Use the data in Table 1 to calculate a value for the electron affinity of chlorine - AQA - A-Level Chemistry - Question 1 - 2020 - Paper 1

To find the electron affinity ( E_a ight) of chlorine using Hess's law, we start with the enthalpy change equation for the formation of an ionic compound:

$$ext{Enthalpy of formation} = ext{Lattice energy} + ext{Enthalpy of atomisation of cations} + ext{Ionisation energies} - ext{Electron affinity}$$

For strontium chloride (SrCl₂), we can express this as:

$$ext{Enthalpy of formation of SrCl₂} = ext{Enthalpy of lattice formation} + ext{Enthalpy of atomisation of Sr} + 2 imes ext{Ionisation energy of Sr} + 2 imes ext{Electron affinity of Cl}$$

Plugging in the known values from Table 1, we can rearrange the equation to isolate the electron affinity:

$$-828 = -2112 + 164 + 2(548) + 2E_a$$

Calculating the left side gives:

$$-828 = -2112 + 164 + 1096 + 2E_a$$ $$-828 = -852 + 2E_a$$

Now solving for $2E_a$:

$$2E_a = -828 + 852$$ $$2E_a = 24$$

Dividing by 2:

$$E_a = 12 ext{ kJ mol}^{-1}$$

Therefore, the electron affinity of chlorine is approximately:

Electron Affinity of Chlorine: 12 kJ mol⁻¹