IChO 2000 · Theory problem 2

Bridge between Denmark and Sweden

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Figure 1

On July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.

Concrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, \(\ce{CaSO4 * 2H2O}\), is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, \(\ce{CaSO4 * 1/2H2O}\). Consider the following reaction:

Figure 2

The following thermodynamic data apply at 25 °C, standard pressure: 1.00 bar:

Compound \(\Delta_{\mathrm f}H^\circ\) (\(\mathrm{kJ\,mol^{-1}}\)) \(S^\circ\) (\(\mathrm{J\,K^{-1}\,mol^{-1}}\))
\(\ce{CaSO4 * 2H2O(s)}\) −2021.0 194.0
\(\ce{CaSO4 * 1/2H2O(s)}\) −1575.0 130.5
\(\ce{H2O(g)}\) −241.8 188.6

Gas constant: \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}=0.08314\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}\)

\(0\ ^\circ\mathrm C=273.15\ \mathrm K.\)

2.1 Calculate \(\Delta H^\circ\) (in kJ) for transformation of 1.00 kg of \(\ce{CaSO4 * 2H2O(s)}\) to hemihydrate \(\ce{CaSO4 * 1/2H2O(s)}\). Is this reaction endothermic or is it exothermic?

2.2 Calculate the equilibrium pressure (in bar) of water vapour in a closed vessel containing \(\ce{CaSO4 * 2H2O(s)}\), \(\ce{CaSO4 * 1/2H2O(s)}\) and \(\ce{H2O(g)}\) at 25 °C.

2.3 Calculate the temperature at which the equilibrium water vapour pressure is 1.00 bar in the system described in problem 2-2. Assume that \(\Delta H^\circ\) and \(\Delta S^\circ\) are temperature independent.

Corrosion of metals is associated with electrochemical reactions. This also applies for the formation of rust on iron surfaces, where the initial electrode reactions usually are:

Figure 3

An electrochemical cell in which these electrode reactions take place is constructed. The temperature is 25 °C. The cell is represented by the following cell diagram:

Figure 4

Standard electrode potentials (at 25 °C):

Figure 5

Nernst factor: \(RT\ln 10/F=0.05916\) volt (at 25 °C)

Faraday constant: \(F=96485\ \mathrm{C\,mol^{-1}}\)

2.4 Calculate the standard electromotive force (the standard cell voltage), \(E^\circ\), at 25 °C.

2.5 Write down the overall reaction which takes place during discharge of the cell under standard conditions.

2.6 Calculate the equilibrium constant at 25 °C for the overall cell reaction.

2.7 The overall reaction referred to above is allowed to proceed for 24 hours under standard conditions and at a constant current of 0.12 A. Calculate the mass of \(\ce{Fe}\) converted to \(\ce{Fe^2+}\) after 24 hours. Oxygen and water may be assumed to be present in excess.

2.8 Calculate \(E\) for the cell at 25 °C for the following conditions :

\[ [\ce{Fe^2+}]=0.015\ \mathrm M,\quad \mathrm{pH}_{\text{right-hand half-cell}}=9.00,\quad p(\ce{O2})=0.700\ \mathrm{bar}. \]

Official English text of the problem.

Answers

  • 2.1: \(\boxed{\Delta H^\circ=+484\ \mathrm{kJ}}\) for 1.00 kg of gypsum; the reaction is endothermic.
  • 2.2: \(\boxed{p(\ce{H2O})=8.15\times10^{-3}\ \mathrm{bar}}\).
  • 2.3: \(\boxed{T=379.7\ \mathrm K=106.5\ ^\circ\mathrm C}\).
  • 2.4: \(\boxed{E^\circ=0.84\ \mathrm V}\).
  • 2.5: \(\boxed{\ce{2Fe(s) + O2(g) + 2H2O(l) -> 2Fe^2+(aq) + 4OH^-(aq)}}\).
  • 2.6: \(\boxed{K=6.2\times10^{56}}\), dimensionless, for the reaction written in 2.5.
  • 2.7: \(\boxed{m(\ce{Fe})=3.0\ \mathrm g}\).
  • 2.8: \(\boxed{E=1.19\ \mathrm V}\).

