Entropy Change, ΔS — Structured Questions (Cambridge International A-Level Chemistry): Exam Questions

The following equation represents the melting of ice.

H₂O (s) → H₂O (l) ∆H^ϴ = +6.03 kJ mol^–1 ∆S^ϴ = +22.1 J K^–1 mol^–1

State the meaning of the symbol ^ϴ in ∆H^ϴ.

Extract:

The following equation represents the melting of ice.

H2O (s) → H2O (l) ∆H^ϴ^ = +6.03 kJ mol^–1^ ∆S^ϴ^ = +22.1 J K^–1^ mol^–1^

Explain why the standard entropy change for this process is positive.

Extract:

The following equation represents the melting of ice.

H₂O (s) → H₂O (l) ∆H° = +6.03 kJ mol⁻¹ ∆S° = +22.1 J K⁻¹ mol⁻¹

Calculate the temperature at which ∆G^ϴ = 0 for this process.

Show your working.

Potassium nitrate is a soluble salt. When it dissolves in water the value of the enthalpy change ΔH^θ = +34.9 kJ mol⁻¹ and the value of the entropy change ΔS^θ = +117 J K⁻¹ mol⁻¹.

Construct an equation, including state symbols, for the process that occurs when potassium nitrate dissolves in water.

Extract:

Potassium nitrate is a soluble salt. When it dissolves in water the value of the enthalpy change ΔH^θ^ = +34.9 kJ mol^−1^ and the value of the entropy change ΔS^θ^ = +117 J K^−1^ mol^−1^.

Construct an equation, including state symbols, for the process that occurs when potassium nitrate dissolves in water.

KNO3 (s) → K^+^ (aq) + NO3^−^ (aq)

Explain why ΔS^θ is positive for this process.

Extract:

Potassium nitrate is a soluble salt. When it dissolves in water the value of the enthalpy change ΔH^θ^ = +34.9 kJ mol^−1^ and the value of the entropy change ΔS^θ^ = +117 J K^−1^ mol^−1^.

Calculate the temperature above which the dissolving of potassium nitrate becomes feasible. Show your working.

Extract:

Potassium nitrate is a soluble salt. When it dissolves in water the value of the enthalpy change ΔH^θ = +34.9 kJ mol^−1^ and the value of the entropy change ΔS^θ = +117 J K^−1^ mol^−1^.

Calculate the temperature above which the dissolving of potassium nitrate becomes feasible.

T = ΔH^θ ÷ ΔS^θ T = (34.9 × 1000) ÷ 117 T = 298 K

i) State the sign of ΔG^θ for this process at a temperature lower than your answer to part (c).

ii) State what this value of ΔG^θ indicates about the dissolving of potassium nitrate at this lower temperature.

Ammonia, NH₃, is produced by the Haber process and is an important chemical in the manufacture of fertilisers and cleaning products.

Ammonia gas can react with oxygen to produce nitrogen monoxide and steam, and is the first step in the Ostwald process which produces nitric acid.

i) Construct an equation for the reaction of ammonia with oxygen to produce nitrogen monoxide and steam.

ii) Standard entropies are shown in Table 2.1.

Table 2.1

Calculate ΔS^θ for this reaction at 298 K.

Extract:

Ammonia gas can react with oxygen to produce nitrogen monoxide and steam, and is the first step in the Ostwald process which produces nitric acid.

i) Construct an equation for the reaction of ammonia with oxygen to produce nitrogen monoxide and steam.

The balanced equation for the reaction of ammonia with oxygen is:

4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (g)

Explain why the standard entropy change for the reaction in (a) is positive.

Extract:

Standard entropies are shown in Table 2.1.

Table 2.1

The second step in the Ostwald process produces nitrogen dioxide as shown in the equation

2NO (g) + O₂ (g) → 2NO₂ (g) ΔH^θ = −112 kJ mol⁻¹

The standard entropy of NO₂ (g) is 240.0 J K⁻¹ mol⁻¹.

i) Use Table 2.1 to calculate ΔS^θ for this reaction at 298 K.

ii) Use your answer to (c)(i) to calculate ΔG^θ for this reaction at 298 K. Show your working.

Fig. 3.1 shows how entropy changes with temperature.

Fig. 3.1

i) The entropy of water is zero when the temperature is zero Kelvin. Explain this in terms of the arrangement and motion of the water molecules.

ii) Explain why the entropy change, ΔS, is larger at temperature T₂ than at temperature T₁.

iii) Label the boiling point, T_b, on the appropriate axis.

Standard entropies can be used to calculate the entropy change of a reaction, ΔS. For example, for the reaction between nitrogen monoxide and ozone.

NO (g) + O₃ (g) → NO₂ (g) + O₂ (g)

Standard entropy values are shown in Table 3.1

Table 3.1

Using Table 3.1, calculate ΔS^θ of the reaction between nitrogen monoxide and ozone at 298 K. Show your working.

The contact process is a method used industrially 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)

Enthalpy of formation values are shown in Table 3.2

Table 3.2

i) Using Table 3.2, calculate the standard enthalpy change of reaction, ΔH^θ, using the data provided. Show your working.

ii) The standard entropy change of this reaction is –189 J K^–1 mol^–1. Use this value and your enthalpy value from part (i) to calculate the standard Gibbs free energy change, ΔG^θ, for this reaction at 298 K. Show your working.

iii) Use your answer to part (ii) to state whether the reaction is feasible at 298 K.