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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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Born Haber Cycle concept. You can view video lessons to learn Born Haber Cycle. Or if you need more Born Haber Cycle practice, you can also practice Born Haber Cycle practice problems.

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Based on our data, we think this problem is relevant for Professor Guerra's class at RUTGERS.

What textbook is this problem found in?

Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl 2nd Edition practice problems.