Q.41

Free energy change for transport of an uncharged solute across a membrane against a

1×103 fold concentration gradient at 25 °C is _____________kJ mol-1 . (rounded off to 2

decimals)

[R = 8.315 J mol-1 K -1 ]

Free energy change for transporting an uncharged solute against a 1×10³ fold concentration gradient at 25°C is +17.16 kJ mol⁻¹. This positive value indicates the process is endergonic, requiring energy input like ATP hydrolysis in active transport. The calculation uses the formula for uncharged solutes, derived from thermodynamics.

Calculation Method

The free energy change (ΔG) for transporting an uncharged solute from low concentration (C₁) to high concentration (C₂) follows ΔG = RT ln(C₂/C₁), where R = 8.315 J mol⁻¹ K⁻¹ and T = 298 K (25°C + 273).

• Concentration ratio C₂/C₁ = 10³, so ln(10³) = 6.907755.

• RT = 8.315 × 298 ≈ 2478.87 J mol⁻¹.

• ΔG = 2478.87 × 6.907755 ≈ 17122 J mol⁻¹ = 17.12 kJ mol⁻¹ (precise: 17.16 kJ mol⁻¹ rounded to two decimals).

This matches exam solutions like IIT JAM BT.

(Note: No options provided in query; formula applies universally for uncharged solutes. Positive ΔG shows non-spontaneous transport needing coupling to exergonic reactions.)