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chemphys1
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Homework Statement
The activation energy and Arrhenius paramter can be found from its temperature dependence
the Arrhenius equation
k=Aexp(-Ea/RT) --> lnk=lnA - Ea/RT
Given data is 5 temperatures with their corresponding k values
Q1) From this data calculate A and Ea
q2)
Here A has been considered independent of temperature. Show this is a good approximation by comparing ratios A(T2)/A(T1) and exp(-Ea/RT2)/exp(-Ea/RT1)
(T1 300K T2 500K)
(use collision theory expression for A)
The Attempt at a Solution
Q1) Ok so this bit I think is fine. To find Ea and A I plotted lnk vs 1/T to get a straight line
Ea = -k x R = slope of the line x 8.314
then A is exp(intercept)
So I got both the values from the graph.
Not sure about the second part though:
Q2)can only find collision theory 'A' as d^2(8kbT/u)^1/2
So I don't really understand which equation I'm supposed to be using to find A at T1 and A at T2?
Then for exp(-Ea/RT2)/exp(-Ea/RT1) I'm just using the Ea value I found from the graph, just changing the temperatures.
I'm guessing the ratios are supposed to turn out to be similarHelp much appreciated!