Question

Using the following data, estimate ∆Hf° for potassium chloride: K(s) + ½ Cl₂(g) → KCl(s).

Answer 1

The estimated standard enthalpy of formation
`(\( \Delta H_f^\circ \))` for potassium chloride (KCl) is approximately -291 kJ/mol, derived from sublimation, ionization, bond dissociation, electron affinity, and lattice formation enthalpies.

To estimate
`\( \Delta H_f^\circ \)` for the formation of potassium chloride KCl, you can use the Hess's law, which states that the total enthalpy change for a reaction is the same, regardless of the number of steps in which the reaction is carried out.

The given reaction is:

`\[ K(s) + (1)/(2) Cl_2(g) \rightarrow KCl(s) \]`

The steps involved in the formation of KCl are:

1. Sublimation of potassium:

`\[ K(s) \rightarrow K(g) \]`

2. Ionization of potassium:

`\[ K(g) \rightarrow K^+(g) + e^- \]`

3. Dissociation of chlorine molecule:

`\[ (1)/(2) Cl_2(g) \rightarrow Cl(g) \]`

4. Electron affinity of chlorine:

`\[ Cl(g) + e^- \rightarrow Cl^-(g) \]`

5. Lattice formation of potassium chloride:

`\[ K^+(g) + Cl^-(g) \rightarrow KCl(s) \]`

Now, apply Hess's law and sum up the enthalpies of these steps to find
`\( \Delta H_f^\circ \)`:

`\[ \Delta H_f^\circ = \text{(Enthalpy of sublimation for K)} + \text{(Ionization energy of K)}`
`+ \text{(Bond dissociation energy of Cl}_2)`
`+ \text{(Electron affinity of Cl)} + \text{(Lattice formation enthalpy of KCl)}`

Substitute the given values:

`\[ \Delta H_f^\circ = 90 \ \text{kJ/mol} + 419 \ \text{kJ/mol} + 239 \ \text{kJ/mol} - 349 \ \text{kJ/mol} - 690 \ \text{kJ/mol} \]\[ \Delta H_f^\circ \approx -291 \ \text{kJ/mol} \]`

So, the estimated
`\( \Delta H_f^\circ \)` for the formation of potassium chloride is approximately -291 kJ/mol.

The complete question is:

Using the following data, estimate ∆Hf° for potassium chloride: K(s) + ½ Cl₂(g) → KCl(s).

Lattice formation enthalpy of KCI = -690 kJ/mol

lonization energy of K = 419 kJ/mol

Electron affinity of CI = -349 kJ/mol

Bond dissociation energy of Cl2 = 239 kJ/mol

Enthalpy of sublimation for K = 90 kJ/mol