Isomerization of glucose 1-phosphate to glucose 6-phosphate is energetically favorable.At 37°C, ΔG° = −1.42 log₁₀K.What is the equilibrium constant for this reaction if ΔG° = −1.74 kcal/mole at 37°C?

Question

Quizplus · Accepted Answer

The equilibrium constant (K) for a reaction can be calculated from the standard free energy change (ΔG°) using the equation ΔG° = -RTlnK, where R is the gas constant (1.987 cal/(mol·K)) and T is the temperature in Kelvin. At 37°C (which is 310.15K), and given ΔG° = -1.74 kcal/mole, converting kcal to cal (1 kcal = 1000 cal), ΔG° = -1740 cal/mole. Plugging these values into the equation gives K = e^(-ΔG°/RT) = e^(1740/(1.987*310.15)) ≈ 16.98.

Isomerization of Glucose 1-Phosphate to Glucose 6-Phosphate Is Energetically Favorable