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 1 of 5: Enthalpy of dehydration
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ΔH∘=−1575.0+32(−241.8)−(−2021.0)=+83.3 kJ mol−1;1.00 kg⇒5.808 mol⇒ΔH=+484 kJ\Delta H^\circ = -1575.0 + \tfrac{3}{2}(-241.8) - (-2021.0) = +83.3\ \text{kJ mol}^{-1};\quad 1.00\ \text{kg}\Rightarrow 5.808\ \text{mol}\Rightarrow \Delta H = +484\ \text{kJ}
Analysis

ΔH∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)=−1575.0−362.7+2021.0=+83.3\Delta H^\circ = \sum\Delta H_f^\circ(\text{products}) - \sum\Delta H_f^\circ(\text{reactants}) = -1575.0 - 362.7 + 2021.0 = +83.3 kJ per mole of gypsum. Since M(CaSOX4 ⋅ 2 HX2O)=172.18M(\ce{CaSO4.2H2O}) = 172.18 g mol−1^{-1}, 1.00 kg is 5.808 mol and the reaction is endothermic, absorbing 484 kJ.