The Arrhenius equation shows the relationship between the rate constant k and th
ID: 495201 • 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 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 In k_2/k_1 = E_a/R (1/T_1 - 1/T_2) which is mathmatically equivalent to In k_1/k_2 = E_a/R (1/T_2 - 1/T_1) where and are the rate constants for a single reaction at two different absolute temperatures (T_1 and T_2). Part A The activation energy of a certain reaction is 48.3 kJ/mol. At 26 degree C, the rate constant is 0.0160s^-1. At what temperature in degrees Celsius would this reaction go twice as fast? Part B Given that the initial rate constant is 0.0160s^-1 at an initial temperature of 26 degree C, what would the rate constant be at a temperature of 200. degree C for the same reaction described in Part A?Explanation / Answer
k = A_1 * e^(-Ea/RT)
where, k - rate constant Ea - activation energy T- temperature
So,
ln(k1/k2) = (Ea/R)*[(1/T2)-(1/T1)]
=> for the reaction to go twice as fast, rate constant should be twice, => k1/k2 = 1/2
Now,
=> ln(1/2) = (48300/8.314)*[(1/T2)-(1/299)]
=> T2 = 309.59 K
=> Temperature = 309.59 K = 36.59 degree C
b)
ln(k1/k2) = (Ea/R)*[(1/T2)-(1/T1)]
=> ln(0.016/k2) = (48300/8.314)*[(1/473)-(1/299)]
=> k2 = 20.18 s^-1
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