A reaction has a rate constant of 1.12×10−2 s−1 at 400.0 K and 0.690 s−1 at 450.0 K. Part 1: Determine the activation barrier for the reaction. Part 2: What is the value of the rate constant at 425 K?
1)
Given:
T1 = 400 K
T2 = 450 K
K1 = 1.12*10^-2 s-1
K2 = 0.69 s-1
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(0.69/1.12*10^-2) = ( Ea/8.314)*(1/400 - 1/450)
4.1208 = (Ea/8.314)*(2.778*10^-4)
Ea = 123337 J/mol
Ea = 123 KJ/mol
Answer: 123 KJ/mol
2)
Given:
T1 = 400 K
T2 = 425 K
K1 = 1.12*10^-2 s-1
Ea = 123337 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/1.12*10^-2) = (123337/8.314)*(1/400 - 1/425)
ln(K2/1.12*10^-2) = 14835*(1.471*10^-4)
K2 = 9.924*10^-2 s-1
Answer: 9.92*10^-2 s-1
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