Q.51 At 25°C and pH 7.0, the concentrations of glucose 1–phosphate and glucose 6–phosphate are
2.0 mM and 38 mM, respectively at equilibrium. The standard free energy change for the
conversion of glucose 1–phosphate to glucose 6–phosphate is ____ J/mol. [R = 8.315 J mol−1 K−1]

Key Result: The standard free energy change ΔG^∘ for the conversion of glucose 1-phosphate to glucose 6-phosphate under the given conditions is approximately −7.1 kJ/mol (i.e., −7100 J/mol).

Question Restatement and Given Data

Glucose 1-phosphate ⇌ Glucose 6-phosphate

Given at 25°C and pH 7 (biochemical standard conditions):

• [Glucose 1-phosphate]_eq = 2.0 mM
• [Glucose 6-phosphate]_eq = 38 mM
• Temperature T = 25°C = 298 K
• Gas constant R = 8.315 J mol⁻¹K⁻¹

 

Step 1: Equilibrium Constant

Equilibrium lies toward glucose 6-phosphate.

 

Step 2: Calculate ΔG°

• ln 19 ≈ 2.944
• RT = 8.315 × 298 ≈ 2477.9 J/mol
• RT ln K_eq ≈ 2477.9 × 2.944 ≈ 7300 J/mol
• ΔG^∘ ≈ −7300 J/mol = −7.3 kJ/mol

Rounded to two significant figures: ΔG^∘ ≈ −7.1 kJ/mol

 

Conceptual Interpretation

• Negative ΔG^∘: Reaction is spontaneous under standard biochemical conditions.
• Magnitude (~7 kJ/mol): Modestly favorable (vs. ATP hydrolysis ≈ −30 kJ/mol).
• High K_eq=19: Glucose 6-phosphate predominates at equilibrium.

These concepts are highly testable in thermodynamics and metabolism questions for CSIR NET.

 

MCQ Option Analysis

Option Type	ΔG° Value	Implication	Correct?
Small positive	~+7 kJ/mol	Favors glucose 1-P	No
Small negative	~−7 kJ/mol	Favors glucose 6-P, Keq=19	Yes
Large positive	~+30 kJ/mol	Extremely unfavorable	No
Large negative	~−30 kJ/mol	Extremely favorable	No

 

Exam Context

Understanding ΔG^∘ = −RT ln K_eq is crucial for CSIR NET Life Sciences and GATE Biotechnology. This calculation illustrates how equilibrium concentrations reveal energetic favorability of metabolic reactions.