Answer : The value of
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)](https://tex.z-dn.net/?f=Na%28g%29%2BBr%28g%29%5Coverset%7B%5CDelta%20H%5Eo%7D%5Crightarrow%20NaBr%28s%29)
= 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)](https://tex.z-dn.net/?f=Na%28g%29%5Coverset%7B%5CDelta%20H_I%7D%5Crightarrow%20Na%5E%7B%2B1%7D%28g%29)
= 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)](https://tex.z-dn.net/?f=Br%28g%29%5Coverset%7B%5CDelta%20H_E%7D%5Crightarrow%20Br%5E-%28g%29)
= 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)](https://tex.z-dn.net/?f=Na%5E%2B%28g%29%2BBr%5E-%28g%29%5Coverset%7B%5CDelta%20H_L%7D%5Crightarrow%20NaBr%28s%29)
= lattice energy of sodium bromide = -736 kJ/mol
To calculate the
for the reaction, the equation used will be:
![\Delta H^o=\Delta H_I+\Delta H_E+\Delta H_L](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5CDelta%20H_I%2B%5CDelta%20H_E%2B%5CDelta%20H_L)
Now put all the given values in this equation, we get:
![\Delta H^o=496kJ/mole+(-325kJ/mole)+(-736kJ/mole)](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D496kJ%2Fmole%2B%28-325kJ%2Fmole%29%2B%28-736kJ%2Fmole%29)
![\Delta H^o=-565kJ/mole](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D-565kJ%2Fmole)
Therefore, the value of
for the reaction is, -565 kJ/mol