Answer

2.303 log10 (K2/K1) = (Ea/R)[(T2 – T1)/T1T2] 

Therefore, 2.303 log10{(4.5 × 107)/(1.5 × 107)} = (Ea/8.314)[(373 – 323)/373 × 323] 

Therefore, Ea = 2.2 × 104 J mol-1 

Now, K = Ae–Ea/RT 

Therefore, 4.5 × 107 = Ae–{(2.2 × 104)/(8.314 × 373)} 

Therefore, A = 5.42 × 1010