What is the lattice energy of NaI? Use the given information below.

Heat of formation for NaI = -288.0 kJ/mol
Heat of sublimation for Na = 107.3 kJ/mol
Ionization energy for Na = 496.0 kJ/mol
Bond dissociation energy for I2 = 149 kJ/mol
Electron affinity of I = -295.0 kJ/mol
Heat of sublimation for I2(s) = 48.30 kJ/mol

dHf = HsubNa + IP + 1/2*BE + EA + 1/2*HsubI + Ecrystal

Solve for Ecryst, then U = lattice energy = -Ecrystal

dHf = heat formation
HsubNa = heat sublimation for Na
IP = ionization energy
BE = bond dissociation energy
EA = electron affinity
HsubI = heat sublimation I2

To calculate the lattice energy (ΔHlattice) of NaI, we can use the Born-Haber cycle. The Born-Haber cycle is a series of steps that represent the formation of an ionic compound from its constituent elements.

The steps involved in the Born-Haber cycle are:
1. Formation of gaseous Na+ ions from solid Na (Na(s) → Na(g))
2. Ionization of gaseous Na atoms (Na(g) → Na+(g) + e-)
3. Sublimation of solid I2 (I2(s) → I2(g))
4. Dissociation of gaseous I2 into gaseous I atoms (I2(g) → 2I(g))
5. Electron affinity of gaseous I atoms (I(g) + e- → I-(g))
6. Formation of solid NaI from gaseous Na+ and I- ions (Na+(g) + I-(g) → NaI(s))

The lattice energy (ΔHlattice) is the calculated energy change for the last step, which represents the addition of opposite charges to form the solid ionic compound.

To calculate ΔHlattice, we can use the following equation:
ΔHlattice = ΔHf(NaI) + ΔHsub(Na) + ΔHion(Na) + 1/2 ΔHdis(I2) + ΔHea(I) - ΔHsub(I2)

Given information:
ΔHf(NaI) = -288.0 kJ/mol (heat of formation for NaI)
ΔHsub(Na) = 107.3 kJ/mol (heat of sublimation for Na)
ΔHion(Na) = 496.0 kJ/mol (ionization energy for Na)
ΔHdis(I2) = 149.0 kJ/mol (bond dissociation energy for I2)
ΔHea(I) = -295.0 kJ/mol (electron affinity of I)
ΔHsub(I2) = 48.30 kJ/mol (heat of sublimation for I2)

Now, let's plug these values into the equation and calculate ΔHlattice:

ΔHlattice = -288.0 kJ/mol + 107.3 kJ/mol + 496.0 kJ/mol + 1/2 (149.0 kJ/mol) + (-295.0 kJ/mol) - 48.30 kJ/mol

Simplifying the equation:

ΔHlattice = -288.0 kJ/mol + 107.3 kJ/mol + 496.0 kJ/mol + 74.5 kJ/mol - 295.0 kJ/mol - 48.30 kJ/mol

ΔHlattice = 46.5 kJ/mol

Therefore, the lattice energy of NaI is 46.5 kJ/mol.

To calculate the lattice energy of NaI, we can use the Born-Haber cycle. The Born-Haber cycle is a series of hypothetical steps used to calculate the lattice energy of an ionic compound.

Step 1: Formation of gaseous sodium atoms (Na):

Na(s) → Na(g) - sublimation
∆H1 = +107.3 kJ/mol (heat of sublimation for Na)

Step 2: Ionization of gaseous sodium atoms:

Na(g) → Na+(g) + e-
∆H2 = +496.0 kJ/mol (ionization energy for Na)

Step 3: Dissociation of gaseous iodine (I2) molecules:

I2(s) → 2I(g)
∆H3 = +48.30 kJ/mol (heat of sublimation for I2)

Step 4: Electron affinity of iodine:

I(g) + e- → I-(g)
∆H4 = -295.0 kJ/mol (electron affinity of I)

Step 5: Formation of solid NaI:

Na+(g) + I-(g) → NaI(s)
∆H5 = -288.0 kJ/mol (heat of formation for NaI)

Now, let's calculate the lattice energy (ΔHlattice) using the Born-Haber cycle:

ΔHlattice = ∆H1 + ∆H2 + ∆H3 + ∆H4 + ∆H5

= (+107.3 kJ/mol) + (+496.0 kJ/mol) + (+48.30 kJ/mol) + (-295.0 kJ/mol) + (-288.0 kJ/mol)

= 68.60 kJ/mol

Therefore, the lattice energy of NaI is approximately 68.60 kJ/mol.