Estimate the Change in Enthalpy and Entropy When Liquid Ammonia Vaporises

Problem: Estimate the change in enthalpy (ΔH) and entropy (ΔS) when liquid ammonia at −33.34°C and 1 atm is converted to gaseous ammonia.

Given data: • Boiling point of NH₃ = −33.34°C = 239.81 K • Enthalpy of vaporisation (ΔH_vap) = 23.35 kJ/mol • Pressure = 1 atm (standard)

Process: NH₃(l) → NH₃(g) at 239.81 K and 1 atm

Step 1: Change in Enthalpy (ΔH) At the boiling point and 1 atm, the enthalpy change equals the enthalpy of vaporisation: ΔH = ΔH_vap = +23.35 kJ/mol

The positive sign indicates the process is endothermic (liquid absorbs heat to become gas).

Step 2: Change in Entropy (ΔS) At the boiling point, liquid and vapour are in equilibrium, so ΔG = 0. Using ΔG = ΔH − TΔS = 0: ΔS = ΔH / T ΔS = 23,350 J/mol ÷ 239.81 K ΔS = +97.4 J/(mol·K)

The positive sign indicates disorder increases (gas is more disordered than liquid).

Answer: • ΔH = +23.35 kJ/mol • ΔS = +97.4 J/(mol·K)

For the vaporisation of liquid ammonia (NH₃(l) → NH₃(g)), both ΔH and ΔS are positive. Here is why:

Why ΔH is positive (endothermic): • Vaporisation requires energy to overcome intermolecular forces • In liquid NH₃, molecules are held together by hydrogen bonds and dipole-dipole forces • Energy must be absorbed to break these attractions and separate molecules into the gas phase • ΔH_vap is always positive for any substance

Why ΔS is positive (entropy increases): • Gas molecules have much more freedom of movement than liquid molecules • In liquid: molecules are close together with limited motion • In gas: molecules are far apart, moving randomly in all directions • More possible arrangements = more disorder = higher entropy • ΔS_vap is always positive for any substance

Thermodynamic spontaneity at the boiling point: • ΔG = ΔH − TΔS = 0 (at boiling point, equilibrium exists) • Above boiling point: TΔS > ΔH → ΔG < 0 → vaporisation is spontaneous • Below boiling point: TΔS < ΔH → ΔG > 0 → condensation is spontaneous • At boiling point: TΔS = ΔH → ΔG = 0 → liquid and gas coexist

The following thermodynamic formulas are essential for solving this problem:

1. Gibbs Free Energy Equation: ΔG = ΔH − TΔS

2. At phase equilibrium (boiling point): ΔG = 0, therefore: ΔS = ΔH / T

3. Clausius Equation for entropy of phase change: ΔS = ΔH_transition / T_transition

This applies to: • Vaporisation: ΔS_vap = ΔH_vap / T_bp • Fusion (melting): ΔS_fus = ΔH_fus / T_mp • Sublimation: ΔS_sub = ΔH_sub / T_sub

4. Temperature conversion: T(K) = T(°C) + 273.15 −33.34°C + 273.15 = 239.81 K

5. Unit conversion: • ΔH is usually in kJ/mol • ΔS is usually in J/(mol·K) • Convert kJ to J by multiplying by 1000 before dividing • 23.35 kJ = 23,350 J

Important: Always use temperature in Kelvin (not °C) in thermodynamic calculations.

Trouton's Rule states that the entropy of vaporisation (ΔS_vap) for most liquids at their normal boiling point is approximately 85–88 J/(mol·K).

ΔS_vap ≈ 88 J/(mol·K) (Trouton's constant)

Ammonia's ΔS_vap = 97.4 J/(mol·K) — this is HIGHER than Trouton's rule predicts.

Why ammonia deviates from Trouton's rule: • NH₃ has strong hydrogen bonding in the liquid phase • More energy is needed to break these H-bonds during vaporisation • Greater change from ordered liquid to disordered gas • Therefore, ΔS_vap is larger than the typical 88 J/(mol·K)

Substances with ΔS_vap > 88 J/(mol·K) — strong intermolecular forces: • Water (H₂O): ΔS_vap = 109.1 J/(mol·K) — very strong H-bonds • Ammonia (NH₃): ΔS_vap = 97.4 J/(mol·K) — strong H-bonds • Ethanol (C₂H₅OH): ΔS_vap = 110.0 J/(mol·K) — H-bonds

