plusminus The Arrhenius Equation The Arrhenius equation shows the relationship b
ID: 957869 • Letter: P
Question
plusminus The Arrhenius Equation 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) which is mathmatically equivalent to ln k_1/k_2 = 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 49.0 kJ/mol. At 30 degree C, the rate constant is 0.0180s^-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 0.0180s^-1 at an initial temperature of 30 degree C, what would the rate constant be at a temperature of 180. degree C for the same reaction described in Part A? Express your answer with the appropriate units.Explanation / Answer
it seems to be you need part-B
K1 = 0.018 s-1
T1 = 30ºC = 273+30 = 303K
T2 = 180 ºC = 273+180 = 453K
Ea = 49.0 kJ/mol convert in to joules = 49000 J/mol
put all these values in the above derived arhenious equation
ln (K2/0.018) = 49000 / 8.314 (1/303 - 1/453)
ln (K2/0.018) = 5893.7 (0.0033 - 0.0022)
ln (K2/0.018) = 6.483307
(K2/0.018) = e -6.483307
(K2/0.018) = 654.13
K2 = 11.77 s-1
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