Chemistry Labs

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, CaSOX4 ⋅ 2 HX2O\ce{CaSO4.2H2O}, is added to improve hardening. Elevated temperatures may form unwanted hemihydrate: CaSOX4 ⋅ 2 HX2O(s)→CaSOX4 ⋅ 12 HX2O(s)+32 HX2O(g)\ce{CaSO4.2H2O(s) -> CaSO4.1/2H2O(s) + 3/2 H2O(g)}. Thermodynamic data at 25 ∘C25\ ^\circ\mathrm{C}, 1.00 bar — CaSOX4 ⋅ 2 HX2O(s)\ce{CaSO4.2H2O(s)}: ΔHf∘=−2021.0\Delta H_f^\circ = -2021.0 kJ mol−1^{-1}, S∘=194.0S^\circ = 194.0 J K−1^{-1} mol−1^{-1}; CaSOX4 ⋅ 12 HX2O(s)\ce{CaSO4.1/2H2O(s)}: −1575.0-1575.0 kJ mol−1^{-1}, 130.5130.5 J K−1^{-1} mol−1^{-1}; HX2O(g)\ce{H2O(g)}: −241.8-241.8 kJ mol−1^{-1}, 188.6188.6 J K−1^{-1} mol−1^{-1}. R=8.314R = 8.314 J mol−1^{-1} K−1^{-1}; 0 ∘C=273.150\ ^\circ\mathrm{C} = 273.15 K. (2.1) Calculate ΔH∘\Delta H^\circ for the transformation of 1.00 kg of CaSOX4 ⋅ 2 HX2O(s)\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 CaSOX4 ⋅ 2 HX2O(s)\ce{CaSO4.2H2O(s)}, CaSOX4 ⋅ 12 HX2O(s)\ce{CaSO4.1/2H2O(s)} and HX2O(g)\ce{H2O(g)} at 25 ∘C25\ ^\circ\mathrm{C}. (2.3) Calculate the temperature at which the equilibrium water-vapour pressure is 1.00 bar, assuming ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ are temperature-independent. Corrosion of iron involves the electrode reactions Fe(s)→FeX2+(aq)+2 eX−\ce{Fe(s) -> Fe^{2+}(aq) + 2e-} and OX2(g)+2 HX2O(l)+4 eX−→4 OHX−(aq)\ce{O2(g) + 2H2O(l) + 4e- -> 4OH-(aq)}. Consider the cell Fe(s)∣FeX2+(aq)∣∣OHX−(aq), OX2(g)∣Pt(s)\ce{Fe(s)|Fe^{2+}(aq)||OH-(aq),O2(g)|Pt(s)} at 25 ∘C25\ ^\circ\mathrm{C} with E∘(FeX2+/Fe)=−0.44E^\circ(\ce{Fe^{2+}/Fe}) = -0.44 V and E∘(OX2/OHX−)=+0.40E^\circ(\ce{O2/OH-}) = +0.40 V; RTln⁡10/F=0.05916RT\ln10/F = 0.05916 V, F=96485F = 96485 C mol−1^{-1}. (2.4) Calculate the standard EMF at 25 ∘C25\ ^\circ\mathrm{C}. (2.5) Write the overall discharge reaction. (2.6) Calculate the equilibrium constant at 25 ∘C25\ ^\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 FeX2+\ce{Fe^{2+}}. (2.8) Calculate EE for [FeX2+]=0.015[\ce{Fe^{2+}}] = 0.015 M, pH of the right half-cell 9.00, p(OX2)=0.700p(\ce{O2}) = 0.700 bar.
Step 3 of 5: Temperature for 1 bar vapour
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p(HX2O)=1.00 bar⇒ΔG∘=0⇒T=ΔH∘ΔS∘=83300219.4=380 K (107 ∘C)p(\ce{H2O}) = 1.00\ \text{bar} \Rightarrow \Delta G^\circ = 0 \Rightarrow T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{83300}{219.4} = 380\ \text{K}\ (107\ ^\circ\mathrm{C})
Analysis

At p(HX2O)=1.00p(\ce{H2O}) = 1.00 bar, K=1K = 1 and ΔG∘=0\Delta G^\circ = 0, so T=ΔH∘/ΔS∘=380T = \Delta H^\circ/\Delta S^\circ = 380 K ≈ 107 ∘C107\ ^\circ\mathrm{C} — above this temperature the hemihydrate forms spontaneously.