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The standard enthalpy of formation of Cr2O3 is to be determined. The following thermodynamic data are given at 27 degrees C: 4Cr(s) + 3O2(g) -> 2Cr2O3(s), delta_r_G° = -2093.4 kJ/mol S°(Cr, s) = 24 J/mol/K, S°(O2, g) = 205 J/mol/K, S°(Cr2O3, s) = 81 J/mol/K Calculate delta_H° (in kJ/mol) for the formation of one mole of Cr2O3.

1. -2258.1 kJ/mol
2. -1129.05 kJ/mol
3. -964.35 kJ/mol
4. None of these

Correct answer: -1129.05 kJ/mol

Solution

The reaction entropy change is 2(81) - 4(24) - 3(205) = -549 J/mol/K. Using delta_H = delta_G + T*delta_S gives delta_H = -2093.4 + 300*(-0.549) = -2258.1 kJ/mol for 2 mol Cr2O3. Per mole of Cr2O3: -2258.1/2 = -1129.05 kJ/mol.

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