Use the data in Table 1 to calculate a value for the electron affinity of chlorine. Enthalpy changes / kJ mol⁻¹ First ionisation energy of strontium +548 Second ionisation energy of strontium +1060 Enthalpy of atomisation of chlorine +121 Enthalpy of atomisation of strontium +164 Enthalpy of formation of strontium chloride -828 Enthalpy of lattice formation of strontium chloride -2112

A-Level AQA Chemistry Properties of Period 3 Elements & their Oxides: Use the data in Table 1 to calcu

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

Question 1

Use the data in Table 1 to calculate a value for the electron affinity of chlorine. Enthalpy changes / kJ mol⁻¹ First ionisation energy of strontium +548 Second io... show full transcript

Worked Solution & Example Answer: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

Step 1

Calculate electron affinity of chlorine

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Answer

To calculate the electron affinity of chlorine, we can use Hess's law. The Born-Haber cycle gives us the following equation:

$\Delta H_f = \Delta H_{atomization(Sr)} + \Delta H_{ionization1(Sr)} + \Delta H_{ionization2(Sr)} + \Delta H_{atomization(Cl)} + \Delta H_{lattice(SrCl2)} + 2 \Delta H_{electron affinity(Cl)}$

By rearranging the equation for electron affinity:

$2 \Delta H_{electron affinity(Cl)} = \Delta H_f - (\Delta H_{atomization(Sr)} + \Delta H_{ionization1(Sr)} + \Delta H_{ionization2(Sr)} + \Delta H_{atomization(Cl)} + \Delta H_{lattice(SrCl2)})$

Plugging in the values from Table 1:

Enthalpy of formation of strontium chloride = -828 kJ/mol
Enthalpy of atomisation of strontium = +164 kJ/mol
First ionisation energy of strontium = +548 kJ/mol
Second ionisation energy of strontium = +1060 kJ/mol
Enthalpy of atomisation of chlorine = +121 kJ/mol
Enthalpy of lattice formation of strontium chloride = -2112 kJ/mol

Thus, substituting the values in:

$2 \Delta H_{electron affinity(Cl)} = -828 - (164 + 548 + 1060 + 121 - 2112)$ $2 \Delta H_{electron affinity(Cl)} = -828 - (-730)$ $2 \Delta H_{electron affinity(Cl)} = -828 + 730$ $2 \Delta H_{electron affinity(Cl)} = -98$

Dividing both sides by 2: $\Delta H_{electron affinity(Cl)} = -49 \, \text{kJ mol}^{-1}$

Step 2

Draw a line from each substance to the enthalpy of lattice formation of that substance

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Answer

Draw a line from MgCl₂ to -2018 kJ/mol.
Draw a line from MgO to -2493 kJ/mol.
Draw a line from BaCl₂ to -3889 kJ/mol.

Step 3

State why there is a difference between the theoretical and experimental values.

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Answer

The difference between the theoretical and experimental values arises primarily due to factors such as covalent character or partial covalent bonding in the actual compounds. The theoretical model assumes purely ionic bonding, while in reality, there may be a degree of electron sharing or polarization, particularly in compounds with larger ions.

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

Worked Solution & Example Answer: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

Calculate electron affinity of chlorine

Draw a line from each substance to the enthalpy of lattice formation of that substance

State why there is a difference between the theoretical and experimental values.

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