4
$\begingroup$

So, we have the Freundlich's Isotherm equation as $\frac{x}{m} = kp^{\frac{1}n}$, where the LHS measures the extent of adsorption with the corresponding pressure $p$ in the RHS with the two constants, $k$ and $n > 1$. But, I got a little confused with the dimensions on both the sides. Since $\frac{x}{m}$ is dimensionless, so should be the RHS. Therefore, I reasoned that the constant $k$ having units should cancel out the units of $p^{\frac{1}n}$. But yet I'm confused, and I don't think its right. What is going on?

$\endgroup$
3
$\begingroup$

Technically speaking, one typically wouldn't write the left hand side unitless, as it is the ratio of the mass of two different components (usually mg of some chemical per gram of the surface).

But, otherwise your intuition is correct and $k$ just needs units that will cancel with $p^{\frac{1}{n}}$ to give the units on the left hand side. So $k$ could have units of $\left(\dfrac{\text{mg chemical}}{\text{g surface}}\right)\left(\mathrm{bar^{\frac{-1}{n}}}\right)$.

We could also reformulate Freundlich's equation in terms of concentration (since we are dealing with an isotherm) as: $$q=KC^{\frac{1}{n}}$$ where $q=m/x$, $C$ is concentration of the chemical and $K$ is a new constant with units of $\left(\dfrac{\text{mg chemical}}{\text{g surface}}\right)\left(\mathrm{{\dfrac{L}{\text{mg chemical}}}}\right)^{\frac{1}{n}}$.

Source

$\endgroup$
0
$\begingroup$

About the units of the constant K of the Freundlich equation in terms of concentration in the answer above. It seems that you have "double thought" it - applied the reciprocal twice - once by inverting mg chemical per L to give L per mg chemical, but then again by writing a negative sign before the 1/n. It seems to me that the units of K should be "(mg chemical/g surface)(L/mg chemical)^(1/n)

$\endgroup$
1
  • $\begingroup$ You are right, I must have been indecisive at the time about how I would show the reciprocal and accidentally did it twice. I will make that edit, thank you for spotting that. In the future, it would be better to suggest an edit/post a comment on the original answer rather than add a new one. $\endgroup$ – Tyberius Aug 14 '20 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.