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

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This question is about silver iodide. 0 1 . 1 Define the term enthalpy of lattice formation. The enthalpy of lattice formation is the standard enthalpy change when... 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 in their standard states.

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 Hess's law. The reaction can be represented as: AgI(s)Ag+(aq)+I(aq)\text{AgI}(s) \rightarrow \text{Ag}^+(aq) + \text{I}^-(aq) Using the values from Table 1:

  • Enthalpy of dissolution of AgI(s): +112 kJ/mol
  • Enthalpy of formation from element Ag(g): -464 kJ/mol
  • Enthalpy of formation from element I(g): -293 kJ/mol

So the enthalpy of lattice formation (ΔH_lattice) can be calculated as: ΔHlattice=ΔHAgI+ΔHAg++ΔHI\text{ΔH}_{lattice} = \text{ΔH}_{AgI} + \text{ΔH}_{Ag^+} + \text{ΔH}_{I^-} Plugging in the numbers: ΔHlattice=(+112)464293=645 kJ/mol\text{ΔH}_{lattice} = (+112) - 464 - 293 = -645 \text{ kJ/mol}

Step 3

Explain this difference.

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Answer

The difference occurs because a perfect ionic model assumes that all ionic interactions are purely electrostatic and does not take into account the covalent character of the bond. In reality, ionic compounds may have some degree of covalent character due to polarization, which results in stronger bonding and hence a more negative enthalpy of lattice formation than predicted by the ideal model.

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

Reagent: Starch solution

Observation: A blue-black coloration is observed when iodide ions are present in the solution due to the formation of a starch-iodine complex.

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