Question

Sucrose C₁₂H₂₂O₁₁ reacts slowly with water in the presence of an acid to form two other sugars, glucose and fructose, both of which have the same molecular formulas, but different structures:

C₁₂H₂₂O₁₁ + H₂O -> C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose)

The reaction is first order and has a rate constant of 6.2 x 10^-5 sec -1 at 35^C under the prevailing solution conditions. Suppose that the initial concentration of sucrose in the solution is 0.40 M.

What will the sucrose concentration be after 2.0 hours? (PLEASE SHOW WORK)

How many minutes will it take for the sucrose concentration to drop to 0.30 M? (PLEASE SHOW WORK)

Answer 1

Since this is a first order reaction so, equation will be  

ln[A]t = -k∙t + ln [A]0

{ [A]t = concentration at time t, k = rate cons. , [A]o = initial conc., t= time in second }

we solve the equation

ln[A]t - ln[A]0 = -kt  

ln( [A]t / [A]0 ) = -k∙t

[A]t / [A]0 = e^(-k∙t)

[A]t = [A]0 ∙ e^(-k∙t)

{ at t = 2.0h = 7200s }

[A]t = 0.40M ∙ e^{(- 6.2x10-5 s⁻¹ ∙ 7200s)}

= 0.40M ∙ e^{(- 0.4464)}

[A]t = 0.256M

b) put the value of [A]t = 0.3 M and find out how much time taken

ln[A]t = -k∙t + ln [A]0

ln[0.3] = -6.2x10^-5 s^-1 x t + ln [0.4]

solve for t

t = 4640 s = 77.33 minutes