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 Tin kelvins and is typically written as k=AeEa/RT where R is the gas constant (8.314 J/molK), Ais a constant called the frequency factor, and Ea is 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 (T1and T2).

a) The activation energy of a certain reaction is 40.7 kJ/mol . At 30  C , the rate constant is 0.0150s1. At what temperature in degrees Celsius would this reaction go twice as fast?

b) Given that the initial rate constant is 0.0150s1 at an initial temperature of 30  C , what would the rate constant be at a temperature of 190  C for the same reaction described in Part A?

Explanation / Answer

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

substituting the values we get

ln(2) = 40700/8.314 * (1/303 - 1/T2)

(1/303 - 1/T2) = 0.00014159276

1/T2 = 0.00315873726

T2 = 316.58K = 43.58C

b)

ln(k2/k1) = 40700/8.314 * (1/303 - 1/463)

ln(k2/k1) = 5.5831

k2 = e^(5.5831) * k1 = 3.988 s^(-1)

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