Question

Fructose-1,6-bisphosphate (FBP) <=====> glyceraldehyde-3-phosphate (GAP) + dihydroxyacetone phosphate (DHAP) The standard free energy change for this reaction is +22.8 kJ/mol. Under actual conditions in the cell the concentrations of GAP,DHAP,and FBP are 0.5 mM, 0.3 mM, and 6.0 mM respectively. Given this information calculate the actual free energy change for this reaction in the cell at a temperature of 37 oC. Provide your answer (just the numerical part) in kJ/mole to two signicant figures

Answer 1

The given reaction is FBP ßà GAP + DHAP is in equilibrium.

The given values are concentration so we can use the below formula for calculating free energy change

ΔG = RT ln Q

Or

ΔG = 2.303 RT log Q

ΔG is change in free energy

R gas constant = 8.314 J mol^-1 K^-1

T is temperature 37 ^oC

Temperature is converted to kelvin scale T = 273 + 37 = 310 K   

Q is equilibrium constant (can be represented as k) and can be calculated by the ratio of molar concentration of products and reactants.

Q = [GAP][DHAP] / [FBP]

[GAP] – 0.5 mmol = 0.0005 mol

[DHAP] – 0.3 mmol = 0.0003 mol

[FBP] – 6.0 mmol = 0.006 mol

On substituting all the values

ΔG = 2.303 X 8.314 J mol^-1 K^-1 X 310 K log (0.0005 mol X 0.0003 mol / 0.006 mol )

ΔG = 2.303 X 8.314 J mol^-1 K^-1 X 310 K log (2.5 X 10­^-5 )

ΔG = -27303.8 J

ΔG = -27.3038 kJ