Consider an ionic compound, MX, composed of generic metal M and generic halogen X.

The enthalpy of formation of MX is ÄHf° = –457 kJ/mol. The enthalpy of sublimation of M is ÄHsub = 121 kJ/mol. The ionization energy of M is IE = 433 kJ/mol. The electron affinity of X is EA = –313 kJ/mol. The bond energy of X2 is BE = 229 kJ/mol.

Determine the lattice energy of MX.

This is a problem on my homework assignment that I really don't understand. Please explain, it would help so much! Thanks!

You need to go through the "cycle" to see this.

M(s) ==> M(g) dH = 121 kJ
M(g) ==> M^+(g) + e dH = 433 kJ
1/2 X2(g) ==> X(g) dH = 229/2 kJ
X(g) + e ==> X^-(g) dH = -313 kJ
M^+(g) + X^-(g) ==> MX(s) dH = Ecrystal
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Add these. M(g) cancels. M^+(g) cancels. e cancels, X^-(g) cancels, and we have left
M(s) + 1/2 X2(g) ==> MX(s) dHf = -457 kJ
-457=121 + 433+(229/2)+(-313)+Ecrystal
And solve for Ecrystal.
Now for the confusing part.
Ecrystal is the lattice energy IF YOU DEFINE lattice energy as the heat liberated by combining the ions to form the solid crystal. But if you define lattice energy as the energy needed to break up the ionic compound into the ions then -Ecrystal = lattice energy. About -813 kJ or +813 kJ for Ecrystal or -Ecrystal. Look in your notes to see which way your prof defines lattice energy.
M^+(g) + X^-(g) ==> MX(s) -813 kJ
MX(s) ==> M^+(g) + X^-(g) +813 kJ

To determine the lattice energy of MX, we need to use the Born-Haber cycle, which relates the enthalpy changes during the formation of an ionic compound from its elements.

The Born-Haber cycle is as follows:

1. Formation of MX from M and X:
M(s) + 1/2 X2(g) → MX(s) ΔH1

2. Ionization of M:
M(s) → M+(g) + e- ΔHsub*(M)

3. Electron affinity of X:
X(g) + e- → X-(g) ΔHsub*(X)

4. Formation of X2 from X:
1/2 X2(g) → X(g) ΔHsub(X2)

5. Lattice energy of MX:
M+(g) + X-(g) → MX(s) ΔHlattice

Using Hess's law, we can express the lattice energy as the sum of the enthalpy changes along the cycle:

ΔHlattice = ΔH1 + ΔHsub*(M) + ΔHsub*(X) + ΔHsub(X2)

Now let's calculate each step of the cycle:

1. Formation of MX from M and X:
ΔH1 = -457 kJ/mol

2. Ionization of M:
ΔHsub*(M) = IE(M) = 433 kJ/mol

3. Electron affinity of X:
ΔHsub*(X) = EA(X) = -313 kJ/mol

4. Formation of X2 from X:
ΔHsub(X2) = -BE(X2) = -229 kJ/mol

Now we can calculate the lattice energy:

ΔHlattice = -457 kJ/mol + 433 kJ/mol - 313 kJ/mol - 229 kJ/mol

ΔHlattice = -566 kJ/mol

Thus, the lattice energy of MX is -566 kJ/mol.

To determine the lattice energy of MX, we need to understand the relationship between the enthalpy of formation, ionization energy, electron affinity, and bond energy.

Lattice energy is the energy released when gaseous ions come together to form a solid lattice. It represents the strength of the ionic bond.

The formation of MX can be broken down into several steps:

1. Ionization of metal M: M -> M+ + e- (requires energy equal to the ionization energy, IE).

2. Electron affinity of halogen X: X + e- -> X- (requires energy equal to the electron affinity, EA).

3. Formation of MX: M+ + X- -> MX (releases energy equal to the enthalpy of formation, ΔHf°).

4. Dissociation of X2: X2 -> 2X (requires energy equal to twice the bond energy of X2, 2BE).

From the given values, we can see that the enthalpy of formation, ΔHf°, is -457 kJ/mol, meaning the formation of MX is exothermic and releases energy.

Now, to calculate the lattice energy, we can use the following relationship:

Lattice Energy = ΔHf° + IE + EA - 2BE

Substituting the given values into the equation:

Lattice Energy = (-457 kJ/mol) + (433 kJ/mol) + (-313 kJ/mol) - 2(229 kJ/mol)

Lattice Energy = -457 + 433 - 313 - 2(229)

Lattice Energy = -457 + 433 - 313 - 458

Lattice Energy = -805 kJ/mol

Therefore, the lattice energy of MX is approximately -805 kJ/mol.

Note: The negative sign indicates that the lattice energy is released (exothermic) during the formation of the compound.