Phosphorylation of glucose is the first step of glycolysis

CHEMISTRY · Multi-part · A Level · Chemistry · Thermodynamics · Thermodynamics

Problem

Phosphorylation of glucose is the first step of glycolysis. The direct reaction is thermodynamically unfavorable, but cells overcome this by coupling it to ATP hydrolysis. Reaction A (unfavorable) Glucose + PiP_i →\rightarrow glucose-6-phosphate + H2OH_2O ΔGA∘=+13.8 kJ mol−1\Delta G^\circ_A = +13.8\,\text{kJ mol}^{-1} Reaction B (favorable) ATP + H2OH_2O →\rightarrow ADP + PiP_i ΔGB∘=−7.3 kcal mol−1\Delta G^\circ_B = -7.3\,\text{kcal mol}^{-1} (Use 1 cal=4.184 J1\,\text{cal} = 4.184\,\text{J}.) Answer the following. (a) Convert the standard Gibbs free energy change for Reaction B into kilojoules per mole. Report your answer to the correct number of significant figures. (b) Add Reactions A and B to obtain the net chemical equation for phosphorylation of glucose that is coupled to ATP hydrolysis. Cancel any species that appear unchanged on both sides. (c) Calculate the standard Gibbs free energy change, ΔGnet∘\Delta G^\circ_{\text{net}}, for the overall reaction obtained in part (b). Show all work, including unit consistency. (d) State whether the coupled reaction is spontaneous under standard conditions and justify your answer based on the sign of ΔGnet∘\Delta G^\circ_{\text{net}}.

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