Heat of displacement and precipitation

This page covers one Form 5 Thermochemistry content standard: the heat of displacement and the heat of precipitation. The two quantities are grouped together because they share the same experimental method and the same Q = mcθ calculation you met in heat of neutralisation. Learn the two definitions precisely, keep the calculation steps in order, and this standard becomes a reliable source of marks.

Heat of displacement

The heat of displacement is the heat released when one mole of a metal is displaced from its salt solution by a more reactive metal. In a displacement reaction, a more reactive metal gives up electrons to the ions of a less reactive metal, pushing that less reactive metal out of solution as the free metal.

A typical example is adding zinc to copper(II) sulfate solution. Zinc is more reactive than copper, so zinc displaces copper:

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

or, as an ionic equation, Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).

The reaction releases heat to the surroundings, so it is exothermic, the temperature of the solution rises, and ΔH is negative. The more reactive the displacing metal is compared with the metal being displaced, the more heat is released, a useful way to link this topic to the reactivity series.

Heat of precipitation

The heat of precipitation is the heat change when one mole of a precipitate is formed from its ions in solution. When two solutions are mixed and an insoluble salt forms, the oppositely charged ions come together into a solid lattice, and this usually releases heat, so precipitation is normally exothermic with a negative ΔH.

A common example is mixing silver nitrate solution with sodium chloride solution to form a white precipitate of silver chloride:

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Here the sodium and nitrate ions are spectator ions; only the silver and chloride ions actually form the precipitate, and the heat of precipitation is defined per mole of AgCl formed.

The calculation, the same three steps

Both quantities use exactly the method from heat of neutralisation.

1. Heat released, Q = mcθ, m is the mass of the solution in grams (density 1 g cm⁻³, so cm³ of solution equals grams), c is the specific heat capacity of the solution (4.2 J g⁻¹ °C⁻¹), and θ is the temperature change in °C.
2. Moles of the metal displaced (displacement) or of the precipitate formed (precipitation), from the limiting reactant.
3. ΔH = −Q ÷ moles, in kJ mol⁻¹, with a negative sign for these exothermic changes.

For displacement, the reacting metal is usually added in excess (as a powder), so the amount of product is fixed by the moles of the salt in solution.

Worked example (displacement)

Question. Excess zinc powder is added to 50 cm³ of 0.40 mol dm⁻³ copper(II) sulfate solution in a polystyrene cup. The temperature rises from 28.0 °C to 48.0 °C. Calculate the heat of displacement of copper. (c = 4.2 J g⁻¹ °C⁻¹; density of solution = 1 g cm⁻³.)

Step 1, Mass of solution. m = 50 × 1 = 50 g

Step 2, Temperature change. θ = 48.0 − 28.0 = 20.0 °C

Step 3, Heat released. Q = mcθ = 50 × 4.2 × 20.0 = 4200 J = 4.2 kJ

Step 4, Moles of copper displaced. Moles of Cu²⁺ = 0.050 dm³ × 0.40 mol dm⁻³ = 0.020 mol From Zn + Cu²⁺ → Zn²⁺ + Cu, 0.020 mol of copper is displaced (zinc is in excess).

Step 5, Heat of displacement. ΔH = −Q ÷ moles = −4.2 kJ ÷ 0.020 mol = −210 kJ mol⁻¹

Answer. The heat of displacement of copper is −210 kJ mol⁻¹; the negative sign shows the reaction is exothermic.

Practice question (precipitation)

Question. 50 cm³ of 0.50 mol dm⁻³ silver nitrate solution is mixed with 50 cm³ of 0.50 mol dm⁻³ sodium chloride solution. A white precipitate forms and the temperature rises from 28.0 °C to 32.0 °C. Calculate the heat of precipitation of silver chloride. (c = 4.2 J g⁻¹ °C⁻¹; density = 1 g cm⁻³.)

Answer. Mass of solution = 50 + 50 = 100 g. Temperature change θ = 32.0 − 28.0 = 4.0 °C. Heat released Q = mcθ = 100 × 4.2 × 4.0 = 1680 J = 1.68 kJ. Moles of Ag⁺ = 0.050 × 0.50 = 0.025 mol and moles of Cl⁻ = 0.025 mol, so 0.025 mol of AgCl is precipitated. ΔH = −1.68 kJ ÷ 0.025 mol = −67.2 kJ mol⁻¹. The negative sign confirms precipitation is exothermic.

Exam tip

The two definitions must both mention one mole, one mole of the metal displaced, or one mole of the precipitate formed, and this is where the definition mark is usually won or lost. For the calculation, work in the same order as heat of neutralisation, and take special care with the moles step: in displacement the metal added is in excess, so the product is fixed by the salt solution, not by the metal; in precipitation, check which ion is limiting before you decide the moles of precipitate. Keep m as the total mass of solution after mixing, keep the negative sign on ΔH, and convert joules to kilojoules before writing kJ mol⁻¹. If a question also asks why the reaction is exothermic, say that bond or lattice formation releases more energy than is absorbed, the same energy reasoning used throughout the chapter.

Where this fits

This is content standard 11.4 of the Thermochemistry chapter. It reuses the exothermic idea and the Q = mcθ method from the earlier standards and connects to the reactivity series you met in Redox and Equilibrium, since displacement depends on one metal being more reactive than another. Consolidate the two definitions with the chapter revision notes and rehearse both calculations with the worked examples. In our online 1-to-1 SPM Chemistry lessons, taught in English, from RM50/hr, our teachers focus on the moles step, because that is where displacement and precipitation questions are most often lost, and this quantity is tested throughout SPM Chemistry.

Quick recap

• Heat of displacement: heat released when one mole of a metal is displaced by a more reactive metal; exothermic, ΔH negative.
• Heat of precipitation: heat change when one mole of a precipitate forms from its ions; usually exothermic, ΔH negative.
• Calculation: Q = mcθ, then ΔH = −Q ÷ moles, in kJ mol⁻¹.
• Watch the moles step, excess metal in displacement, limiting ion in precipitation.