Question

1.16 Ti(a) transforms to Ti(B) at 882°C (1155 K), with a heat of transforma- tion equal to 830 cal/mole (3,473 J/mol). Calculate the heat of reaction when Ti(B) is oxidized by pure oxygen gas to form T10, at 1400°C (1673 K). The First Law of Thermodynamics 27 Given: Ti(a) + Oy - T102; 298 -225.5 kcal/mole (-943.49 kJ/mol). AHO Cp, Ti(a) 5.28 + 2.4 x 10-3 T cal/deg/mole (22.09 - 10.04 * 10-2 T J/K/mol). Cp Ti(B) = 6.91 cal/deg/mole (28.91 J/K/mol). Cp, TiO2 18.0 + 0.28 x 10-31 - 4.35 x 105 1-2 cal/deg/mole (75.31 + 1.17 x 10- 1. - 18.2 x 10 J/K/mol). = 7.16 + 1.0 x 10-31 - 0.4 x 105 7-2 cal/deg/mole (29.98 + 4.184 x 10-7 - 1.87 * 108 7-2 J/K/mol). Cp,02

Ti(a) transforms toTi(3) at 882°C (1155 K), with a heat of transforma-
tion equal to 830 cal/mole (3,473 J/mol). Calculate the heat of reaction
when Ti(3) is oxidized by pure oxygen gas to form TiO^ at 1400°C (1673 K).

Answer 1

Given: Ti(a) + Oz дн? 298 Ti0z; -225.5 kcal/mole (-943.49 kJ/mol). = Cp, Ti(a) 5.28 + 2.4 x 10-3 I cal/deg/mole (22.09 + 10.04 X 10 10-3 T J/K/mol). Cp,Ti(B) 6.91 cal/deg/mole (28.91 J/K/mol). Cp, TiO2 - 3 18.0 + 0.28 x 10 T - 4.35 x 1-2 cal/deg/mole -3 (75, 31 + 1.17 X 10 T- 18.2 X 1097-2 J/K/mol). 105 Cp,02 7.16 + 1.0 x 10-3 T - 0.4 x 105,-2 cal/deg/mole (29.96 + 4,184 X 10 1-3 - 1.67 x 105 J/K/mol). -2 T T