physical chemistry - Does the Maxwell-Boltzmann distribution apply to gases only?

The Maxwell-Boltzmann distribution can be used to determine the fraction of particles with sufficient energy to react. I know that the curve applies to gaseous reactants and would like to know whether solids and/or liquids are also described by a similar distribution.

In other words, can I use a Maxwell-Boltzmann distribution to interpret reaction rates at the molecular level in liquids and/or solids?

Answer

I'm going to go against the grain here: the Maxwell distribution does describe the distribution of molecule speeds in any (3-D) matter, regardless of phase.

Suppose we have a system of $N$ molecules with masses $m_i$, positions $\vec{r}_i$, and velocities $\vec{v}_i$ (with $i = 1, ..., N$). Assume that the total energy of this system is of the form $$ E(\vec{r}_1, \dots, \vec{r}_N; \vec{p}_1, \dots, \vec{p}_N) = U(\vec{r}_1, \dots, \vec{r}_N) + \sum_{i = 1}^N \frac{\vec{p}_i^2}{2 m_i }, $$ i.e., a potential energy depending on the molecules' positions and a kinetic energy. According to Boltzmann statistics, the probability of finding this system within a small volume of phase space $d^{3N} \vec{r} \, d^{3N} \vec{p}$ is $$ \mathcal{P}(\vec{r}_1, \dots, \vec{r}_N; \vec{p}_1, \dots, \vec{p}_N) d^{3N} \vec{r} \, d^{3N} \vec{p} \propto e^{-E/kT} d^{3N} \vec{r} \, d^{3N} \vec{p} \\= e^{-U(\vec{r}_1, \dots, \vec{r}_N)/kT} \left[ \prod_{i=1}^N e^{-p_i^2/2m_i k T}\right]d^{3N} \vec{r} \, d^{3N} \vec{p} $$ To find the probability distribution of finding molecule #1 with a particular momentum $\vec{p}_1$, we integrate over all other configuration variables (i.e., $\vec{r}_1$ through $\vec{r}_N$ and $\vec{p}_2$ through $\vec{p}_N$). Because the $E$ is the sum of a contribution from the positions and a contribution from the momenta, the Boltzmann factors can be split between these integrals, with the result that $$ \mathcal{P}(\vec{p}_1) \, d^3\vec{p}_1 \propto e^{-p_1^2/2 m_1 kT} d^3\vec{p}_1 \left[ \int e^{-U(\vec{r}_1, \dots, \vec{r}_N)/kT} d^{3N}\vec{r}\right] \left\{ \prod_{i = 2}^N \int e^{-p_i^2/2 m_i kT} d^3 \vec{p}_i \right\} $$ These integrals are nasty (especially the one over all $N$ position vectors), but they're just constants with respect to $\vec{p}_1$, which means that the can be folded into the proportionality constant: $$ \mathcal{P}(\vec{p}_1) \, d^3\vec{p}_1 \propto e^{-p_1^2/2 m_1 kT} d^3\vec{p}_1. $$ A similar logic applies to every other molecule in my system. In other words, the probability of finding a molecule with a momentum $\vec{p}$ doesn't depend at all on how they interact with each other, assuming that their interaction energy is dependent only on their collective positions. Since these molecules obey the same momentum distribution, they must also have the same velocity distribution, and in particular they obey the Maxwell speed distribution.

So the answer to the question asked in your title, "Does the Maxwell-Boltzmann distribution apply to gases only?" is "no"; it applies to all phases of matter, in the sense that it describes the distribution of particle speeds and energies. However, the question you ask at the bottom of your post, "Can I use a Maxwell-Boltzmann distribution to interpret reaction rates at the molecular level in liquids and/or solids?" may also be "no"; the connection between reaction rate and activation energy is not straightforward if the medium is dense, as was pointed out by @porphyrin in their answer.