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: $$ \begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)+2 \mathrm{CO}_{2}(g) \end{array} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

Short answer

The equilibrium constants for the two reactions are: \(K_1 = e^{-\dfrac{ΔG°_1}{RT}}\) \(K_2 = e^{-\dfrac{ΔG°_2}{RT}}\) 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: \(W_{max1}=-ΔG°_1\) \(W_{max2}=-ΔG°_2\) 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

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 (ΔG°) by the following formula: $$ K = e^{-\dfrac{ΔG°}{RT}} $$ Where: \(K\) = equilibrium constant \(ΔG°\) = standard Gibbs free energy change in J/mol \(R\) = universal gas constant, 8.314 J/(K·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: $$ C_6H_{12}O_6(s) \rightleftharpoons 2C_2H_5OH(l) + 2CO_2(g) $$

Calculate the ΔG° for both reactions

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

Calculate the equilibrium constants

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

Compare the equilibrium constants

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

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: $$ W_{max}=-ΔG° $$ For reaction 1: $$ W_{max1}=-ΔG°_1 $$ For reaction 2: $$ W_{max2}=-ΔG°_2 $$

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.

Key concepts

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: \[ \Delta G° = \Delta H - T\Delta S \]Here, \( \Delta H \) represents the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy. A negative \( \Delta G° \) value indicates that the reaction is spontaneous.
For chemical reactions, the equilibrium constant \( K \) is related to the standard Gibbs free energy change \( \Delta G° \) by the formula: \[ K = e^{- \dfrac{ \Delta G°}{RT} } \]Where \( R \) is the universal gas constant and \( T \) is the temperature.
By converting \( \Delta G° \) 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 \( \Delta G° \), 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.