At present, fossil fuels are the most important energy sources for humankind. Their use is generatin — Physical Chemistry — Thermodynamics Chemistry Question
Theoretical Problem 22
At present, fossil fuels are the most important energy sources for humankind. Their use is generating two major concerns. First, energy production from fossil fuels releases a lot of carbon dioxide into the atmosphere, which is now understood to contribute to global warming. In addition, natural supplies of fossil fuels are expected to be exhausted at the present rate of use in a relatively short time on a historical scale. Many experts believe that alternative sources, like hydrogen or methanol could find widespread use as environmentally friendly substitutes for fossil fuels.
Hydrogen is not a primary energy source; it would have to be produced using another source of power, e.g. nuclear or solar power. The best way would be to produce hydrogen from water, a process popularly called water splitting.
Thermodynamic parameters at 298 K:
H2O(l) ∆fHº = –286 kJ mol–1 O2(g) Sº = 205 J mol–1 K–1
H2O(g) ∆fHº = –242 kJ mol–1 CO2(g) ∆fGº = –394.4 kJ mol–1
H2O(l) Sº = 70 J mol–1 K–1 C8H18(l) ∆fGº = 6.4 kJ mol–1
H2O(g) Sº = 189 J mol–1 K–1 CH3OH(l) ∆fGº= –166.3 kJ mol–1
H2(g) Sº = 131 J mol–1 K–1
Calculate how many kWh of electricity is needed to produce 1 kg hydrogen if the electrolysis operates at a voltage of 1.6 V and 90 % efficiency. Evaluate this process economically based on current industrial electricity and hydrogen prices (use average prices of 0.10 euro/kWh for electricity and 2 euro/kg for H2).
Model Answer
Water splitting by electrolysis:
Anode: 2 H2O → 4 H+ + O2 + 4 e−
Cathode: 4 H2O + 4 e− → 2 H2 + 4 OH−
Therefore a charge of 4 mol e− is necessary to produce 2 mol H2.
For 1 kg hydrogen: 2×1000 g/ M(H2) = 992 mol e−,
9.55×107 C charge is needed. E = 1.6 V × 9.55×107 C = 1.5×108 J
The electricity necessary taking the efficiency into account:
1.5×108 J / (0.90 × 3.6×106 J kWh−1) = 47 kWh
The cost of electricity for producing 1 kg hydrogen is therefore:
47 kWh × 0.10 euro / kWh = 4.7 euro, which is more than the present industrial price of 1 kg hydrogen (2 euro), so electrolysis does not seem economical.
Calculate the volumetric and gravimetric energy density of hydrogen at atmospheric pressure and 298 K. (Assume that hydrogen follows the ideal gas law under these conditions.)
Model Answer
The free energy of hydrogen burning at 298 K: H2(g) + 0.5 O2(g) → H2O(l)
∆rGº = ∆fHº(H2O) − T(Sº(H2O) − Sº (H2) − 0.5 Sº(O2)) = −237 kJ mol−1
The gravimetric energy density: 237 kJ mol−1 / 0.00202 kg mol−1 = 1.17×105 kJ kg−1
The molar volume of hydrogen can be estimated from the ideal gas law:
Vm = RT / p = R × 298 K / 101325 Pa = 0.02445 m3 mol−1
The volumetric energy density: 237 kJ mol−1 / 0.02445 m3 mol−1 = 9.69×103 kJ m−3
Hydrogen is often transported in cylinders which are normally filled to 200 bar. A typical big gas cylinder, made of steel (density 7.8 g cm–3) has a useful volume of 50 dm3 and weighs 93 kg when empty. At this high pressure, hydrogen no longer follows the ideal gas law. A better description can be obtained from the van der Waals equation:
(p + a/Vm^2)(Vm - b) = RT
where p is the pressure, Vm is the molar volume, R is the gas constant, T is the thermodynamic temperature, a and b are gas-specific constants. For hydrogen, a = 2.48×10−2 Pa m6mol−2 and b = 2.66·10−5 m3mol−1 (printed as 2×66·10−5 m3mol−1 in the text). Compressed hydrogen cannot be transported without a cylinder.
