Question

Given the lattice energy of NaCl = 787 kJ/mol, the ionization energy of Na = 496 kJ/mol and the electron affinity of Cl = −349 kJ/mol, calculate the ΔH o for the reaction: Na(g) + Cl(g) → NaCl(s)

Answer 1

Answer :  The value of $\Delta H_f^o$ for the reaction is, -640 KJ/mole

Explanation :  

The steps involved in the formation of $NaCl$ :

(1) Conversion of gaseous sodium atoms into gaseous sodium ions.

$Na(g)\overset{\Delta H_I}\rightarrow Na^{+1}(g)$

$\Delta H_I$ = ionization energy of sodium = 496 kJ/mol

(2) Conversion of gaseous chlorine atoms into gaseous chlorine ions.

$Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)$

$\Delta H_E$ = electron affinity energy of chlorine  = -349 kJ/mol

(3) Conversion of gaseous cations and gaseous anion into solid sodium chloride.

$Na^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow NaCl(s)$

$\Delta H_L$ = lattice energy of sodium chloride  (always negative) = -787 kJ/mol

To calculate the overall energy the equation used will be:

$\Delta H_f^o=\Delta H_I+\Delta H_E+\Delta H_L$

Now put all the given values in this equation, we get:

$\Delta H_f^o=496KJ/mole+(-349KJ/mole)+(-787KJ/mole)$

$\Delta H_f^o=-640KJ/mole$

Therefore, the value of $\Delta H_f^o$ for the reaction is, -640 KJ/mole