Standard Free Energy Change (ΔG°′) for the Conversion of Glucose-1-Phosphate to Glucose-6-Phosphate | Complete Numerical Solution with Formula

Correct Answer

ΔG°′ = −7.3 × 10³ J mol⁻¹

or

ΔG°′ ≈ −7295 J mol⁻¹

Introduction

The standard free energy change (ΔG°′) is one of the most important thermodynamic parameters in biochemistry. It indicates whether a biochemical reaction is energetically favorable under standard biochemical conditions (pH 7.0). The relationship between the equilibrium constant and free energy is fundamental to understanding metabolic pathways, enzyme-catalyzed reactions, and energy transfer in living organisms.

The conversion of glucose-1-phosphate (G1P) into glucose-6-phosphate (G6P) is catalyzed by the enzyme phosphoglucomutase, which plays an important role in glycogen metabolism. Since the concentrations given are at equilibrium, we can determine the equilibrium constant (K) and then calculate the standard free energy change (ΔG°′).

This type of numerical is frequently asked in CSIR NET, GATE Biotechnology, IIT JAM, CUET PG, NEET PG, and M.Sc. entrance examinations.

Concept Used

The standard free energy change is related to the equilibrium constant by the Gibbs free energy equation.

Standard Gibbs Free Energy Equation

ΔG°′ = −RT ln K

Where:

• ΔG°′ = Standard free energy change (J mol⁻¹)
• R = Gas constant = 8.315 J mol⁻¹ K⁻¹
• T = Absolute temperature (Kelvin)
• K = Equilibrium constant

Step 1: Write the Reaction

Glucose-1-phosphate ⇌ Glucose-6-phosphate

Step 2: Calculate the Equilibrium Constant (K)

The equilibrium constant is calculated as:

K = [Products] / [Reactants]

Substitute the equilibrium concentrations:

K = [Glucose-6-phosphate] / [Glucose-1-phosphate]

K = 38 / 2

K = 19

Therefore,

Equilibrium Constant (K) = 19

Step 3: Apply the Gibbs Free Energy Equation

ΔG°′ = −RT ln K

Substitute the values:

ΔG°′ = −(8.315 J mol⁻¹ K⁻¹) × (298 K) × ln(19)

Step 4: Calculate the Natural Logarithm

ln(19) = 2.944

Step 5: Perform the Multiplication

First,

8.315 × 298 = 2477.87

Now,

2477.87 × 2.944 = 7294.9

Therefore,

ΔG°′ = −7294.9 J mol⁻¹

Rounded to two significant figures,

ΔG°′ = −7.3 × 10³ J mol⁻¹

or

ΔG°′ = −7.29 kJ mol⁻¹

Final Answer

Standard Free Energy Change (ΔG°′) = −7.3 × 10³ J mol⁻¹

Why Is the Standard Free Energy Change Negative?

Products Are Favored at Equilibrium

The equilibrium constant is:

K = 19

Since K > 1, the concentration of products is much greater than the concentration of reactants.

According to the equation:

ΔG°′ = −RT ln K

• When K > 1, ln K is positive.
• Multiplying by −RT gives a negative value of ΔG°′.

Therefore, the reaction is thermodynamically favorable under standard biochemical conditions.

Biological Significance of This Reaction

Glycogen Breakdown

Glucose-1-phosphate is produced during glycogen degradation and is converted into glucose-6-phosphate before entering other metabolic pathways.

Glycolysis

Glucose-6-phosphate serves as an important intermediate in glycolysis for ATP production.

Pentose Phosphate Pathway

Glucose-6-phosphate can also enter the pentose phosphate pathway to produce NADPH and ribose-5-phosphate.

Metabolic Regulation

This reversible conversion allows cells to regulate carbohydrate metabolism efficiently according to their energy requirements.

Relationship Between ΔG°′ and the Equilibrium Constant

When K > 1

• Products are favored.
• ΔG°′ is negative.
• Reaction is spontaneous under standard conditions.

When K < 1

• Reactants are favored.
• ΔG°′ is positive.
• Reaction is non-spontaneous under standard conditions.

When K = 1

• Reactants and products are present in equal concentrations.
• ΔG°′ = 0

Using Temperature in Celsius

Always convert:

25°C = 298 K

Using the Wrong Equilibrium Expression

For this reaction:

K = [Glucose-6-phosphate] / [Glucose-1-phosphate]

Products should always be placed in the numerator.

Using log Instead of ln

The Gibbs free energy equation uses the natural logarithm (ln), not log₁₀.

Forgetting the Negative Sign

The equation begins with:

ΔG°′ = −RT ln K

Missing the negative sign leads to an incorrect final answer.

High-Yield Exam Points

Standard Free Energy Equation

ΔG°′ = −RT ln K

Temperature

Always use Kelvin.

Equilibrium Constant

K = Products / Reactants

Interpretation

• K > 1 → ΔG°′ is negative
• K < 1 → ΔG°′ is positive
• K = 1 → ΔG°′ = 0

Frequently Asked Questions

Why is ΔG°′ negative?

Because the equilibrium constant (K = 19) is greater than one, indicating that product formation is favored.

Why do we use Kelvin instead of Celsius?

Thermodynamic equations require absolute temperature.

Therefore,

25°C + 273 = 298 K

Why do we use ln instead of log?

The Gibbs free energy equation is derived using the natural logarithm (ln), which is based on the mathematical constant e.

Key Takeaways

• The equilibrium constant is:

  K = 38 / 2 = 19

• The Gibbs free energy equation is:

  ΔG°′ = −RT ln K

• Substituting the values gives:

  ΔG°′ = −7295 J mol⁻¹

• Rounded value:

  ΔG°′ = −7.3 × 10³ J mol⁻¹

• A negative ΔG°′ indicates that the conversion of glucose-1-phosphate to glucose-6-phosphate is thermodynamically favorable under standard biochemical conditions.

Final Answer

Correct Answer: −7.3 × 10³ J mol⁻¹

Explanation

At equilibrium, the equilibrium constant for the reaction is K = 19, calculated from the ratio of 38 mM glucose-6-phosphate to 2 mM glucose-1-phosphate. Using the Gibbs free energy equation, ΔG°′ = −RT ln K, with R = 8.315 J mol⁻¹ K⁻¹, T = 298 K, and ln(19) = 2.944, the standard free energy change is −7295 J mol⁻¹, which is approximately −7.3 × 10³ J mol⁻¹. The negative value indicates that the reaction proceeds spontaneously toward the formation of glucose-6-phosphate under standard biochemical conditions.