Estimate the volumetric and gravimetric energy density of compressed hydrogen.
Model Answer
Rearranging the van der Waals equation gives:
(p + a/Vm^2)(Vm - b) = RT → Vm = b + RT/p - a/(p Vm) + a b/(p Vm^2)
Using this as an iterative formula to find Vm for p = 200 bar at T = 298 K:
Vm(i+1) = 1.50×10−4 m3 mol−1 – 1.24×10−9 / Vm(i) + 3.30×10−14 / Vm(i)^2
If one starts the iteration with Vm(0) = 1×10−2 m3 mol−1:
Vm(1) = 1.50×10−4 m3 mol−1; Vm(2) = 1.43×10−4 m3 mol−1; Vm(3) = 1.43×10−4 m3 mol−1
Therefore Vm = 1.43×10−4 m3 mol−1 under these conditions.
The volume of the cylinder is 0.050 m3, so it contains 0.050 m3 / 1.43×10−4 m3 mol−1 = 350 mol H2.
The total mass of the cylinder is therefore 93 kg + 350 mol × 0.00202 kg mol−1 = 93.7 kg.
The gravimetric energy density: 350 mol × 237 kJ mol−1 / 93.7 kg = 8.85×102 kJ kg−1
The volume of the filled cylinder is 50 dm3 + 93 kg / 7.8 kg dm−3 = 62 dm3
The volumetric energy density: 350 mol × 237 kJ / mol / 0.062 m3 = 1.3×106 kJ m−3
Hydrogen can also be transported in the form of metal hydrides. NaBH4 is a promising substance in this respect, as it reacts with water in the presence of a catalyst to give hydrogen.
How many moles of hydrogen can be produced from 1 mol of NaBH4?
Model Answer
1 mol NaBH4 gives 4 mol H2. A greatly simplified chemical equation of the reaction is:
NaBH4 + 2 H2O → NaBO2 + 4 H2
(The actual chemical form of borate ions is much more complicated.)
As water is a ubiquitous substance, it does not have to be transported together with the metal hydride. Estimate the volumetric and gravimetric energy density of NaBH4 as a hydrogen source. Its density is 1.07 g cm–3.
Model Answer
The gravimetric energy density: 4 mol × 237 kJ mol−1 / 0.03784 kg = 2.51×104 kJ kg−1
The volumetric energy density: 2.51×104 kJ kg−1 × 1070 kg m−3 = 2.68×107 kJ m−3
To put the previously calculated energy densities in perspective, determine the volumetric and gravimetric energy densities of the following energy sources:
i) Graphite as a model of coal. Calculate the density based on the fact that the bond length in graphite is 145.6 pm and the interlayer distance is 335.4 pm.
ii) n-Octane (C8H18) as a model of gasoline. Its density is 0.70 g cm–3.
iii) Methanol, the use of which instead of hydrogen was proposed by the 1994 Nobel laureate György Oláh. Its density is 0.79 g cm–3.
iv) A Ni-MH rechargeable AA battery with a capacity of 1900 mAh and voltage of 1.3 V, which is shaped like a cylinder (diameter: 14.1 mm, height: 47.3 mm, mass: 26.58 g).
v) Water as a source of hydrogen in an imaginary fusion reactor simply converting 1H into 4He. The relative atomic masses are: Ar(1H) = 1.00782, Ar(4He) = 4.00260
Model Answer
i. Burning of graphite:
C(s, gr) + O2(g) → CO2(g) ∆rGº = ∆fGº(CO2) = −394.4 kJ mol−1
The gravimetric energy density: 394.4 kJ/mol / 0.01201 kg mol−1 = 3.28×104 kJ kg−1
The structure of graphite consists of regular hexagons in one layer. The area of one hexagon is: 6 × √3 × (1.456×10−10 m)2 / 4 = 5.508×10−20 m2
The volume of a hexagon-based prism found between two layers: 5.508×10−20 m2 × 3.354×10−10 m = 1.847×10−29 m3
In this cell, there is one carbon atom in each vertex, and each vertex is common to six cells. The cell thus contains 12 × 1/6 = 2 C atoms.
