Question

The hydrolysis of sucrose to fructose is first order in sucrose with a half-life of 3.30e+03 s at some temperature. What fraction of sucrose, expressed as a percent, would hydrolyze in 2.90 hours? % How long (in hours) would it take to hydrolyze 79.0% of the sucrose? hours

Answer 1

Given

1) Reaction is First order

2) Half life t _1/2 = 3.30 x 10 ⁰³ seconds

3) t = 2.9 hours= 2.9 hour x ( 60 min / 1 hour ) x ( 60 s / 1 min ) = 10440 seconds.

For first order reaction we know that ,

K = 0.693 /  t _1/2

K = 0.693 / 3.30 x 10 ⁰³ seconds

K = 2.1 x 10 ^-04 s ^-01

We have relation K= (2.303 / t ) log (a /a-x)

where a= initial concentration of reactant

a-x = concentration after time t

t = time for concentration change

K = ( 2.303 / 10440 seconds.) log ( 100 / 100-x)

2.1 x 10 ^-04 s ^-01 = 2.2059 x 10 ^-04(  log 100 - log 100-x )

= 2.2059 x 10 ^-04 ( 2 - log 100-x )

( 2 - log 100-x ) =  2.1 x 10 ^-04 s ^-01/2.2059 x 10 ^-04 s ^-01

= 0.952

2-0.952 =  log 100-x

1.048 =  log 100-x

100-x = 10 ^1.048

= 11.17

x = 100 - 11.17

= 88.83

ANSWER : 88.83 % sucrose decomposed in 2.9 hours.

2) time for 79 % decomposition

K= (2.303 / t ) log (a /a-x)

   2.1 x 10 ^-04 s ^-01 =2.303 / t log 100 / 21

= 2.303 / t x 0.6778

t= 2.303 x 0.6778 / 2.1 x 10 ^-04 s ^-01

t = 7432 seconds