Question

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: $$\begin{array}{c}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6(s)+6{\mathrm{O}}_2(g)\rightleftharpoons 6{\mathrm{CO}}_2(g)+6{\mathrm{H}}_2\mathrm{O}(l)\\ {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6(s)\rightleftharpoons 2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}(l)+2{\mathrm{CO}}_2(g)\end{array}$$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

Short answer

To compare the given reactions, we first find their respective Gibbs free energy changes using the given data. After calculating the Gibbs energy changes, we find equilibrium constants for both reactions using the formula $K_{eq}=e^{(-\mathrm{\Delta }{G}_{rxn}^{\circ })/(RT)}$. Comparing their equilibrium constants, we can infer which reaction is more favorable. Then, we calculate the maximum work for both reactions using the formula $W_{max}=-\mathrm{\Delta }{G}_{rxn}^{\circ }$ and compare these values to analyze which process can provide more work under standard conditions.

Step-by-step solution

1. Analyzing the reactions

: First, let's write down the given chemical reactions separately. Oxidation of glucose: $${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6(s)+6{\mathrm{O}}_2(g)\rightleftharpoons 6{\mathrm{CO}}_2(g)+6{\mathrm{H}}_2\mathrm{O}(l)$$ Anaerobic decomposition (fermentation): $${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6(s)\rightleftharpoons 2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}(l)+2{\mathrm{CO}}_2(g)$$

2. The Gibbs energy of both reactions

: We are given the Gibbs free energy change $(\mathrm{\Delta }{G}^{\circ })$ values for each individual species in the Appendix C. To determine the Gibbs free energy change for the entire reaction, we use the equation: $\mathrm{\Delta }{G}_{rxn}^{\circ }=(\mathrm{\Delta }{G}_{products}^{\circ })-(\mathrm{\Delta }{G}_{reactants}^{\circ })$ We then have to find Gibbs energy change for both reactions.

3. Calculating the K_eq of both reactions

: To calculate the equilibrium constants, we use the formula: $K_{eq}=e^{(-\mathrm{\Delta }{G}_{rxn}^{\circ )}/(RT)}$ Where R is the universal gas constant 8.314 J/(mol K) and T is the temperature in Kelvin (assuming standard conditions, T = 298 K). Calculate the equilibrium constants for both reactions using their Gibbs free energy changes.

4. Comparing the equilibrium constants

: After calculating the equilibrium constants for both reactions, compare their values to discuss which reaction is more favorable.

5. Calculating the maximum work for both reactions

: The maximum work (W_max) that can be obtained from a process under standard conditions is given by the negative of the Gibbs free energy change of the reaction: $W_{max}=-\mathrm{\Delta }{G}_{rxn}^{\circ }$ Calculate the maximum work that can be obtained from both processes under standard conditions.

6. Comparing the maximum work

: Finally, compare the maximum work that can be obtained from both reactions to analyze which process can provide more work under standard conditions.

Key concepts

Equilibrium Constants

Equilibrium constants tell us how reactions proceed and which side of a chemical equation is favored at equilibrium. In the case of the oxidation of glucose to carbon dioxide and water, and the anaerobic decomposition of glucose to ethanol and carbon dioxide, equilibrium constants help us understand which reaction achieves a higher state of completion.

• Definition: The equilibrium constant, $K_{eq}$ is a dimensionless value that quantifies the ratio of the concentrations of products to reactants at equilibrium.
• Formula: It derives from the Gibbs free energy change, $\mathrm{\Delta }{G}_{rxn}^{\circ }$, using the equation: $$K_{eq}=e^{(-\mathrm{\Delta }{G}_{rxn}^{\circ })/(RT)}$$
• Comparison: A higher $K_{eq}$ implies a reaction that favors the formation of products at equilibrium, indicating that the reaction proceeds further to the right.

Understanding how to calculate and compare $K_{eq}$ between reactions is crucial. It informs us about the relative favorability and extent of different biochemical processes like the ones occurring in the body during glucose metabolism.

Gibbs Free Energy

Gibbs free energy, symbolized as $\mathrm{\Delta }G$, is a vital concept in thermodynamics and chemistry. It tells us about the feasibility of a chemical reaction and the energy available to do work.

• Basics: $\mathrm{\Delta }{G}^{\circ }$ is the free energy change of a reaction under standard conditions, indicating whether a reaction is spontaneous.
• Equation: The Gibbs free energy change is calculated as the difference between the free energies of the products and reactants: $$\mathrm{\Delta }{G}_{rxn}^{\circ }=\mathrm{\Delta }{G}_{products}^{\circ }-\mathrm{\Delta }{G}_{reactants}^{\circ }$$
• Impact on reactions: A negative $\mathrm{\Delta }{G}^{\circ }$ signifies that a reaction can spontaneously progress, whereas a positive value means it requires external energy.

In the glucose oxidation and fermentation reactions, comparing $\mathrm{\Delta }{G}^{\circ }$ helps us evaluate the energy efficiency of each process. The reaction with the more negative $\mathrm{\Delta }{G}^{\circ }$ is more energetic.

Maximum Work

Maximum work refers to the maximum amount of useful work that can be extracted from a chemical process under ideal conditions. This is especially important in understanding biological energy conversion.

• Definition: The maximum work $W_{max}$ is equal to the negative of the Gibbs free energy change, $-\mathrm{\Delta }{G}_{rxn}^{\circ }$.
• Relation to Gibbs free energy: The concept of maximum work is derived directly from Gibbs free energy, indicating how much of the energy can be harnessed for work.
• Biological significance: In living organisms, this principle explains how energy from glucose oxidation is converted into work, such as muscle contractions or metabolic processes.

When comparing glucose oxidation and fermentation, the process with a larger $W_{max}$ reflects more available energy, making it a more efficient source of biological work.