Does the term ##ln(k)## have units in reaction based equations?

[phymath7]

TL;DR Summary

I encounteted a chemical equation as follows: $$ln{x/t}=lnk +aln[S_2O^{2-}_8] $$ where I wonder whether the $ln{x/t}$ and $ln[S_2O^{2-}_8]$ terms are unitless?

$$ln{x/t}=lnk +aln[S_2O^{2-}_8] $$ From the above equation,I made a table with the experimental data of the two terms $ln{x/t}$ and $ln[S_2O^{2-}_8]$Bur I didn't mention units for them as I had known that even if k had unit,$lnk$ wouldn't have had.But my teacher pointed out that both terms had unit.Is it true?If yes,then is it only for these type of cases or for all $lnk$ when k has unit?

 

[Borek]

log is unitless and requires an unitless argument.

Whenever you see something like log of concentration it is actually log of unitless activity. Activity is f×c, where c - concentration - has units of mol/L, and f - activity coefficient - has units selected in such a way the product is unitless.

The more diluted solution the close the value of activity coefficient to 1, which is why in typical situation we omit it, but it is no longer true for the more concentrated solutions. Compare https://www.chembuddy.com/pH-calculation-definitions-of-ionic-strength-and-activities

 

[phymath7]

Is there any proof or sth like that that $lnk$ is unitless?It would be very helpful if you provide the proof as well cause I didn't find any on the Internet.

 

[Borek]

Check for example https://doi.org/10.1021/ed1000476 (sadly, behind paywall).

 

[berkeman]

Is there any proof or sth like that ln(k) that is unitless?

https://en.wikipedia.org/wiki/Logarithm

 

[Borek]

https://en.wikipedia.org/wiki/Logarithm

Unfortunately most of the sources (wiki included) don't say a word about units nor dimensions.

This is probably intended to mean they are not part of the definition so the definition as such applies only to unitless numbers. Sadly, this produces a common confusion that hits back now and again and most discussions I have seen in the past were quite handwavy and never gave any solid, strict answer.

 

[TeethWhitener]

This is kind of a deep question.

Not only are the outputs of logarithms and exponentials unitless, but their inputs must be as well. Here's an easy way to see this. Assume that you have two constants, $k_1$ and $k_2$ , and they have the same units $[u_a]$. That means you can add them together such that $k_1+k_2$ also has units $[u_a]$. If you multiply them, you get $k_1k_2$ with units $[u_a]^2$. So far so good. But what happens if we exponentiate them? The exponential will change the units, and presumably, the exponential always changes the same units to the same thing. So let's say the units of $e^{[u_a]}$ are $[u_b]$. But the problem now is that $e^{k_1+k_2}=e^{k_1}e^{k_2}$. This means that $e^{[u_a]}=e^{[u_a]}e^{[u_a]}$, or $[u_b]=[u_b]^2$. That's not very useful.

Logarithms operate in a similar manner, just backwards: we have that $\ln{k_1}+\ln{k_2}=\ln{k_1k_2}$. We end up having $\ln{[u_a]}+\ln{[u_a]}=\ln{[u_a]^2}$, which means we have to assert that the log function transforms $[u_a]$ and $[u_a]^2$ into the same units for the two sides to equal. That's also not very useful.

So we end up saying that exponentials and logarithms take unitless inputs and spit out unitless outputs. It gets even more complicated when you look at @Borek 's example. You need a unitless input for things like the Arrhenius equation $\ln{k}=\ln{A}-\frac{-E_a}{RT}$, so really, the arguments of the logs should be $k/k_0$ and $A/A_0$. But then you get into trouble because mathematically: $\ln{k/k_0}=\ln{k}-\ln{k_0}$. The common consensus is simply to say "you can't do that," which is admittedly a little dissatisfying, but more accurate is to say that only unitless numbers can be inputs, and the unitless number can't be split into unitful numbers and have the properties of logarithms still apply.

