A certain reaction has an activation energy of 33.70 kJ/mol. At what Kelvin temperature will the...

Question

Question

Answer 1

Answer #1

Ea = 33.70KJ/mole = 33700J/mole

T1 = 355K

K1 = 1

K2 = 4.50 times

T2 =

logK2/K1 = Ea/2.303R [ 1/T1 - 1/T2]

log4.5/1 = 33700/2.303*8.314[ 1/355 - 1/T2]

0.6532 = 1760.05(0.00282-1/T2)

(0.00282-1/T2) = 0.6532/1760.05

(0.00282-1/T2) = 0.000371

-1/T2 = 0.000371-0.00282

-1/T2 = -0.002449

T2 = 408.33K

T = 408.33K >>>>answer

Answer 2

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