Question

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) hydrolyzes into glucose and fructose. The hydrolysis is a first-order reaction. The half-life for the hydrolysis of sucrose is 64.2 min at \(25^{\circ} \mathrm{C}\). How many grams of sucrose in \(1.25 \mathrm{~L}\) of a \(0.389 \mathrm{M}\) solution are hydrolyzed in 1.73 hours?

Short answer

Answer: Approximately 109.54 grams of sucrose are hydrolyzed in 1.73 hours.

Step-by-step solution

Convert the time to minutes

First, we need to convert the time given (1.73 hours) into minutes. There are 60 minutes in an hour, so we can use the conversion factor: time (minutes) = 1.73 hours × 60 minutes/hour = 103.8 minutes Now we have the time in minutes to use with the half-life.

Determine the fraction of sucrose remaining

We are given that the half-life of sucrose is 64.2 minutes, and we know that the relationship between the first-order reaction rate constant k and half-life t₍₁/₂₎ is: t₍₁/₂₎ = ln(2) / k With the half-life and the elapsed time, we can determine the fraction of sucrose remaining after 103.8 minutes. Let x be the fraction of sucrose remaining after 103.8 minutes. Then, at time t = 103.8 minutes, the following relationship holds: x = (1/2)^(time / half-life) = (1/2)^(103.8 / 64.2) Calculating this value results in: x ≈ 0.341 So, approximately 34.1% of the sucrose remains after 1.73 hours.

Calculate the amount of sucrose hydrolyzed

Now that we know the fraction of sucrose remaining, we can calculate the amount of sucrose hydrolyzed. Since the solution was initially 0.389 M and had a volume of 1.25 L, the initial moles of sucrose can be calculated as: moles of sucrose (initial) = 0.389 mol/L × 1.25 L = 0.48625 mol Now, we can calculate the amount of sucrose hydrolyzed by finding the difference between the initial moles and the moles remaining: moles of sucrose (hydrolyzed) = moles of sucrose (initial) × (1 - fraction remaining) = 0.48625 mol × (1 - 0.341) ≈ 0.320 mol

Convert moles of sucrose to grams

Finally, we need to convert the moles of sucrose hydrolyzed into grams. The molar mass of sucrose is 342.3 g/mol, so we can use the conversion factor: grams of sucrose (hydrolyzed) = 0.320 mol × 342.3 g/mol ≈ 109.54 g Therefore, approximately 109.54 grams of sucrose are hydrolyzed in 1.73 hours.