Question

Pasteurization of milk takes 30 minutes if performed at 600C, and 20 seconds at 700C. Noting that rate is inversely proportional to the reaction time, determine the activation energy for the pasteurization.

Answer 1

Solution :-

T1 = 600 C + 273 = 873 K

T2 = 700 C + 273 = 973 K

rate 1 = 1/(30min * 60 s / 1 min) = 5.56*10^-4 s-1

rate 2 = 1/20 s = 0.05 s-1

Arhhenius equation

ln (K2/K1)= Ea/R[(1/T1)-(1/T2)]

lets put the values in the formula

ln[0.05/5.56*10^-4] = Ea/8.314 J per mol K * [(1/873)-(1/973)]

4.499 = Ea/8.314 J per mol K * 0.0001177

4.99* 8.314 J per mol K / 0.0001177 K = Ea

3.18*10^5 J per mol = Ea

lets convert it to kJ

3.18*1065 J per mol * 1 kJ / 1000 J = 318 kJ/ mol

So the activation energy of the pasteurization is 318 kJ/mol