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 | Reaction | Enthalpy change | kJ mol⁻¹ | |---------------------------------|------------------|----------| | AgI(s) → Ag⁺(aq) + I⁻(aq) | +112 | | Ag(g) → Ag⁺(aq) | −464 | | I(g) → I⁻(aq) | −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 | Reacti... 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 defined as the amount of energy released when one mole of an ionic compound is formed from its gaseous ions in their standard states. It is a measure of the stability of the ionic lattice structure.

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 AgI, we need to apply Hess's law using the reactions in Table 1:

Starting from the formation of the ionic compound from its gaseous ions:

$ext{Ag}^+(aq) + ext{I}^-(aq) \rightarrow ext{AgI}(s)$

Using the enthalpy values:

$\Delta H = \text{Enthalpy (products)} - \text{Enthalpy (reactants)}$

Thus, we have:

$\Delta H = [\Delta H_{AgI(s)} + \Delta H_{Ag^+(aq)} + \Delta H_{I^-(aq)}]$

Substituting the known values:

$\Delta H = [0 + (-464) + (-293)] - (+112) = -730 \, kJ \, mol^{-1}$

Therefore, the enthalpy of lattice formation of silver iodide, AgI, is -730 kJ mol⁻¹.

Step 3

Explain this difference.

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Answer

The difference arises because a perfect ionic model assumes that all ions are point charges with no polarity or covalent character. However, in reality, silver iodide may exhibit some degree of covalent character due to the polarization of the iodide ion by the silver ion. This results in a stronger interaction than predicted by the perfect ionic model, yielding a higher lattice enthalpy.

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 to identify iodide ions (I⁻) in solution is lead(II) nitrate (Pb(NO₃)₂). When lead(II) nitrate is added to a solution containing iodide ions, a bright yellow precipitate of lead(II) iodide (PbI₂) is formed. The observation would be the appearance of a distinct yellow color in the solution.

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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