Q.65 For the coupled reactions given below
Glucose 6-phosphate + H2O → Glucose + Pi (Reaction 1)
ATP + Glucose → ADP + Glucose 6-phosphate (Reaction 2)
the standard free energy change of ATP hydrolysis at 25 °C is ____ kJ/mol.
The equilibrium constants for Reaction 1 and Reaction 2 are 360 and 800,
respectively; Gas constant R = 8.314 J mol-1 K-1.
(Round off to two decimal places)
Reaction Overview
Reaction 1 (hydrolysis of glucose 6-phosphate): Glucose 6-phosphate + H2O → Glucose + Pi, with K1 = 360.
Reaction 2 (ATP phosphorylation): ATP + Glucose → ADP + Glucose 6-phosphate, with K2 = 800.
ATP hydrolysis is the reverse of Reaction 2 combined with Reaction 1: ATP + H2O → ADP + Pi.
Calculation Method
ΔG° = -RT ln K, where R = 8.314 J mol-1 K-1 (0.008314 kJ mol-1 K-1) and T = 298 K.
For ATP hydrolysis, K = K1 × K2 = 360 × 800 = 288,000.
ΔG° = -(0.008314)(298) ln(288,000) = -30.66 kJ/mol (rounded to two decimal places).
Step-by-Step Derivation
First, compute ln(288,000) ≈ 12.571.
RT = (0.008314 kJ mol-1 K-1)(298 K) ≈ 2.478 kJ/mol.
Thus, ΔG° = -2.478 × 12.571 ≈ -30.66 kJ/mol.
This negative value indicates the reaction is spontaneous under standard conditions.
Biological Relevance
ATP hydrolysis drives endergonic reactions like glucose phosphorylation, coupling a favorable ΔG° (~ -30 kJ/mol) to overcome positive ΔG° for glucose 6-phosphate formation (~ +13-16 kJ/mol).
Overall, such coupling maintains cellular energy balance in glycolysis and metabolism.
