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=AeEa/RT

where R is the gas constant (8.314 J/molK), 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(1T11T2)

which is mathmatically equivalent to

lnk1k2=EaR(1T21T1)

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 42.2 kJ/mol . At 25  C , the rate constant is 0.0140s1. At what temperature in degrees Celsius would this reaction go twice as fast?

Part B

Given that the initial rate constant is 0.0140s1 at an initial temperature of 25  C , what would the rate constant be at a temperature of 130.  C for the same reaction described in Part A?

Explanation / Answer

PART A

ln(k1/k2) = (Ea/R) · (1/T2 - 1/T1)

here, ln(k1/k2) = 2 and T2 = 25 C = 298 K and Ea = 42200 J

2 = 42200/8.314 * ( 1/298- 1/T1)

T1 = 337.64 K

T1 = 64.65 C ............. answer

PART B

ln(k1/k2) = (Ea/R) · (1/T2 - 1/T1)

here,

k1 = 0.014 ,T1 = 25 C = 298 K ,

T2 = 130 C = 403 K then K2 = ?

ln( 0.014/K2) = (42200/8.314)(1/403-1/298)

K2 = 1.184 s^-1 ........... answer

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