Chapter 5, Problem 5.4PS

Problem Set

Enthalpy, Entropy, and Free Energy. The oxidation of glucose to carbon dioxide and water is represented by the following reaction, whether the oxidation occurs by combustion in the laboratory or by biological oxidation in living cells:

${\text{C}}_6{\text{H}}_{12}{\text{O}}_6+6{\text{O}}_2\rightleftharpoons 6{\text{O}}_2+6{\text{H}}_2\text{O}$    (5-27)

This reaction is highly exothermic, with an enthalpy change (ΔH) of –673 kcal/mol. As you know from Figure 5-9, ΔG for this reaction at 25°C is –686 kcal/mol, so the reaction is also highly exergonic.

(a)    Explain in your own words what the ΔH and ΔG values mean. What do the negative signs mean in each case?

(b)    What does it mean to say that the difference between the ΔG and ΔH values is due to entropy?

(c)    Without doing any calculations, would ΔS for this reaction be positive or negative? Explain your answer.

(d)    Now calculate ΔS for this reaction at 25°C. Does the calculated value agree in sign with your prediction in part c?

(e)    What are the values of ΔG, ΔH, and ΔS for the reverse of the above reaction as carried out by a photosynthetic algal cell that is using CO₂ and H₂O to make C₆H₁₂O₆?

Figure 5-9 Changes in Free Energy for the Oxidation and Synthesis of Glucose. The exergonic oxidation of glucose shown in (a) has a large negative ΔG that is exactly equal in magnitude but opposite in sign to the large positive ΔG for the endergonic synthesis of glucose shown in (b).