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What is logarithms (chemistry help)

Hi I'm in year 2 chem and we used lograthims in the Arhlenius equation

I know how to use it but what actually is it? I don't even know why I'm doing it and what I'm doing

edit: ty for everyone’s response I think I’ve gotten the grasp of it now 🙂
(edited 1 month ago)
Reply 1
Original post by 7889224
Hi I'm in year 2 chem and we used lograthims in the Arhlenius equation
I know how to use it but what actually is it? I don't even know why I'm doing it and what I'm doing

Its simply the inverse of an exponential function, so
log_2(8) = 3
because 2^3 = 8. So it tells you how many times the base (2 in this case) is imultiplied by itself to get the argument (8) which is obviously 3 as 2^3=8.

There are plenty of introductory tutorials, have a quick google? But the key idea is that youre reasoning about exponents (of the base), rather than the actual number.
(edited 1 month ago)
Reply 2
The Arrhenius equation is used to predict activation energy from rate constants and temperature
IMG_2360.jpeg
IMG_2361.jpeg
The Arrhenius equation can be linearised by taking the natural log ( log of e Euler’s number) this is then fitted to a straight line
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Credits to the creators of images
Original post by 7889224
Hi I'm in year 2 chem and we used lograthims in the Arhlenius equation
I know how to use it but what actually is it? I don't even know why I'm doing it and what I'm doing

For the sake of A level chemistry, you don’t need to know what a logarithm is conceptually- just that ln(x) and e^x are inverse functions (as are log10(x) and 10^x, which you will encounter when you get to acids and bases).

As above, logarithms are a convenient means of linearising exponential equations as well. Since Arrhenius’ equation shows that the rate constant varies exponentially with the negative reciprocal of the temperature (-1/T) and by taking logarithms (and using the appropriate laws of logarithms) gives you the logarithmic form of the Arrhenius equation:

k = A exp(-Ea/RT) (where exp(x) = e^x)

So ln(k) = ln[-A exp(-Ea/RT)

= ln(A) + ln[exp(-Ea/RT)] (by the log multiplication law)

= ln(A) - Ea/RT (since exp(x) and ln(x) are inverse functions, they “cancel” each other out)

So we can say that

ln(k) = -Ea/R x 1/T + ln(A)

Which looks like y = mx + c (since Ea, R and A are all constants)

As such, if k is measured for the same reaction over a range of temperatures, a graph of ln(k) vs 1/T can be plotted since the graph should form a straight line, one should easily be able to calculate Ea from the gradient (simply by multiplying it by -R, or roughly -8.314 and to optionally convert it to kJ/mol, divide the result by 1000). Similarly, the pre-exponential factor A can be measured by determining the vertical intercept of the graph and exponentiating it.

The form of the Arrhenius equation shown in the post right above mine is the form for two different temperatures. It is rarely encountered at A level, but can be derived from the logarithmic form.

Suppose that when T = T1, k = k1 and that when T = T2, k = k2. We can say

(1) ln(k1) = ln(A) - Ea/RT1 and (2) ln(k2) = ln(A) - Ea/RT2

If we subtract the equation (1) from equation (2), we have

ln(k2) - ln(k1) = -Ea/RT2 + Ea/RT1 (beware the double negatives!)

Using the log division law, we can say the LHS = ln(k2/k1):

ln(k2/k1) = -Ea/RT2 + Ea/RT1

Taking a common factor of Ea/R out of the RHS:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

This form of the Arrhenius equation is useful if you only know the rate constant at two different temperatures and want to calculate the activation energy (since two data points aren’t really much good for plotting a reliable straight line).
(edited 1 month ago)
Reply 4
You also need to know about logs for calculating pH and pKa though. But it will makes if you understood the example from mqb2766.
Original post by CoalL
You also need to know about logs for calculating pH and pKa though. But it will makes if you understood the example from mqb2766.

Also true - I have stated this myself (albeit without identifying pH and pKa calculations specifically).

If the OP is doing the Edexcel A level course, they will also need to be familiar with using logs for ΔG as the equation ΔG = -RTlnK is expected. However, to my knowledge no other exam boards require this equation to be memorised and so they will just give it to you in an exam if you have to use it.

To add to my previous response, logs (this time of base 10 rather than euler’s number e) are used with pH as a means of indicating how high the ”concentration” of “hydrogen ions” is (in reality, concentrations are not used at all with logs as you cannot put anything with units into a logarithmic function and instead you use a related, unitless quantity called the activity which is well approximated by concentration if the concentration is relatively low and water is the solvent. Also, technically hydrogen ions don’t exist- you have H3O^+ adducts, but I digress).

Using pH is helpful as statements like “the concentration of {insert acid} is 0.50 mol dm^-3” often don’t give much of an indicator as to how acidic the solution actually is (for a multitude of reasons) and whether it is fit for a purpose e.g as a disinfectant.

pKas are helpful as they indicate how easily the acidic proton is removed from the conjugate acid. The more negative the pKa, the stronger the acid. Since some acids are literally millions of times more acidic than others, it is more appropriate to use a logarithmic scale to make comparisons.
(edited 1 month ago)

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