Part A
The activation energy of a certain reaction is 36.4 kJ/mol . At 21 ∘C , the rate constant is 0.0160s−1. At what temperature in degrees Celsius would this reaction go twice as fast?
Express your answer with the appropriate units.
Hints

Part B
Given that the initial rate constant is 0.0160s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temperature of 130. ∘C for the same reaction described in Part A?
Express your answer with the appropriate units.
A)
T1 = 21.0 oC
=(21.0+273)K
= 294.0 K
K2/K1 = 2
Ea = 36.4 KJ/mol
= 36400.0 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1  1/T2)
ln(2) = (36400.0/8.314)*(1/294.0  1/T2)
0.693147 = 4378.157325*(1/294.0  1/T2)
T2 = 308
= (308273) oC
= 35 oC
Answer: 35 oC
B)
T1 = 21.0 oC
=(21.0+273)K
= 294.0 K
T2 = 130.0 oC
=(130.0+273)K
= 403.0 K
K1 = 1.6*10^2 S1
Ea = 36.4 KJ/mol
= 36400.0 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1  1/T2)
ln(K2/1.6*10^2) = (36400.0/8.314)*(1/294.0  1/403.0)
ln(K2/1.6*10^2) = 4378*(9.2*10^4)
K2 = 0.898 S1
Answer: 0.898 S1
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