Abstract
The enthalpy of the process 2RbBr(c)+TaBr 4(c) = Rb 2TaBr 6(c) has been determined by solution calorimetry. The standard enthalpy of formation of Rb 2TaBr 6 was found to be −316.6 kcal mol −1. Based upon a Kapustinskii calculation of lattice energy, the standard enthalpy of formation of TaBr 2− 6 (g) was found to be −194.4 kcal mol −1. The double bromide affinity of TaBr 4(g) and the lattice energy of Rb 2TaBr 6(c) were compared with values of corresponding compounds of Ti, Zr, Hf, W, Re and Pt.
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