13
$\begingroup$

I couldn't find a source that explicitly say this, but given the gas constant R is used, is the Arrhenius equation only valid when all reactants are gases? Do they have to be ideal gases?

If the above is true, is there a similar equation for reactions where at least one reactant is in liquid/solid form?

Improve this question
$\endgroup$
5
  • $\begingroup$ No, it has no such limits, obviously R being named "gas constant" is misleading. $\endgroup$
    – Mithoron
    Jun 14, 2015 at 22:56
  • 1
    $\begingroup$ $R$ is called the gas constant because it was first determined in relation to gases. However, $R$ is related to two other constants that have wide usage outside the gas world: Boltzmann's constant $k_B$ and Avogadro's number $N_A$: $$R=N_A \cdot k_B$$ $\endgroup$
    – Ben Norris
    Jun 15, 2015 at 10:59
  • $\begingroup$ @BenNorris From what I understand the exponential is a probability function, is that the same regardless of phase? If the equation has no phase restriction then I am curious about the theoretical reasoning behind using the gas constant R as opposed to some other constant (i.e. Boltzmann constant)? $\endgroup$
    – Yandle
    Jun 15, 2015 at 15:54
  • $\begingroup$ @Yandle - calling the 'gas' constant such is a misnomer. The constant has applications all over thermodynamics at the molar scale. It just happened to be discovered during the study of gases. We choose not to use $k_B$ very often because it is more convenient for small numbers of particles. $\endgroup$
    – Ben Norris
    Jun 15, 2015 at 22:32
  • $\begingroup$ @BenNorris My textbook describes the $e^{\frac{-E_a}{RT}}$ as an energy factor that expresses the frequency of collisions that occur with an energy above threshold for reaction ($E_a$). Am I correct to interpret this expression as valid regardless of what phase or combination of phases the reactants are composed of? $\endgroup$
    – Yandle
    Jun 16, 2015 at 1:49

1 Answer 1

Reset to default
4
$\begingroup$

As Ben Norris explained, $R$ can be just more convenient so that one can deal with quantities per mole.

I.e. one could write either $e^{(-E_a/RT)}$ or $e^{(-E_a/K_BT)}$, where in the first instance one would use $E_a$ in $\ce{J~mol^{-1}}$, in the second instance just in $\ce{J}$. The term in the exponent should of course be dimensionless.

Since the numbers in $\ce{kJ~mol^{-1}}$ are much more familiar for chemists, one would often use the first one to avoid having to use very small numbers.

The crux is that the Boltzmann factor in the Arrhenius expression is related to the chance for particles having enough energy at a given temperature, $T$, to surmount the activation barrier.

Improve this answer
$\endgroup$
2
  • $\begingroup$ So is the exponential (to my understanding is expresses the probability of a collision resulting in successful reaction) in the Arrhenius equation valid for all phases and not just gas? Is there an explanation/proof as to why this is the case? $\endgroup$
    – Yandle
    Jun 19, 2015 at 5:05
  • $\begingroup$ Indeed, but not exactly. The exponential relates to the probability of having sufficient energy to surmount the barrier upon collision. Whether or not, given sufficient energy, the collision leads to reaction depends on other factors. E.g. in gas phase theory, the reaction probability depends on the collision cross section and a 'geometrical' factor depending on whether the reaction partners are oriented in the right way to react. (For reactions involving hydrogen atoms and/or reactions at low T, also collisions with insufficient E can lead to reactions: tunneling). $\endgroup$
    – Fedor
    Jun 28, 2015 at 18:49

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.