/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 99 The oxidation of glucose \(\left... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 99

The oxidation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in body tissue produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2}\). (a) Using data given in Appendix \(\mathrm{C}\), compare the equilibrium constants for the following reactions: $$ \begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)+2 \mathrm{CO}_{2}(g) \end{array} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

Short Answer

Expert verified
The equilibrium constants for the two reactions are: \(K_1 = e^{-\dfrac{螖G掳_1}{RT}}\) \(K_2 = e^{-\dfrac{螖G掳_2}{RT}}\) Compare the calculated equilibrium constants (K_1 and K_2). The larger equilibrium constant corresponds to the more spontaneous reaction. The maximum work that can be obtained from each process under standard conditions are: \(W_{max1}=-螖G掳_1\) \(W_{max2}=-螖G掳_2\) Compare the maximum works (W_max1 and W_max2) calculated for both reactions. The reaction with higher maximum work represents the process that can produce more work under standard conditions.

Step by step solution

01

Write down the equilibrium constant formula for both reactions

The equilibrium constant (K) for a reaction can be expressed in terms of standard Gibbs free energy change (螖G掳) by the following formula: $$ K = e^{-\dfrac{螖G掳}{RT}} $$ Where: \(K\) = equilibrium constant \(螖G掳\) = standard Gibbs free energy change in J/mol \(R\) = universal gas constant, 8.314 J/(K路mol) \(T\) = temperature, 298K (since, standard conditions) Now, we can write the equilibrium constants for the given reactions: Reaction 1: $$ C_6H_{12}O_6(s) + 6O_2(g) \rightleftharpoons 6CO_2(g) + 6H_2O(l) $$ Reaction 2: $$ C_6H_{12}O_6(s) \rightleftharpoons 2C_2H_5OH(l) + 2CO_2(g) $$
02

Calculate the 螖G掳 for both reactions

To calculate the 螖G掳 for both reactions, we'll use the following equation: $$ 螖G掳 = 螖G掳_{(products)} - 螖G掳_{(reactants)} $$ For reaction 1: $$ 螖G掳_1 = [6螖G掳(CO_2(g)) + 6螖G掳(H_2O(l))] - [螖G掳(C_6H_{12}O_6(s)) + 6螖G掳(O_2(g))] $$ For reaction 2: $$ 螖G掳_2 = [2螖G掳(C_2H_5OH(l)) + 2螖G掳(CO_2(g))] - 螖G掳(C_6H_{12}O_6(s)) $$ Using the values of 螖G掳 provided in Appendix C, calculate 螖G掳 for both reactions.
03

Calculate the equilibrium constants

Now, we can use the calculated 螖G掳 for both reactions and the previously mentioned equation to find the equilibrium constants for both reactions: For reaction 1: $$ K_1 = e^{-\dfrac{螖G掳_1}{RT}} $$ For reaction 2: $$ K_2 = e^{-\dfrac{螖G掳_2}{RT}} $$
04

Compare the equilibrium constants

Compare the calculated equilibrium constants (K_1 and K_2). The larger equilibrium constant corresponds to the more spontaneous reaction.
05

Calculate the maximum work

The maximum work (W_max) that can be obtained from a process under standard conditions can be calculated using the following equation: $$ W_{max}=-螖G掳 $$ For reaction 1: $$ W_{max1}=-螖G掳_1 $$ For reaction 2: $$ W_{max2}=-螖G掳_2 $$
06

Compare the maximum work

Finally, compare the maximum works (W_max1 and W_max2) calculated for both reactions. The reaction with higher maximum work represents the process that can produce more work under standard conditions.

