Question

The hydrolysis of sucrose to fructose is first order in sucrose with a half-life of 3.60e+03 s at some temperature.

a) What fraction of sucrose, expressed as a percent, would hydrolyze in 3.50 hours?

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

Answer 1

C_sucrose initial concentration = 1M
For first-order reaction, t_1/2 = 3.60*10³ sec
At t_1/2 = 3.6*10³ sec = 1hour C_sucrose = 0.5 M
t_1/2 = 0.693 / K = 3.6*10³ sec
K = 1.925*10^-4 sec^-1
[C_sucrose] = [ C_sucrose ]_o e^-kt   
t = 3.50 hrs = 12600 sec
[C_sucrose] = 1 * e^{-(1.925*10-4) 12600}
[C_sucrose] = 0.088433 M ( left out after 3.5 hrs)
fraction of sucrose % =[ (1-0.088433) / 1 ] * 100 = 91.156 %

b) 65 % =[ (1 - 0.35) / 1 ] * 100
   [C_sucrose] = 0.35 M
0.35 = 1 * e^{-(1.925*10-4)t}  
apply ln on both sides,
ln (0.35) = - 1.925 *10^-4 * t
t = 5453.621 sec = 1.514 hrs