Arhenius equation chem a-level

Does anyone know a good well explained video or website for arhenius equation inside chemistry a-level and any tips on how to rearrange it would be useful?

Reply 1

BankaiGintoki

Original post by Chickenunicorn17

Does anyone know a good well explained video or website for arhenius equation inside chemistry a-level and any tips on how to rearrange it would be useful?

https://www.chemguide.co.uk/physical/basicrates/arrhenius.html
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.02%3A_Temperature_Dependence_of_Reaction_Rates/6.2.03%3A_The_Arrhenius_Law
https://www.savemyexams.com/a-level/chemistry/ocr/17/revision-notes/5-physical-chemistry--transition-elements-a-level-only/5-1-rates-orders--arrhenius/5-1-7-the-arrhenius-equation/

Reply 2

UtterlyUseless69

Original post by Chickenunicorn17

Does anyone know a good well explained video or website for arhenius equation inside chemistry a-level and any tips on how to rearrange it would be useful?

If you really can’t rearrange the equations, you can always just plug the numbers in and work from there.

In general, you can expect to be given the Arrhenius equation in one of two forms in the question (maybe unless you do OCR, where both forms are given in the data book).

The exponential form, k = A e^(-Ea/RT) is harder to rearrange if you aren’t doing A level maths. This form is most useful if you have the values of A and Ea and are trying to calculate k or if you have k and Ea and are trying to calculate A.

The logarithmic form, ln(k) = ln(A) - Ea/RT is easier to work with. Plotting ln(k) (y-axis) against 1/T (x-axis) gives a straight line with a gradient of -Ea/R and a y-intercept of ln(A). As for rearranging it, usually you only ever have to rearrange it for Ea.

In which case:

ln(k) = ln(A) - Ea/RT

Subtracting ln(A) from both sides: ln(k) - ln(A) = -Ea/RT

Multiplying both sides by -RT: -RTln(k) + RTln(A) = Ea

You can clean this up with factorisation and the laws of logs if you want to (i.e Ea = RTln(A/k) or -RTln(k/A)). Note that the laws of logs are an A level maths topic that is not on the chemistry syllabus and so you will not be expected to use them in A level chemistry.

(edited 1 year ago)

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