Skip to content
MathsGenie logo
Open app

Course home

  1. A Level
  2. Chemistry AQA
  3. Question bank

Thermodynamics (A-level only)

EasyMediumHard
12345678910111213141516171819202122232425
Question 14

This question is about enthalpy changes and thermodynamic feasibility.

1.

Theoretical values for enthalpies of lattice dissociation can be calculated using a perfect ionic model. State the meaning of the term perfect ionic model.

[2]
2.

Table 1 shows some enthalpy changes:

Enthalpy changeΔH/kJ mol−1\Delta H / \text{kJ mol}^{-1}ΔH/kJ mol−1
Enthalpy of atomisation of sulfur+279+279+279
Enthalpy of atomisation of sodium+107+107+107
Enthalpy of formation of sodium sulfide−365-365−365
First ionisation energy of sodium+496+496+496
First electron affinity of sulfur−200-200−200
Second electron affinity of sulfur+640+640+640

Use the data in Table 1 to calculate the enthalpy of lattice dissociation of sodium sulfide, Na2S\text{Na}_2\text{S}Na2​S.

[3]
3.

Explain why the enthalpy of lattice dissociation for lithium chloride (LiCl\text{LiCl}LiCl) is significantly greater than the enthalpy of lattice dissociation for sodium chloride (NaCl\text{NaCl}NaCl).

[3]
4.

Calculate the temperature, in ∘C^\circ\text{C}∘C, above which the following reaction becomes feasible:

NaBr(s)→Na(s)+12Br2(g) \text{NaBr(s)} \rightarrow \text{Na(s)} + \frac{1}{2}\text{Br}_2\text{(g)} NaBr(s)→Na(s)+21​Br2​(g)

where

ΔH=+361 kJ mol−1 \Delta H = +361\text{ kJ mol}^{-1} ΔH=+361 kJ mol−1 ΔS=+115.2 J K−1 mol−1 \Delta S = +115.2\text{ J K}^{-1}\text{ mol}^{-1} ΔS=+115.2 J K−1 mol−1
[3]

Thermodynamics (A-level only) Questions

  1. A Level
  2. /Chemistry
  3. /Thermodynamics (A-level only)