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Use the Born Haber cycle (see equations and enthalpy values below) to determine the lattice energy...

Use the Born Haber cycle (see equations and enthalpy values below) to determine the lattice energy for BeI2 (s) (∆H LE (BeI2 (s))= ?) Show your work. Box your final answer.

A. Be(g)→Be1+ (g) + 1 e–∆H = + 899.5kJ

B. Be1+ (g) →Be2+ (g) + 1 e–∆H = +1757 kJ

C. Be(s)→Be(g)∆H= +302kJ

D. I2(s)→I2(g)∆H= + 62.4kJ

E. I(g) + e–→I–(g)∆H= –295kJ

F. I2(g)→2I(g)∆H= + 151 kJ

G. Be(s) + I2(s) →BeI2(s)∆H= –208 kJ

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Answer #1

Be (s) + I2 (s) --------------> BeI2

According Born Haber cycle OR Hess law,

Enthalpy change of overall reaction = sum of enthalpy changes of all individual reactions

deltaH = deltaH0sub(Be) + deltaH0ie1 + deltaH0ie2 + deltaH0Sub(I2) + deltaH0b + 2 x deltaH0ea + deltaH0lat

- 208 = + 302 + 899.5 + 1757 + 62.4 + 151 + 2 ( - 295 ) + deltaH0lat

- 208 = 2581.9 + deltaH0lat

deltaH0lat = - 208 - 2581.9

deltaH0lat = - 2789.9 kJ

Therefore,

Lattice enthalpy of BeI2 = - 2789.9 kJ


answered by: ANURANJAN SARSAM
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