Chemistry 102: Equilibrium: Value of Equilibrium constant

Question

The value of ΔG° for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol.

What is the value of the equilibrium constant for the reaction at 25.0°C?

Answer 1

∆Gº = -RT ln K

∆Gº = 13.8 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ / Kmol (changed units to kJ)

T = temp in Kelvin = 25.0C + 273 = 298K

K = ?

13.8 = -(0.008314)(298) ln K

13.8 = -2.478 ln K

ln K = -5.569

K = 3.81x10^-3