MOFs as sponges for gas: hydrogen, carbon dioxide and methane

A MOF is a rigid sponge that can soak up gas and give it back

Intuition

A kitchen sponge soaks up water through its holes and gives it back when squeezed. A MOF does the same with gas molecules. Its holes are so many and so open that the lecture quotes surface areas of up to 6000 m² g⁻¹, and gas sticks to all of that surface.

Highly porous MOFs are host-guest systems. MOF-5 has Zn₄O corner clusters, 1,4-benzenedicarboxylate linkers and 19 Å cavities, and is stable to 300 °C. Gas sorption measures how much gas goes in and out. Nanoporous materials do this reversibly and leave the host lattice intact. Table 1 of the cited paper shows desolvated MOF-5 taking up for example 1492 mg g⁻¹ of Ar and 831 mg g⁻¹ of N₂ (both at −194 °C), with free volume 0.61 to 0.54 cm³ cm⁻³. The N₂ isotherm plateaus near 830 mg g⁻¹ above $P/P_0 \approx 0.2$.

The lecture covers three gases: H₂, CO₂ and CH₄. Tuning the MOF’s structure optimises how each is adsorbed and released.

Hydrogen is a good fuel that is hard to pack tightly

Hydrogen has a high energy density per weight. It is “green” if made with renewable energy, burns cleanly in a fuel cell ($H_2 + \tfrac12 O_2 \to H_2O$), converts efficiently from chemical to electrical energy, and may help decarbonise hard industries such as aviation. The problems are a low energy density per volume at room temperature and pressure, and energy lost in producing, compressing and storing it. Boiling point and critical temperature (numbers for interest only): hydrogen −253 °C and −240 °C, methane −161 °C and −83 °C, propane −42 °C and 97 °C, hexane 69 °C and 254 °C. Hydrogen’s critical temperature is far below room temperature, so it cannot be liquefied at room temperature by pressure alone.

Targets for storage: high gravimetric and volumetric densities, suitable thermodynamics, fast kinetics, long cycle life, safety and low cost. For interest only: liquid H₂ is held at 21 K, cylinders withstand 700 atm, and liquefaction needs about 30% of the energy. Chemisorption forms bonds with the surface but is energy-intensive and slow. Physisorption uses weak intermolecular forces. MOFs and other nanoporous materials are competitive for physisorption.

For physisorption you want two things. First, maximise the surface area per mass and volume, ideally above 4000 m² g⁻¹. Second, maximise the interaction strength: ideally $\Delta H \approx 14$ kJ mol⁻¹, whereas typical values are 4 to 6 kJ mol⁻¹. Frameworks have very high areas per mass. H₂ forms a monolayer at low temperature and uptake scales with surface area. At 77 K the plot gives about 75 mg g⁻¹ for MOF-177 (about 5600 m² g⁻¹), falling through IRMOF-20, IRMOF-1, IRMOF-6, IRMOF-11 and HKUST-1 to about 22 mg g⁻¹ for MOF-74 (about 1070 m² g⁻¹). Those figures are read from the plot, so they are approximate.

In MOF-5 the α binding site is a cup-shaped site with three Zn atoms. A quantum simulation at 0 K gives $\Delta H_{ads} = -6.4$ kJ mol⁻¹. Experiment at 77 K gives −3.8 to −4.8 kJ mol⁻¹.

The best binding energy balances filling at high pressure against emptying at low pressure

Intuition

Think of a tap and a plug. Bind too weakly and the gas will not stay in even at high pressure. Bind too strongly and you cannot get it out again without pulling a vacuum. The ideal MOF holds the gas firmly at the filling pressure and lets go at the delivery pressure.

The lecture does a back-of-envelope Langmuir calculation at 298 K. Fractional sorption is $\theta = KP/(1+KP)$ with $K = e^{\Delta S^\circ/R}e^{-\Delta H^\circ/RT}/P^\circ$ and $\Delta S^\circ \cong -8R$. Modelled isotherms for $\Delta H = -20$, $-15$, $-10$ and $-5$ kJ mol⁻¹ show that at $-5$ kJ mol⁻¹ uptake is small even at high pressure. At $-20$ kJ mol⁻¹ pressures below 1 bar are needed to release the gas. The best release at room temperature, within practical pressure limits of 1.5 to 100 bar, is for about 15 kJ mol⁻¹. The release peaks for $-5$, $-10$, $-15$ and $-20$ kJ mol⁻¹ are near 110, 220, 330 and 440 K. The optimum energy also depends on pressure: about 18 kJ mol⁻¹ near 0 bar, 13.6 at 100 bar and 12.3 at 300 bar.

The formula for the optimum is

$\Delta H^\circ_{opt} = T\Delta S^\circ + \frac{RT}{2}\ln\!\left(\frac{P_1 P_2}{P^{\circ 2}}\right) = -13.6\ \text{kJ mol}^{-1} \quad (T = 298\ \text{K})$

Try it: worked example

Reproduce delta H_opt = -13.6 kJ/mol at T = 298 K, with delta S = -8R, P1 = 1.5 bar and P2 = 100 bar.

Show answer and steps

AnswerAbout -13.6 kJ/mol. The slide gives the equation and the result. The 1 bar value for P0 is an assumption that reproduces the slide's number.

