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.
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' Keq'=e-∆G'°RT=e--25.1*10008.315*298=2.51×104