The density is therefore 2 M(C) / (NA × 1.847×10−29 m3) = 2160 kg m−3
The volumetric energy density: 3.28×104 kJ kg−1 × 2160 kg m−3 = 7.08×107 kJ m−3
ii. Burning of n-octane:
C8H18(l) + 12.5 O2(g) → 8 CO2(g) + 9 H2O(l)
∆rGº = 8 ∆fGº(CO2) + 9 ∆fGº(H2O) − ∆fGº(C8H18) = −5295 kJ mol−1
The gravimetric energy density: 5295 kJ mol−1 / 0.11426 kg mol−1 = 4.63×104 kJ kg−1
The volumetric energy density: 4.63×104 kJ kg−1 × 700 kg m−3 = 3.24×107 kJ m−3
iii. Burning of methanol:
CH3OH(l) + 1.5 O2(g) → CO2(g) + 2 H2O(l)
∆rGº = ∆fGº(CO2) + 2 ∆fGº(H2O) − ∆fGº(CH3OH) = −702 kJ mol−1
The gravimetric energy density: 702 kJ mol−1 / 0.03205 kg mol−1 = 2.19×104 kJ kg−1
The volumetric energy density: 2.19×104 kJ kg−1 × 790 kg m−3 = 1.73×107 kJ m−3
iv. The energy of the battery:
E = 1.9 A × 3600 s × 1.3 V = 8.9×103 J
The gravimetric energy density: 8.9 kJ / 0.02658 kg = 3.3×102 kJ kg−1
The volume of the battery: (7.05 mm)2 × π × 47.3 mm = 7386 mm3
The volumetric energy density: 8.9 kJ / 7.386×10−6 m3 = 1.2×106 kJ m−3
v. The mass change associated with the nuclear reaction is:
∆m = 4 × 1.00782 − 4.00260 = 0.02868 g mol−1
∆E = ∆m × c2 = 2.868×10−5 kg mol−1 × (3.000×108 m s−1)2 = 2.581×1012 J mol−1
Each molecule of water contains 2 hydrogen nuclei.
The gravimetric energy density: 2.578×1012 J mol−1 / (2 × 0.01802 kg mol−1) = 7.15×1010 kJ kg−1
The volumetric energy density: 7.15×1010 kJ kg−1 × 1000 kg m−3 = 7.15×1013 kJ m−3
Hydrogen could also be stored as a cryogenic liquid at very low temperatures. The density of liquid hydrogen at its boiling point (–253 °C) is 0.071 g cm –3.
Estimate the volumetric and gravimetric energy density of liquid hydrogen.
Model Answer
The gravimetric energy density of liquid hydrogen is the same as that of gaseous hydrogen: 1.17×108 J kg−1
The volumetric energy density: 1.17×108 J kg−1 × 71 kg m−3 = 8.3×109 J m−3
However, it should be noted that cooling involves a lot of extra cost not reflected directly by the energy density.
What is the advantage of using liquid methanol instead of hydrogen in a hypothetical future economy?
Model Answer
The volumetric and gravimetric energy densities of methanol are high. There is no need for liquification.
Methanol can also be used in methanol fuel cells. The net reaction of the fuel cell is:
CH3OH(l) + 1.5 O2 (g) → CO2(g) + 2 H2O(l)
Write down the cathode and the anode reactions.
Model Answer
CH3OH + H2O → CO2 + 6 H+ + 6 e– on the anode and
1.5 O2 + 6 H+ + 6 e– → 3 H2O on the cathode.
Calculate the maximum voltage of the methanol fuel cell at 25 °C.
Model Answer
∆rGº = –z F Eº, ∆rGº = – 702 kJ mol−1, z = 6, Eºcell = 1.213 V
The methanol fuel cell operates best at 120 °C. At this temperature, the cell reaction potential is 1.214 V. Compare this number with your calculated data.
Model Answer
It is practically the same, but the efficiency of the fuel cell is higher at higher temperature.