The activation energy of a certain reaction is 30.1 kJ/mol . At 28 ∘C , the rate constant is 0.0120s−1 . At what temperature in degrees Celsius would this reaction go twice as fast?
Given:
T1 = 28 oC
=(28+273)K
= 301 K
K1 = 1.2*10^-2 s-1
K2 = 2.4*10^-2 s-1
Ea = 30.1 KJ/mol
= 30100 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(2.4*10^-2/1.2*10^-2) = (30100.0/8.314)*(1/301 - 1/T2)
0.6931 = 3620.3993*(1/301 - 1/T2)
T2 = 319 K
= (319-273) oC
= 46 oC
Answer: 46 oC
The activation energy of a certain reaction is 30.1 kJ/mol . At 28 ∘C , the...
The activation energy of a certain reaction is 30.1 kJ/mol . At 28 ∘C , the rate constant is 0.0120s−1 . At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units.
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