Part A: The activation energy of a certain reaction is 37.2kJ/mol . At 20 degree
ID: 851010 • Letter: P
Question
Part A: The activation energy of a certain reaction is 37.2kJ/mol . At 20 degrees celcius , 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. T2 =
Part B: Given that the initial rate constant is 0.0190s-1 at an initial temperature of 20 degrees Celcius , what would the rate constant be at a temperature of 130 degrees Celcius for the same reaction described in Part A? Express your answer with the appropriate units. k2 =
Explanation / Answer
1) First, the answer should be approximately 10 degrees higher because a decent rule of thumb is that for every 10 degree increase a reaction rate will double.
Now to solve it, use the Arrhenius equation in this form:
ln(k2/k1) = Ea/R(1/T1-1/T2)
where k2 and k1 are the reaction rate constants, Ea is energy of activation in units of J/mon, R =8.314 J/mol K, and T1 and T2 are the two temperatures that correspond to the rate constants (in Kelvin). For this problem, I'll call k1 and T1 the given rate constant at 20 degrees C and we'll solve for T2 when k2 should be double of k1. Since you know k2 is twice as big as k1, the ratio of k2/k1 = 2 so you don't even need to know the initial rate constant (0.0130 per sec).
ln(2) = 35,300/8.314(1/293 - 1/T2)
1.633 x10^-4 = 1/293 - 1/T2
3.25 x10^-3 = 1/T2
T2 = 307.7 Kelvin = 308 with sig figs = 35 degrees celsius.
We predicted 30 degrees using the rule of thumb, so this is in the ballpark. 35 degrees is the answer.
2) k = k?