The Arrhenius equation shows the relationship between the rate constant k and th

Question

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k=Ae^-Ea/RT where R is the gas constant (8.314 J/mol*K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is lnk2/k1=Ea/R(1/T1-1/T2) which is mathmatically equivalent to ln k1/k2=Ea/R(1/T2-1/T1) where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2).

Part A

The activation energy of a certain reaction is 37.2kJ/mol . At 20 ?C , the rate constant is 0.0190s?1. At what temperature in degrees Celsius would this reaction go twice as fast?

Part B

Given that the initial rate constant is 0.0190s?1 at an initial temperature of 20 ?C , what would the rate constant be at a temperature of 130 ?C for the same reaction described in Part A?

Explanation / Answer

log(K2/K1) = Ea/2.303R [1/T1 - 1/T2]
log(2K/K) = 37.2/2.303*8.314*10^-3 [1/293 - 1/T2]
T2 = 306.93 K
or
T2 = 33.93 0C or 34 0C

PartB
log(K2/0.0190) = 37.2/2.303*8.314*10^-3 [1/293 - 1/303]
K2 = 0.03144 sec^-1

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