What is the value of K for this aqueous reaction at 298 K

Question

A + B ⇌ C + D

Delta G Deegree = 23.53 kJ/mol

Answer 1

∆Gº = -RT ln K

∆Gº = 23.53 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol

T = temperature in K = 298K

K = ?

23.53 kJ/mol = -(0.008314 kJ/Kmol)(298K) ln K

23.53 kJ/mol = -2.478 kJ/mol ln K

ln K = -9.496

K = 7.515x10^-5

(be sure to check all of the math)