At 25°C, an acid has a pKa value of 5,73; this corresponds to a standard ΔGsys value (in kcalmol-1) of...?

Question

At 25°C, an acid has a pK_a value of 5,73; this corresponds to a standard ΔG_sys value (in kcalmol^-1) of...?

Answer 1

pKa = 5.73

Ka = 1x10^-5.73

Ka = 1.86x10^-6

∆G = -RT ln K

∆G = -(8.314 J/Kmol)(298K) ln 1.86x10^-6

∆G = -2478 J/mol x -13.19

∆G = +32697 J/mol

Converting to cal / mol and then Kcal / mol we have...

32697 J / mol x 1 cal / 4.184 J = 7818 cal / mol = 7.82 kcal / mol