Problem 15 - Chemistry and Chemical Reactivity

Question

The rate equation for the hydrolysis of sucrose to fructose and glucose $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})$$ is \(-\Delta[\text { sucrose }] / \Delta t=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] .\) After 27 minutes at \(27^{\circ} \mathrm{C},\) the sucrose concentration decreased from \(0.0146 \mathrm{M}\) to \(0.0132 \mathrm{M} .\) Find the rate constant, \(k.\)

Step-by-Step Solution

Answer

The rate constant \(k\) is approximately \(5.93 \times 10^{-5} \text{ s}^{-1}.\)

Step 1: Understand the Rate Equation

The rate equation given is \ -\Delta[\text{sucrose}]/\Delta t=k[\text{C}_{12}\text{H}_{22}\text{O}_{11}]. \ This implies that the rate of reaction depends on the concentration of sucrose and is directly proportional to it, with \(k\) being the rate constant.

Step 2: Identify Known Quantities

We know that the initial concentration of sucrose \([\text{C}_{12}\text{H}_{22}\text{O}_{11}]_0\) is 0.0146 M, and the final concentration \([\text{C}_{12}\text{H}_{22}\text{O}_{11}]_f\) is 0.0132 M, after 27 minutes. \(\Delta t = 27\) minutes needs to be converted to seconds for our rate, so \(\Delta t = 27 \times 60 = 1620\) seconds.

Step 3: Calculate Change in Concentration

The change in concentration \(-\Delta[\text{sucrose}]\) is the difference between the initial and final concentrations: \ -\Delta[\text{sucrose}] = [\text{C}_{12}\text{H}_{22}\text{O}_{11}]_0 - [\text{C}_{12}\text{H}_{22}\text{O}_{11}]_f \ = 0.0146 \text{ M} - 0.0132 \text{ M} = 0.0014 \text{ M}.

Step 4: Apply Values to Rate Equation

Substitute the values into the rate equation: \ \frac{0.0014 \text{ M}}{1620 \text{ s}} = k \times 0.0146 \text{ M}. \ This simplifies the expression to find \(k\).

Step 5: Solve for Rate Constant \(k\)

Rearrange the equation to solve for \(k\): \ k = \frac{0.0014 \text{ M}}{1620 \text{ s} \times 0.0146 \text{ M}} \ = \frac{0.0014}{1620 \times 0.0146}. \ Perform the calculation for \(k\).

Step 6: Calculate and Verify

Calculate \(k\) using the formula from Step 5: \ k = \frac{0.0014}{1620 \times 0.0146} \approx 5.93 \times 10^{-5} \text{ s}^{-1}. \ Verify by checking the dimensional consistency and recalculating as needed.

Key Concepts

Rate Equation

The rate equation is a fundamental concept in chemical kinetics that describes how the rate of a reaction depends on the concentration of one or more reactants. In our example, the hydrolysis of sucrose is governed by the equation: \[-\frac{\Delta [\text{sucrose}]}{\Delta t} = k[\text{C}_{12}\text{H}_{22}\text{O}_{11}].\]This indicates that the reaction rate is directly proportional to the concentration of sucrose present. The higher the concentration of sucrose, the quicker the reaction will proceed. This is often expressed with a negative sign representing the decrease in the concentration of sucrose over time.

• The rate expression involves the change in concentration \( \Delta [\text{sucrose}] \) over change in time \( \Delta t \).
• The proportionality constant, \( k \), is known as the rate constant, which is crucial in connecting concentration and reaction rate.

Using rate equations, we can predict how fast reactions occur based on reactant concentrations, which helps in chemical manufacturing and environmental science.

Rate Constant

The rate constant \( k \) is a pivotal element in the rate equation. It is a measure of the speed of a chemical reaction. In the exercise, the rate constant for the hydrolysis of sucrose was calculated to be approximately \( 5.93 \times 10^{-5} \) s\(^{-1}\). This number, often specific to the reaction and conditions such as temperature and pressure, helps chemists understand the reaction's characteristics.

• The value of \( k \) tells us about the reaction's speed. A higher \( k \) means a faster reaction, whereas a lower \( k \) indicates a slower reaction.
• Units of \( k \) vary depending on the order of the reaction. Here it's s\(^{-1}\), suitable for first-order reactions.

In practice, determining \( k \) involves experiments where reactant concentrations and reaction rates are measured, ensuring accurate mathematical descriptions of chemical processes.

Hydrolysis Reaction

A hydrolysis reaction is a chemical process where a compound reacts with water, leading to the breakdown of that compound. In the context of the given exercise, sucrose undergoes hydrolysis to yield two simpler sugars: fructose and glucose.

• This reaction is significant in both biological systems, for instance in digestion, and industrial applications like sugar refining.
• Hydrolysis often involves a water molecule contributing a hydrogen ion to one fragment and a hydroxide ion to another. This facilitates the breakdown and formation of new substances.

Understanding hydrolysis reactions has vast implications, such as improving efficient production processes and enhancing our comprehension of biological mechanisms.