Arrhenius equation (IB Chemistry HL)

Hi everyone!

I am looking for someone, who is willing to teach me how to use the Arrhenius equation by using data given on a graph (1/T versus ln k). Is anyone willing to explain this to me via Skype? I would be very grateful as my IB exams are in two weeks.

I have got a question that we can use as a sample for the session.

Here's a screenshot of the question:

Here's the link for the paper (Question no: 7)d)iv) ) : http://teacherweb.com/NS/CitadelHighSchool/KathyReid/E_Chemistr_HL_paper-2_TZ2_M09_E.pdf

And lastly my skype username: ege.okyar

It's not that hard.

The equation itself is given to you in the databooklet which you get for paper 2

You just need to be able to use it:

Arrhenius equation:

k = Ae^-Ea/RT

and the natural log form:

lnk = lnA - Ea/RT

where:

A is a constant related to the orientation of collisions occurring between the particles in the reaction. You can think of it as the proportion of collisions with adequate activation energy that successfully react.

k is the rate constant

R is the universal gas constant

T is the absolute temperature

Determination of the activation energy from practical results

Rates experiments are carried out at different temperatures. The results of values of k at different temperatures can be plotted on a graph to obtain a value for the activation energy for a specific reaction.

Using the natural log form:

ln k = lnA - Ea/RT

A plot of natural log of k (y-axis) against (1/T) (x-axis) will give a straight line of gradient = -Ea/R

You just need to measure the gradient, paying close attention to the values of the axes. Divide it by R and you have the activation energy. Remember that this MUST be positive.

If you are asked for the Arrhenius constant, then it's the intercept of the line with the 'y' axis when 1/T = 0

And your question
-----------------

(iv) Calculate the activation energy, Ea, for the reaction, using Table 1 of the Data Booklet. [4]

Difference in y values (–5.8 - - 9.8) / difference in x values (0.00191-0.00212)
Gradient = 4/-0.00021 = 19047.6
This is equal to Ea/R = 19047.6
Ea = 19047.6 x 8.314 = 158362 J = 158 kJ