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

Short answer

By calculating the Gibbs free energy change for both reactions, we find that the aerobic oxidation of glucose has a much larger negative value than the anaerobic decomposition. Therefore, \(K_1\) (the equilibrium constant for aerobic oxidation) is significantly larger than \(K_2\) (for anaerobic decomposition), indicating that the aerobic oxidation is thermodynamically more favored than the anaerobic process. The maximum work obtained under standard conditions is determined by the magnitude of the Gibbs free energy change. Since the aerobic oxidation has a larger negative Gibbs free energy change, the maximum work that can be obtained from this process (\(W_{\text{max}_1}\)) is significantly greater than that of anaerobic decomposition (\(W_{\text{max}_2}\)).

Step-by-step solution

Calculate the Gibbs free energy change for both reactions

For this, we will use the standard Gibbs free energy change values (\(\Delta G^\circ_f\)) of each compound, which can be found in Appendix C. The standard Gibbs free energy change for each reaction is given by: \(\Delta G^\circ = \sum \Delta G_{1}^\circ_{products} - \sum \Delta G_{1}^\circ_{reactants}\) For the aerobic oxidation of glucose: \(\Delta G_1^\circ = [6\Delta G_{\mathrm{CO}_2}^\circ + 6\Delta G_{\mathrm{H}_2\mathrm{O}}^\circ] - [\Delta G_{\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6}^\circ + 6\Delta G_{\mathrm{O}_2}^\circ]\) For the anaerobic decomposition of glucose: \(\Delta G_2^\circ = [2\Delta G_{\mathrm{C}_2\mathrm{H}_5\mathrm{OH}}^\circ + 2\Delta G_{\mathrm{CO}_2}^\circ] - [\Delta G_{\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6}^\circ]\) Now, substitute the standard Gibbs free energy change values for each compound and calculate \(\Delta G_1^\circ\) and \(\Delta G_2^\circ\).

Calculate the equilibrium constants for both reactions

Now that we have the Gibbs free energy change for both reactions, we can determine the equilibrium constants using the following formula: \(K = e^{\frac{-\Delta G^\circ}{RT}}\) where R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin (298.15 K for standard conditions). For the aerobic oxidation: \(K_1 = e^{\frac{-\Delta G_1^\circ}{RT}}\) For the anaerobic decomposition: \(K_2 = e^{\frac{-\Delta G_2^\circ}{RT}}\) Calculate \(K_1\) and \(K_2\), then compare their values.

Calculate the maximum work obtained for both processes under standard conditions

We can calculate the maximum work obtained for both processes using the Gibbs free energy change: \(W_\text{max} = -\Delta G^\circ\) For the aerobic oxidation: \(W_{\text{max}_1} = -\Delta G_1^\circ\) For the anaerobic decomposition: \(W_{\text{max}_2} = -\Delta G_2^\circ\) Calculate \(W_\text{max}\) for both processes and compare their values. By following these steps, you'll determine the equilibrium constants for both reactions, compare them to see which process is more favored, and determine the maximum work that can be obtained for both processes under standard conditions.