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 use Hess's law and the following thermodynamic equation:

$\Delta H_f = \Delta H_{atom}(Sr) + \Delta H_{ion}(Sr) + \Delta H_{atom}(Cl) + \Delta H_{aff} + \Delta H_{lattice}$

Substituting the values from Table 1:

• Enthalpy of formation of strontium chloride, (\Delta H_f = -828 \text{ kJ mol}^{-1})
• Enthalpy of atomization of strontium, (\Delta H_{atom}(Sr) = +164 \text{ kJ mol}^{-1})
• First ionization energy of strontium, (\Delta H_{ion}(Sr) = +548 \text{ kJ mol}^{-1})
• Enthalpy of atomization of chlorine, (\Delta H_{atom}(Cl) = +121 \text{ kJ mol}^{-1})
• Enthalpy of lattice formation of strontium chloride, (\Delta H_{lattice} = -2112 \text{ kJ mol}^{-1})

The equation can be rearranged to solve for (\Delta H_{aff}):

$-828 = 164 + 548 + 121 + \Delta H_{aff} - 2112$

Calculating:

$-828 = 164 + 548 + 121 - 2112 + \Delta H_{aff}$ $-828 = -1279 + \Delta H_{aff}$

Thus, we get:

$\Delta H_{aff} = -828 + 1279 = 451 \text{ kJ mol}^{-1}$

Therefore, the electron affinity of chlorine is approximately (-451) kJ mol⁻¹.