Question

presente Organic Chemistry oudon The standard free energy of activation of one reaction A is 93.80 kJ mol-'(22.42 kcal mol-'). The standard free energy of activation of another reaction B is 74.80 kJ mol-'(17.88 kcal mol). Assume a temperature of 298 K and 1 M concentration. By what factor is one reaction faster than the other? Number Tools 10 Which reactior O Reaction A is faster. Reaction B is faster. Cannot be determined. O

Answer 1

k=rate constant Ea = activation energy R = has constant - 8.314 J/malik I= Temperature for reaction A- T = 298k {a = 93.80 kJ imel = 93.80 X 1000 = 93800 J/mol 938005/ma K = A e 8.3143/malik X 298K Ik = Ae Ik = A x 3.618 10-17 for reaction B- T. 298k Ea = 74. 80 kJ/mol =74800 J/mol - 74800 J/mol k= & 314 J/molk X 298k A e k - A 30.14 -30-19 k = A x 7.74 X 10-14 Divide kolka A X 7.74 x 10-14 A X 3.61 x 10-17 Tko = 2144 IKA - KB 2144 ka reaction B is 2144 times faster than reaction A