7. (a) Determine the crystal lattice energy for LiF(s) given that the ∆Hsubl for Li = 155kJ/mol,

∆H0f for F(g) = 78.99kJ/mol, IE1 of Li = 520kJ/mol, EA for Fluorine = -322kJ/mol and the ∆H0f LiF(s) = -589.5kJ/mol
(b) For the following reactions:
NH4NO3(s) + H2O(l) �¨ NH4+(aq) + NO3-(aq). Given that
∆H0f of NO3- = ∆H0f = 205.0kJ/mol, and S0 = 146J/mol .k,
∆H0 f of NH4+ = -132.57kJ/mol, and S0 = 113.4 J/mol .k,
∆H0 f of NH4NO3 = -365.56kJ/mol, and S0 = 151.08 J/mol .k

DHsubl + IP + DHF + EA(negative) + U = -589.5

Solve for U.

(a) To determine the crystal lattice energy for LiF(s), we can use the Born-Haber cycle which relates the lattice energy to various enthalpy changes.

The steps involved in the Born-Haber cycle are:
1. Formation of LiF from its elements (Li and F)
2. Sublimation of Li(s) to form Li(g)
3. Dissociation of F2(g) to form F(g)
4. Ionization of Li(g) to form Li+(g)
5. Electron affinity of F(g) to form F-(g)
6. Formation of LiF(s) from Li+(g) and F-(g)

We are given the following enthalpy changes:
∆Hsubl for Li = 155 kJ/mol
∆H0f for F(g) = 78.99 kJ/mol
IE1 of Li = 520 kJ/mol (ionization energy)
EA for Fluorine = -322 kJ/mol (electron affinity)
∆H0f LiF(s) = -589.5 kJ/mol

Now let's determine the crystal lattice energy using the Born-Haber cycle:
1. ΔH0f(LiF) = ΔH0f(Li) + 0.5 * ΔHf(F2) + ΔHsubl(Li) + Lattice Energy
- ΔH0f(LiF) = -589.5 kJ/mol (given)
- ΔH0f(Li) = 0 kJ/mol (elemental form)
- ΔHf(F2) = 0 kJ/mol (elemental form)
- ΔHsubl(Li) = 155 kJ/mol (given)
- Let's solve for the Lattice Energy.

2. ΔH0f(LiF) = ΔHsubl(Li) + Lattice Energy
Lattice Energy = ΔH0f(LiF) - ΔHsubl(Li)
Lattice Energy = -589.5 kJ/mol - 155 kJ/mol
Lattice Energy = -744.5 kJ/mol

Therefore, the crystal lattice energy for LiF(s) is -744.5 kJ/mol.

(b) To determine the enthalpy change (ΔH) for the given reaction NH4NO3(s) + H2O(l) → NH4+(aq) + NO3-(aq), we can use Hess's Law.

Hess's Law states that the overall enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps that make up the reaction.

Given the enthalpy changes:
ΔH0f of NO3- = 205.0 kJ/mol
S0 = 146 J/mol·K
ΔH0f of NH4+ = -132.57 kJ/mol
S0 = 113.4 J/mol·K
ΔH0f of NH4NO3 = -365.56 kJ/mol
S0 = 151.08 J/mol·K

The enthalpy change (∆H) for the reaction can be calculated using the formula:
∆H = Σ∆H0f(products) - Σ∆H0f(reactants)

∆H = [ΔH0f(NH4+) + ΔH0f(NO3-)] - [ΔH0f(NH4NO3)]

∆H = [-132.57 kJ/mol + 205.0 kJ/mol] - [-365.56 kJ/mol]

∆H = 72.43 kJ/mol + 365.56 kJ/mol

∆H = 437.99 kJ/mol

Therefore, the enthalpy change (∆H) for the given reaction is 437.99 kJ/mol.