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Q.

The ionization enthalpy of Na+ formation from Nag is 495.8  kJ  mol−1 , while the electron gain enthalpy of Br is -325.0 kJ mol-1. Given the lattice enthalpy of NaBr is −728.4  kJ mol−1 . The energy for the formation of  NaBr ionic solid is −_______×10−1  kJ mol−1 .

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answer is -5576.

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Detailed Solution

=495.8+(-32.5)+(-728.4)=-557.6=-5576×10-1KJ/mol.Note : The above calculation is not for ΔHformation but for ΔHReactionBut on the basis of given data it is the best ans
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