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Ionization enthalpy of sodium = 492 kJ/mole;
Electron gain enthalpy of chlorine = -349 kJ/mole;
Sublimation enthalpy of sodium = +108 kJ/mole ;
Bond dissociation enthalpy of chlorine = +243 kJ/mole ;
Change in enthalpy for the process, $Na (g) + Cl (g) \to {Na}^{+} (g) + {Cl}^{-} (g)$ will be

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+251 kJ

+372.5 kJ

+143 kJ

-241 kJ

answer is C.

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Detailed Solution

Change in enthalpy for the process, $Na (g) + Cl (g) \to {Na}^{+} (g) + {Cl}^{-} (g)$ will be +143 kJ/mole ;

$Na (g) \to {Na}^{+} (g) + e, ∆ H = + 492 kJ . . . . . (1) Cl (g) + e \to {Cl}^{-} (g), ∆ H = - 349 kJ . . . . . (2) Adding (1) and (2) Na (g) + Cl (g) \to {Na}^{+} (g) + {Cl}^{-} (g), ∆ H = + 143 kJ;$

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