The Arrhenius Equation The Arrhenius equation shows the relationship between the

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 Ea is the activation energy for the reaction. However, a more practical form of this equation is which is mathematically equivalent to 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 32.7kJ/mol . At 27 degree C , the rate constant is 0.0120s^-1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. T2 = Part B Given that the initial rate constant is 0.0120s^-1 at an initial temperature of 27 degree C , what would the rate constant be at a temperature of 100 degree C for the same reaction described in Part A? Express your answer with the appropriate units. k2 =

Explanation / Answer

A)

we know that

according to arhenius equation

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

given

Ea = 32.7 kJ/mol

R= 8.314

T1 = 27 C = 27 + 273 = 300 K

K1 = 0.012

K2 = 2K1

so

using those values

we get

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

ln(2K1/K1) = ( 32.7 x 1000/8.314) ( 1/300 - 1/T2)

ln2 = (32.6 x 1000 / 8.314 ) ( 1/300 - 1/T2)

solving we get

T2 = 316.74 K

so

the reaction twice as fast at 316.74 K = 43.74 C

Part B)

given

k1 = 0.012

T1 = 300 K

T2 = 100 C = 100 + 273 = 373 K

using those values we get

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

ln(K2/0.012) = ( 32.7 x 1000 / 8.314 ) ( 1/300 - 1/373)

solving we get

K2 = 0.156

so

so the rate constant at 100 C is 0.156 s-1

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