Answer to Question #292399 in General Chemistry for bbs

Question

The energy for the following reaction was measured to be -1033.0 kJ/mol.

Mg(g) + 2 Cl(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) + 2Cl⁻(g).

Answer 1

The enthalpy of formation of MgCl₂ is given as

Mg(g)+Cl₂(g)------>MgCl₂(s) ∆H=-1033kJ/mol.........(i)

From ionisation energy data;

Mg(g)---->Mg⁺ + e I.E-1=737.8kJ/mol

Mg⁺(g)---->Mg²⁺ + e I.E-2=1450.7kJ/mol

Net ionisation energy is;

Mg(g)----->Mg²⁺ + 2e I.E=(737.8+1450.7)kJ/mol=2188.5kJ/mol......(ii)

Cl electron affinity is given by;

Cl(g)+e---->Cl^- Ea=-349kJ/mol

For 2 Cl atoms

2Cl +2e---->2Cl^- Ea=(-349*2)=-698kJ/mol.......(iii)

We now rearrange equation (i)

MgCl₂----->Mg(g)+2Cl(g) ∆H=1033kJ/mol......(iv)

Now putting equations (ii), (iii) and (iv) together we get;

MgCl₂(s)+2Cl+2e+Mg(g)----->Mg(g)+2Cl(g)+2Cl^-+Mg²⁺+2e(g) ∆H=1033+(-698)+2188.5kJ/mol

We can simplify that to

MgCl₂(s)---->Mg²⁺(g)+2Cl^-(g) ∆H=2522.5kJ/mol

Hence the enthalpy required to separate Mg²⁺ and 2Cl^- ions from MgCl₂ is 2522.5kJ/mol