Question 16.9: The lactic acid (C3H6O3(aq), ΔG°f = -559 kJ) produced in mus...
The lactic acid (C_{3}H_{6}O_{3}(aq), ΔG°_{f} = -559 kJ) produced in muscle cells by vigorous exercise eventually is absorbed into the bloodstream, where it is metabolized back to glucose (ΔG°_{f} = -919 kJ) in the liver. The reaction is
2C_{3}H_{6}O_{3}(aq) → C_{6}H_{12}O_{6}(aq)
(a) Calculate ΔG° at 25°C for this reaction, using free energies of formation.
(b) If the hydrolysis of ATP to ADP is coupled with this reaction, how many moles of ATP must react to make the process spontaneous?
| ANALYSIS | |
| equation for the reaction (2C_{3}H_{6}O_{3}(aq) → C_{6}H_{12}O_{6}(aq))
ΔG°_{f} values: C_{3}H_{6}O_{3}(aq) (-559 kJ); C_{6}H_{12}O_{6}(aq)(-919 kJ) T(25°C) |
Information given: |
| (a) ΔG° (b) mol ATP for spontaneity |
Asked for: |
STRATEGY
(a) Find ΔG° using the ΔG°_{f} values given in Appendix 1.
ΔG° = ΣΔG°_{f_{ products}} – ΣΔG°_{f_{ reactants}}
(b) Convert the energy obtaned in (a) to moles ATP by using the conversion factor: 31 kJ/mol ATP
| ΔG° = ΔG°_{f} C_{6}H_{12}O_{6}(aq) – 2ΔG°_{f} C_{3}H_{6}O_{3}(aq)
= -919 kJ + 2(559 kJ) = +199 kJ 199 kJ × \frac{1 mol ATP}{31 kJ} = 6.4 mol ATP |
(a) ΔG°
(b) mol ATP |