IChO 2011 · Theory problem 4

Hydrogen storage in sodium borohydride

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For sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride (\(\ce{NaBH4}\)), being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases \(\ce{H2}\) gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.

Figure 1

Colloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete \(\ce{H2}\) release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of \(\ce{NaBH4}\) is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is \(92\ \mathrm{mol}\,\ce{H2}\cdot(\mathrm{mol}\,\ce{Ru})^{-1}\cdot\mathrm{min}^{-1}\) at \(25\,^\circ\mathrm{C}\).

4.1 Calculate the amount of ruthenium catalyst (in mg) which must be added to \(0.100\ \mathrm{dm^3}\) of \(\ce{NaBH4}\) solution with a concentration of \(1.0\ \mathrm{mol\,dm^{-3}}\) to supply the hydrogen gas at a rate of \(0.100\ \mathrm{dm^3\,min^{-1}}\) at \(25\,^\circ\mathrm{C}\) and \(1.0\ \mathrm{atm}\), that is required for a portable fuel cell.

4.2 For how many minutes will this system supply hydrogen gas at this rate?

4.3 The Arrhenius activation energy for this catalytic hydrolysis of sodium borohydride is \(E_{\mathrm a}=42.0\ \mathrm{kJ\,mol^{-1}}\). Calculate the temperature required to achieve the same rate of hydrogen evolution by using a half of the amount of ruthenium catalyst used at \(25.0\,^\circ\mathrm{C}\).

4.4 A fuel cell (see figure) is made up of three segments sandwiched together: the anode, the electrolyte, and the cathode. Hydrogen is used as fuel and oxygen as oxidant. Two chemical reactions occur at the interfaces of the three different segments:

Figure 2
Figure 3
Figure 4

The net result of the two reactions is

Figure 5

The hydrogen for the fuel cell is supplied from the hydrolysis of sodium borohydride. Calculate the standard potential for the cathode half reaction if the standard reduction potential for the anode half reaction is \(-0.83\ \mathrm{V}\) and \(\Delta_{\mathrm f}G^\circ(\ce{H2O(l)})\) is \(-237\ \mathrm{kJ\,mol^{-1}}\).

4.5 Calculate the volume of air at a temperature of \(25\,^\circ\mathrm{C}\) and a pressure of \(1.0\ \mathrm{atm}\) needed to generate a constant current of \(2.5\ \mathrm{A}\) for \(3.0\) hours in this fuel cell. Assume that air contains \(20\,\%\) by volume \(\ce{O2(g)}\).

4.6 The efficiency of a fuel cell is given by the ratio of the work produced to the heat dissipated by the cell reaction. Thus, the maximum efficiency for a fuel cell is given by:

\[ \eta_{\text{fuel cell}}=\frac{\text{work}}{\text{heat}} \]

Calculate the maximum efficiency for the fuel cell using the data given below at \(25\,^\circ\mathrm{C}\) and a standard pressure.

\(S^\circ\ (\mathrm{J\,mol^{-1}\,K^{-1}})\)
\(\ce{H2(g)}\) 130.7
\(\ce{O2(g)}\) 205.2
\(\ce{H2O(l)}\) 70.0

The second law of thermodynamics states that it is impossible to convert all of the heat, \(q_{\mathrm H}\), from a high-temperature reservoir at \(T_{\mathrm H}\) into work. At least, some of the energy, \(q_{\mathrm C}\), must be transferred to a low-temperature reservoir at \(T_{\mathrm C}\). Thus, a heat engine with 100% efficiency is thermodynamically impossible. When the heat engine is working reversibly, as in a Carnot cycle, the efficiency will be a maximum.

Figure 6

For a heat engine working reversibly between two reservoirs the following relations are applied:

\[ q_{\mathrm H}=w+q_{\mathrm C} \]

and

\[ \frac{q_{\mathrm H}}{T_{\mathrm H}} = \frac{q_{\mathrm C}}{T_{\mathrm C}} \]

4.7 What should be the temperature of the hot reservoir, \(T_{\mathrm H}\), of a Carnot heat engine to maintain the efficiency of the fuel cell calculated in part 4.6, if the temperature of cold reservoir \(T_{\mathrm C}\) is \(40\,^\circ\mathrm{C}\)?

(If you have been unable to calculate the value for the efficiency then use the value 0.80 for the calculation.)

Official English text of the problem.

Answers

  • 4.1: \(m(\ce{Ru})=\boxed{4.5\ \mathrm{mg}}\).
  • 4.2: \(\boxed{98\ \mathrm{min}}\).
  • 4.3: \(\boxed{T=310.9\ \mathrm{K}=37.7\,^\circ\mathrm{C}}\).
  • 4.4: \(\boxed{E^\circ_{\text{cathode}}=+0.40\ \mathrm{V}}\), as a reduction potential.
  • 4.5: \(\boxed{8.6\ \mathrm{dm^3}}\) of air at \(25\,^\circ\mathrm{C}\) and \(1.0\ \mathrm{atm}\).
  • 4.6: \(\boxed{\eta_{\max}=0.830=83.0\%}\).
  • 4.7: \(\boxed{T_{\mathrm H}=1.84\times10^3\ \mathrm{K}}\), approximately \(\boxed{1.56\times10^3\,^\circ\mathrm{C}}\).

