Question

Consider the interconversion shown, which occurs in glycolysis.

Fructose 6-phosphate ⇌glucose 6-phosphate
The equilibrium constant, K'eq, is 1.97 at 25.0 °C. Calculate ∆G°' for the reaction.
∆G° = ▭kJ/mol
If the concentration of fructose 6-phosphate is adjusted to 1.5 M and that of glucose 6-phosphate is adjusted to 0.50 M, what is
∆G°?
∆G°=▭kJ/mol

Answer 1

To calculate the standard free energy change (ΔG°') for the interconversion of fructose 6-phosphate to glucose 6-phosphate in glycolysis, you use the equilibrium constant and the equation ΔG°' = -RTlnK'eq.

Step-by-step explanation:

The student asked how to calculate the standard free energy change (ΔG°') of the reaction where fructose 6-phosphate is converted to glucose 6-phosphate, which is a step in glycolysis. Given the equilibrium constant (K'eq) of 1.97 at 25.0 °C, the ΔG°' can be calculated using the equation ΔG°' = -RTlnK'eq, where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K'eq is the equilibrium constant.

To calculate ΔG under non-standard conditions, we use the modified Gibbs free energy equation: ΔG = ΔG°' + RTln(Q), where Q is the reaction quotient. In this example, when the concentrations of fructose 6-phosphate and glucose 6-phosphate are 1.5 M and 0.50 M respectively, Q = [glucose 6-phosphate]/[fructose 6-phosphate].For non-standard conditions, the modified Gibbs free energy equation applies, incorporating the reaction quotient Q based on actual reactant and product concentrations.