Use the data in Table 1 to calculate a value for the electron affinity of chlorine. Enthalpy change / 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 Energetics: 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 change / 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 the Electron Affinity of Chlorine

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Answer

To calculate the electron affinity ( ext{EA}) of chlorine, we can use Hess's law based on the Born-Haber cycle:

The equation for the formation of SrCl₂(s) can be written as:

$\Delta H_f = \Delta H_{atomization (Sr)} + \Delta H_{atomization (Cl)} + \Delta H_{IE(Sr)} + \Delta H_{IE(Sr)} + \Delta H_{EA} + \Delta H_{lattice}$

Inserting the values from the table:

$-828 = 164 + 121 + 548 + 1060 + \Delta H_{EA} - 2112$

Rearranging the equation to solve for \Delta H_{EA}:

$\Delta H_{EA} = -828 - 164 - 121 - 548 - 1060 + 2112$

Calculating:

$\Delta H_{EA} = 730 \, \text{kJ mol}^{-1}$

Step 2

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

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Answer

MgCl₂ → -2018

MgO → -2493

BaCl₂ → -3889

Step 3

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

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Answer

Theoretical values assume purely ionic bonding, while experimental values account for covalent character or partial covalent bonding.

Step 4

Explain why the enthalpy of hydration becomes less exothermic from Li⁺ to K⁺.

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Answer

As the size of the cation increases from Li⁺ to K⁺, the charge density decreases, leading to reduced attraction between the cation and water molecules, thus resulting in a less exothermic enthalpy of hydration.

Step 5

Use the data in Table 4 to calculate the enthalpy of hydration, in kJ mol⁻¹, of bromide ions.

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Answer

Given:

Enthalpy of solution of calcium bromide = -110 kJ mol⁻¹
Enthalpy of lattice formation of calcium bromide = -2176 kJ mol⁻¹
Enthalpy of hydration of calcium ions = -1850 kJ mol⁻¹

Using the relationship:

$\Delta H_{solution} = \Delta H_{lattice} + \Delta H_{hydration}$

We rearrange it to find \Delta H_{hydration}:

$\Delta H_{hydration} = \Delta H_{solution} - \Delta H_{lattice} - \Delta H_{hydration (Ca^{2+})}$

Calculating:

$\Delta H_{hydration} = -110 - (-2176) - (-1850)$

This results in:

$\Delta H_{hydration} = 2176 - 110 - 1850 \, kJ mol^{-1}$

After calculation: $\Delta H_{hydration} = -3716 \, kJ mol^{-1}$

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 the 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.

Explain why the enthalpy of hydration becomes less exothermic from Li⁺ to K⁺.

Use the data in Table 4 to calculate the enthalpy of hydration, in kJ mol⁻¹, of bromide ions.

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