Dihydrogen is a promising fuel for the future, notably for power production or mobility purposes. It — Physical Chemistry — Thermodynamics Chemistry Question
Hydrogen storage
Dihydrogen is a promising fuel for the future, notably for power production or mobility purposes. It is an attractive alternative to the use of fossil fuels (hydrocarbons), which release carbon dioxide during their combustion, thus contributing to global warming. Unfortunately, storing efficiently large amounts of H2 is not easy.
Storing dihydrogen as a complex
In 1984, using measurements obtained from neutron diffraction, G. J. Kubas and his collaborators (G. J. Kubas et al., J. Am. Chem. Soc., 1984) identified a tungsten complex within the field of other ligands. The dehydrogenated complex is assumed to be a square-based pyramid, which the dihydrogen molecule is added to.
Metallic central atom
Kubas complex
Since the complex is considered as a square-based pyramid to which the H2 molecule is added, we have to take into account the influence of other ligands. The splitting thus obtained is given in the diagram below.
Storing hydrogen in form of formic acid
In 2006, a research team of EPFL (Switzerland) (C. Fellay et al., Angew. Chem. Int. Ed., 2008) proposed to store H2 in form of formic acid. The main idea is to use formic acid...
Formic acid (2.3 g) is added to a 1 dm3 container with 0.1 g of ruthenium catalyst, under constant atmospheric pressure and at an initial temperature of 25 °C. The container initially contains dinitrogen.
Thermodynamic data at normal conditions of temperature and pressure (20 °C, 1 atm)
compound HCOOH(g) HCOOH(l) CO2(g) H2(g) N2(g)
ΔfH° kJ mol‒1 ‒378.60 ‒425.09 ‒393.51 0.00 0.00
Sm° J mol‒1 K‒1 248.70 131.84 213.79 130.68 191.61
Densities
Gaseous dihydrogen, standard conditions: 0.08988 g L‒1
Liquid dihydrogen, ‒252.78 °C: 70.849 g dm‒3
Specific latent heat of fusion (at standard pressure): ΔfusH°m = 58.089 kJ kg‒1
Specific latent heat of vaporization (at standard pressure): ΔvapH°m = 448.69 kJ kg‒1
16 K
25 K
77 K
293 K
Model Answer
Correct statements :
16 K, 25 K
Using the Clausius-Clapeyron relation, calculate the pressure needed to liquefy ideal gaseous dihydrogen at 27.15 K.
Model Answer
Using the Clausius-Clapeyron relation and the boiling point under a pressure of 1 atm:
∆vapH°m = 448.69 kJ kg–1, so that ∆vapH° = 897.38 J mol–1
Calculate the mass of the dehydrogenated complex needed to store 1 kg of dihydrogen. Calculate ρH (the density of hydrogen in the complex, defined as the mass of hydrogen atoms per volume unit of complex).
Model Answer
Dehydrogenated complex: W(CO)3(P(iPr)3)2 = WC21O3P2H42, M = 588.4 g mol–1
Each complex can store one molecule of dihydrogen. In 1 kg of dihydrogen, there are 500 mol of dihydrogen. Hence, m = 294.2 kg of dehydrogenated complex are needed to store 1 kg of dihydrogen.
Once bound to 1 kg of H2, the complex thus weighs mKubas = 295.2 kg
hence ρH = 6.6×10-6 kg m-3
Give the electronic configuration of atomic tungsten. Specify the number of valence electrons.
Model Answer
[Xe] (6s)2(4f)14(5d)4 so 6 valence electrons (4f layer is full)
Fill in the table with the name of each depicted atomic orbital (s, dyz, dz2, d(x2 – y2), dxz, dxy).
Draw and fill the molecular orbital diagram of dihydrogen.
Give the two planes of symmetry of the Kubas complex (using the axes of Figure 2).
Model Answer
xz and yz planes
Indicate for each orbital d of the metallic central atom if they are symmetric or anti-symmetric with respect to each of the symmetry planes (using the axes of figure 2).
Model Answer
An orbital is symmetric with respect to a symmetry element if it remains the same when the symmetry operation is applied. An orbital is antisymmetric with respect to a symmetry element if it changes to its opposite when the symmetry operation is applied. The results are gathered in the diagram below.
Fill in the diagram in figure 1 with electrons.
Model Answer
As a general rule, interactions between two orbitals implying two electrons with the
Knowing that only orbitals with the same symmetry interact, enumerate the possible interactions for each conformation. Which conformation is the most stable one?
Model Answer
For all conformations, dz² and dx²-y² interacts with σH2.
For conformation (1), only dxz interacts with σH2.
For conformation (2), only dyz interacts with σH2.
Calculate ρH (the density of hydrogen at 25 °C defined as the mass of hydrogen atoms per volume unit of formic acid). Compare this value to those obtained for gaseous dihydrogen at 500 bar and for liquid dihydrogen.
Model Answer
The hydrogen density is higher in formic acid than for high pressure (500 bars) dihydrogen (31 kg m‒3) but lower than for cryogenic liquid hydrogen (70.85 kg m‒3). If one can extract efficiently H2 molecule from formic acid, it constitutes a good alternative to pure H2 storage.
Calculate the standard enthalpy and entropy of reaction at 20 °C for reaction (R1).
calculate the equilibrium constant at 20 °C for reaction (R1).
Model Answer
ΔrGo(T) = ΔrHo – T ΔrSo
ΔrGo(T) = 31.58 – 0.213 x 293 = – 30.8 kJ mol-1
Thus: Ko = 3.1×105
Determine the final composition of the mixture.
Model Answer
n(N2) = n(N2O) = pV / RT
n(N2) = (1.013×105 × 1.0×10-3) / (8.314 × 298.15) = 0.04 mol