Question

In the Born-Haber cycle for the formation of solid common salt$(NaCl)$, the largest contribution comes from
A.The low ionization potential of $Na$
B.The high electron affinity of $Cl$
C.The low $\Delta {H_{vap}}$of $Na(s)$
D.The lattice energy

Answer 1

As we know that the lattice enthalpy is related to the formation of crystalline solid from its ions. The Born-Haber cycle is a stepwise process to calculate lattice enthalpy. Now, try to answer this question accordingly.

Complete step by step answer:
Now we discuss about the concept of lattice enthalpy as,
The lattice enthalpy of a crystalline solid may be a measure of the energy released when the ions are combined to sort this compound.
The lattice enthalpy is indirectly determined by the utilization of the Born-Haber cycle. This procedure is based on Hess’s law. Now let’s find the lattice enthalpy of sodium chloride by using Born-Haber cycle. The formation reaction, which is defined from the elemental states at 25 degrees and 1 atm.
Solid sodium atom sublimes to gaseous atom by absorbing heat energy.
$Na(s) \to Na(g)$, $\Delta {H_{sub,Na(s)}}$= +108 KJ
Gaseous sodium atoms absorb the ionization energy to release one electron and ort gaseous sodium ions. Diatomic gaseous chlorine breaks into two individual atoms by absorbing bond energy, such each chlorine atom absorbs half the bond energy of chlorine molecules.
$\dfrac{1}{2}C{l_{2(g)}} \to Cl(g),$ $\dfrac{1}{2}\Delta {H_{dissociation,C{l_2}}}_{(g)} = \dfrac{1}{2} \times 244 KJ = + 122KJ$
Gaseous sodium ion and gaseous chloride ion combine to sort a solid sodium-chloride molecule and release energy equivalent to lattice energy.

$N{a^ + }(g) + C{l^ - }(g) \to NaCl(s),$ $\Delta {H_{lattice}} = ?$
Summation of enthalpy of all processes from all steps given the net enthalpy of formation of solid crystalline sodium chloride from sodium and chlorine in their standard conditions of solid and gas respectively. For one mole of NaCl

$Na(s) + \dfrac{1}{2}C{l_{2(g)}}\xrightarrow{{\Delta {H^ \circ }_f}}NaCl(s)$
$\Delta {H_{sub}} = + 108kJ \downarrow$ , $\downarrow \Delta {H_{diss}} = + 122KJ$
$Na(g)$ $Cl(g)$ \uparrow $$$$U = - 788KJ
$IE = + 496KJ \downarrow$ , $\downarrow EA = - 349KJ$
$N{a^ + }$ $C{l^ - }$
$\Delta {H_{{f^ \circ }}} = \Delta {H_{sub}} + IE + \Delta {H_{diss}} + EA + U$
$\Delta {H_{f \circ }} = 108 + 496 + 122 - 349 - 788 = - 411KJ/mole$

From the above details we can summarise as,
The low ionization potential of Na, IE = 496KJ
The electron affinity of Cl, EA = -349 KJ
The low $\Delta {H_{vap}}$of Na(s), $\Delta {H_{sub}} = 108kJ$
the lattice energy, U = -788 KJ
From the options, the most important contribution comes from the lattice energy is that the Born-Haber cycle.

So, the correct answer is Option D.

Note:
Now we discuss Hess's law statement as,‘‘The enthalpy change of a reaction is that the same constant volume and pressure whether it takes place in single or multiple steps long because the initial reactant and the final products remain the same’’.