Solution

The molecular and ionic species used in the solution have the following names and SMILES:

Species Name SMILES
\(\ce{NaBH4}\) Sodium borohydride [Na+].[BH4-]
\(\ce{H2}\) Hydrogen [H][H]
\(\ce{H2O}\) Water O
\(\ce{Na+}\) Sodium ion [Na+]
\(\ce{BO2^-}\) Metaborate ion [O-]B=O
\(\ce{O2}\) Oxygen O=O
\(\ce{OH^-}\) Hydroxide ion [OH-]

The metaborate SMILES is a formal monomeric representation of the formula used in the problem. These SMILES do not specify phases, solvation, or the structure of the ruthenium nanoclusters.

Use the standard constants \[ R=0.082057\ \mathrm{dm^3\,atm\,mol^{-1}\,K^{-1}} =8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}, \] \[ F=96485\ \mathrm{C\,mol^{-1}}, \qquad M(\ce{Ru})=101.07\ \mathrm{g\,mol^{-1}}. \]

4.1

The balanced hydrolysis reaction is \[ \ce{NaBH4(aq) + 2H2O(l) ->[catalyst] Na+(aq) + BO2^-(aq) + 4H2(g)}. \]

At \(25\,^\circ\mathrm{C}=298.15\ \mathrm{K}\), the required molar hydrogen-production rate follows from the ideal-gas equation: \[ \dot n_{\ce{H2}} = \frac{P\dot V}{RT} = \frac{(1.0\ \mathrm{atm})(0.100\ \mathrm{dm^3\,min^{-1}})} {(0.082057\ \mathrm{dm^3\,atm\,mol^{-1}\,K^{-1}})(298.15\ \mathrm K)} = 4.0874\times10^{-3}\ \mathrm{mol\,min^{-1}}. \]

Each mole of ruthenium produces \(92\) moles of hydrogen per minute, so \[ n_{\ce{Ru}} = \frac{4.0874\times10^{-3}\ \mathrm{mol\,min^{-1}}} {92\ \mathrm{min^{-1}}} = 4.4429\times10^{-5}\ \mathrm{mol}. \]

Therefore, \[ m_{\ce{Ru}} = (4.4429\times10^{-5}\ \mathrm{mol}) (101.07\ \mathrm{g\,mol^{-1}}) = 4.4904\times10^{-3}\ \mathrm g = \boxed{4.5\ \mathrm{mg}}. \]

4.2

The initial amount of sodium borohydride is \[ n_{\ce{NaBH4}} = cV = (1.0\ \mathrm{mol\,dm^{-3}})(0.100\ \mathrm{dm^3}) = 0.100\ \mathrm{mol}. \]

The hydrolysis produces four moles of hydrogen per mole of sodium borohydride: \[ n_{\ce{H2},\text{total}}=4(0.100)=0.400\ \mathrm{mol}. \]

Because the reaction is zero order in the substrate, the stated model gives a constant production rate until the substrate is exhausted. Thus, \[ t = \frac{n_{\ce{H2},\text{total}}}{\dot n_{\ce{H2}}} = \frac{0.400\ \mathrm{mol}} {4.0874\times10^{-3}\ \mathrm{mol\,min^{-1}}} = 97.86\ \mathrm{min} \approx \boxed{98\ \mathrm{min}}. \]

4.3

The hydrogen-production rate is proportional to the amount of catalyst: \[ \dot n_{\ce{H2}}=k(T)n_{\ce{Ru}}. \]

To maintain the same molar production rate with half the catalyst, \[ k(T_2)\frac{n_{\ce{Ru}}}{2}=k(T_1)n_{\ce{Ru}}, \qquad \frac{k(T_2)}{k(T_1)}=2. \]

The Arrhenius equation gives \[ \ln\frac{k(T_2)}{k(T_1)} = \frac{E_{\mathrm a}}{R} \left(\frac1{T_1}-\frac1{T_2}\right). \]

Consequently, \[ T_2 = \left[ \frac1{298.15\ \mathrm K} - \frac{(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}})\ln2} {42000\ \mathrm{J\,mol^{-1}}} \right]^{-1} = 310.867\ \mathrm K. \]

Hence, \[ \boxed{T_2=310.9\ \mathrm K=37.7\,^\circ\mathrm C}. \]

4.4

Oxygen is reduced at the cathode because it accepts electrons: \[ \ce{O2(g) + 2H2O(l) + 4e^- -> 4OH^-(aq)}. \]

