0.3010

t= 58.05 min

Explanation: