Question

Given the lattice energy of NaBr = 736 kJ/mol, the ionization energy of Na = 496 kJ/mol and the electron affinity of Br = −325 kJ/mol, calculate the ΔH o for the reaction Na (g) + Br (g) = NaBr (s).

Answer 1

Answer : The value of $\Delta H^o$ for the reaction is, -565 kJ/mol

Explanation :  

The formation of sodium bromide is,

$Na(g)+Br(g)\overset{\Delta H^o}\rightarrow NaBr(s)$

$\Delta H^o$ = enthalpy of the reaction

The steps involved in the reaction are:

(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 bromine atoms into gaseous bromine ions.

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

$\Delta H_E$ = electron affinity energy of bromine = -325 kJ/mol

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

$Na^+(g)+Br^-(g)\overset{\Delta H_L}\rightarrow NaBr(s)$

$\Delta H_L$ = lattice energy of sodium bromide = -736 kJ/mol

To calculate the $\Delta H^o$ for the reaction, the equation used will be:

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

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

$\Delta H^o=496kJ/mole+(-325kJ/mole)+(-736kJ/mole)$

$\Delta H^o=-565kJ/mole$

Therefore, the value of $\Delta H^o$ for the reaction is, -565 kJ/mol