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

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

Step by step solution

01

- 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+].
02

- 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.
03

- 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.
04

- 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.
05

- 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.

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Key Concepts

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

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.

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Most popular questions from this chapter

The effect of substrate concentration on the first-order growth rate of a microbial population follows the Monod equation: \(\mu=\frac{\mu_{\max } S}{K_{\mathrm{s}}+S}\) where \(\mu\) is the first-order growth rate \(\left(\mathrm{s}^{-1}\right), \mu_{\max }\) is the maximum growth rate \(\left(\mathrm{s}^{-1}\right), S\) is the substrate concentration \(\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) and \(K_{\mathrm{s}}\) is the value of \(S\) that gives one-half of the maximum growth rate (in \(\mathrm{kg} / \mathrm{m}^{3}\) ). For \(\mu_{\max }=1.5 \times 10^{-4} \mathrm{~s}^{-1}\) and \(K_{\mathrm{s}}=0.03 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Plot \(\mu\) vs. \(S\) for \(S\) between 0.0 and \(1.0 \mathrm{~kg} / \mathrm{m}^{3}\). (b) The initial population density is \(5.0 \times 10^{3}\) cells \(/ \mathrm{m}^{3}\). What is the density after \(1.0 \mathrm{~h}\), if the initial \(S\) is \(0.30 \mathrm{~kg} / \mathrm{m}^{3} ?\) (c) What is it if the initial \(S\) is \(0.70 \mathrm{~kg} / \mathrm{m}^{3}\) ?

Archaeologists can determine the age of an artifact made of wood or bone by measuring the amount of the radioactive isotope \({ }^{14} \mathrm{C}\) present in the object. The amount of this isotope decreases in a first-order process. If \(15.5 \%\) of the original amount of \({ }^{14} \mathrm{C}\) is present in a wooden tool at the time of analysis, what is the age of the tool? The half- life of \({ }^{14} \mathrm{C}\) is \(5730 \mathrm{yr}\).

In the lower troposphere, ozone is one of the components of photochemical smog. It is generated in air when nitrogen dioxide, formed by the oxidation of nitrogen monoxide from car exhaust, reacts by the following mechanism: Assuming the rate of formation of atomic oxygen in step 1 equals the rate of its consumption in step \(2,\) use the data below to calculate (a) the concentration of atomic oxygen [O] and (b) the rate of ozone formation. $$ \begin{array}{lr} k_{1}=6.0 \times 10^{-3} \mathrm{~s}^{-1} & {\left[\mathrm{NO}_{2}\right]=4.0 \times 10^{-9} \mathrm{M}} \\ k_{2}=1.0 \times 10^{6} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s} & {\left[\mathrm{O}_{2}\right]=1.0 \times 10^{-2} \mathrm{M}} \end{array} $$

16.103 Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]\). The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) In the 1960 s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism: (1) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) [fast] (2) \(\mathrm{H}_{2}(g)+\mathrm{I}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{I}(g)\) [fast] (3) \(\mathrm{H}_{2} \mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) \quad[\) slow \(]\) Show that this mechanism is consistent with the rate law.

Arrhenius proposed that each reaction has an energy threshold that must be reached for the particles to react. The kinetic theory of gases proposes that the average kinetic energy of the particles is proportional to the absolute temperature. How do these concepts relate to the effect of temperature on rate?

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