Question

The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate law. In neutral solution, k = 2.1 X 10^(-11) s^(-1) at 27 °C.

a) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 X 10^(-7) M.
How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst?
b) Why does assuming that the reaction is irreversible simplify the calculation in part (a)?

Answer 1

The time required for sucrose to reach equilibrium concentration can be calculated using the first-order rate law and the provided rate constant. Since the equilibrium concentration of sucrose is very low, the reaction can reasonably be assumed to be irreversible, simplifying the calculation by excluding the reverse reaction rate.

Step-by-step explanation:

The hydrolysis of sucrose into glucose and fructose follows a first-order rate law, which allows for the calculation of how long it will take for the concentration of sucrose to change from an initial concentration to a concentration at equilibrium using the first-order rate equation. Given the rate constant k = 2.1 x 10-11 s-1 at 27 °C and the initial concentration of 0.150 M reaching an equilibrium concentration of 1.65 x 10-7 M, we use the formula:

ln([A]0/[A]) = kt

Where:

• [A]0 is the initial concentration of sucrose (0.150 M)
• [A] is the concentration of sucrose at equilibrium (1.65 x 10-7 M)
• k is the rate constant (2.1 x 10-11 s-1)
• t is the time to reach equilibrium

Solving for t will give the time required to reach equilibrium concentration. Since the equilibrium concentration of sucrose is considerably low, it is reasonable to assume the reaction is irreversible. This assumption simplifies the calculation because we do not need to consider the reverse reaction rate, making the mathematics less complex.