Q.12 The pH of gastric juice in the stomach is 2.0. However the pH inside the cells that line the stomach
is 7.0. For transport of protons from inside the cell to the stomach, the free energy change (∆G) in
kJmol –1 at 37 o
C is _____
[Assume Universal Gas constant R = 8.314 Jmol –1 K–1 ]

Free Energy Change for Proton Transport Across Gastric pH Gradient (pH 7.0 to 2.0) at 37 °C

The free energy change for transporting protons from cytosol (pH 7.0) to gastric juice (pH 2.0) at 37 °C is about 29.6 kJ mol⁻¹, meaning energy must be supplied to move H⁺ against this steep pH gradient.

Introduction

Transport of protons from stomach lining cells (pH 7.0) into gastric juice (pH 2.0) is a classic example used to calculate the free energy change across a biological membrane. Understanding this bioenergetic requirement explains why gastric H⁺ pumping needs ATP and how cells maintain strong pH gradients in physiology.

Given data and concept

pH inside stomach cells (cytosol): 7.0

pH of gastric juice (lumen): 2.0

Temperature: 37 °C = 310 K

Universal gas constant R: 8.314 J mol⁻¹K⁻¹

Movement considered: from inside cell (pH 7) to stomach lumen (pH 2)

For transport of an uncharged solute, the chemical part of Gibbs free energy change across a membrane is:

ΔG = RT ln (C_final/C_initial)

For protons expressed with pH, this becomes:

ΔG = 2.303 RT ΔpH

where ΔpH = pH_final − pH_initial.

Step-by-step solution

Identify initial and final sides

Initial side: cell interior (where H⁺ starts) → pH_initial = 7.0

Final side: stomach lumen (destination of H⁺) → pH_final = 2.0

Here the proton moves from a region of lower [H⁺] (pH 7) to higher [H⁺] (pH 2), so the process is uphill and requires positive ΔG.

Calculate pH difference

ΔpH = pH_final − pH_initial = 2.0 − 7.0 = −5.0

Often in bioenergetics, the magnitude of work required is taken as:

|ΔG| = 2.303 RT |ΔpH|

so |ΔpH| = 5.0.

Substitute numerical values

|ΔG| = 2.303 × 8.314 J mol⁻¹K⁻¹ × 310 K × 5.0

First combine constants:

2.303 × 8.314 ≈ 19.15 J mol⁻¹K⁻¹

19.15 × 310 ≈ 5936.5 J mol⁻¹

5936.5 × 5.0 ≈ 29682.5 J mol⁻¹

Convert to kJ mol⁻¹

|ΔG| ≈ 29682.5 / 1000 ≈ 29.7 kJ mol⁻¹

Rounded as in standard keys:

ΔG ≈ +29.6 kJ mol⁻¹

The positive sign shows that energy input (from ATP hydrolysis in the H⁺/K⁺-ATPase) is required to pump protons into the acidic stomach lumen.

Explanation of possible options (if MCQ)

If the original question appeared as a multiple-choice item, typical options might be:

• 0 kJ mol⁻¹
• 5.9 kJ mol⁻¹
• 29.6 kJ mol⁻¹
• 59.0 kJ mol⁻¹

Explanation:

• 0 kJ mol⁻¹: Incorrect; would mean no pH gradient, but the gradient of 5 pH units corresponds to a 10⁵-fold difference in [H⁺], so significant energy is needed.
• 5.9 kJ mol⁻¹: Too small; this approximates energy for only a 1-unit pH gradient at 37 °C, not 5 units.
• 29.6 kJ mol⁻¹: Correct; matches the calculation 2.303RT × 5 at 310 K for one mole of protons.
• 59.0 kJ mol⁻¹: Roughly double the correct value; might result from mistakenly multiplying by 10 pH units or by 2 moles of H⁺ instead of 1 mole.

Hence, the correct answer is 29.6 kJ mol⁻¹ for the free energy change of transporting one mole of protons from pH 7.0 inside the cells to pH 2.0 in the stomach at 37 °C.