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 \(0.500 \mathrm{M}\) 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 approximately 22.34 hours for glucose and fructose to reach half the sucrose concentration.

Step 1: Understand the Reaction

The reaction given is a hydrolysis of sucrose to form glucose and fructose. It is described by the equation:\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6 \]In dilute solution, this can be treated as a pseudo first-order reaction since the concentration of water remains constant.

Step 2: Initial Conditions and Rate Law

The problem states that 25% of a 0.500 M solution is hydrolyzed in 9.00 hours. Using the pseudo-first-order kinetics, the integrated rate law is:\[ [A] = [A]_0 e^{-kt} \]where - \([A]_0\) is the initial concentration,- \(t\) is time,- \(k\) is the first-order rate constant.

Step 3: Calculate the Rate Constant

If 25% of the sucrose is hydrolyzed, 75% remains. Thus, \([A] = 0.75 \times 0.5 \, \text{M} = 0.375 \, \text{M}\).Substitute this into the rate law: \[ 0.375 = 0.500 e^{-9k} \]Solve for \(k\):\[ e^{-9k} = \frac{0.375}{0.500} = 0.75 \]\[ -9k = \ln(0.75) \]\[ k = -\frac{\ln(0.75)}{9} \approx 0.03101 \, \text{hr}^{-1} \]

Step 4: Determine Time for Half Concentration of Sucrose

We need the time \(t\) when the concentrations of glucose and fructose are equal to one-half the concentration of the remaining sucrose. This means:\[ [\text{sucrose}] = 2\times[\text{glucose}] = 2\times[\text{fructose}] \]As glucose and fructose form equally from sucrose, they must each have concentrations half that of sucrose. So, substitute into:\[ [A] = [A]_0 e^{-kt} \]We want:\[ [A] = \frac{1}{2}[A]_0 \text{ (as glucose concentration) = 0.500 e^{-kt}} \]From this, solve:\[ 0.5 = e^{-kt} \]\[ kt = \ln(2) \]\[ t = \frac{\ln(2)}{k} \approx \frac{0.693}{0.03101} \approx 22.34 \, \text{hours} \]

Step 5: Conclusion: Find Total Time

Thus, it will take approximately 22.34 hours for the concentrations of glucose and fructose to be equal to one-half of the concentration of the remaining sucrose.

Key Concepts

Pseudo First-Order Reaction

In the world of chemical kinetics, reactions are typically classified by their order, which describes how the rate depends on the concentration of reactants. A pseudo first-order reaction is a convenient classification for cases when a reaction seems to behave as a first-order reaction, even if it's not inherently first-order. This can happen when one reactant is in large excess, thus maintaining a relatively constant concentration during the reaction.

For our exercise, the hydrolysis of sucrose into glucose and fructose is originally a second-order reaction because two reactants (sucrose and water) are involved. However, in a dilute aqueous solution, the concentration of water remains constant as it is present in excess. Under these conditions, the reaction can be simplified to a pseudo first-order reaction, allowing the use of first-order kinetics to describe it efficiently. This simplification is a key to solving problems involving large solvent excess and is useful in simplifying kinetic calculations.

Integrated Rate Law

In chemical kinetics, the integrated rate law connects the concentration of reactants or products to time explicitly. For a first-order reaction, the integrated rate law provides a simple mathematical expression that relates these variables, allowing determination of either the concentration at a given time or the time taken to reach a particular concentration.

The integrated rate law for first-order kinetics is written as: \([A] = [A]_0 e^{-kt}\)

• \([A]\) is the concentration of the reactant at time \(t\).
• \([A]_0\) is the initial concentration of the reactant.
• \(k\) is the first-order rate constant.

This equation allows us to determine how much reactant remains at any time \(t\) given the rate constant \(k\). It is particularly useful for reactions that follow pseudo first-order kinetics, like the sucrose hydrolysis in our problem. By knowing the initial and final concentrations and the time interval, we can calculate the rate constant and predict future concentrations, demonstrating the power of this integrated approach.

Rate Constant Calculation

Calculating the rate constant \(k\) is a fundamental step in analyzing reaction kinetics because it quantifies how quickly a reaction proceeds.

In pseudo first-order reactions, where one reactant concentration is constant, the rate constant can be derived from the integrated rate law by rearranging: \([A] = [A]_0 e^{-kt}\)
Given that a certain percentage of a reactant is consumed, we can rearrange the formula to solve for \(k\): \( -kt = ext{ln}igg(rac{[A]}{[A]_0}igg)\)
Subsequently, the rate constant \(k\) becomes: \(k = -rac{ ext{ln}([A]/[A]_0)}{t}\). For example, using our problem where 25% of sucrose is hydrolyzed in 9 hours, with 75% remaining, we can substitute into the formula to find \(k\): \(k = -rac{ ext{ln}(0.75)}{9}\).

Understanding this calculation allows us to determine how fast a reaction unfolds and is crucial for predicting how systems behave over time or designing optimal reaction conditions in laboratory and industrial settings.