CBSE
Gujarat Board
Haryana Board
Class 10
Class 12
K1 = 4.5 x 103 S–1 , T1 = 273 + 10 = 283 KEa = 60 kJ mol–1 T2 = ?K2 = 1.5 x 104S–1We know thatlogK2K1 = Ea(T2-T1)2.303R T1T2log1.5 × 1044.5×103 = 60(T2-283)2.303 × 8.314 × 283 × T2log 15045 = 60(T2×283)5418.64 T2log 150 - log 45 = 60T2-169805418.64 T22.1761 - 1.6532 = 60T2-169805418.64 T20.5229 = 60T2-169805418.64 T22833.4 T2 = 60T2-169802833.4 T2 = 60T2 = -169802773.4 T2 = 16980T2 = 169802733.4 = 6.1 K