The oxidation of glucose (C₆H₁₂O₆) in body tissue produces CO₂ and H₂O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C₂H₅OH) and CO2.
(b) Compare the maximum work that can be obtained from these processes under standard conditions.
C₆H₁₂O₆(s) + 6 O₂(g) ⇌ 6 CO₂(g) + 6 H₂O(l)
C₆H₁₂O₆(s) ⇌ 2 C₂H₅OH(l) + 2 CO₂(g)

Question

The oxidation of glucose (C₆H₁₂O₆) in body tissue produces CO₂ and H₂O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C₂H₅OH) and CO2.

(b) Compare the maximum work that can be obtained from these processes under standard conditions.

C₆H₁₂O₆(s) + 6 O₂(g) ⇌ 6 CO₂(g) + 6 H₂O(l)

C₆H₁₂O₆(s) ⇌ 2 C₂H₅OH(l) + 2 CO₂(g)

Accepted Answer

Welcome back everyone in this example we have the lactic acid fermentation of glucose which produces lactic acid below in the following equilibrium reaction. And we're told that the anaerobic fermentation of glucose produces ethanol and carbon dioxide. By this below equilibrium reaction we need to make a comparison between maximum work produced by these processes. Under standard conditions were also given the gibbs free energy of formation of each of our re agents and both reactions. So we wanna first recall to calculate our standard change of gibbs free energy of both reactions which we can do by taking the standard gibbs free energy of formation of our products subtracted from from the standard gibbs free energy of formation of our reactant. So we're going to follow this formula for both of our equations where for our first reaction we can say that the standard gibbs free energy change of our reaction is equal to beginning with the gibbs free energy of formation of our product. Where we have two moles of lactic acid, multiplied by its gibbs free energy of formation given in the prompt as negative 711.62 kg joules per mole. This is then subtracted from. So this completes the sum of the entropy of formation of our products. And now we can subtract from the some of the entropy of formation of our only other reactant or sorry, of our single reactant which is our glucose where we have one mole of glucose multiplied by its standard entropy of formation given in the prompt as negative 910.4 kg joules per mole. And just so everything is visible. I'll scoot this over. So this is kilo joules per mole as our units here. So just for purposes of room will actually move all of this down below, we'll label this reaction one. So this is all for reaction number one. This is going to yield a gibbs free or standard gibbs free energy of reaction one equal to a value of negative 512.84 kg joules. We're left with kilograms because we can cancel our units of moles because it's in the numerator and the denominator. So now moving on to reaction number two, we want to calculate the same or calculate by the same process. So we have our standard gibbs free energy of our reaction which is equal to we begin with our products where we have two products for the second reaction, beginning with our two moles of our ethanol, multiplied by ethanol's um. Standard gibbs free energy of formation which is given in the prompt as the value negative 174.8 kg joules per mole. This is then added to our second product where we have two moles of carbon dioxide multiplied by its gibbs. Free energy of formation given in the prompt as negative 394.4 kg jewels Permal, completing our some of our gifts. Free energy of formation of our products. And now we can subtract from our some of our gibbs free energy of formation of our single reactant in reaction to where we have Glucose again in this case, one mole of glucose multiplied by its entropy of formation. Given in the prompt as the value negative 910. kg joules per mole. And so for this second reaction we would get a gibbs free standard gibbs free energy of our reaction equal to a value of negative 228 kg joules. Because again, we can cancel our units of moles and we can see that between these two standard gibbs free energies of these two reactions, we have a more negative standard gibbs free energy or delta G. From reaction number one. And this means that therefore reaction one reaction number one does more work than reaction to and thus has a greater energy release again due to this more negative gives free energy value. So, based on what we've outlined here in these statements, we can say that the lactic acid fermentation of glucose will produce more work than the anaerobic fermentation of glucose because it has a more negative change in gibbs free energy delta G. And actually, that won't make sense that we actually need to write that out. So in gibbs free energy and this statement here is going to summarize our final answer for this example, which compares the maximum work produced by the two different reactions under standard conditions. I hope everything is clear If you have any questions, please leave them down below and I'll see everyone in the next practice video.

Verified Solution

Key Concepts

Oxidation and Reduction Reactions

Standard Gibbs Free Energy Change (ΔG°)

Cellular Respiration vs. Fermentation