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The ionization enthalpy of Na^+ formatio...

The ionization enthalpy of `Na^+` formation from Nad is `495.8 kJ mol^(-1)`, while the electron gain enthalpy of Bris `-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 `(-)"______" xx 10^(-1)kJ "mol"^(-1)`

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The correct Answer is:
5576
`Na(s) to Na^(+) (s) DeltaH = 495.8`
`1/2 Br_2(l) + e^(-) to Br^(-) (g) DeltaH = 325`
`Na^(+) (g)+ Br^(-) (g) to NaBr(s) DeltaH = -728.4`
`Na(s) + 1/2 Br_2 (l) to NaBr(s) DeltaH = ?`
`DeltaH = 495.8 - 325 - 728.4`
` - 557.6 kJ = - 5576 xx 10^(-1) kJ`
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