Final answer:
The standard free energy change, ΔG′⁰, for the reaction Fructose 6-phosphate = Glucose 6-phosphate is approximately -191 J. If the concentrations of fructose 6-phosphate and glucose 6-phosphate are adjusted, the actual free energy change, ΔG, is approximately -103 J.
Step-by-step explanation:
The question asks to calculate the standard free energy change, ΔG′⁰, for the reaction: Fructose 6-phosphate = Glucose 6-phosphate.
From the information provided, the standard free energy change, ΔG°, for the reaction G6P = F6P is given as 1.7 kJ/mol.
To calculate ΔG′⁰, we can use the equation ΔG′⁰ = -RTlnK′⁰, where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (25°C = 298 K), and K′⁰ is the equilibrium constant.
ΔG′⁰ = -(8.314 J/mol·K)(298 K)ln(1.97) = -190.606 J ≈ -191 J
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, we can calculate the actual free energy change, ΔG, using the equation ΔG = ΔG′⁰ +RTln(Q), where Q is the reaction quotient.
Assuming the reaction is at equilibrium, Q = K′⁰ = 1.97.
ΔG = -191 J + (8.314 J/mol·K)(298 K)ln(1.97) = -103.378 J ≈ -103 J.
Therefore, ΔG′⁰ for the reaction is approximately -191 J and ΔG is approximately -103 J.