Solution

The chemical identities used throughout are:

Substance or species Name SMILES
\(\ce{CaSO4 * 2H2O}\) Calcium sulfate dihydrate (gypsum) [Ca+2].[O-]S(=O)(=O)[O-].O.O
\(\ce{CaSO4 * 1/2H2O}\) Calcium sulfate hemihydrate [Ca+2].[O-]S(=O)(=O)[O-].[Ca+2].[O-]S(=O)(=O)[O-].O
\(\ce{H2O}\) Water O
\(\ce{O2}\) Dioxygen O=O
\(\ce{Fe^2+}\) Iron(II) ion [Fe+2]
\(\ce{OH^-}\) Hydroxide ion [OH-]
\(\ce{Fe}\) Iron [Fe]
\(\ce{Pt}\) Platinum [Pt]

The hydrate SMILES specify components, not crystal packing or coordination geometry. The hemihydrate entry represents two formula units, giving one water molecule. Atom SMILES do not describe metallic lattices, and SMILES do not specify physical phases.

2.1

The dehydration reaction is

\[ \ce{CaSO4 * 2H2O(s) -> CaSO4 * 1/2H2O(s) + 3/2 H2O(g)}. \]

Using formation enthalpies,

\[ \begin{aligned} \Delta_{\mathrm r}H^\circ &=\sum\nu\Delta_{\mathrm f}H^\circ_{\text{products}} -\sum\nu\Delta_{\mathrm f}H^\circ_{\text{reactants}}\\ &=(-1575.0)+\frac32(-241.8)-(-2021.0)\\ &=+83.3\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Using conventional atomic molar masses,

\[ M(\ce{CaSO4 * 2H2O}) =40.08+32.06+4(16.00)+2[2(1.008)+16.00] =172.172\ \mathrm{g\,mol^{-1}}. \]

Therefore,

\[ n=\frac{1000\ \mathrm g}{172.172\ \mathrm{g\,mol^{-1}}} =5.808\ \mathrm{mol}, \]

and

\[ \Delta H^\circ =(5.808\ \mathrm{mol})(83.3\ \mathrm{kJ\,mol^{-1}}) =\boxed{+484\ \mathrm{kJ}}. \]

The positive enthalpy change means that heat is absorbed: the reaction is endothermic.

2.2

For the same reaction,

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=130.5+\frac32(188.6)-194.0\\ &=219.4\ \mathrm{J\,K^{-1}\,mol^{-1}}. \end{aligned} \]

At \(T=298.15\ \mathrm K\),

\[ \begin{aligned} \Delta_{\mathrm r}G^\circ &=\Delta_{\mathrm r}H^\circ-T\Delta_{\mathrm r}S^\circ\\ &=83300-(298.15)(219.4)\\ &=17885.9\ \mathrm{J\,mol^{-1}}. \end{aligned} \]

Hence the dimensionless equilibrium constant is

\[ K_p=\exp\left(-\frac{\Delta_{\mathrm r}G^\circ}{RT}\right) =\exp\left[-\frac{17885.9}{(8.314)(298.15)}\right] =7.35\times10^{-4}. \]

Both pure solids have unit activity. Treating the vapour as an ideal gas,

\[ K_p=\left(\frac{p(\ce{H2O})}{p^\circ}\right)^{3/2}, \qquad p^\circ=1.00\ \mathrm{bar}. \]

Thus,

\[ p(\ce{H2O}) =p^\circ K_p^{2/3} =(1.00\ \mathrm{bar})(7.35\times10^{-4})^{2/3} =\boxed{8.15\times10^{-3}\ \mathrm{bar}}. \]

2.3

When the equilibrium water vapour pressure equals the standard pressure,

\[ K_p=\left(\frac{1.00\ \mathrm{bar}}{1.00\ \mathrm{bar}}\right)^{3/2}=1. \]

Consequently, \(\Delta_{\mathrm r}G^\circ=-RT\ln K_p=0\). With temperature-independent enthalpy and entropy,

\[ 0=\Delta_{\mathrm r}H^\circ-T\Delta_{\mathrm r}S^\circ, \]

so

\[ T=\frac{\Delta_{\mathrm r}H^\circ}{\Delta_{\mathrm r}S^\circ} =\frac{83300\ \mathrm{J\,mol^{-1}}} {219.4\ \mathrm{J\,K^{-1}\,mol^{-1}}} =379.67\ \mathrm K. \]

Therefore,

\[ \boxed{T=379.7\ \mathrm K=106.5\ ^\circ\mathrm C}. \]

