Question

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate \(=k\left[\mathrm{H}^{+}\right][\) sucrose \(] .\) The initial rate of sucrose breakdown is measured in a solution that is \(0.01 M \mathrm{H}^{+}\), \(1.0 \mathrm{M}\) sucrose, \(0.1 \mathrm{M}\) fructose, and \(0.1 \mathrm{M}\) glucose. How does the rate change if (a) [sucrose] is changed to \(2.5 \mathrm{M} ?\) (b) [sucrose], [fructose], and [glucose] are all changed to \(0.5 \mathrm{M?}\) (c) \(\left[\mathrm{H}^{+}\right]\) is changed to \(0.0001 \mathrm{M} ?\) (d) [sucrose] and \(\left[\mathrm{H}^{+}\right]\) are both changed to \(0.1 \mathrm{M?}\)

Short answer

(a) Increase by 2.5 times (b) Decrease by two times (c) Decrease by 100 times (d) No change.

Step-by-step solution

- Understanding the Given Problem

The problem provides a rate law for the breakdown of sucrose in acidic solution: rate = k[H+][sucrose]. The initial conditions are given as 0.01 M H+, 1.0 M sucrose, 0.1 M fructose, and 0.1 M glucose. We need to determine how the rate changes with different variations in [sucrose] and [H+].

- Analyze Part (a)

In part (a), [sucrose] is changed to 2.5 M. According to the rate law, rate = k[H+][sucrose]. Since [H+] remains constant, the change in rate due to [sucrose = 2.5 M] is calculated as: rate_new = k[H+](2.5). The change in rate is calculated as: rate_new / rate_initial = (k[H+][2.5]) / (k[H+][1.0]) = 2.5/1.0 = 2.5. Therefore, the rate will increase by a factor of 2.5.

- Analyze Part (b)

In part (b), [sucrose], [fructose], and [glucose] are all changed to 0.5 M. However, the rate law only depends on [H+] and [sucrose]. Thus, rate = k[H+][sucrose]. The new rate with [sucrose = 0.5 M] is calculated as: rate_new = k[H+][0.5]. The change in rate is: rate_new / rate_initial = (k[H+][0.5]) / (k[H+][1.0]) = 0.5/1.0 = 0.5. Therefore, the rate will decrease by a factor of 0.5, or it will be halved.

- Analyze Part (c)

In part (c), [H+] is changed to 0.0001 M while [sucrose] remains 1.0 M. The new rate can be calculated as: rate_new = k[0.0001][sucrose]. The change in rate is: rate_new / rate_initial = (k[0.0001][1.0]) / (k[0.01][1.0]) = 0.0001/0.01 = 0.01. Therefore, the rate will decrease by a factor of 0.01, or it will be reduced to 1% of the initial rate.

- Analyze Part (d)

In part (d), both [sucrose] and [H+] are changed to 0.1 M each. The new rate can be calculated as: rate_new = k[0.1][0.1]. The change in rate is: rate_new / rate_initial = (k[0.1][0.1]) / (k[0.01][1.0]) = (0.01)/(0.01) = 1. Therefore, the rate will not change; it will remain the same.

Key concepts

rate law

A rate law is an equation that links the rate of a chemical reaction to the concentration of the reactants. For the breakdown of sucrose in an acidic solution, the rate law is given as: rate = k[H+][sucrose]. Here, k is the rate constant, [H+] is the concentration of hydrogen ions, and [sucrose] is the concentration of sucrose.

The rate law is essential because it helps predict how the rate of reaction changes when the concentrations of the reactants change. By understanding the rate law, we can better control reactions in a laboratory or industrial setting.

It's important to note that the rate law only includes the reactants whose concentrations significantly affect the reaction rate. In this example, [fructose] and [glucose] do not appear in the rate law, suggesting their concentrations do not affect the reaction rate.

sucrose breakdown

The breakdown of sucrose into glucose and fructose is an important biochemical reaction. Sucrose is a disaccharide, composed of one glucose molecule and one fructose molecule. When sucrose breaks down, it forms these two simpler sugars, which are more easily used by organisms.

The process of sucrose breakdown can be triggered in different environments. In this exercise, it occurs in an acidic solution, often assisted by an enzyme called invertase. The step-by-step breakdown involves the following:

• Protonation of sucrose by the acidic solution, making it more reactive.
• Breaking of the glycosidic bond linking glucose and fructose.
• Formation of glucose and fructose as end products.

Understanding this breakdown process is important in fields like food science and biochemistry, where controlling the production of glucose and fructose is crucial.

chemical kinetics

Chemical kinetics is the study of reaction rates and the steps involved in chemical reactions. By analyzing how different conditions impact the speed of reactions, we gain insights into reaction mechanisms and can optimize conditions for desired outcomes.

In this exercise, various aspects of chemical kinetics are exemplified:

• Dependence of reaction rates on reactant concentrations (as reflected in the rate law).
• Role of catalysts and environmental factors, such as acidic solutions in facilitating the reaction.
• Quantitative analysis to determine how changes in concentration affect the rate of sucrose breakdown.

These principles help us design better experiments and industrial processes, ensuring reactions proceed at optimal rates.

acidic solution

An acidic solution is one that has a high concentration of hydrogen ions (H+). These ions play a crucial role in many chemical reactions by affecting the reactivity of other molecules.

In the context of sucrose breakdown, an acidic solution provides a conducive environment for the reaction by protonating sucrose. This makes it easier for the glycosidic bond in sucrose to break, leading to the formation of glucose and fructose.

Understanding how acidic conditions impact reactions can help chemists control reaction conditions more effectively. This knowledge is especially important in industries such as pharmaceuticals, where precise reaction conditions are necessary to produce desired compounds efficiently and safely.