Question

If an enzyme rate increases 10-fold for every 5.7 kJ/mol decrease in the energy of activation, what rate increase would be observed if three H-bonds were stabilizing the transition state with each H-bond having a strength of 21 kJ/mol

Answer 1

Total stabilization in the transition state due to 3 hydrogen bonds = 3 x 21 kJ/mol = 63 kJ/mol

Now, for every 5.7 kJ/mol decrease in energy, rate increases 10-fold

Therefore, for 63 kJ/mol decrease in energy, rate will increase by = (10 x 63 kJ/mol)/5.7 kJ/mol

                                                                                                        = 111-fold