Answer : The value of [tex]\Delta H^o[/tex] for the reaction is, -565 kJ/mol
Explanation :
The formation of sodium bromide is,
[tex]Na(g)+Br(g)\overset{\Delta H^o}\rightarrow NaBr(s)[/tex]
[tex]\Delta H^o[/tex] = enthalpy of the reaction
The steps involved in the reaction are:
(1) Conversion of gaseous sodium atoms into gaseous sodium ions.
[tex]Na(g)\overset{\Delta H_I}\rightarrow Na^{+1}(g)[/tex]
[tex]\Delta H_I[/tex] = ionization energy of sodium = 496 kJ/mol
(2) Conversion of gaseous bromine atoms into gaseous bromine ions.
[tex]Br(g)\overset{\Delta H_E}\rightarrow Br^-(g)[/tex]
[tex]\Delta H_E[/tex] = electron affinity energy of bromine = -325 kJ/mol
(3) Conversion of gaseous cations and gaseous anion into solid sodium bromide.
[tex]Na^+(g)+Br^-(g)\overset{\Delta H_L}\rightarrow NaBr(s)[/tex]
[tex]\Delta H_L[/tex] = lattice energy of sodium bromide = -736 kJ/mol
To calculate the [tex]\Delta H^o[/tex] for the reaction, the equation used will be:
[tex]\Delta H^o=\Delta H_I+\Delta H_E+\Delta H_L[/tex]
Now put all the given values in this equation, we get:
[tex]\Delta H^o=496kJ/mole+(-325kJ/mole)+(-736kJ/mole)[/tex]
[tex]\Delta H^o=-565kJ/mole[/tex]
Therefore, the value of [tex]\Delta H^o[/tex] for the reaction is, -565 kJ/mol
