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 = A e^ -E_ a/RT where R is the gas constant (8.314 J/mol K), A is a constant called the frequency factor, and E_ a is the activation energy for the reaction. However, a more practical form of this equation is ln k_2/k_1 = E_ a/R (1/T_1 - 1/T_2) which is mathmatically equivalent to ln k_1/k_2 = E_ a/R (1/T_2 - 1/T_1) where k_1 and k_2 are the rate constants for a single reaction at two different absolute temperatures (T_1 and T_2). The activation energy of a certain reaction is 49.8 kJ/mol. At 20 Degree C, the rate constant is 0.0170s^-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 0.0170s^-1 at an initial temperature of 20 Degree C, what would the rate constant be at a temperature of 110.Degree C for the same reaction described in Part A? Express your answer with the appropriate units.

Explanation / Answer

Given Ea= 49.8 Kj/mol= 49.8*1000 jolues/mol, T1=20+273.15= 293.15K

K1= 0.0170sec-1 K2= 2*0.0170sec-1 ( since at T2 the reaction will have to goes twice as fast at T1)

K2/K1=2, R =8.314 j/mol.K

Ln2= (49.8*1000/8.314)*(1/293.15-1/T2)

1/293.15-1/T2= .000116

1/T2= 0.003296   and T2=303.4438 K=30.29 deg.c

This proves to the fact for every 10 deg.c rise in temperature, the reaction rate gets doubled.

b)

given K1= 0.017sec-1 at 293.15K T2= 110deg.c = 110+273.15= 383.15 K

ln (K2/K1)= (49.8*1000/8.314)*(1/293.15-1/383.15)=4.8

K2/K1= 121.5 K2=2.0655 sec-1

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