Given: Li(s) + 1/2I2 (s) → LiI(s) ΔE = –272 kJ/mol
Born-Haber Cycle for LiI:
Consider the following:
Li(s) + 1/2I2 (s) → LiI(s) ΔE = -272 kJ/mol
LiI(s) has a lattice energy of -753 kJ/mol. The ionization energy of Li(g) is 520. kJ/mol, the bond energy of I2(g) is 151kJ/ mol, and the electron affinity of I(g) is -295 kJ/mol. Use these data to determine the energy of sublimation of Li(s).
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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.