Arrhenius equation help

Why when rearranging the equation do you sometimes use logarithms but sometimes keep the exponential, I'm very confused.

Reply 1

Nitrotoluene

The Arrhenius equation is: k = Ae^(-Ea/RT)

Use of logarithms:
You can isolate the activation energy ==> log(k) = log(A) - Ea/RT. This allows you to solve for Ea by rearranging the equation.
You can simplify the equation ==>log(k) = log(A) - Ea/R x 1/T
You can change the base of the logarithms ==> (e.g. logarithm base 10) ==> log10(k) = log10(A) - (Ea/RT) x log10(e)

Use of the exponential:
When you need to analyse the temperature dependence of the rate constant.
When you need to simplify the equation for specific conditions, e.g. to determine the rate constant at a specific temperature.

I hope I have remembered everything.🤨

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Sandro

Reply 2

Slugzie(:

How do you know when to use logarithms and when to keep the exponential in the equation?

Reply 3

UtterlyUseless69

Original post by Slugzie(:

How do you know when to use logarithms and when to keep the exponential in the equation?

In general, you can use either. However, if you aren’t aware with how to rearrange equations to “get rid” of exponentials or logs (i.e you don’t take A level maths), you may wish to pick the easier form.

Exponential form:

k = Ae^-Ea/RT

Logarithmic form:

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

Suppose you want to calculate k or A easily. If you try to use the logarithmic form, it is generally more cumbersome as it requires more rearrangement due to the quantities of interest being within log functions. As such, the exponential form is better.

If you want to calculate Ea or T easily, it is better to use the logarithmic form. In this case, the exponential form is more problematic as Ea and T are within the exponential function and this takes even more rearranging.

As a sidenote, if rearranging equations confuses you due to the algebra, you are not penalised in an exam for taking the equation as is, plugging in all the quantities (with the correct units) and simplifying to calculate the variable of interest. Whilst I don’t do this myself, the A level chemistry students I used to mentor who didn’t take A level maths often found this to make the calculations easier.

(edited 9 months ago)

Reply 4

UtterlyUseless69

Original post by Nitrotoluene

The Arrhenius equation is: k = Ae^(-Ea/RT)

Sandro

My understanding is the OP is still studying A level chemistry and so understanding of how to change the base of the logarithm isn’t required. As such, they will either use Arrhenius’ equation in the natural logarithmic form or in the exponential form.

It is worth noting that A level chemistry doesn’t actually teach students how logarithms work as there is no maths for chemists part of the course. In my opinion, this is a rather unfortunate error on the part of those who designed the course, given that calculations involving both exponentials and logs (and rearranging them) are assessed and do come up in the exams, which does cause some problems for students without A level maths.

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