Heat of solution and mixing, enthalpy-concentration charts

Why it matters

Dissolving NaOH, absorbing HCl or NH₃ in water and diluting sulfuric acid release enough heat to boil the solution if it is not removed. Heats of solution and mixing size the coolers on absorbers and dilution tanks, explain why acid is added to water and not the reverse, and enthalpy–concentration charts make energy balances on evaporators and absorbers for such non-ideal solutions quick and reliable.

Key ideas

Ideal vs non-ideal mixing. For an ideal solution the enthalpy of the mixture is the sum of the pure-component enthalpies – mixing at constant T causes no heat effect. Real solutions of electrolytes, acids, bases and polar liquids show a heat of mixing, positive (endothermic, the solution cools) or negative (exothermic, it heats up).

Heat of solution. The enthalpy change when a solute (often a solid or gas) dissolves in a solvent at constant T and P. The term heat of mixing is used when two liquids are mixed; physically both are the same quantity – the ΔH of an isothermal mixing process.

Integral heat of solution ΔĤ_s(r). The enthalpy change when 1 mol of solute dissolves in r mol of solvent at 25 °C, starting from pure solute and pure solvent. It is tabulated (from your data book) against r. As r → ∞ it approaches the heat of solution at infinite dilution, ΔĤ_s(∞).

Heat of dilution. Adding solvent to an existing solution changes r from r₁ to r₂; the heat effect per mol solute is ΔĤ_s(r₂) − ΔĤ_s(r₁). Once the solution is already dilute, further dilution has almost no heat effect.

Enthalpy of a solution. Taking pure solute and pure solvent at 25 °C as references, the enthalpy of a solution with r mol solvent per mol solute at 25 °C is simply ΔĤ_s(r) per mol solute. At another temperature add the sensible heat of the solution, n·C_p·(T − 25) or m·c_p·(T − 25).

Energy balances. Use the solution enthalpy just defined in Q = ΔH. For an adiabatic dissolution the heat of solution goes into raising the solution temperature.

Enthalpy–concentration (H–x) charts. For strongly non-ideal binaries (H₂SO₄–H₂O, NaOH–H₂O, NH₃–H₂O) the specific enthalpy of solution is plotted against mass fraction, with isotherms, using stated reference states (often pure water at 0 °C and pure solute at 25 °C – check the chart). They are used for:

• Adiabatic mixing: the mixture point lies on the straight line joining the two feed points, at the mass-weighted average concentration (lever rule). The isotherm through that point gives the outlet temperature.
• Non-adiabatic processes: Q = m_out·H_out − Σ m_in·H_in, reading each H at its concentration and temperature.
• Evaporators handling NaOH solutions, where the chart (or a Dühring plot for boiling-point elevation) replaces C_p data.

Formulas

• ΔĤ_s(r) – integral heat of solution, kJ per mol solute in r mol solvent at 25 °C (tabulated).
• Q = n_solute·ΔĤ_s(r) – isothermal dissolution at 25 °C (Q < 0 → heat must be removed).
• ΔĤ_dil = ΔĤ_s(r₂) − ΔĤ_s(r₁) – per mol solute, dilution from r₁ to r₂.
• Ĥ_solution(T) = ΔĤ_s(r) + (m_solution / n_solute)·c_p·(T − 25 °C) – per mol solute, references pure components at 25 °C.
• Adiabatic: m_solution·c_p·(T − 25) = −n_solute·ΔĤ_s(r) – feeds at 25 °C.
• Adiabatic mixing on H–x chart: x_mix = (m₁·x₁ + m₂·x₂) / (m₁ + m₂) ; H_mix = (m₁·H₁ + m₂·H₂) / (m₁ + m₂).
• Q = m_out·H_out − Σ m_in·H_in – with chart enthalpies (kJ/kg).

Worked examples

Integral heats of solution at 25 °C (kJ/mol solute, data-book values): HCl(g): r = 5, −64.05; r = 10, −69.49. NaOH(s): r = 10, −42.51; r = 50, −42.72.

Example 1 (standard). 100 mol/h of HCl gas at 25 °C is absorbed in water at 25 °C to make a solution with r = 5 mol H₂O/mol HCl, leaving at 25 °C. Find the cooling duty and the strength of the acid.

1. Water needed = 5 × 100 = 500 mol/h.
2. Q = n_HCl·ΔĤ_s(5) = 100 × (−64.05) = −6405 kJ/h, so 6405 kJ/h (1.78 kW) must be removed.
3. Mass fraction HCl = 36.46 / (36.46 + 5 × 18.02) = 0.288, i.e. 28.8 wt %.

Answer: Remove 6405 kJ/h; product ≈ 28.8 wt % HCl

Example 2 (GATE level). 1 mol of solid NaOH at 25 °C is dissolved in 10 mol water at 25 °C in an insulated vessel. Take the solution c_p = 3.8 J/(g·K). (a) Find the final temperature. (b) The solution is cooled back to 25 °C and diluted to r = 50 with water at 25 °C. What is the heat effect of the dilution?

1. Solution mass = 40.0 + 10 × 18.02 = 220.2 g.
2. Adiabatic: 220.2 × 3.8 × (T − 25) = −1 × (−42 510 J).
3. T − 25 = 42 510 / 836.8 = 50.8 K, so T = 75.8 °C.
4. Dilution at 25 °C: ΔĤ_dil = ΔĤ_s(50) − ΔĤ_s(10) = −42.72 − (−42.51) = −0.21 kJ/mol NaOH.
5. Only 0.21 kJ is released: almost all the heat of solution appears in the first 10 mol of water; the r = 10 solution is already close to infinite dilution.

Answer: (a) ≈ 75.8 °C; (b) 0.21 kJ released per mol NaOH

Common mistakes

• Treating non-ideal solutions as ideal and ignoring the heat of solution.
• Using ΔĤ_s(∞) for a concentrated solution.
• Forgetting that ΔĤ_s is per mol of solute, not per mol of solution.
• Mixing reference states between an H–x chart and a C_p calculation.
• Applying the lever rule with mole fractions on a mass-based chart.
• Adding water to concentrated acid (local boiling) instead of acid to water.

For GATE CH

Expect NAT questions on heat removed during absorption or dissolution using a table of integral heats, the heat of dilution between two r values, adiabatic temperature rise, and adiabatic mixing on an H–x chart. Practise reading r carefully and keeping the solute basis.

Quick check

1. Is the heat of solution of NaOH in water positive or negative?
2. Define r in an integral heat of solution table.
3. What is the heat effect per mol HCl when diluting from r = 5 to r = 10?
4. On an H–x chart, where does the adiabatic mixture of two streams lie?
5. Why does further dilution of an already dilute solution release little heat?

Answers: 1. negative (exothermic); 2. mol solvent per mol solute; 3. −5.44 kJ/mol (released); 4. on the straight line joining them, at the mass-weighted mean composition; 5. ΔĤ_s(r) has nearly reached its infinite-dilution value.