Question

The conversion of glucose-6-phosphate to fructose-6-phosphate has a K′eq of 0.5. Based on that information, which of the following sets of changes in the terms of the Gibbs equation is correct?

A) ΔG > 0, ΔH > 0, TΔS < 0
B) ΔG > 0, ΔH < 0, TΔS > 0
C) ΔG < 0, ΔH < 0, TΔS < 0
D) ΔG < 0, ΔH > 0, TΔS > 0

Answer 1

The conversion of glucose-6-phosphate to fructose-6-phosphate with a ‘K’eq’ of 0.5 suggests a non-spontaneous reaction under standard conditions due to a positive ΔG. ΔH and TΔS must align such that TΔS is greater for the reaction to be spontaneous under non-standard cellular conditions, where substrates and products concentrations vary.

Step-by-step explanation:

When analyzing the conversion of glucose-6-phosphate (G6P) to fructose-6-phosphate (F6P) with a Gibbs free energy change (ΔG°) of 1.7 kJ and a K'eq of 0.5, we can ascertain certain thermodynamic characteristics of the reaction under standard conditions. Since a K'eq less than 1 indicates that the reaction favors the reactants at equilibrium, it suggests an endergonic reaction. However, the ΔG° being positive signifies that the reaction is non-spontaneous under standard conditions, wherein a positive ΔG° means that free energy is absorbed.

The particularities of ΔH (enthalpy change) and TΔS (temperature times entropy change) are not explicitly given, but they must satisfy the equation ΔG = ΔH - TΔS. For ΔG to be < 0, as in this case when the reaction proceeds in a cell under non-standard conditions, TΔS must be greater than ΔH. This would mean the reaction may proceed spontaneously under specific cellular conditions where the concentration of substrates and products are different from standard conditions. Furthermore, the reaction's spontaneity in a typical cell may vary depending on the actual concentration of reactants and products, as well as cellular temperature (which is usually around 37°C in human cells), in contrast to standard conditions.