Question

A protein has a binding site for a single ligand with a A.G-16.7 KJ/mol at 298 K. What is Keg for this reaction?

Answer 1

Explanation:

The given data is as follows.

          $\Delta_{r} G$ = -16.7 kJ/mol = $-16.7 \times 10^{3}$,       T = 298 K

          R = 8.314 J/mol K,       $K_{eq}$ = ?

Relation between $\Delta_{r} G$ and $K_{eq}$ is as follows.

                $\Delta_{r} G$ = $-RT ln K_{eq}$

Hence, putting the values into the above equation as follows.

                $\Delta_{r} G$ = $-RT ln K_{eq}$

                        $-16.7 \times 10^{3} J/mol$ = $-8.314 J/mol K \times 298 K ln K_{eq}$

                     $ln K_{eq}$ = $\frac{-16.7 \times 10^{3} J/mol}{-8.314 J/mol K \times 298 K}$    

                                         = 6.740

                            $K_{eq}$ = antilog (6.740)

                                            = 846

Thus, we can conclude that $K_{eq}$ for given values is 846.