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Key Concepts

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

Gibbs Free Energy
Gibbs Free Energy is a concept used to determine whether or not a reaction can occur spontaneously. The formula used to calculate it is: \[ \Delta G掳 = \Delta H - T\Delta S \]Here, \( \Delta H \) represents the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy. A negative \( \Delta G掳 \) value indicates that the reaction is spontaneous.
For chemical reactions, the equilibrium constant \( K \) is related to the standard Gibbs free energy change \( \Delta G掳 \) by the formula: \[ K = e^{- \dfrac{ \Delta G掳}{RT} } \]Where \( R \) is the universal gas constant and \( T \) is the temperature.
By converting \( \Delta G掳 \) into an equilibrium constant, scientists can predict which way a reaction will proceed. In biological systems, this is critical for understanding processes such as cellular respiration and fermentation.
Cellular Respiration
Cellular respiration is a biological process by which cells convert glucose into energy. It takes place mainly in the mitochondria of cells and can be divided into several stages, including glycolysis, the Krebs cycle, and oxidative phosphorylation.
The overall reaction for cellular respiration is:\[ \text{C}_{6}\text{H}_{12}\text{O}_{6} + 6\text{O}_{2} \rightarrow 6\text{CO}_{2} + 6\text{H}_{2}\text{O} + \text{Energy (as ATP)} \]
Key aspects of cellular respiration:
  • Produces a large amount of energy (ATP)
  • Involves the oxidation of glucose molecules
  • Oxygen is required for the process (aerobic)

In this process, glucose is broken down and energy is transferred to ATP, the energy currency of the cell. This reaction is highly efficient due to its large negative \( \Delta G掳 \), which makes it a highly spontaneous process under normal cellular conditions.
Anaerobic Fermentation
Anaerobic fermentation is a process that allows cells to convert glucose into energy without the need for oxygen. It typically occurs in environments lacking oxygen, and in certain organisms like yeast or in muscle cells when oxygen levels are low.
The general reaction for anaerobic fermentation (as in yeast) is:\[ \text{C}_{6}\text{H}_{12}\text{O}_{6} \rightarrow 2\text{C}_{2}\text{H}_{5}\text{OH} + 2\text{CO}_{2} \]
Important points about anaerobic fermentation:
  • Less efficient than cellular respiration
  • Produces ethanol and carbon dioxide
  • Occurs in the absence of oxygen (anaerobic)

This process is used extensively in industrial applications like brewing and bread-making. Its inefficiency compared to cellular respiration means less ATP is produced per glucose molecule. However, it is a critical survival mechanism for some organisms in low-oxygen environments.

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Most popular questions from this chapter

Consider what happens when a sample of the explosive TNT is detonated under atmospheric pressure. (a) Is the detonation a reversible process? (b) What is the sign of \(q\) for this process? (c) Is w positive, negative, or zero for the process?

A standard air conditioner involves a \(r\) frigerant that is typically now a fluorinated hydrocarbon, such as \(\mathrm{CH}_{2} \mathrm{~F}_{2}\). An air- conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram). During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (a) What is the sign of \(q\) for the expansion? (b) What is the sign of \(q\) for the compression? (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why? (d) Imagine that a sample of liquid refrigerant undergoes expansion followed by compression, so that it is back to its original state. Would you expect that to be a reversible process? (e) Suppose that a house and its exterior are both initially at \(31^{\circ} \mathrm{C}\). Some time after the air conditioner is turned on, the house is cooled to \(24^{\circ} \mathrm{C}\). Is this process spontaneous of nonspontaneous?

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .(\mathbf{b})\) How much heat is produced in burning \(1 \mathrm{~mol}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) under standard conditions if both reactants and products are brought to \(298 \mathrm{~K} ?\) (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction?

When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here: Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on!). What do you observe? Are your observations consistent with your answer to part (b)?

Consider the reaction $$ \mathrm{PbCO}_{3}(s) \rightleftharpoons \mathrm{PbO}(s)+\mathrm{CO}_{2}(g) $$ Using data in Appendix \(\mathrm{C}\), calculate the equilibrium pressure of \(\mathrm{CO}_{2}\) in the system at $$ \text { (a) } 400^{\circ} \mathrm{C} \text { and } $$ $$ \text { (b) } 180^{\circ} \mathrm{C} \text { . } $$
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