Question

The energy for the following reaction was measured to be -1033.0 kJ/mol. Mg(g) + Cl₂(g) → MgCl₂(s). Using this fact and the data in the table below, calculate the enthalpy (in kJ/mol) required to separate the ions from the lattice for this reaction: MgCl₂(s) → Mg₂(g) + 2CH(g). . Substance lonization Energies Electron Affinities Mg(g) 737.8 0 Mg*(g) 1450.7 - Cl(g) 1251.2 -349.0

Answer 1

The enthalpy required to separate the ions from the lattice for the given reaction MgCl₂(s) → Mg₂(g) + 2Cl(g) is -1018.1 kJ/mol.

Step-by-step explanation:

The enthalpy required to separate the ions from the lattice for the reaction MgCl₂(s) → Mg₂(g) + 2Cl(g) can be calculated using a thermochemical cycle and the given data. First, we can write a series of reactions and use them to cancel out the compounds we don't need:

MgCl₂(s) → Mg(g) + Cl₂(g) (reverse of the given reaction)

Mg(g) → Mg²+(g) + 2e-

2Cl(g) + 2e- → 2Cl-(g) (balance the charges)

We can now sum up the enthalpies of these three reactions:

ΔH = -1033.0 kJ/mol + (1450.7 kJ/mol - 737.8 kJ/mol) + 2(-349.0 kJ/mol)

ΔH = -1033.0 kJ/mol + 712.9 kJ/mol - 698.0 kJ/mol

ΔH = -1018.1 kJ/mol

Therefore, the enthalpy required to separate the ions from the lattice for the given reaction is -1018.1 kJ/mol.

Answer 2

The enthalpy required to separate the ions from the lattice in the reaction MgCl₂(s) → Mg²⁺(g) + 2Cl⁻(g) is 2938.2 kJ/mol.

To calculate the enthalpy required to separate the ions from the lattice for the reaction MgCl₂(s) → Mg²⁺(g) + 2Cl⁻(g), you can use Hess's Law, which states that the overall enthalpy change of a reaction is the sum of the enthalpy changes of its individual steps. Here are the steps to calculate it:

1. Calculate the enthalpy change for the formation of Mg²⁺(g) from Mg(g):

ΔH₁ = Ionization energy of Mg(g) - Electron affinity of Mg²⁺(g)

ΔH₁ = (737.8 kJ/mol) - (0 kJ/mol) = 737.8 kJ/mol

2. Calculate the enthalpy change for the formation of Cl⁻(g) from Cl(g):

ΔH₂ = Ionization energy of Cl(g) - Electron affinity of Cl⁻(g)

ΔH₂ = (1251.2 kJ/mol) - (-349.0 kJ/mol) = 1600.2 kJ/mol

3. Now, calculate the overall enthalpy change for the desired reaction:

ΔH_reaction = ΔH₁ + 2ΔH₂

ΔH_reaction = 737.8 kJ/mol + 2(1600.2 kJ/mol) = 2938.2 kJ/mol

So, the enthalpy required to separate the ions from the lattice in the reaction MgCl₂(s) → Mg²⁺(g) + 2Cl⁻(g) is 2938.2 kJ/mol.