The phosphorylation of glucose to glucose-6-phosphate has

CHEMISTRY · Single correct · AP · Chemistry · Thermodynamics and Electrochemistry · Thermodynamics and Electrochemistry

Problem

The phosphorylation of glucose to glucose-6-phosphate has ΔG∘=+13.8 kJ mol−1\Delta G^{\circ}=+13.8\,\text{kJ}\,\text{mol}^{-1}. When this reaction is coupled to ATP hydrolysis under standard conditions, which statement best explains why the overall process becomes spontaneous? (a) The coupled ΔG∘\Delta G^{\circ} equals +13.8 kJ mol−1+30.5 kJ mol−1=+44.3 kJ mol−1+13.8\,\text{kJ mol}^{-1}+30.5\,\text{kJ mol}^{-1}=+44.3\,\text{kJ mol}^{-1}, leaving the reaction non-spontaneous. (b) The coupled ΔG∘\Delta G^{\circ} equals +13.8 kJ mol−1+(−30.5 kJ mol−1)=−16.7 kJ mol−1+13.8\,\text{kJ mol}^{-1}+(-30.5\,\text{kJ mol}^{-1})=-16.7\,\text{kJ mol}^{-1}, so the overall reaction is spontaneous. (c) Coupling exactly offsets the energy cost, making the overall ΔG∘\Delta G^{\circ} effectively 0 kJ mol−10\,\text{kJ mol}^{-1}. (d) ATP hydrolysis increases entropy so greatly that the sign of ΔG∘\Delta G^{\circ} no longer determines spontaneity.

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