Vapour pressure of H$_2$ in equilibrium with MgH$_2$

Let us study a specific example, namely equilibrium between the MgH$_2$ crystal with H$_2$ gas and the Mg crystal:

$${\rm MgH_2 \leftrightarrow Mg + H_2}.$$ (6.30)

The chemical potentials of the two sides of  must be equal:

$$\mu_{\rm MgH_2} = \mu_{\rm Mg} + \mu_{\rm H_2}.$$ (6.31)

We have:

$$\mu_{\rm H_2} = \frac{1}{N_{\rm H_2}} \left [ F_{\rm H_2} + PV_{\rm H_2}\right ],$$ (6.32)

where $N_{\rm H_2}$ is the number of molecules, $V_{\rm H_2}$ the volume of the gas and $F_{\rm H_2}$ the Helmholtz free energy ^6.2:

$$F_{\rm H_2} = -k_{\rm B}T \ln Z_{\rm H_2},$$ (6.33)

Figure: Crystal structure of MgH$_2$. The Mg and H atoms are represented by light blue and dark red colors, respectively.

with the partition function given by:

$$Z_{\rm H_2} = \frac{1}{N_{\rm H_2}!} \left ( Z_{tr} Z_{int} \right )^{N_{\rm H_2}}.$$ (6.34)

The internal partition function is given by the product of the partition functions associated to the rotational and the vibrational degrees of freedom of the H$_2$ molecule:

$$Z_{int} = Z_{rot} Z_{vib},$$ (6.35)

given by

$$Z_{vib} = \sum_{n=0}^\infty e^{(n+1/2)\hbar \omega / k_{\rm B}T},$$ (6.36)

with $\omega$ the H$_2$ stretching frequency ^6.3 and

$$Z_{rot} = \frac{1}{2}\sum_{j=0}^\infty (2 j + 1 ) e^{-\frac{\hbar... ... B}T} = \frac{1}{2}\sum_{j=0}^\infty (2 j + 1 ) e^{-j (j+1) \frac{\Theta_r}{T}}$$ (6.37)

where $I = m_{\rm H} R_0^2 / 2$ is the moment of inertia of the molecule, with $m_{\rm H}$ the mass of the hydrogen atom, $R_0$ the distance between the two atoms in the molecule ^6.4, and $\Theta_r = \frac{\hbar^2}{2I k_{\rm B}}\simeq 88$ K. The factor $1/2$ in front of the r.h.s of Eq.  is due to the identity of the two atoms in the molecule, which reduces by a factor of two the number of distinct available microstates.

We can therefore write the Helmholtz free energy of the H$_2$ molecule as:

$$F_{\rm H_2} = -N_{\rm H_2} k_{\rm B}T \left [ \ln \frac{eV_{\rm H... ...2\pi mk_{\rm B}T}{h^2}\right)^\frac{3}{2} + \ln Z_{rot} + \ln Z_{vib} \right ],$$ (6.38)

and so

$$\mu_{\rm H_2} = - k_{\rm B}T \left [ \ln \frac{k_{\rm B}T}{P}\lef... ...2\pi mk_{\rm B}T}{h^2}\right)^\frac{3}{2} + \ln Z_{rot} + \ln Z_{vib} \right ].$$ (6.39)

Using this in Eq.  we have:

$$\ln P = \ln k_{\rm B}T \left ( \frac{2\pi mk_{\rm B}T}{h^2}\right... ...t} + \ln Z_{vib} + \left ( \mu_{\rm MgH_2} - \mu_{\rm Mg}\right ) / k_{\rm B}T.$$ (6.40)

If the temperature is large compared to the separation between the rotational energy levels we can approximate the sum in Eq.  using the Euler-MacLaurin identity ^6.5, and we obtain:

$$Z_{rot} = \frac{1}{2} \left [ \frac{T}{\Theta_r} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_r}{T} + \dots \right ].$$ (6.41)

The stretching frequency of the hydrogen molecule is $\nu \simeq 130$ THz, meaning that at room temperature only the ground state has significant probability of being occupied ^6.6, and therefore we can approximate the vibrational partition function as:

$$Z_{vib} \simeq e^{-\hbar \omega / 2 k_{\rm B}T},$$ (6.42)

and so the vapour pressure is:

$$P \simeq k_{\rm B}T \left ( \frac{2\pi mk_{\rm B}T}{h^2}\right)^\... ...heta_r} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_r}{T} + \dots \right ] \times$$

$$\times e^{-\hbar \omega / 2 k_{\rm B}T} e^{\left ( \mu_{\rm MgH_2} - \mu_{\rm Mg}\right ) / k_{\rm B}T}.$$ (6.43)

If we ignore the $p V$ terms for the solid phases, which contribute very little to the total chemical potential at reasonable conditions, then their chemical potentials are simply given by Helmholtz free energies per formula unit, and we can write our final expression for the vapour pressure:

$$P \simeq k_{\rm B}T \left ( \frac{2\pi mk_{\rm B}T}{h^2}\right)^\... ...heta_r} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_r}{T} + \dots \right ] \times$$

$$\times e^{\left ( f_{\rm MgH_2} - f_{\rm Mg} -\hbar \omega / 2 \right) / k_{\rm B}T}.$$ (6.44)

This expression shows how the vapour pressure decreases exponentially with the free energy of the MgH$2$ crystal, reduced by the free energy of the Mg crystal and the zero point energy of the H$_2$ molecule. Therefore, if MgH$_2$ is more stable than Mg + H$_2$, this provides a mechanism to store H$_2$ at a reduced pressure compared with pure H$_2$ gas. The vapour pressure can then be tuned by changing the temperature, on which it depends as $T^{7/2}$, and so H$_2$ gas can be released as needed by increasing $T$ to sufficiently high values.