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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, \(k\).

Short Answer

Expert verified
The rate constant, \(k\), is approximately \(5.92 \times 10^{-5}\, \text{L/s}\).

Step by step solution

01

Understand the Rate Equation

The rate of disappearance of sucrose is given by the equation \(-\Delta[\text{sucrose}] / \Delta t=k\left[\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}\right]\). This means the rate is directly proportional to the concentration of sucrose.
02

Determine the Change in Concentration

Calculate the change in sucrose concentration: \(\Delta[\text{sucrose}] = [\text{initial sucrose}] - [\text{final sucrose}] = 0.0146\,M - 0.0132\,M = 0.0014\,M\).
03

Calculate the Change in Time

Convert the time duration from minutes to seconds. \(27\, \text{minutes} = 27 \times 60 = 1620\, \text{seconds}\). Success is in scientific conversions, so make sure you are meticulous when converting time units.
04

Insert Values Into Rate Equation

Insert the values into the rate equation: \(-\frac{0.0014}{1620} = k(0.0146)\). Here, \(-\Delta[\text{sucrose}] / \Delta t = -0.0014 / 1620\) and \([\text{sucrose}] = 0.0146\,M\).
05

Solve for the Rate Constant \(k\)

Rearrange the equation to solve for \(k\): \(k = \frac{0.0014}{1620 \times 0.0146}\). Calculate \(k\) to find the rate constant.
06

Perform Calculation

Calculate the value of \(k\). This results in \(k \approx 5.92 \times 10^{-5}\, \text{L/s}\). It's often useful to check your work calculator if available to ensure accuracy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Constant
The rate constant, often represented by the symbol \( k \), is a fundamental parameter in chemical kinetics. It establishes the relationship between the reaction rate and the concentration of reactants. In essence, \( k \) determines how quickly a reaction proceeds under specific conditions, such as temperature and pressure.

For the sucrose hydrolysis reaction, the rate constant can be derived from the rate equation \(-\Delta[\text{sucrose}] / \Delta t = k[\text{C}_{12} \text{H}_{22} \text{O}_{11}]\). Here, \( k \) is directly linked to how fast the concentration of sucrose decreases over time.
  • The larger the value of \( k \), the faster the reaction rate.
  • The rate constant is sensitive to temperature changes. As temperature increases, \( k \) usually increases, resulting in a faster reaction.
  • Units of \( k \) can vary depending on the order of the reaction. For a first-order reaction like sucrose hydrolysis, the units are \( \text{L/s} \).
Reaction Rate
The reaction rate refers to how quickly a reaction occurs. It is defined as the change in concentration of reactants or products per unit time.

For the hydrolysis of sucrose, the reaction rate is observed in terms of how fast sucrose concentration decreases. Mathematically, it is given by \(-\Delta[\text{sucrose}] / \Delta t\), which signifies the rate of disappearance of sucrose over time.
  • Reaction rates can be affected by several factors, including concentrations, temperature, and the presence of catalysts.
  • A higher concentration of reactants generally leads to a higher reaction rate.
  • Monitoring the reaction rate helps in understanding the kinetics of a reaction and predicting the time required for the reaction.
Hydrolysis
Hydrolysis is a type of chemical reaction where a molecule is split into two parts by the addition of a water molecule. In the context of sucrose, hydrolysis involves breaking down the disaccharide sucrose into its monomer components, glucose, and fructose.

This process is essential in biological systems and industrial applications. During hydrolysis, enzymes can act as catalysts, speeding up the reaction that would otherwise occur slowly.
  • Sucrose, a common sugar, undergoes hydrolysis in the presence of water to yield glucose and fructose.
  • The reaction equation for hydrolysis is \( \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} + \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \).
  • Hydrolysis is crucial in digestion where complex molecules are broken down into simpler ones for absorption.
Sucrose
Sucrose, commonly known as table sugar, is a disaccharide composed of glucose and fructose. It's a natural carbohydrate that plants use for energy and storage.

In chemistry, sucrose plays a vital role in studying reaction kinetics, especially in understanding hydrolysis reactions. When sucrose undergoes hydrolysis, it breaks down into glucose and fructose, making it an excellent model for kinetic studies.
  • Sucrose is highly soluble in water, making it easy to study its concentration changes in aqueous reactions.
  • It is used as a benchmark in studies of enzymatic reactions, particularly with sucrase, the enzyme that catalyzes sucrose hydrolysis.
  • Sucrose hydrolysis is a first-order reaction, where the rate depends on the concentration of sucrose alone.

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Most popular questions from this chapter

Compare the lock-and-key and induced-fit models for substrate binding to an enzyme.

The decomposition of ammonia on a metal surface to form \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) is a zero-order reaction (Figure \(14.7 \mathrm{c}) .\) At \(873^{\circ} \mathrm{C},\) the value of the rate constant is \(1.5 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\). How long it will take to completely decompose 0.16 g of \(\mathrm{NH}_{3}\) in a \(1.0-\mathrm{L}\) flask?

To which species should an enzyme bind best: the substrate, transition state, or product of a reaction?

\(\mathrm{NO}_{x^{\prime}}\) a mixture of \(\mathrm{NO}\) and \(\mathrm{NO}_{2},\) plays an essential role in the production of pollutants found in photochemical smog. The \(\mathrm{NO}_{x}\) in the atmosphere is slowly broken down to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) in a first-order reaction. The average half-life of \(\mathrm{NO}_{x}\) in the smokestack emissions in a large city during daylight is 3.9 hours. (a) Starting with \(1.50 \mathrm{mg}\) in an experiment, what quantity of \(\mathrm{NO}_{x}\) remains after 5.25 hours? (b) How many hours of daylight must have elapsed to decrease \(1.50 \mathrm{mg}\) of \(\mathrm{NO}_{x}\) to \(2.50 \times\) \(10^{-6} \mathrm{mg} ?\)

In the reaction \(2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g}),\) the rate of for- mation of \(\mathrm{O}_{2}\) is \(1.5 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\). What is the rate of decomposition of \(\mathrm{O}_{3} ?\)

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