IChO 2006 · Theory problem 7

Hydrogen Economy

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Hydrogen is more energy-dense than carbon, by mass. Thus, historically there has been a move toward fuel with higher hydrogen content: coal → oil → natural gas → hydrogen. Cost-effective production and safe storage of hydrogen are two major hurdles to the successful inauguration of a hydrogen economy.

7.1 Consider hydrogen in a cylinder of 80 MPa at 25 °C. Using the ideal gas law, estimate the density of hydrogen in the cylinder in kg m\(^{-3}\).

7.2 Calculate the ratio between heat generated when hydrogen is burned and heat generated when the same weight of carbon is burned. The difference comes to a large extent from the fact that the most abundant isotope of hydrogen has no neutron and hydrogen has no inner electron shell. \(\Delta H_{\mathrm f}^{\circ}[\ce{H2O(l)}] = -286\ \mathrm{kJ/mol}\), \(\Delta H_{\mathrm f}^{\circ}[\ce{CO2(g)}] = -394\ \mathrm{kJ/mol}\).

7.3 Calculate the theoretical maximum work produced by the combustion of 1 kg hydrogen (a) from the electric motor using hydrogen fuel cell and (b) from the heat engine working between 25 °C and 300 °C. The efficiency (work done/heat absorbed) of an ideal heat engine working between \(T_{\mathrm{cold}}\) and \(T_{\mathrm{hot}}\) is given by \([1-T_{\mathrm{cold}}/T_{\mathrm{hot}}]\).

\[ S_{298}^{\circ}[\ce{H2(g)}]=131\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

\[ S_{298}^{\circ}[\ce{O2(g)}]=205\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

\[ S_{298}^{\circ}[\ce{H2O(l)}]=70\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

If the fuel cell is working at 1 W and the standard potential difference, how long will the electric motor run at what current?

Official English text of the problem.

Answers

  • 7.1: \(\boxed{\rho\approx65.1\ \mathrm{kg\,m^{-3}}}\), using the ideal gas law.
  • 7.2: \(\boxed{q_{\mathrm{hydrogen}}/q_{\mathrm{carbon}}\approx4.32}\) for equal fuel masses.
  • 7.3(a): \(\boxed{W_{\max}\approx1.18\times10^{8}\ \mathrm J=118\ \mathrm{MJ}}\).
  • 7.3(b): \(\boxed{W_{\max}\approx6.81\times10^{7}\ \mathrm J=68.1\ \mathrm{MJ}}\).
  • 7.3, operation at 1 W: \(\boxed{t\approx1.18\times10^{8}\ \mathrm s\approx3.73\ \text{years}}\), at \(\boxed{I\approx0.813\ \mathrm A}\), with standard cell voltage \(\boxed{E^\circ\approx1.23\ \mathrm V}\).

These values use \(M(\ce{H2})=2.016\ \mathrm{g\,mol^{-1}}\) and \(M(\ce{C})=12.01\ \mathrm{g\,mol^{-1}}\).

Solution

Use \[ R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}, \qquad F=96485\ \mathrm{C\,mol^{-1}}, \qquad T=298.15\ \mathrm K. \]

The molecular species involved are molecular hydrogen, \(\ce{H2}\), [H][H]; molecular oxygen, \(\ce{O2}\), O=O; water, \(\ce{H2O}\), O; and carbon dioxide, \(\ce{CO2}\), O=C=O. Elemental carbon is taken as graphite, whose extended lattice has no single molecular SMILES.

7.1 — Density of compressed hydrogen

From the ideal gas law, \[ PV=nRT=\frac{m}{M}RT, \] so \[ \rho=\frac{m}{V}=\frac{PM}{RT}. \]

Substituting in SI units, \[ \rho= \frac{(80\times10^{6}\ \mathrm{Pa}) (2.016\times10^{-3}\ \mathrm{kg\,mol^{-1}})} {(8.314\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}) (298.15\ \mathrm K)} =\boxed{65.1\ \mathrm{kg\,m^{-3}}}. \]

This is the ideal-gas estimate requested.

7.2 — Heat released by equal masses

The balanced combustion reactions are \[ \ce{H2(g) + 1/2 O2(g) -> H2O(l)}, \qquad \Delta_{\mathrm r}H^\circ=-286\ \mathrm{kJ\,mol^{-1}}, \] and \[ \ce{C(s, graphite) + O2(g) -> CO2(g)}, \qquad \Delta_{\mathrm r}H^\circ=-394\ \mathrm{kJ\,mol^{-1}}. \]

The reaction enthalpies equal the given product formation enthalpies because each reactant is an element in its standard state.

