The hydrolysis of sucrose (C12H22O11) into glucose and fructose in acidic water has a rate constant of 1.8*10-4s-1 at 25 ∘C. Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.55 L of a 0.160 M sucrose solution is allowed to react for 200 minutes.

Mass of sucrose that is hydrolyzed = 123.55 g

Explanation:

Since the reaction is a first ofree reaction,

Let the initial concentration of sucrose be C₀

And the concentration of sucrose left at any time be C.

r = (dC/dt) = - kC (Minus sign because it's a rate of reduction)

K = rate constant

(dC/dt) = - kC

(dC/C) = - kdt

∫ (dC/C) = - k ∫ dt

Solving the two sides as definite integrals by integrating the left hand side from C₀ to C and the Right hand side from 0 to t.

We obtain

In (C/C₀) = - kt

(C/C₀) = (e⁻ᵏᵗ)

C = C₀ e⁻ᵏᵗ

So, we can find the concentration of sucrose left at the given time

k = 1.8 * 10⁻⁴ s⁻¹

t = 200 minutes = 200 * 60 = 12000 s

kt = 12000 * 1.8 * 10⁻⁴ = 2.16

C₀ = 0.160 M

C = 0.160 e⁻²•¹⁶ = 0.01845 M

So, the concentration that has been used up = 0.160 - 0.01845 = 0.142 M

To calculate the mass of sucrose that has reacted, we'll convert the concentration used up into number of moles.

Concentration = (number of moles) / (volume in Litres)

number of moles = concentration * volume in Litres

Volume in Litres = 2.55L

Number of moles = 0.142 * 2.55 = 0.361 moles

Then, mass = number of moles * Molar mass

Molar mass of sucrose = 342.3 g/mol

Mass of sucrose used up = 0.361 * 342.3 = 123.55 g