The Arrhenius Equation is typically written as k = A e E a/ R T However, the fol
ID: 592700 • Letter: T
Question
The Arrhenius Equation is typically written as
k=AeEa/RT
However, the following more practical form of this equation also exists:
lnk2k1=EaR(1T11T2)
where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1and T2).
Part A
The activation energy of a certain reaction is 32.1 kJ/mol . At 20 C, the rate constant is 0.0130 s1. At what temperature would this reaction go twice as fast?
Express your answer numerically in degrees Celsius
Part B
Given that the initial rate constant is 0.0130 s1 at an initial temperature of 20 C, what would the rate constant be at a temperature of 100 C?
Express your answer numerically in inverse seconds.
Explanation / Answer
A)
we have:
T1 = 20 oC
=(20+273)K
= 293 K
K2/K1 = 2/1
Ea = 32.1 KJ/mol
= 32100 J/mol
we have below equation to be used:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(2/1) = (32100.0/8.314)*(1/293.0 - 1/T2)
0.6931 = 3860.9574*(1/293.0 - 1/T2)
T2 = 309 K
= (309-273) oC
= 36 oC
Answer: 36 oC
B)
we have:
T1 = 20 oC
=(20+273)K
= 293 K
T2 = 100 oC
=(100+273)K
= 373 K
K1 = 1.3*10^-2 s-1
Ea = 32.1 KJ/mol
= 32100 J/mol
we have below equation to be used:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/1.3*10^-2) = (32100.0/8.314)*(1/293.0 - 1/373.0)
ln(K2/1.3*10^-2) = 3861*(7.32*10^-4)
K2 = 0.2195 s-1
Answer: 0.220 s-1
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