The ionization enthalpy of Na+ formation from Na(g) 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 $-$728.4 kJ mol$-$1. The energy for the formation of NaBr ionic solid is ($-$) ____________ $\times$ 10$-$1 kJ mol$-$1.

Question

The ionization enthalpy of Na⁺ formation from Na_(g) 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 $-$728.4 kJ mol^$-$1. The energy for the formation of NaBr ionic solid is ($-$) ____________ $\times$ 10^$-$1 kJ mol^$-$1.

Correct Answer

5576

Solution

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