Hydrogen is oxidized at the anode because it releases electrons: \[ \ce{H2(g) + 2OH^-(aq) -> 2H2O(l) + 2e^-}. \]

Multiplying the anode reaction by two and adding it to the cathode reaction gives \[ \ce{2H2(g) + O2(g) -> 2H2O(l)}. \]

Four moles of electrons are transferred per mole of this overall reaction. The standard Gibbs energies of formation of elemental hydrogen and oxygen in their standard states are zero, so \[ \Delta_{\mathrm r}G^\circ = 2\Delta_{\mathrm f}G^\circ(\ce{H2O(l)}) = 2(-237) = -474\ \mathrm{kJ\,mol^{-1}}. \]

Using \(\Delta_{\mathrm r}G^\circ=-nFE^\circ_{\text{cell}}\), \[ E^\circ_{\text{cell}} = \frac{474000\ \mathrm{J\,mol^{-1}}} {4(96485\ \mathrm{C\,mol^{-1}})} = 1.22817\ \mathrm V. \]

Both electrode potentials in \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode,red}} - E^\circ_{\text{anode,red}} \] are reduction potentials. Therefore, \[ E^\circ_{\text{cathode,red}} = 1.22817+(-0.83) = 0.39817\ \mathrm V \approx \boxed{+0.40\ \mathrm V}. \]

4.5

The total charge delivered is \[ Q=It = (2.5\ \mathrm A)(3.0\times3600\ \mathrm s) = 27000\ \mathrm C. \]

The cathode reaction consumes one mole of oxygen per four moles of electrons: \[ n_{\ce{O2}} = \frac{Q}{4F} = \frac{27000}{4(96485)} = 0.069959\ \mathrm{mol}. \]

For ideal gases at the same temperature and pressure, volume fraction equals mole fraction. Air containing \(20\%\) oxygen therefore has the required total amount \[ n_{\text{air}} = \frac{n_{\ce{O2}}}{0.20} = 0.349795\ \mathrm{mol}. \]

Its volume is \[ V_{\text{air}} = \frac{n_{\text{air}}RT}{P} = \frac{(0.349795)(0.082057)(298.15)}{1.0} \ \mathrm{dm^3} = 8.5578\ \mathrm{dm^3}. \]

Thus, \[ \boxed{V_{\text{air}}=8.6\ \mathrm{dm^3}}. \]

4.6

At constant temperature and pressure, the maximum electrical work obtainable is \(-\Delta_{\mathrm r}G^\circ\). Using the total reaction enthalpy release as the available chemical-energy input, \[ \eta_{\max} = \frac{-\Delta_{\mathrm r}G^\circ}{-\Delta_{\mathrm r}H^\circ}. \]

For \[ \ce{2H2(g) + O2(g) -> 2H2O(l)}, \] the standard entropy change is \[ \begin{aligned} \Delta_{\mathrm r}S^\circ &= 2S^\circ(\ce{H2O(l)}) - 2S^\circ(\ce{H2(g)}) - S^\circ(\ce{O2(g)})\\ &= 2(70.0)-2(130.7)-205.2\\ &= -326.6\ \mathrm{J\,mol^{-1}\,K^{-1}}. \end{aligned} \]

Since \[ \Delta_{\mathrm r}G^\circ = \Delta_{\mathrm r}H^\circ-T\Delta_{\mathrm r}S^\circ, \] we obtain \[ \begin{aligned} \Delta_{\mathrm r}H^\circ &= \Delta_{\mathrm r}G^\circ+T\Delta_{\mathrm r}S^\circ\\ &= -474+(298.15)(-0.3266)\\ &= -571.3758\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Therefore, \[ \eta_{\max} = \frac{474}{571.3758} = 0.8295766 \approx \boxed{0.830=83.0\%}. \]

4.7

For the reversible heat engine, \[ \eta = \frac{w}{q_{\mathrm H}} = \frac{q_{\mathrm H}-q_{\mathrm C}}{q_{\mathrm H}} = 1-\frac{q_{\mathrm C}}{q_{\mathrm H}}. \]

The reversible relation \(q_{\mathrm H}/T_{\mathrm H}=q_{\mathrm C}/T_{\mathrm C}\) gives \[ \eta=1-\frac{T_{\mathrm C}}{T_{\mathrm H}}. \]

Using absolute temperatures, \[ T_{\mathrm C}=40+273.15=313.15\ \mathrm K. \]

With the unrounded efficiency from part 4.6, \[ T_{\mathrm H} = \frac{T_{\mathrm C}}{1-\eta} = \frac{313.15}{1-0.8295766} = 1837.48\ \mathrm K. \]

Thus, \[ \boxed{T_{\mathrm H}\approx1.84\times10^3\ \mathrm K \approx1.56\times10^3\,^\circ\mathrm C}. \]

The solution was checked against the official solution.