The Arrhenius equation shows the relationship between the rate constant k and th
ID: 1008206 • 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^-Ea/RT where R is the gas constant (8.314 J/mol middot 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 k2/k1 = E_a/R (1/T_1 - 1/T_2) which is mathematically equivalent to ln k1/k2 = E_a/R (1/T_2 - 1/T_1) 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 43.6 kJ/mol. At 20 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 20 degree C, what would the rate constant be at a temperature of 200. degree C for the same reaction described in ? Express your answer with the appropriate units.Explanation / Answer
logK2/K1 = Ea/2.303R [1/T1-1/T2]
K1 = 0.017Sce-1 T1 = 20C0 = 20+273 = 293K
K2 = 2*0.017sec-1 T2 =
Ea = 43.6kj/mole = 43600joule/mole
R = 8.314j/mole-k
logK2/K1 = Ea/2.303R [1/T1-1/T2]
log2*0.017/0.017 = 43600/2.303*8.314 [1/293 -1/T2]
0.3020 = 2277 [0.00341-1/T2]
0.3020/2277 = [0.00341-1/T2]
0.000132 = 0.00314-1/T2
0.000132-0.00314 = -1/T2
-0.003008 = -1/T2
T2 = 332.5K = 332.5-273 = 59.5C0
2.
logK2/K1 = Ea/2.303R [1/T1-1/T2]
K1 = 0.017Sce-1 T1 = 20C0 = 20+273 = 293K
K2 = T2 = 200C0 = 200+273 = 473K
Ea = 43.6kj/mole = 43600joule/mole
logK2/K1 = Ea/2.303R [1/T1-1/T2]
logK2/0.017 = 43600/2.303*8.314 [ 1/293- 1/473]
logK2/0.017 = 2277 [0.00341-0.00211]
logK2/0.017 = 2.9601
K2/0.017 = 102.9601
K2/0.017 = 912.22
K2 = 912.22*0.017 = 15.5Sec-1 >>> answer
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