Problem 14 - Analytical Chemistry

Question

Sucrose is hydrolyzed to glucose and fructose: $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ In dilute aqueous solution, the water concentration remains essentially constant. and so the reaction is pseudo first order and follows first-order kinetics. If \(25.0 \%\) of a \(0.500 M\) sucrose solution is hydrolyzed in \(9.00 \mathrm{~h}\), in how much time will the glucose and fructose concentration be equal to one-half the concentration of the remaining sucrose?

Step-by-Step Solution

Answer

It takes about 20.4 hours for glucose and fructose concentrations to equal half the remaining sucrose.

Step 1: Understanding the Reaction and Initial Conditions

The reaction involves the hydrolysis of sucrose into glucose and fructose. The initial concentration of sucrose is given as \(0.500\, \text{M}\), and \(25.0\%\) of it is hydrolyzed in \(9.00 \, \text{h}\). The constant concentration of water makes the reaction pseudo first-order.

Step 2: Write the First-Order Kinetics Equation

For a first-order reaction, the rate equation is \( \ln (\text{[A]}_0/\text{[A]}_t) = kt \), where \( \text{[A]}_0 \) is the initial concentration, \( \text{[A]}_t \) is the concentration at time \( t \), and \( k \) is the rate constant.

Step 3: Calculate the Rate Constant \( k \)

If \(25.0\%\) of sucrose is hydrolyzed, then \( \text{[A]}_t = 0.750 \times 0.500 \, \text{M} = 0.375 \, \text{M} \). Substitute into the first-order equation:\[ \ln \left( \frac{0.500}{0.375} \right) = k \times 9.00 \]Solving this, we find \( k \approx 0.0339 \, \text{h}^{-1} \).

Step 4: Determine Time for Half Concentration Remaining

To find when glucose and fructose concentrations are equal to one-half the remaining sucrose concentration, we need \( \text{[A]} = 0.250 \, \text{M} \) (since original \(0.500\, \text{M}/2=0.250 \, \text{M}\) after half reaction). Set up: \[ \ln \left( \frac{0.500}{0.250} \right) = k \times t \]\[ \ln 2 = 0.0339 \times t \]Solving gives \( t \approx 20.4 \, \text{h} \).

Key Concepts

Pseudo First-Order Reactions

Pseudo first-order reactions are a fascinating concept in chemical kinetics. They occur when the concentration of one reactant remains constant throughout the reaction. This is typical in cases where one of the reactants is present in a large excess or doesn’t change much. In such scenarios, a reaction that might normally follow a higher order kinetic, behaves like a first-order reaction. This makes the mathematics and calculations more straightforward.
For example, in the hydrolysis of sucrose, water is a reactant but its concentration is so large that it stays nearly constant. As a result, we treat it as a pseudo-first-order reaction. We analyze it using the first-order kinetics formula which simplifies our calculations and understanding of the reaction kinetics.

Rate Constant Calculation

The rate constant, represented by the symbol \( k \), is a crucial factor in determining how fast a reaction proceeds. To calculate it for a first-order reaction, we use experimental data such as the amount of reactant decomposed over a certain time.

• Use the equation: \( \ln \left( \frac{[A]_0}{[A]_t} \right) = kt \)
• Where \([A]_0\) is the initial concentration and \([A]_t\) is the concentration at time \( t \).
• The expression simplifies to \( k = \frac{\ln([A]_0/[A]_t)}{t} \)

In the sucrose hydrolysis problem, we knew that \( 25\% \) of sucrose hydrolyzed in \( 9.00 \text{ h} \) to a concentration of \( 0.375\, \text{M} \). Using these values, we calculate \( k \) to be approximately \( 0.0339\, \text{h}^{-1} \). This value tells us how quickly the reaction occurs.

Chemical Hydrolysis

Chemical hydrolysis is a common chemical reaction involving the breaking down of a compound due to the addition of water. It plays an essential role in various biological and industrial processes. In the case of sucrose, hydrolysis converts it into glucose and fructose.
Hydrolysis reactions tend to be slow. Therefore, understanding the kinetics, especially when enhanced to pseudo first-order in water-rich environments, is key to controlling and utilizing such reactions. Recognizing that the reaction mimics a first-order process due to constant water concentration helps in streamlining both study and application.

Reaction Rate Equations

Reaction rate equations describe how the concentration of reactants and products change over time. For pseudo first-order reactions, the equation \( \ln([A]_0/[A]_t) = kt \) is often applied for quick calculations.

• This equation links initial and remaining concentrations after time \( t \).
• Use it to predict concentrations at future time points given a rate constant.

In our exercise, knowing \( 0.500 \, \text{M} \) is the starting concentration, and calculating the time needed for the product concentration to reach a particular threshold (such as half the starting amount) is straightforward with this setup.
These equations allow scientists to forecast how changes in conditions affect reaction rates.