This question is about silver iodide. 0.1.1 Define the term enthalpy of lattice formation. 0.1.2 Some enthalpy change data are shown in Table 1. Table 1 AgI(s) → Ag+(aq) + I−(aq) ∆H = +112 Ag+(g) → Ag+(aq) ∆H = -464 I−(g) → I−(aq) ∆H = -293 Use the data in Table 1 to calculate the enthalpy of lattice formation of silver iodide. 0.1.3 A calculation of the enthalpy of lattice formation of silver iodide based on a perfect ionic model gives a smaller numerical value than the value calculated in Question 1.2. Explain this difference. 0.1.4 Identify a reagent that could be used to indicate the presence of iodide ions in an aqueous solution and describe the observation made.

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This question is about silver iodide - AQA - A-Level Chemistry - Question 1 - 2017 - Paper 1

Question 1

This question is about silver iodide. 0.1.1 Define the term enthalpy of lattice formation. 0.1.2 Some enthalpy change data are shown in Table 1. Table 1 AgI(s) →... show full transcript

Worked Solution & Example Answer:This question is about silver iodide - AQA - A-Level Chemistry - Question 1 - 2017 - Paper 1

Step 1

Define the term enthalpy of lattice formation.

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Answer

The enthalpy of lattice formation is the standard enthalpy change when one mole of an ionic solid is formed from its gaseous ions under standard conditions. It is a measure of the strength of the forces between the ions in the lattice.

Step 2

Use the data in Table 1 to calculate the enthalpy of lattice formation of silver iodide.

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Answer

To calculate the enthalpy of lattice formation for silver iodide (AgI), we can use the provided data from Table 1. The enthalpy of lattice formation can be derived from the following equation:

$ext{Lattice Energy} = ext{Energy of ions in aqueous state} - ext{Energy of solid}$

Start from AgI(s) to ions:
- AgI(s) → Ag+(aq) + I−(aq) ∆H = +112 kJ/mol
Add the enthalpy change for converting gaseous ions to aqueous ions:
- Ag+(g) → Ag+(aq) ∆H = -464 kJ/mol
- I−(g) → I−(aq) ∆H = -293 kJ/mol
Overall calculation:

$ext{Enthalpy of lattice formation} = +112 - 464 - 293 = -645 ext{ kJ mol}^{-1}$

Step 3

Explain this difference.

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Answer

The difference arises because the perfect ionic model assumes that the ions are point charges and that there are no interactions, such as polarization between ions. In reality, the ions experience some degree of polarization due to their electron clouds, which results in stronger interactions than those predicted by the simplistic model. As a result, the calculated enthalpy using the perfect ionic model is a smaller numerical value than that obtained from the experimental data.

Step 4

Identify a reagent that could be used to indicate the presence of iodide ions in an aqueous solution and describe the observation made.

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Answer

A suitable reagent for detecting iodide ions (I−) in an aqueous solution is silver nitrate (AgNO₃). When AgNO₃ is added to a solution containing iodide ions, a bright yellow precipitate of silver iodide (AgI) forms, indicating the presence of iodide ions. This reaction can be represented as:

ightarrow ext{AgI}(s) ext{ (yellow precipitate)}$$

This question is about silver iodide - AQA - A-Level Chemistry - Question 1 - 2017 - Paper 1

Worked Solution & Example Answer:This question is about silver iodide - AQA - A-Level Chemistry - Question 1 - 2017 - Paper 1

Define the term enthalpy of lattice formation.

Use the data in Table 1 to calculate the enthalpy of lattice formation of silver iodide.

Explain this difference.

Identify a reagent that could be used to indicate the presence of iodide ions in an aqueous solution and describe the observation made.

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