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In a Born-Haber cycle for compound A2B3(s): - A2B3(s) dissolves to give 2A³+(aq) + 3B²-(aq): delta_H_solution = -100 J/mol - Lattice energy of A2B3(s) = +90 J/mol - Hydration enthalpy of A³+(g) = -50 J/mol Calculate the hydration enthalpy of B²-(g).

1. +30 J/mol
2. -30 J/mol
3. -140 J/mol
4. +140 J/mol

Correct answer: +30 J/mol

Solution

delta_H_solution = -Lattice Energy + 2*HE(A³+) + 3*HE(B²-). -100 = -90 + 2*(-50) + 3*HE(B²-). -100 = -90 - 100 + 3*HE(B²-). -100 = -190 + 3*HE(B²-). 3*HE(B²-) = 90. HE(B²-) = +30 J/mol.

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