GATE 2022 BT – Question 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.
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Correct answer: (D) 100.5
Explanation
The standard binding free energy is related to the equilibrium constant by
$$\Delta G^\circ=-RT\ln K\;\Rightarrow\;K=\exp\!\left(-\frac{\Delta G^\circ}{RT}\right).$$
**Numbers.** $\Delta G^\circ=-11.5$ kJ/mol $=-11\,500$ J/mol, $R=8.314$ J mol⁻¹ K⁻¹, $T=300$ K:
$$RT=8.314\times300=2494.2\ \text{J/mol}$$
$$K=\exp\!\left(\frac{11\,500}{2494.2}\right)=e^{4.611}=\mathbf{100.5}.$$