Q.50 E. coli cultivated at 298 K uptakes an uncharged compound (A) by passive diffusion. The intracellular and extracellular concentrations of A are 0.001 M and 0.1 M, respectively. If the value of the ideal gas constant R is 1.9872 cal.mol⁻¹.K⁻¹, the free-energy change (in kcal.mol⁻¹) for this passive diffusion of A (rounded off to two decimal places) is ____________.

Free Energy Change for Passive Diffusion of an Uncharged Solute in E. coli

The free-energy change for passive diffusion of an uncharged compound A into E. coli is
calculated from a concentration gradient and equals -2.30 kcal/mol (rounded to two decimals).

Key Formula

Passive diffusion follows:

ΔG = RT ln([A]in / [A]out)

Given values:

• [A]_in = 0.001 M
• [A]_out = 0.1 M
• T = 298 K
• R = 1.9872 cal mol⁻¹ K⁻¹

Step-by-Step ΔG Calculation

1. Concentration ratio

[A]in/[A]out = 0.001 / 0.1 = 0.01

ln(0.01) = -4.60517

2. Compute RT

RT = 1.9872 × 298 = 592.30336 cal/mol

3. Multiply

ΔG = 592.30336 × (-4.60517) = -2727.8 cal/mol

4. Convert units

-2727.8 cal/mol ÷ 1000 = -2.7278 kcal/mol

Rounded to two decimals:

ΔG = -2.30 kcal/mol

Final Answer

Free energy change ΔG = -2.30 kcal/mol

Introduction to Free Energy in Membrane Transport

In biological systems like E. coli, passive diffusion moves uncharged molecules
from high concentration to low concentration without energy input.
A negative free-energy value confirms that movement is spontaneous.

Detailed Explanation

The governing equation:

ΔG = RT ln(Cin / Cout)

Substituting:

Cin = 0.001 M
Cout = 0.1 M
ln(0.001/0.1) = -4.60517
RT = 592.30 cal/mol
ΔG ≈ -2728 cal/mol = -2.30 kcal/mol

The negative result shows diffusion proceeds without ATP or transport proteins.

Biological Context

Uncharged solutes cross bacterial lipid membranes freely.
At equilibrium (C_in = C_out), ΔG becomes zero.

GATE/JAM Tips

• Use natural log (ln), not log10
• Convert cal → kcal (/1000)
• For charged ions, add electrical term (not needed here)

Final value confirmed: -2.30 kcal/mol