Question

Calculate the △g∘' for the reaction fructose-6-phosphate → glucose-6-phosphate given the equilibrium constant is 1.97 and the physiological relevant temperature is 37∘c.

Answer 1

The standard free energy change (ΔG°') for the isomerization of fructose-6-phosphate to glucose-6-phosphate at 37°C is calculated to be -2.14 kJ/mol, indicating the reaction is spontaneous under standard conditions.

Step-by-step explanation:

To calculate the standard free energy change (ΔG°') for the isomerization of fructose-6-phosphate (F6P) to glucose-6-phosphate (G6P) at a physiological temperature of 37°C (310.15 K), we use the equation that relates ΔG°' to the equilibrium constant (Keq):

ΔG°' = -RTln(Keq)

Here, R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Keq is the equilibrium constant for the reaction.

First, we convert the temperature to Kelvin:

T = 37°C + 273.15 = 310.15 K

Next, we can plug the values into the equation:

ΔG°' = -8.314 J/mol·K × 310.15 K × ln(1.97)

After performing the calculations, we find:

ΔG°' = -2.14 kJ/mol

ΔG°' is negative, indicating that the reaction is spontaneous under standard thermodynamic conditions. Moreover, since glycolysis is a biochemical pathway, the free energy change provides insight into the metabolic processes and the role of ATP in driving them.