0 votes 0 votes The binding free energy of a ligand to its receptor protein is $-11.5 \; \text{kJ mol}^{-1}$ at $300 \; \text{K.}$ What is the value of the equilibrium binding constant? Use $\text{R = 8.314 J mol}^{-1} \text{K}^{-1}.$ $0.01$ $1.0$ $4.6$ $100.5$ Others gatebt-2022 + – Lakshman Bhaiya asked Mar 20, 2022 • edited Mar 27, 2022 by Lakshman Bhaiya Lakshman Bhaiya 1.4k points answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Ans: 1.0 delta G= -RTlnKeq Given, deltaG= -11.5KJ/mol; R=8.314 J/mol/K; T=300K Subtitute in formula -11.5= -8.314* 300* ln Keq -11.5/-8.314*300= ln Keq ln Keq= 0.0046 Keq= e^0.0046 Keq= 1.0046 Keq = 1.0 shilpasathish answered Mar 8, 2023 shilpasathish 140 points comment Share ask related question Follow See all 0 reply Please log in or register to add a comment.