Problem: Consider the following:Li(s) + 1/2I2 (s) → LiI(s)      ΔE = -272 kJ/molLiI(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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Given: Li(s) + 1/2I2 (s) → LiI(s)      ΔE = 272 kJ/mol


Born-Haber Cycle for LiI:

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Problem Details

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.