Question

How does the bond energy for the formation of NaCl(s) take place?

Answer 1

In the above question, we have to find out how the bond energy for the formation of NaCl takes place. We have to write a series of balanced chemical equations along with heat released in each step to get the desired lattice enthalpy.

Complete step by step solution:
Sodium is metal and chlorine is a diatomic gas in room temperature and pressure but in NaCl, sodium and chlorine ions are present. So, we have to basically find out the following reaction:
${\text{Na(s)}}\xrightarrow{{{\text{sublimation}}}}{\text{Na(l)}}\xrightarrow{{{\text{1st ionization}}}}{\text{N}}{{\text{a}}^{\text{ + }}}{\text{(g)}}$
$\dfrac{{\text{1}}}{{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}}\xrightarrow{{\dfrac{{\text{1}}}{{\text{2}}}{\text{ bond breaking}}}}{\text{Cl(g)}}\xrightarrow{{{\text{1st electron affinity}}}}{\text{C}}{{\text{l}}^{\text{ - }}}{\text{(g)}}$
The standard heat of formation of NaCl is given as:
${\text{NaCl(s)}} \to {\text{Na(s) + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}{{\Delta H}}_{{\text{f,NaCl}}}^{\text{0}}{\text{ = 411}}$ kJ …………..(1)
The enthalpy change during sublimation of sodium is given by:
${\text{Na(s)}} \to {{Na(g),\Delta }}{{\text{H}}_{{\text{sub,Na}}}}{\text{ = 107}}$ kJ …………..(2)
The first ionization energy of Na, that is, removal of one electron from sodium atom is given by:
${\text{Na(g)}} \to {\text{N}}{{\text{a}}^{\text{ + }}}{\text{(g) + }}{{\text{e}}^{\text{ - }}}{\text{,I}}{{\text{E}}_{{\text{1,Na(g)}}}}{\text{ = 502}}$ KJ…………..(3)
Now let us look at the enthalpy change when chlorine molecule breaks down to chlorine atom which is given by:
$\dfrac{{\text{1}}}{{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}{\text{(g)}} \to {\text{Cl(g),}}\dfrac{{\text{1}}}{{\text{2}}}{{\Delta }}{{\text{H}}_{{\text{bond,C}}{{\text{l}}_{\text{2}}}{\text{(g)}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{{ \times 242 = 121}}$ kJ…………..(4)
The enthalpy change when chlorine gains an electron is given by:
${\text{Cl(g) + }}{{\text{e}}^{\text{ - }}} \to {\text{C}}{{\text{l}}^{\text{ - }}}{\text{(g),E}}{{\text{A}}_{{\text{1,Cl(g)}}}}{\text{ = - 355}}$ kJ…………..(5)
We have to find the enthalpy change ( ${{\Delta }}{{\text{H}}_{{\text{lattice}}}}$ ) during-
${\text{N}}{{\text{a}}^{\text{ + }}}{\text{(g) + C}}{{\text{l}}^{\text{ - }}}{\text{(g)}} \to {\text{NaCl(s)}}$ …………..(6)
If we add equation 1, 2, 3,4 and 5, we will get:
${\text{NaCl(s) + Na(s) + Na(g) + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}{\text{(g) + Cl(g) + }}{{\text{e}}^{\text{ - }}} \to {\text{Na(s) + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}(g) + {\text{Na(g)}} + {\text{N}}{{\text{a}}^{\text{ + }}}{\text{(g) + }}{{\text{e}}^{\text{ - }}} + {\text{Cl(g)}} + {\text{C}}{{\text{l}}^{\text{ - }}}{\text{(g)}}$
Simplifying the above equation, we get:
${\text{NaCl(s)}} \to {\text{N}}{{\text{a}}^{\text{ + }}}{\text{(g) + C}}{{\text{l}}^{\text{ - }}}{\text{(g)}}$
${{\Delta H = 411 + 107 + 502 + 121 - 355 = 786}}$ kJ
Hence, for the reaction
${\text{N}}{{\text{a}}^{\text{ + }}}{\text{(g) + C}}{{\text{l}}^{\text{ - }}}{\text{(g)}} \to {\text{NaCl(s)}}$
${{\Delta }}{{\text{H}}_{{\text{lattice}}}}{{ = - \Delta H = - 786}}$ kJ
Hence, the bond energy for the formation of NaCl(s) is ${\text{ - 786}}$ kJ.

Note:
The lattice energy of a crystalline solid is a measure of the energy released when ions are combined to make a compound. It is a measure of the cohesive forces that bind ions. Lattice energy is relevant to many practical properties which includes solubility, hardness, and volatility.