Question

9. Use the data in the table below to calculate the activation energy and the value of the pre-exponential factor. Experiment 2 3 4 Temperature in Kelvin Rate constant (L/mol s) 1125 1053 11.59 1.67 0.380 0.0011 1001 838

Physical Chemisty Question* #9

*integrate if possible

*Show work please

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Answer #1

use,
lnK2/K1 = (Ea/R)*{1/T1 - 1/T2}
ln(1.67/11.59) = (Ea/8.314)*{1/1125 - 1/1053}
ln0.144 = (Ea/8.314)*{8.89 - 9.50}*10^-4
-1.94 = (Ea/8.314)*(-0.61*10^-4)
Ea = (264*10^3)J/mol
= 264 KJ/mol

use, 4th data
lnk = lnA - Ea/(R*T)
ln0.0011 = lnA - (264*10^3)/(8.314*838)
-6.8 = lnA - 37.9
lnA = 31.1
A = (3.2*10^13) L/mol.s

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