Problem 2

Bridge between Denmark and Sweden. The combined tunnel and bridge between Copenhagen and Malmö uses concrete and steel; this problem concerns reactions in their production and degradation. Concrete is made from cement, water, sand and stones. Cement consists mainly of calcium silicates and aluminates; in the final step some gypsum,

\ce{CaSO4.2H2O}

, is added to improve hardening. Elevated temperatures may form unwanted hemihydrate:

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

. Thermodynamic data at

25\ ^\circ\mathrm{C}

, 1.00 bar —

\ce{CaSO4.2H2O(s)}

\Delta H_f^\circ = -2021.0

kJ mol

^{-1}

S^\circ = 194.0

J K

^{-1}

mol

^{-1}

;

\ce{CaSO4.1/2H2O(s)}

-1575.0

kJ mol

^{-1}

130.5

J K

^{-1}

mol

^{-1}

;

\ce{H2O(g)}

-241.8

kJ mol

^{-1}

188.6

J K

^{-1}

mol

^{-1}

R = 8.314

J mol

^{-1}

^{-1}

;

0\ ^\circ\mathrm{C} = 273.15

K. (2.1) Calculate

\Delta H^\circ

for the transformation of 1.00 kg of

\ce{CaSO4.2H2O(s)}

into hemihydrate; is it endothermic or exothermic? (2.2) Calculate the equilibrium water-vapour pressure (bar) in a closed vessel containing

\ce{CaSO4.2H2O(s)}

\ce{CaSO4.1/2H2O(s)}

and

\ce{H2O(g)}

25\ ^\circ\mathrm{C}

. (2.3) Calculate the temperature at which the equilibrium water-vapour pressure is 1.00 bar, assuming

\Delta H^\circ

and

\Delta S^\circ

are temperature-independent. Corrosion of iron involves the electrode reactions

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

and

\ce{O2(g) + 2H2O(l) + 4e- -> 4OH-(aq)}

. Consider the cell

\ce{Fe(s)|Fe^{2+}(aq)||OH-(aq),O2(g)|Pt(s)}

25\ ^\circ\mathrm{C}

with

E^\circ(\ce{Fe^{2+}/Fe}) = -0.44

V and

E^\circ(\ce{O2/OH-}) = +0.40

RT\ln10/F = 0.05916

F = 96485

C mol

^{-1}

. (2.4) Calculate the standard EMF at

25\ ^\circ\mathrm{C}

. (2.5) Write the overall discharge reaction. (2.6) Calculate the equilibrium constant at

25\ ^\circ\mathrm{C}

. (2.7) The cell runs 24 h at a constant 0.12 A under standard conditions; calculate the mass of Fe converted to

\ce{Fe^{2+}}

. (2.8) Calculate

E

for

[\ce{Fe^{2+}}] = 0.015

M, pH of the right half-cell 9.00,

p(\ce{O2}) = 0.700

bar.

Step 2 of 5: Equilibrium vapour pressure

← Previous Next →

\Delta S^\circ = 130.5 + \tfrac{3}{2}(188.6) - 194.0 = 219.4\ \text{J K}^{-1}\text{mol}^{-1};\quad K = p(\ce{H2O})^{3/2} = e^{-\Delta G^\circ/RT} = 7.35\times10^{-4}\Rightarrow p(\ce{H2O}) = 8.15\times10^{-3}\ \text{bar}

Analysis

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 83300 - 298.15\times219.4 = 17886$ J mol $^{-1}$ . With both solids in standard state, $K = p(\ce{H2O})^{3/2}$ , so $\ln K = -\Delta G^\circ/RT$ gives $K = 7.35\times10^{-4}$ and $p(\ce{H2O}) = K^{2/3} = 8.15\times10^{-3}$ bar.