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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 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.
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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.