1. Take $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ and $\Delta S^\circ = -8R = -66.5\ \text{J mol}^{-1}\text{K}^{-1}$.
2. First term: $T\Delta S^\circ = 298 \times (-66.5) = -19\,820\ \text{J mol}^{-1} \approx -19.8\ \text{kJ mol}^{-1}$.
3. The slide extract does not give a number for $P^\circ$. Taking the standard pressure as 1 bar, $P_1P_2/P^{\circ 2} = 1.5 \times 100 = 150$, and $\ln 150 = 5.01$.
4. Second term: $\dfrac{RT}{2}\ln 150 = \dfrac{8.314 \times 298}{2} \times 5.01 \approx 1239 \times 5.01 \approx 6210\ \text{J mol}^{-1} = +6.2\ \text{kJ mol}^{-1}$.
5. Add them: $-19.8 + 6.2 = -13.6\ \text{kJ mol}^{-1}$, matching the slide.

Optional aside

A reading of the formula, not stated on the slide: setting $K\sqrt{P_1P_2} = 1$ in the Langmuir expression and solving for $\Delta H^\circ$ gives exactly the equation above. In plain words, the best binding is the one where the material is half full at the middle (geometric mean) of the fill and release pressures.

Exam detail

Know the equation for $\Delta H^\circ_{opt}$, that it uses $\Delta S^\circ \cong -8R$, $P_1 = 1.5$ bar and $P_2 = 100$ bar, and that it gives $-13.6$ kJ mol⁻¹. Know what too-weak and too-strong binding do to uptake and release.

Open metal sites give hydrogen something stronger to hold

Metal hydrides bind H₂ by chemisorption with interaction energies that are too high. The lecture asks whether the idea of metal sites can raise the interaction energy of MOFs. Its table of binding enthalpies (kJ mol⁻¹) includes Li⁺(g) 27, Na⁺(g) 10, K⁺(g) 6, Ti⁺(g) 37, Cu⁺ in chabazite 56, a CuCl surface 93, Cr(CO)₅ 78 and OsH₂(CO)(PiPr₃)₂ 82.

In HKUST-1, Cu₃(btc)₂, made from btc (1,3,5-benzenetricarboxylate) and $[Cu_2(Ac)_4]$, water is coordinated to Cu as made. Heating removes the water, leaving undercoordinated Cu, a bare metal site. Neutron diffraction shows H₂ binds the bare metal sites first, because the interaction energy is more favourable, and then fills the pores. In the pores, 48 D₂ sit in cage B (a face-capped truncated rhombic dodecahedron) and 32 D₂ in cage A (a face-capped truncated octahedron). A distance of 2.39 Å is labelled at the bare metal site, but the slide does not name the two atoms it spans.

CO₂ capture needs a surface that picks CO₂ out of a messy gas

The flue-gas detail is marked “for interest only” on the slides. The MOF design point is part of the lecture’s learning outcomes. Flue gas is a mixture (N₂ 73 to 77%, CO₂ 15 to 16%, H₂O 5 to 7%, O₂ 3 to 4%, plus small amounts of SO₂, SO₃, NOₓ, HCl, CO, hydrocarbons and Hg) at 50 to 75 °C and 1 atm. That means high gas volume and low CO₂ concentration. Carbon capture and storage increases costs by 25 to 40%. Conventional wet scrubbing with an alkanolamine such as monoethanolamine absorbs at 40 to 80 °C and regenerates at 100 to 140 °C, which is expensive because of water’s high heat capacity.

MOFs offer high surface area, and large pore voids with small pore openings for size-selective adsorption (zeolitic imidazolate frameworks, ZIFs, take up much CO₂ and little N₂). Surface design helps further. In $H_3[(Cu_4Cl)_3(BTTri)_8]$, an amine such as ethylenediamine is tethered to the metal sites, and the free end reacts with CO₂ to form carbonates. Amine grafting reduces the BET surface area (non-grafted 1770 m² g⁻¹, then en 345, pip 380, men 634, mmen 870, dimen 854). That is a disadvantage. But it raises CO₂ uptake and lowers N₂ uptake, an overall advantage.

Grafting also raises the enthalpy of adsorption, $-Q_{st}$, which shows a change from physisorption to chemisorption. Zero-coverage $-Q_{st}$ (kJ mol⁻¹): non-grafted 21, en 90, pip 96, men 82, mmen 96, dimen 46. The slide’s plot has grafted samples starting near 95 and falling to about 40 kJ mol⁻¹ at about 3.2 mmol g⁻¹ of CO₂, with non-grafted flat near 24 and N₂ flat near 15 (read from the plot).

Methane sticks best where windows, metal sites and neighbours all help

For methane the lecture wants high surface area and a favourable interaction energy with the right local structure. In HKUST-1, CH₄ binds to the bare metal site, creating an ion-induced dipole on the guest, which strengthens the interaction. The local structure puts CH₄ molecules next to each other, and CH₄–CH₄ interactions strengthen it further. Binding enthalpies: primary adsorption in the windows $-21.8$ kJ mol⁻¹, metal only $-9.4$ kJ mol⁻¹, and with CH₄–CH₄ interactions about $-15$ kJ mol⁻¹. Cu₃(btc)₂ has the highest known room-temperature volumetric methane uptake: 230 cm³(STP) cm⁻³ at 35 bar and 270 cm³(STP) cm⁻³ at 65 bar, meeting the volumetric target set by DOE. PCN-14, with Cu–Cu dimers, appears as another example.

Pitfall

The slide numbers $-21.8$, $-9.4$ and $-15$ kJ mol⁻¹ are three different binding situations in the same material, not three materials. And do not mix up the H₂ optimum ($-13.6$ kJ mol⁻¹, from the Langmuir equation) with the CH₄ windows value ($-21.8$ kJ mol⁻¹, measured for HKUST-1).

Recall

Why does the best hydrogen-storage MOF not have the strongest possible binding?

Because you have to get the gas back out. Very strong binding holds hydrogen so tightly that you need pressures below 1 bar to release it. Very weak binding gives almost no uptake at practical pressures. The best value sits between them, about 14 to 15 kJ/mol.