The Arrhenius equation shows the relationship between the rate constant k and th
ID: 929262 • Letter: T
Question
The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k = Ae^-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) 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 38.9 kJ/mol At 27 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 27 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
Arrhenus equation k = A e-Ea/RT where k = rate of reaction
Arrhenius equation can be written as
In (k2/k1) = Ea/R (1/T1 - 1/T2) -------- Eq (1)
1) We have to find the temperature at which reaction go twice fast.
Hence, Initial rate of reaction = k1
Final rate of reaction k2 = 2k1
Initial temperature T1 = 27oC = 27 + 273 K = 300 K
Final temperature T2 = ?
Activation enegry Ea = 38.9 kJ/mol = 38900 J/mol
R = 8.314 J/K/mol
Substitute all these values in eq (1),
In (k2/k1) = [Ea/R] (1/T1 - 1/T2)
In (2k1/k1) = [38900/8.314 ] [1/300 - 1/T2]
[1/300 - 1/T2] = 1.481 x 10-4
1/T2 = (1/300) - 1.481 x 10-4
1/T2 = 3.185 x 10-3
T2 = 1/ ( 3.185 x 10-3 )
= 313.95 K
= 40.95 oC
T2 = 40.95 oC
Therefore, at 40.95 oC , the reaction go twice fast.
2)
Initial rate of reaction k1 = 0.017 s-1
Final rate of reaction k2 = ?
Initial temperature T1 = 27oC = 27 + 273 K = 300 K
Final temperature T2 = 110oC = 110 + 273 K = 383 K
Activation enegry Ea = 38.9 kJ/mol = 38900 J/mol
R = 8.314 J/K/mol
Substitute all these values in eq (1),
In (k2/k1) = [Ea/R] (1/T1 - 1/T2)
In (k2/0.017) = [38900/8.314 ] [1/300 - 1/383]
k2 = 0.017 x e [38900/8.314 ] [1/300 - 1/383]
= 0.5 s-1
k2 = 0.5 s-1
Therefore, rate constant at 110oC = 0.5 s-1
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