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 of chlorine, we can utilize Hess's law along with the enthalpy changes provided in Table 1. The Born-Haber cycle equation can be expressed as:

$\Delta H_f^\circ = \Delta H_{atom} + \Delta H_{ion} + \Delta H_{lattice}$

Rearranging this gives us:

$\Delta H_{EA} = \Delta H_f^\circ + \Delta H_{lattice} - (\Delta H_{atom, Sr} + \Delta H_{ion, 1} + \Delta H_{ion, 2})$

Substituting the values:

• Enthalpy of formation of strontium chloride ((\Delta H_f^\circ)): -828 kJ mol⁻¹
• Enthalpy of lattice formation of strontium chloride ((\Delta H_{lattice})): -2112 kJ mol⁻¹
• Enthalpy of atomisation of strontium ((\Delta H_{atom, Sr})): +164 kJ mol⁻¹
• First ionisation energy of strontium ((\Delta H_{ion, 1})): +548 kJ mol⁻¹
• Second ionisation energy of strontium ((\Delta H_{ion, 2})): +1060 kJ mol⁻¹

Substituting these values into the rearranged equation yields:

$\Delta H_{EA} = -828 + (-2112) - (164 + 548 + 1060)$

Calculating step-by-step:

1. Calculate the sum of ionization energies: (\Delta H_{ion} = 164 + 548 + 1060 = 1772)
2. Substitute this into the equation: (\Delta H_{EA} = -828 - 2112 - 1772 = -3712)

Thus,

Electron affinity of chlorine is approximately -3712 kJ mol⁻¹.