Question

± The Arrhenius Equation

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 Eais the activation energy for the reaction.

However, a more practical form of this equation is

lnk2k1=EaR(1T1−1T2)

which is mathmatically equivalent to

lnk1k2=EaR(1T2−1T1)

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 30.9 kJ/mol . At 22  ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

Hints

T2 =

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Part B

Given that the initial rate constant is 0.0190s−1 at an initial temperature of 22  ∘C , what would the rate constant be at a temperature of 140.  ∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

Hints

k2 =

Answer 1

A)

T1 = 22.0 oC

=(22.0+273)K

= 295.0 K

K1 = 1 S-1

K2 = 2 S-1

Ea = 30.9 KJ/mol

= 30900 J/mol

use:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(2/1) = (30900/8.314)*(1/295.0 - 1/T2)

0.693147 = 3716.622564*(1/295.0 - 1/T2)

T2 = 312

= (312-273) oC

= 39 oC

Answer: 39 oC

B)

T1 = 22.0 oC

=(22.0+273)K

= 295.0 K

T2 = 140.0 oC

=(140.0+273)K

= 413.0 K

K1 = 1.9*10^-2 S-1

Ea = 30.9 KJ/mol

= 30900.0 J/mol

use:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(K2/1.9*10^-2) = (30900.0/8.314)*(1/295.0 - 1/413.0)

ln(K2/1.9*10^-2) = 3717*(9.685*10^-4)

K2 = 0.695 s-1

Answer: 0.695 s-1