Question

The hydrolysis of sucrose to fructose is first order in sucrose with a half-life of 4.60e+03 s at some temperature.

What fraction of sucrose, expressed as a percent, would hydrolyze in 3.40 hours?

_____ %

How long (in hours) would it take to hydrolyze 65.0% of the sucrose?

______hours

Answer 1

first calculate the rate constant

K = 2.303 / t_1/2   = 2.303 / 4.60 x 10³

K = 1.5 x 10^-4  s^-1

now after 3.40 h = 12240 s

K = (2.303 / t) log [A₀] / [A]

1.5 x 10^-4  s^-1 = (2.303 /12240 ) log (100 / x)

1.5 x 10^-4  s^-1 = (1.9 x 10^-4) log (100 / x)

log (100 / x) = 0.79

2 - log x = 0.79

log x = 2 - 0.79

log x = 1.21

x = 16.22

100 - 16.22 = 83.78 % sucrose hydrolyze.

part 2)

65.0% hydrolyze means

[A₀] = 100

[A] = 100 - 65 = 35

K = (2.303 / t) log [A₀] / [A]

1.5 x 10^-4  s^-1 = (2.303 / t) log [100]/[35]

  1.5 x 10^-4  s^-1 = (2.303 / t) x 0.46

(2.303 / t) = 3.26 x 10^-4

t = 7064.42 s

t = 117.7 m

t = 1.96 h