part A k=AeEa/RT where k is the rate constant, A is the frequency factor, Ea is
ID: 565899 • Letter: P
Question
part A
k=AeEa/RT where k is the rate constant, A is the frequency factor, Ea is the activation energy, R=8.314 J/(molK) is the universal gas constant, and T is the absolute temperature. A certain reaction has an activation energy of 70.0 kJ/mol and a frequency factor of A1 = 4.00×1012 M1s1 . What is the rate constant, k, of this reaction at 24.0 C ?
part B
An unknown reaction was observed, and the following data were collected: T (K) k (M1s1) 352 109 426 185 Determine the activation energy for this reaction.
Explanation / Answer
A)
Given:
T = 24.0 oC
=(24.0+273)K
= 297.0 K
A = 4*10^12 M-1s-1
Ea = 70.0 KJ/mol
= 70000 J/mol
R = 8.314 J/mol.K
use:
K = A*e^(-Ea/RT)
= 4*10^12*e^(-70000.0/(8.314*297.0))
= 4*10^12*e^(-28.3486)
= 4*10^12*4.879*10^-13
= 1.95 M-1s-1
Answer: 1.95 M-1s-1
B)
Given:
T1 = 352 K
T2 = 426 K
K1 = 1.09*10^2 M-1s-1
K2 = 1.85*10^2 M-1s-1
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(1.85*10^2/1.09*10^2) = ( Ea/8.314)*(1/352.0 - 1/426.0)
0.529 = (Ea/8.314)*(4.935*10^-4)
Ea = 8912 J/mol
Ea = 8.91 KJ/mol
Answer: 8.91 KJ/mol
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