The activation energy of a certain reaction is 47.9 kJ/molkJ/mol . At 29 ∘C ∘C , the rate constant is 0.0180s−10.0180s−1 . At what temperature in degrees Celsius would this reaction go twice as fast?
Express your answer with the appropriate units.
T2=
Given that the initial rate constant is 0.0180s−10.0180s−1 at an initial temperature of 29 ∘C ∘C , what would the rate constant be at a temperature of 170. ∘C ∘C for the same reaction described in Part A?
k2=
1)
Given:
T1 = 29 oC
=(29+273)K
= 302 K
K1 = 1.8*10^-2 s-1
K2 = 2*1.8*10^-2 s-1 = 3.6*10^-2 s-1
Ea = 47.9 KJ/mol
= 47900 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(3.6*10^-2/1.8*10^-2) = (47900.0/8.314)*(1/302 - 1/T2)
0.6931 = 5761.3664*(1/302 - 1/T2)
T2 = 313 K
= (313-273) oC
= 40 oC
Answer: 40 oC
2)
Given:
T1 = 29 oC
=(29+273)K
= 302 K
T2 = 170 oC
=(170+273)K
= 443 K
K1 = 1.8*10^-2 s-1
Ea = 47.9 KJ/mol
= 47900 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/1.8*10^-2) = (47900.0/8.314)*(1/302 - 1/443.0)
ln(K2/1.8*10^-2) = 5761*(1.054*10^-3)
ln(K2/1.8*10^-2) = 6.072
(K2/1.8*10^-2) = e^(6.072)
(K2/1.8*10^-2) = 4.336*10^2
K2 = 7.804 s-1
Answer: 7.80 s-1
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