Question

consider the following interconversion, which occurs in glycolysis: fructose 6-phosphate glucose 6-phosphate k’eq = 1.97 (a) what is ∆g’0 for the reaction at 25 c?

Answer 1

The free energy change (ΔG´₀) for the conversion of fructose 6-phosphate to glucose 6-phosphate, with an equilibrium constant of 1.97 at 25°C, is calculated as approximately -1.6823 kJ/mol, indicating an exergonic and spontaneous reaction under standard conditions.

Step-by-step explanation:

The student asked about the free energy change (ΔG´₀) for the interconversion of fructose 6-phosphate (F6P) to glucose 6-phosphate (G6P) in glycolysis, given the equilibrium constant (K'eq) is 1.97 at 25°C. The ΔG´₀ for a reaction can be calculated using the formula ΔG´₀ = -RT ln(K'eq), where R is the universal gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin. Since the equilibrium constant is greater than 1, we can infer that the ΔG´₀ will be negative, indicating a spontaneous reaction under standard conditions. At 25°C (or 298.15K), the ΔG´₀ can be calculated as:

ΔG´₀ = - (8.314 J/(mol·K))(298.15 K) ln(1.97) = - (8.314 ⋅ 298.15 ⋅ 0.6802) = -1,682.3 J/mol = -1.6823 kJ/mol

The reaction is slightly exergonic, signifying a release of energy and feasibility under standard conditions.