The positive magnitudes of heat released per gram are \[ q_{\mathrm{hydrogen}} =\frac{286}{2.016} =141.9\ \mathrm{kJ\,g^{-1}}, \] \[ q_{\mathrm{carbon}} =\frac{394}{12.01} =32.81\ \mathrm{kJ\,g^{-1}}. \]

Thus, for equal masses, \[ \boxed{ \frac{q_{\mathrm{hydrogen}}}{q_{\mathrm{carbon}}} =\frac{286/2.016}{394/12.01} =4.32 }. \]

7.3(a) — Maximum work from the fuel cell and electric motor

At constant temperature and pressure, the maximum electrical work delivered by a reversible fuel cell is the decrease in Gibbs energy: \[ W_{\mathrm{electrical,max}}=-n\Delta_{\mathrm r}G^\circ. \] An ideal electric motor converts all this electrical work into mechanical work.

For the hydrogen combustion reaction, \[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=S^\circ(\ce{H2O(l)}) -S^\circ(\ce{H2(g)}) -\frac12S^\circ(\ce{O2(g)})\\ &=70-131-\frac12(205)\\ &=-163.5\ \mathrm{J\,mol^{-1}\,K^{-1}}. \end{aligned} \]

Using the supplied \(298\ \mathrm K\) entropy values at \(298.15\ \mathrm K\), \[ \begin{aligned} \Delta_{\mathrm r}G^\circ &=\Delta_{\mathrm r}H^\circ-T\Delta_{\mathrm r}S^\circ\\ &=-286-(298.15)(-0.1635)\\ &=-237.252\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

The amount of hydrogen in \(1\ \mathrm{kg}\) is \[ n=\frac{1000\ \mathrm g}{2.016\ \mathrm{g\,mol^{-1}}} =496.03\ \mathrm{mol}. \]

Therefore, \[ W_{\max} =(496.03)(237.252)\ \mathrm{kJ} =\boxed{1.18\times10^8\ \mathrm J}. \]

7.3(b) — Maximum work from the heat engine

The temperatures must be expressed in kelvin: \[ T_{\mathrm{cold}}=298.15\ \mathrm K, \qquad T_{\mathrm{hot}}=573.15\ \mathrm K. \]

The Carnot efficiency is \[ \eta_{\max} =1-\frac{298.15}{573.15} =0.4798. \]

Using the supplied combustion enthalpy, the heat available from \(1\ \mathrm{kg}\) of hydrogen is \[ Q=n|\Delta_{\mathrm r}H^\circ| =(496.03)(286) =1.41865\times10^5\ \mathrm{kJ}. \]

In the idealized heat-engine model specified by the problem, \[ W_{\max}=\eta_{\max}Q =(0.4798)(1.41865\times10^8\ \mathrm J) =\boxed{6.81\times10^7\ \mathrm J}. \]

7.3 — Running time and current at 1 W

Each hydrogen atom changes oxidation state from \(0\) in \(\ce{H2}\) to \(+1\) in water. Consequently, combustion of one molecule of \(\ce{H2}\) transfers two electrons through the external circuit.

The standard cell voltage therefore follows from \[ \Delta_{\mathrm r}G^\circ=-2FE^\circ: \] \[ E^\circ =\frac{237252\ \mathrm{J\,mol^{-1}}} {2(96485\ \mathrm{C\,mol^{-1}})} =1.2295\ \mathrm V. \]

Since electrical power is \(P=E^\circ I\), \[ I=\frac{1\ \mathrm W}{1.2295\ \mathrm V} =\boxed{0.813\ \mathrm A}. \]

At a constant power of \(1\ \mathrm W=1\ \mathrm{J\,s^{-1}}\), the maximum running time is \[ t=\frac{W_{\mathrm{electrical,max}}}{P} =\frac{1.17685\times10^8\ \mathrm J} {1\ \mathrm{J\,s^{-1}}} =\boxed{1.18\times10^8\ \mathrm s}. \]

Using \(365.25\) days per year, \[ \boxed{t\approx3.73\ \text{years}}. \]

If rounded molar masses \(M(\ce{H2})=2.00\) and \(M(\ce{C})=12.0\ \mathrm{g\,mol^{-1}}\) are used instead, the corresponding results are \(64.5\ \mathrm{kg\,m^{-3}}\), \(4.36\), \(119\ \mathrm{MJ}\), \(68.6\ \mathrm{MJ}\), and \(3.76\) years. The voltage and current are unchanged.

The solution was checked against the official solution.