Química·Murcia·2016·OrdinariaEjercicio4Opción A2 puntosCalcule a partir de qué temperatura la siguiente reacción será espontánea:2 FeX2OX3(s)+3 C(s)→4 Fe(s)+3 COX2(g)\ce{2Fe2O3(s) + 3C(s) -> 4Fe(s) + 3CO2(g)}2FeX2OX3(s)+3C(s)4Fe(s)+3COX2(g)DatosΔHf0(COX2(g))=−393,5 kJ mol−1\Delta H_f^0(\ce{CO2(g)}) = -393{,}5\,\text{kJ mol}^{-1}ΔHf0(COX2(g))=−393,5kJ mol−1ΔHf0(FeX2OX3(s))=−824 kJ/mol\Delta H_f^0(\ce{Fe2O3(s)}) = -824\,\text{kJ/mol}ΔHf0(FeX2OX3(s))=−824kJ/molS0(COX2(g))=213,7 J mol−1 K−1S^0(\ce{CO2(g)}) = 213{,}7\,\text{J mol}^{-1}\text{ K}^{-1}S0(COX2(g))=213,7J mol−1 K−1S0(Fe(s))=27,2 J mol−1 K−1S^0(\ce{Fe(s)}) = 27{,}2\,\text{J mol}^{-1}\text{ K}^{-1}S0(Fe(s))=27,2J mol−1 K−1S0(FeX2OX3(s))=87,4 J mol−1 K−1S^0(\ce{Fe2O3(s)}) = 87{,}4\,\text{J mol}^{-1}\text{ K}^{-1}S0(FeX2OX3(s))=87,4J mol−1 K−1S0(C(s))=5,7 J mol−1 K−1S^0(\ce{C(s)}) = 5{,}7\,\text{J mol}^{-1}\text{ K}^{-1}S0(C(s))=5,7J mol−1 K−1