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 In k2/k1 = Ea/R(1/T1 -1/T2) which is mathmatically equivalent to In k1/k2 = Ea/R(1/T2-1/T1) Where k1 and k2 are the rate constants for a single reaction at two different absolute temparatures(T1 and T2) The activation energy of a certain reaction is 43.8kJ/mol. At 30 degree 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. Given that the initial rate constant is 0,0190s-1 at an initial temperature of 30 degree C , what would the rate constant be at a temperature of 180 degree C for the same reaction described in Part A? Express your answer with the appropriate units

Explanation / Answer

part A

Ea=43.8 * 1000J/kmol

T1=30o=273k+30=303k

k1=0.0190s-1

k1/k2=1/2=0.5

ln k1/k2=Ea/R(1/T1-1/T2)

ln 0.5 =43.8 * 1000 J/mol/8.314J/Kmol(1/303-1/T2)

-0.6931=5268.22/K(0.0033K-1 - 1/T2)

-131.5625*10-6 K =0.0033K-1 - 1/T2

-1.3156 *10-4 K=33.0 10-4 K- 1/T2

31.6844*10-4 K=1/T2

T2=0.031561*10000K=315.61K=315.61-273=42.6o

part B

ln k1/k2=Ea/R(1/T1-1/T2)

T2=180+273=453K

ln 0.0190s-1/k2 =5268.22/K(0.0033K-1 - 1/453k)

                       =5268.22/K*150/137259 K-1

       ln 0.0190s-1/k2 =316.46

taking antilog of both sides,

k2=0.6004 * 10-4s-1

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