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11. Bonding & Molecular Structure - Part 4 of 4

11. Bonding & Molecular Structure - Part 4 of 4

11. Bonding & Molecular Structure / Born Haber Cycle / Problem 10

The lattice energy of CuCl₂(s) is 2824 kJ/mol. Use the Born-Haber cycle to determine the ΔH°_f for CuCl₂(s) from its elements. Compare this with the ΔH°_f for CuCl(s) (–137.2 kJ/mol) and determine which compound is more stable.

ANSWERS OPTIONS

A

ΔH°_f of CuCl₂(s) = 139.1 kJ/mol; CuCl(s) is more stable.

B

ΔH°_f of CuCl₂(s) = –195.7 kJ/mol; CuCl₂(s) is more stable.

C

ΔH°_f of CuCl₂(s) = 103.2 kJ/mol; CuCl(s) is more stable.

D

ΔH°_f of CuCl₂(s) = –237.8 kJ/mol; CuCl₂(s) is more stable.

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