 

[pasmith]

Polynomials and transcendental functions

Scalar arguments to transcendental functions such as exponential, trigonometric and logarithmic functions, or to inhomogeneous polynomials, must be dimensionless quantities. (Note: this requirement is somewhat relaxed in Siano's orientational analysis described below, in which the square of certain dimensioned quantities are dimensionless.)

While most mathematical identities about dimensionless numbers translate in a straightforward manner to dimensional quantities, care must be taken with logarithms of ratios: the identity $\log(a/b) = \log a − \log b$, where the logarithm is taken in any base, holds for dimensionless numbers $a$ and $b$, but it does not hold if $a$ and $b$ are dimensional, because in this case the left-hand side is well-defined but the right-hand side is not.

See https://en.wikipedia.org/wiki/Dimensional_analysis, which sadly cites no source for these assertions.

 

[Lord Jestocost]

See entry "Can you take the logarithm of a dimensioned quantity?" in "The Original Usenet Physics FAQ" (https://math.ucr.edu/home/baez/physics/General/logs.html)

 

[TeethWhitener]

But then you get into trouble because mathematically: ln⁡k/k0=ln⁡k−ln⁡k0. The common consensus is simply to say "you can't do that," which is admittedly a little dissatisfying, but more accurate is to say that only unitless numbers can be inputs, and the unitless number can't be split into unitful numbers and have the properties of logarithms still apply.

To elaborate on this point a little bit, supposing you have a unitless number $n$, there are infinite ways to arbitrarily write it as the product of two unitful numbers. For instance, if $n=5$, we can just as easily say that $n=25\text{ meters} \times\frac{1}{5}\text{ meters}^{-1}$ as we can $n=\sqrt{5}\text{ seconds}\times\sqrt{5}\text{ Hz}$. Mathematically it's all the same, so once you've arrived at a unitless argument for a logarithm, it really doesn't make much sense to separate pieces of the argument back out into unitful numbers because any choice of units will be arbitrary.

 

[Orodruin]

Any function of a single dimensionful quantity is necessarily a power of that quantity. This follows from the Buckingham pi theorem. Assume that the dimensionful quantity is $x$ and that the result of some function of $x$ is $y$. We are looking for a physical relationship between $x$ and $y$, meaning that we must construct a dimensionless quantity $\pi = x^a y^b$ and the physical relationship necessarily takes the form $f(\pi)=1$ for some function $f$. (Without loss of generalization, we can pick $b=1$.) It follows that $\pi$ is necessarily a constant and thus $y = \pi x^{-a}$.

Now with that said, people will many times cheat with logarithms due to the logarithm law $\ln(ab) = \ln(a) + \ln(b)$. If $a$ and $b$ are dimensional, the above is technically nonsense since $\ln$ requires a dimensionless argument. However, the way to make sense of it is to squint and assume that whoever writes such a thing really means that we must assume to use the same base units for any included quantities. If $ab$ is dimensionless and we use a unit $u_a$ for $a$, then the unit of $b$ is assumed to be $u_b = 1/u_a$ and therefore
$$
\ln(ab) = \ln(\lambda_a \lambda_b) = \ln(\lambda_a) + \ln(\lambda_b)
$$
where $\lambda_x = x/u_x$ is dimensionless. Thus, interpreting $\ln(ab) = \ln(a)+\ln(b)$ to really be an equation expressing relation between the values of $a$ and $b$ in a particular unit does make some sense.

Unfortunately, the Buckingham pi theorem is often not covered as well as it should be.

 

[Orodruin]

Just to take a little more explicit example of the above. The Green’s function of the Laplace operator in two dimensions is of the form $G(r) = \ln(r) - k$ (modulo an overall multiplicative constant). However, $r$ has dimension length and what is really intended is $\ln(r) - k = \ln(r)-\ln(r_0) = \ln(r/r_0)$ with $r_0$ being some constant normalising lenth scale.