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Question

The ionization enthalpy of Na+ formation from Na(g) is 495.8 kJ mol1 while the electron gain enthalpy of Br is 325.0 kJ mol1. Given the lattice enthalpy of NaBr is 728.4 kJ mol1. The energy for the formation of NaBr ionic solid is (–) _____ ×101 kJ mol1 .

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Solution

Na(s)Na+(g)+e ΔH=495.8 kJ mol1

12Br2(l)+eBr(g) ΔH=325.0 kJ mol1

Na+(g)+Br(g)NaBr(s) ΔH=728.4 kJ mol1

On adding the above three equations,
Na(s)+12Br2(l)NaBr(s) ΔHf=?
ΔHf=495.8325.0728.4=557.6 kJ mol1ΔHf=5576×101 kJ mol1


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