1
$\begingroup$

I've recently seen partial derivatives in the velocity definition but in most cases is with total derivatives (infinitesimal change of concentration over infinitesimal change of time). Maybe is more accurate to use partial derivatives because it shows that is only differentiation with respect time and not the other variables (which ones?). Is this even remarkable or is just some change of notation?

Figure that shows how partial derivatives are used in CINÉTICA QUÍMICA, A. G. UREÑA, Ed. Síntesis, 2005

Improve this question
$\endgroup$
8
  • $\begingroup$ The other variables can be the temperature, the pressure, the viscosity, the concentration of substances not taking part to the reaction, etc. $\endgroup$
    – Maurice
    Commented Aug 23, 2022 at 14:09
  • 1
    $\begingroup$ Viscosity may affect the rate of molecular collisions, as there are stronger intermolecular forces. $\endgroup$
    – Poutnik
    Commented Aug 23, 2022 at 14:32
  • 1
    $\begingroup$ A chemical reaction has a rate that tends to zero if the "solution" is like honey, then like glue, then like a solid. $\endgroup$
    – Maurice
    Commented Aug 23, 2022 at 14:34
  • 1
    $\begingroup$ I searched for examples of $$\frac{\partial [A]}{\partial t}$$ and could not find them. Where did you see these partial derivatives? Could you edit your question to provide the sources. $\endgroup$
    – Karsten ♦
    Commented Aug 23, 2022 at 15:30
  • 1
    $\begingroup$ Now that I see the source, I think it is just unusual notation without significance. In the example, there is only one variable used for differentiation, time. I don't think it is wrong but rather unexpected and therefor puzzling. $\endgroup$
    – Karsten ♦
    Commented Aug 24, 2022 at 20:28

1 Answer 1

Reset to default
1
$\begingroup$

If we consider the ideal gas state equation as an example, in the form $V=f(n,p,T)$, then we use a partial derivative $\left(\frac{\partial V}{\partial T}\right)_{n,p}$.

But if we use it in a closed, isobaric scenario as $V=f(T)$, with $n, p$ implicitly constant, then we can use normal/total derivative $\left(\frac{\mathrm{d} V}{\mathrm{d} T}\right)$

The distinguishing of both cases is kind of formal, depending on if you formally consider the other independent variables explicitly or not.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Ok, but in this case the function was concentration, so all the variables that @Maurice said (p,T, viscosity, etc...) are what concentration depends on? $\endgroup$
    – Jorge Bonifaz
    Commented Aug 23, 2022 at 14:59
  • $\begingroup$ I suppose it's concentration because is in the velocity definition $\endgroup$
    – Jorge Bonifaz
    Commented Aug 23, 2022 at 15:00
  • $\begingroup$ And about your answer, you mean that when total derivatives in the velocity defifnition are used is because the other variables are constant, don't you? $\endgroup$
    – Jorge Bonifaz
    Commented Aug 23, 2022 at 15:02
  • 1
    $\begingroup$ Then apply the example on concentration or generally on any independent and dependent variables. The topic is rather mathematical than chemical. $\endgroup$
    – Poutnik
    Commented Aug 23, 2022 at 15:05

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