The ionization enthalpy of $Na^{+}$ formation from $Na_{(g)}$ is $495.8 \ kJ \ mol^{-1},$ while the electron gain enthalpy of $Br_{(g)}$ is $-325.0 \ kJ \ mol^{-1}$. Given the lattice enthalpy of $NaBr_{(s)}$ is $-728.4 \ kJ \ mol^{-1}$. The energy for the reaction $Na_{(g)} + Br_{(g)} \rightarrow NaBr_{(s)}$ is $(-) \dots \dots \dots \dots \dots \times 10^{-1} \ kJ \ mol^{-1}$.

• A
  $5581$
• B
  $4856$
• C
  $5596$
• D
  $5576$