Substances that follow Trouton's rule (~88 J/(mol·K)): • Benzene: ΔS_vap = 87.2 J/(mol·K) • Carbon tetrachloride: ΔS_vap = 85.8 J/(mol·K) • Hexane: ΔS_vap = 84.4 J/(mol·K)

Substances with ΔS_vap < 88 J/(mol·K) — weak forces or already disordered: • Hydrogen (H₂): ΔS_vap = 44.4 J/(mol·K)

Key thermodynamic properties of ammonia (NH₃):

Basic properties: • Molecular formula: NH₃ • Molar mass: 17.03 g/mol • Boiling point: −33.34°C (239.81 K) at 1 atm • Melting point: −77.73°C (195.42 K) • Critical temperature: 132.25°C (405.40 K) • Critical pressure: 111.3 atm

Thermodynamic data: • ΔH_vap = 23.35 kJ/mol (at boiling point) • ΔH_fus = 5.66 kJ/mol (at melting point) • ΔS_vap = 97.4 J/(mol·K) • ΔS_fus = 28.96 J/(mol·K) • Specific heat capacity (liquid): 4.70 J/(g·K) • Specific heat capacity (gas): 2.06 J/(g·K)

Intermolecular forces in NH₃: • Hydrogen bonding (N–H···N) — strongest force • Dipole-dipole interactions (NH₃ is a polar molecule, μ = 1.47 D) • London dispersion forces (weakest)

NH₃ has a trigonal pyramidal shape (sp³ hybridised nitrogen with one lone pair), giving it a significant dipole moment and the ability to form hydrogen bonds.

Example 1: Entropy of vaporisation of water Given: ΔH_vap = 40.7 kJ/mol, T_bp = 100°C = 373.15 K ΔS_vap = 40,700 ÷ 373.15 = 109.1 J/(mol·K) Water has a very high ΔS_vap due to extensive hydrogen bonding.

Example 2: Entropy of fusion (melting) of ammonia Given: ΔH_fus = 5.66 kJ/mol, T_mp = −77.73°C = 195.42 K ΔS_fus = 5,660 ÷ 195.42 = 28.96 J/(mol·K) Fusion has a smaller ΔS than vaporisation (liquid and solid are both condensed phases).

Example 3: Using Trouton's rule to estimate ΔH_vap of benzene Given: T_bp = 80.1°C = 353.25 K, ΔS_vap ≈ 88 J/(mol·K) ΔH_vap = TΔS = 353.25 × 88 = 31,086 J/mol = 31.1 kJ/mol Actual value: 30.8 kJ/mol — Trouton's rule gives a good estimate!

Example 4: ΔG for vaporisation of NH₃ at 25°C (298 K) ΔG = ΔH − TΔS = 23,350 − (298 × 97.4) = 23,350 − 29,025 = −5,675 J/mol = −5.68 kJ/mol ΔG is negative → vaporisation is spontaneous at 25°C (which makes sense because 25°C > boiling point of −33.34°C).

Example 5: At what temperature is ΔG = 0 for NH₃ vaporisation? ΔG = 0 when T = ΔH/ΔS = 23,350 ÷ 97.4 = 239.7 K ≈ −33.4°C This confirms the boiling point!

Summary of ΔH and ΔS signs for all phase changes:

Phase Change | ΔH | ΔS | ΔG at equilibrium Melting (solid→liquid) | + (endothermic) | + (more disorder) | 0 at melting point Vaporisation (liquid→gas) | + (endothermic) | + (more disorder) | 0 at boiling point Sublimation (solid→gas) | + (endothermic) | + (more disorder) | 0 at sublimation point Freezing (liquid→solid) | − (exothermic) | − (less disorder) | 0 at freezing point Condensation (gas→liquid) | − (exothermic) | − (less disorder) | 0 at boiling point Deposition (gas→solid) | − (exothermic) | − (less disorder) | 0 at deposition point

Key relationships: • ΔH_vap = −ΔH_condensation • ΔS_vap = −ΔS_condensation • ΔH_sub = ΔH_fus + ΔH_vap (Hess's law) • At any phase transition temperature: ΔG = 0, so ΔS = ΔH/T

For ammonia: • Vaporisation: ΔH = +23.35 kJ/mol, ΔS = +97.4 J/(mol·K) • Condensation: ΔH = −23.35 kJ/mol, ΔS = −97.4 J/(mol·K) • Melting: ΔH = +5.66 kJ/mol, ΔS = +28.96 J/(mol·K) • Freezing: ΔH = −5.66 kJ/mol, ΔS = −28.96 J/(mol·K)