# Kinetics, Arrhenius

1. Apr 5, 2014

### chemphys1

1. The problem statement, all variables and given/known data

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)

3. 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 similar

Help much appreciated!

2. Jan 27, 2017

### TeethWhitener

You've already found Ea and A at 300K from the graph that you plotted. The thing to focus on from collision theory is that A is proportional to the square root of T:
$$A \propto \sqrt{T}$$
The proportionality constant isn't going to change with temperature, so you're just going to get
$$A = k_{prop}T$$
You can plug in the value you found for A in part 1 of your question, and plug in T=300K to find $k_{prop}$. Then you can put this back in the above equation along with T=500K to get the value for what A would be at 500K.

Another option is simply to notice that
$$\frac{A_{500K}}{A_{300K}} = \sqrt{\frac{500K}{300K}} \approx 1.29$$
since the other constants in your collision theory expression for A all cancel. So even over a range of 200 degrees, the prefactor only changes by less than 30%. Thus constant A turns out to be a decent approximation.