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Hainan

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Hainan
6. a) ∆G'°=-RTlnKeq'=-8.315Jmol∙K×298K×ln1.97=-1.68 kJ/mol b) ∆G'=∆G'°+RTlnglucose-6-phosphate fructose-6-phosphate=-1.68kJmol+8.315Jmol∙K×298K×ln0.51.5
=-4.40kJ/mol
c) ∆G'° is the standard energy change at the given temperature and fixed concentration of chemicals in the reaction. (For this reaction, both glucose-6-phosphate and fructose-6-phosphate are at 1M) . In contrast, ∆G' is a variable energy change for different sets of reactant and product concentrations, which can be calculated from the equation used in b). As the condition is different for ∆G'° and ∆G', the energy change to reach the equilibrium is comprehensibly different.
10.
∆Gp=∆G'°+RTlnADPPiATP ∆G'°=-30.5kJmol ATP/mM | 5 | 3 | 1 | 0.2 | 5 | ADP/mM | 0.2 | 2.2 | 4.2 | 5 | 25 | Pi/mM | 10 | 12.1 | 14.1 | 14.9 | 10 | Q/M | 0.0004 | 0.008873 | 0.05922 | 0.3725 | 0.05 | lnQ | -7.82405 | -4.7247 | -2.8265 | -0.98752 | -2.99573 | ΔG | -49.887 | -42.2072 | -37.5037 | -32.9469 | -37.923 |

A higher [ATP]/[ADP] ratio cause a higher ΔG that can be used by the cell, so metabolism is regulated to keep the ratio [ATP]/[ADP] high.
23.
a) NAD+/NADH, of which the standard reduction potential is lower and represents a greater tendency to lose electrons. b) Pyruvate/lactate, of which the standard reduction potential is higher and represents a greater tendency to gain electrons. c) E=-0.19V-(-0.32V)=0.13V So the direction is as from left to right, in another word, lactate formation. d) ∆G'°=-nFE=-2×9.65×104×0.13=-25.1kJ/mol e) ∆G'°=-RTlnKeq'

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