The Gibbs free energy for Step 3 of glycolysis, is ∆G = –5.3 kcal/mole. In terms of energy changes, how would you classify this reaction?

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BlackTom AI · Accepted Answer

A negative Gibbs free energy (∆G = –5.3 kcal/mol) indicates the reaction can proceed spontaneously and releases free energy, so it is exergonic. 1. Relevant knowledge points and concepts - Gibbs free energy (∆G) determines spontaneity: ∆G < 0 means spontaneous (favorable), ∆G > 0 means non-spontaneous (requires input). - Exergonic vs. endergonic: exergonic reactions release free energy (∆G negative); endergonic reactions consume free energy (∆G positive). - Exothermic vs. endothermic refer to heat/enthalpy (∆H), not directly to free energy; a reaction can be exergonic but not necessarily exothermic. - Anabolic processes build molecules and are often endergonic overall; catabolic processes break down molecules and are often exergonic. 2. Step-by-step reasoning - Given ∆G = –5.3 kcal/mol. The sign is negative, so by the Gibbs criterion the reaction is thermodynamically favorable and spontaneous under standard conditions. - Spontaneous release of free energy corresponds to an exergonic classification. 3. Calculation (intermediate step) - Converting to kJ: –5.3 kcal/mol × 4.184 kJ/kcal = –22.2 kJ/mol. The negative value remains, reaffirming spontaneity in SI units. 4. Why the correct answer is correct - Exergonic is defined by ∆G < 0. Since ∆G is negative, the reaction releases free energy and is therefore exergonic. The magnitude (around –22.2 kJ/mol) indicates a modestly favorable step in glycolysis. 5. Why the other options are incorrect - Anabolic: describes biosynthetic (building) pathways; classification refers to function, not directly to ∆G sign. Anabolic steps are often endergonic, so this is not the best choice. - Endergonic: would require ∆G > 0 (positive); here ∆G is negative, so this is incorrect. - Endothermic: refers to heat absorption (∆H > 0). ∆G negative does not imply ∆H positive; the question gives ∆G, so endothermic is not the correct classification. 6. Conclusion Because ∆G = –5.3 kcal/mol is negative, the reaction is exergonic. This matches the provided answer: Exergonic.

The Gibbs free energy for Step 3 of glycolysis, is ∆G = –5.3 kcal/mole. In terms of energy changes, how would you classify this reaction?单项选择题

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The Gibbs free energy for Step 3 of glycolysis, is ∆G = –5.3 kcal/mole. In terms of energy changes, how would you classify this reaction?单项选择题

类似问题

The ∆Gº is a _______________ value often written in textbooks. The _______________ is used in cellular conditions because varying temperature and a high amount of _______________ may make even a ______________ ∆Gº reaction spontaneous. (overthrowing the natural favorability/non-favorability).

During a laboratory experiment, you discover that an enzyme-catalyzed reaction has a deltaG of -20kcal/mol. If you double the amount of enzyme in the reaction, what will be the deltaG for the new reaction?

The transaminase transfer of an amine group from alanine to α-ketoglutarate to form pyruvate and glutamate has a change in Gibb’s free energy (ΔG) of 1.9 kJ/mol. What does this ΔG value tell you about the reaction?

The de-phosphorylation of phosphoenolpyruvate (PEP) to form pyruvate has a change in Gibb’s free energy (ΔG) of -61.9 kJ/mol. How would you describe the reaction?

The free energy change of hydrolysis of sucrose into glucose and fructose is ΔG = - 7 kcal/mol. When the enzyme sucrase is added to the reaction mixture, what happens to the ΔG of the reaction?

The transaminase transfer of an amine group from alanine to α-ketoglutarate to form pyruvate and glutamate has a change in Gibb’s free energy (ΔG) of 1.9 kJ/mol. What does this ΔG value tell you about the reaction?

更多留学生实用工具

希望你的学习变得更简单

The ∆Gº is a _______ value often written in textbooks. The _ is used in cellular conditions because varying temperature and a high amount of _ may make even a ______ ∆Gº reaction spontaneous. (overthrowing the natural favorability/non-favorability).