2.4

The oxygen reduction couple has the more positive standard reduction potential, so oxygen is reduced at the cathode:

\[ \ce{O2(g) + 2H2O(l) + 4e^- -> 4OH^-(aq)}, \qquad E^\circ_{\text{cathode}}=+0.40\ \mathrm V. \]

Iron is oxidized at the anode:

\[ \ce{Fe(s) -> Fe^2+(aq) + 2e^-}. \]

The tabulated reduction potential for the iron couple is \(-0.44\ \mathrm V\). Thus,

\[ E^\circ_{\text{cell}} =E^\circ_{\text{cathode}}-E^\circ_{\text{anode, reduction}} =0.40-(-0.44) =\boxed{0.84\ \mathrm V}. \]

Electrode potentials are not multiplied when half-reactions are multiplied to balance electrons.

2.5

Multiplying the iron oxidation half-reaction by two gives

\[ \ce{2Fe(s) -> 2Fe^2+(aq) + 4e^-}. \]

Adding the oxygen reduction half-reaction cancels four electrons:

\[ \boxed{\ce{2Fe(s) + O2(g) + 2H2O(l) -> 2Fe^2+(aq) + 4OH^-(aq)}}. \]

Both atoms and total charge are balanced. The positive standard cell voltage establishes this as the spontaneous discharge direction under standard conditions.

2.6

For the overall reaction as written in 2.5, \(n=4\) mol of electrons are transferred per mole of reaction. Combining

\[ \Delta_{\mathrm r}G^\circ=-nFE^\circ \quad\text{and}\quad \Delta_{\mathrm r}G^\circ=-RT\ln K \]

gives

\[ \log_{10}K =\frac{nFE^\circ}{RT\ln10} =\frac{nE^\circ}{0.05916\ \mathrm V}. \]

Therefore,

\[ \log_{10}K =\frac{4(0.84)}{0.05916} =56.795, \]

and

\[ \boxed{K=10^{56.795}\approx6.2\times10^{56}}. \]

This dimensionless value corresponds specifically to the equation with two iron atoms in 2.5.

2.7

The charge passed is

\[ q=It =(0.12\ \mathrm{C\,s^{-1}})(24\times3600\ \mathrm s) =10368\ \mathrm C. \]

Each mole of iron converted to iron(II) releases two moles of electrons, so

\[ n(\ce{Fe}) =\frac{q}{2F} =\frac{10368\ \mathrm C} {2(96485\ \mathrm{C\,mol^{-1}})} =0.0537286\ \mathrm{mol}. \]

Using \(M(\ce{Fe})=55.85\ \mathrm{g\,mol^{-1}}\),

\[ m(\ce{Fe}) =(0.0537286)(55.85)\ \mathrm g =3.0007\ \mathrm g. \]

To the precision of the current,

\[ \boxed{m(\ce{Fe})=3.0\ \mathrm g}. \]

2.8

At 25 °C, using \(\mathrm{p}K_{\mathrm w}=14.00\),

\[ \mathrm{pOH}=14.00-\mathrm{pH}=14.00-9.00=5.00, \]

so

\[ [\ce{OH^-}]=10^{-5.00}\ \mathrm{mol\,L^{-1}}. \]

For the overall reaction in 2.5, the reaction quotient is

\[ Q= \frac{a(\ce{Fe^2+})^2a(\ce{OH^-})^4} {p(\ce{O2})/p^\circ}. \]

Pure iron and liquid water have unit activity. Approximating dissolved-ion activities by their concentrations divided by \(c^\circ=1\ \mathrm{mol\,L^{-1}}\),

\[ Q=\frac{(0.015)^2(10^{-5})^4}{0.700} =3.2143\times10^{-24}. \]

The Nernst equation, with \(n=4\), gives

\[ \begin{aligned} E &=E^\circ-\frac{0.05916\ \mathrm V}{4}\log_{10}Q\\ &=0.84\ \mathrm V -\frac{0.05916\ \mathrm V}{4} \log_{10}(3.2143\times10^{-24})\\ &=0.84\ \mathrm V+0.34746\ \mathrm V\\ &=1.18746\ \mathrm V. \end{aligned} \]

Thus,

\[ \boxed{E=1.19\ \mathrm V}. \]

The solution was checked against the official solution.