5.1 Thermodynamics (A Level only)

Draw a fully labelled Born-Haber cycle for the formation of sodium selenide, Na₂Se (s), from its elements. Include state symbols in all equations.

If sulfur is used as opposed to selenium in the lattice, what would you expect to happen to the value of the enthalpy of lattice dissociation. Explain your answer.

Use the data in Table 1 to calculate the value for the enthalpy of lattice formation of aluminium oxide.

Table 1

Aluminium oxide is insoluble in water, but sodium oxide is soluble. Explain why there is no enthalpy of solution data for sodium oxide.

The equations for two separate reversible reactions are as follows:

Reaction A      2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)

Reaction B      CO (g) + H₂O (g) ⇌ CO₂ (g) + H₂ (g)

Use the information in Table 1 to calculate the temperature at which the free energy change for reaction A becomes feasible

Table 1

By using the data in Table 1, deduce if reaction B is feasible at a low temperature.

Magnesium carbonate decomposes at a relatively high temperature. Using the data in Table 2, determine if this reaction is feasible at 280 C.

Table 2

Using your answer to part (c), determine the temperature in C at which the decomposition of magnesium carbonate becomes feasible.

This question is about enthalpy changes. 

Strontium is used as a red colouring agent in fireworks as it provides a very intense red colour.

Use the data in Table 1 to calculate the atomisation energy for chlorine in strontium chloride.

Using information in part (a) suggest a value for the lattice enthalpy of formation of rubidium chloride. Explain your choice of value.

A different Group 1 metal, lithium, can be used in rechargeable batteries for mobile phones, laptops, digital cameras and electric vehicles. Lithium is also used in some non-rechargeable batteries for things like heart pacemakers, toys and clocks.     

Lithium carbonate will decompose on heating.

Figure 1 shows the graph of ΔG plotted against temperature, K for the decomposition of lithium carbonate.

Figure 1

Which of the following compounds would have the greatest difference between the theoretical and experimental value for lattice enthalpy. Explain your answer.

Silver fluoride

Silver iodide

Sodium fluoride

Butane can be converted into ethene and hydrogen which can be used in hydrogenation of vegetable oil in the presence of a nickel catalyst.

The free energy change for this reaction varies with temperature.

Hydrazoic acid is a colorless, volatile explosive liquid under standard conditions. The decomposition of the acid in a gaseous state is shown in the following equation.

HN₃ (g) →  H₂ (g) + N₂ (g)

Figure 1 shows a graph of graph of free energy against temperature on the grid below.

Figure 1

Use the data give in Table 2 to explain why the reaction for the formation of ethene from its elements is never feasible at any temperature. You must show your working.

Table 2

This question is about the energy changes represented in a Born-Haber cycle when forming silver chloride.

Figure 1

This question focuses on the difference between the theoretical enthalpy values and experimental values of ionic substances. A perfect ionic model can be used to calculate a theoretical value for the enthalpy of lattice dissociation.

This question looks at the enthalpy steps of calcium sulfide and its Born-Haber cycle.

Figure 2

This question is about enthalpy changes in solution.

This question focuses on the enthalpy of hydration of fluoride compounds.

This question is about the enthalpy of hydration of halide ions.

Table 1

Using the data from Table 1 above, explain why the value for the enthalpy of hydration for the fluoride ion is more negative than that for the chloride ion. You must refer to the attractive forces involved in your answer.

Use the enthalpy values below in Table 2 to suggest why there is a difference between the hydration values of calcium ions and sodium ions.

Table 2

This question looks at how the entropy change of water varies with temperature.

Figure 1

Standard entropies can be used to calculate the entropy change of a reaction, ΔS.

For example, for the reaction between nitrogen monoxide and oxygen which is shown below.

2NO (g) + O₂ (g) → 2NO₂ (g)

Table 2

The contact process is a method used by industries to form sulfur trioxide, by reacting sulfur dioxide and oxygen together over a vanadium (V) oxide catalyst.

The equation for this reaction is shown below:

2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)

Table 3

The standard entropy change of this reaction is –189 JK^-1 mol^-1. Use this value and your enthalpy value from part (i) calculate a value for the free energy change for this reaction at 45 C.
If you did not calculate an answer for part (i), use 70 C as your value. Note that this is not the correct answer to part (i). 

Use your answer to part (ii) to explain whether the reaction is feasible at 45 C.

This question is about the enthalpy of solution of sodium chloride.

If the value of enthalpy of solution of sodium chloride is +4 kJ mol^-1, explain why the free energy change for dissolving sodium chloride in water is negative, despite the enthalpy change being a positive value.

Calcium carbonate thermally decomposes to form calcium oxide and carbon dioxide, as shown below:

CaCO₃ (s) → CaO (s) + CO₂ (g)

The enthalpy change of the above reaction is ΔH = +178 kJ mol^-1 and the entropy change is ΔS = +161 J K^-1 mol^-1.

Calculate the temperature at which the free-energy change, ΔG, for this process is zero.

Some ionic compounds such as potassium chloride, KCl, will dissolve in water at room temperature in an endothermic process.

KCl (s) → K⁺ (aq) + Cl^- (aq)                 ΔH = +16 kJ mol^-1      

Table 1

Using the data from Table 1, prove that this process is feasible at 25 C.

Diamond and graphite are both allotropes of carbon.

Table 2 below shows the data for the conversion of graphite into diamond. Use this data to calculate values for ΔH and ΔS for the reaction. Use these values to explain why this reaction is not feasible under standard pressure at any temperature.

Carbon (graphite) → Carbon (diamond)

Table 2