A) We need to use the Arrhenius equation; the integrated form is given as ln k 2

Question

A) We need to use the Arrhenius equation; the integrated form is given as

ln k2/k1 = Ea/R*(1/T1 – 1/T2)

where k1 is the rate of the reaction at absolute temperature T1 and k2 is the rate of the reaction at absolute temperature T2.

Given Ea = 48.0 kJ/mol = (48.0 kJ/mol)*(1000 J/1 kJ) = 48000 J/mol, T1 = 27°C (27 + 273) K = 300 K, k1 = 0.0120 s-1 and k2 = 2*0.0120 s-1 = 0.0240 s-1, plug in values and obtain

ln (0.0240 s-1)/(0.0120 s-1) = (48000 J/mol)/(8.314 J/mol.K)*(1/300 – 1/T2) K-1

====> ln (2) = (5773.394275)*(0.003333 – 1/T2)

====> 0.69315 = (5773.394275)*(0.003333 – 1/T2)

====> 0.003333 – 1/T2 = 0.69315/5773.394275 = 0.000120

====> 1/T2 = 0.003213

====> T2 = 1/0.003213 = 311.2356 311.23

The desired temperature is 311.23 K (311.23 – 273)°C = 38.23°C (ans).

B) We shall use the Arrhenius equation again.

Given Ea = 48.0 kJ/mol = (48.0 kJ/mol)*(1000 J/1 kJ) = 48000 J/mol, T1 = 27°C (27 + 273) K = 300 K, T2 = 170°C (170 + 273) K = 443 K and k1 = 0.0120 s-1, plug in values and obtain

ln k2/k1 = Ea/R*(1/T1 – 1/T2)

====> ln k2/(0.0120 s-1) = (48000 J/mol)/(8.314 J/mol.K)*(1/300 – 1.443) K-1

====> ln k2/(0.0120 s-1) = (5773.394275)*(0.001076) = 6.2122

====> k2/(0.0120 s-1) = exp^(6.2122) = 498.7974

====> k2 = 498.7974*(0.0120 s-1) = 5.98556 s-1 5.9856 s-1 (ans).

Explanation / Answer

A) The activation energy of a certain reaction is 48.0 kJ/mol . At 27 C , the rate constant is 0.0120s1. At what temperature in degrees Celsius would this reaction go twice as fast?

B) Given that the initial rate constant is 0.0120s1 at an initial temperature of 27 C , what would the rate constant be at a temperature of 170. C for the same reaction described in Part A?

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