In: Chemistry
Part A:
The activation energy of a certain reaction is 34.4 kJ/mol . At 24 ∘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.
Part B:
Given that the initial rate constant is 0.0190s−1 at an initial temperature of 24 ∘C , what would the rate constant be at a temperature of 130. ∘C for the same reaction described in Part A? Express your answer with the appropriate units.
Part A
By using Arrhenius equation:
ln(k2/k1) = (Ea / R)(1/T1 - 1/T2)
Ea, activation energy = 34.4 kJ/mol or 34.4 *103 J/mol
Temperature, T1 = 24oC = (24+273) K,
T2 = ?
Rate constant k2 = 2k1
Gas constant, R = 8.314 J/mol-K
Substituting all these values in the above equation:
ln(2k1/k1) = (34.4*103 / 8.314)(1/(24 + 273) -
1/T2)
0.693147 = 4137.59923 (1/297 - 1/T2)
1.68 *10-4 = 1/297 - 1/T2
1/T2 = -0.0032
T2 = 312.5 K
T2 = 312.5 – 273 K = 39.5 oC
Part B
By using Arrhenius equation:
ln(k2/k1) = (Ea / R)(1/T1 - 1/T2)
Ea, activation energy = 34.4 kJ/mol or 34.4 *103 J/mol
Temperature, T1 = 24oC = (24+273) K,
T2 = 130 oC = (130+273)K
Rate constant k1 = 0.0190 /s
k2 = ?
Gas constant, R = 8.314 J/mol-K
Substituting all these values in the above equation:
ln(k2/0.0190) = (34.4*103 / 8.314)[1/(24 + 273) -
1/(130+273)]
ln(k2/0.0190) = 4137.59923 (1/297 - 1/403)
ln(k2/0.0190) = 4137.59923 * 0.000885614
ln(k2/0.0190) = 3.664314931
(k2/0.0190) = e3.664314931
k2 = 39.02938918 * 0.019
rate constant k2 = 0.74156 s-1