Q.15 The binding free energy of a ligand to its receptor protein is −11.5 kJ mol⁻¹ at 300 K.
What is the value of the equilibrium binding constant?

Use R = 8.314 J mol⁻¹ K⁻¹.
(A) 0.01
(B) 1.0
(C) 4.6
(D) 100.5

Correct Answer

Option (D) 100.5

Explanation

The equilibrium binding constant is calculated using:

ΔG° = −RT lnK_a

Rearranging:

K_a = e^−ΔG°/RT

• ΔG° = −11.5 kJ/mol = −11500 J/mol
• R = 8.314 J mol⁻¹ K⁻¹
• T = 300 K

So:

RT = 8.314 × 300 = 2494.2 J/mol
−ΔG° / RT = 11500 / 2494.2 ≈ 4.610

Therefore:

K_a = e^4.610 ≈ 100.5

Option Explanations

(A) 0.01: Assumes reversed sign or uses dissociation constant logic. Incorrect because favorable binding makes ΔG negative and K_a > 1.

(B) 1.0: Would only be correct if ΔG° = 0 (no binding). Not valid for negative free energy.

(C) 4.6: Misinterprets exponent argument as final K_a. Requires exponentiation.

(D) 100.5: Correct. Represents strong ligand–receptor binding affinity.