If an elementary reaction was conducted in the gaseous phase, say $$\ce{X (g) -> Y(g)},$$ would the formal definition of the rate law be $\mathrm{rate} = kP_\ce{X}$ or $\mathrm{rate} = k[\ce{X}]$?


The ambiguity doesn't lie so much in the definition of the rate, because as long as the constants $a$ and $b$ in $ap_\ce{X}$ and $b[\ce{X}]$ are chosen appropriately, the two expressions are entirely equivalent. So you can define the rate either way, and there is no physical difference, not even a mathematical difference. It is somewhat akin to asking whether you should factorise the number $12$ as $2 \times 6$ or $4 \times 3$. It doesn't matter how you factorise $12$, it's still the same number.

The ambiguity is actually in the definition of the rate coefficient, $k$. The constants $a$ and $b$ above, though directly proportional to each other, are obviously different: and only one of them can be the "true" rate coefficient $k$.

Before launching into the answer proper I should mention that there is no physical difference between using the pressure and the concentration. At the end of the day, it doesn't change the actual rate at which your system is reacting: it's just an arbitrary choice as to what physical constants you want to include in the rate coefficient $k$.

That said, the IUPAC Gold Book seems to come down quite strongly on the side of the concentration, meaning that $k$ is the same as the second constant $b$. They do not make any reference to the phase of the reaction, so it should probably be assumed that they intended it to hold for any general phase. (If anyone knows of a more specialised definition, please point it out!)

order of reaction, $n$ (DOI: 10.1351/goldbook.O04322)

If the macroscopic (observed, empirical or phenomenological) rate of reaction ($nu$) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $k[\ce{A}]^\alpha[\ce{B}]^\beta\cdots$ (expressing in full the dependence of the rate of reaction on the concentrations $[\ce{A}]$, $[\ce{B}], \cdots$)
The proportionality factor $k$ above is called the ($n$th order) 'rate coefficient'. Rate coefficients referring to (or believed to refer to) elementary reactions are called 'rate constants' or, more appropriately 'microscopic' (hypothetical, mechanistic) rate constants.

This approach is quite consistent throughout. For example, the rate of reaction is also expressed in terms of concentrations, at least under constant-volume conditions:

For the general chemical reaction: $$\ce{aA + bB -> pP + qQ + \cdots}$$ occurring under constant-volume conditions, without an appreciable build-up of reaction intermediates, the rate of reaction $\nu$ is defined as: $$\nu = -\frac{1}{a}\frac{\mathrm{d}[\ce{A}]}{\mathrm{dt}} = -\frac{1}{b}\frac{\mathrm{d}[\ce{B}]}{\mathrm{dt}} = \frac{1}{p}\frac{\mathrm{d}[\ce{P}]}{\mathrm{dt}} = \frac{1}{q}\frac{\mathrm{d}[\ce{Q}]}{\mathrm{dt}}$$

The quantity $$\dot{\xi} = \frac{\mathrm{d}\xi}{\mathrm{d}t}$$ defined by the equation: $$\dot{\xi} = -\frac{1}{a}\frac{\mathrm{d}n_\ce{A}}{\mathrm{dt}} = -\frac{1}{b}\frac{\mathrm{d}n_\ce{B}}{\mathrm{dt}} = \frac{1}{p}\frac{\mathrm{d}n_\ce{P}}{\mathrm{dt}} = \frac{1}{q}\frac{\mathrm{d}n_\ce{Q}}{\mathrm{dt}}$$ (where $n_\ce{A}$ designates the amount of substance $\ce{A}$, conventionally expressed in units of mole) may be called the 'rate of conversion' and is appropriate when the use of concentrations is inconvenient, e.g. under conditions of varying volume.

One tangential question I can anticipate is: what's the difference between the expressions $r = -(1/a)(\mathrm{d}[\ce{A}]/\mathrm{d}t)$ and $r = k[\ce{A}]$? The answer is: the former is a definition. The rate is literally defined by how fast the reactants disappear, which is a universal truth, applicable to all reactions. The latter is an empirically observed expression, which only holds true for a small subset of reactions (i.e. first-order ones), and it says that the rate at which the reactant disappears is directly proportional to its concentration.

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    $\begingroup$ @Orthocresol Can we conclude that $r = -(1/a)(\mathrm{d}[\ce{A}]/\mathrm{d}t) = k[\ce{A}]$ $\endgroup$ – Adnan AL-Amleh Apr 29 at 23:55
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    $\begingroup$ @AdnanAL-Amleh Well, if (and only if) that first-order dependence is indeed observed, then yes you can equate them: and then you can integrate with respect to $t$ to get $[\ce{A}]_t = [\ce{A}]_0 \mathrm{e}^{-akt}$. If $\ce{A}$ is the only reactant, then it's quite typical to set its stoichiometric coefficient $a$ to be just $1$. The resulting equation should be familiar to you if you have studied the integrated rate equations. $\endgroup$ – orthocresol Apr 30 at 0:06
  • $\begingroup$ @orthocresol Thank you for the answer! More specifically, if the reaction rate is given to the reader as bar/hour, would this imply that the rate would be written in terms of partial pressure? I'm confused because P and n/V are just a factor of RT away. $\endgroup$ – ilovechemistry Apr 30 at 0:08
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    $\begingroup$ @ilovechemistry You could always convert between pressure and concentration as you say. But if the rate is given in terms of pressures, then it's by far easier to just stick to pressures throughout. It won't be the IUPAC-sanctioned "rate coefficient", but it's just a proportionality factor, so not exactly a huge deal (and doesn't affect the physical interpretation of the system). $\endgroup$ – orthocresol Apr 30 at 0:11

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