Required Practical 2: Measurement of an Enthalpy Change

The aim of Required Practical 2 is to find the enthalpy change of thermal decomposition of potassium hydrogencarbonate, a reaction that cannot be measured directly:

2KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l) ΔH₃ (the target)

Direct measurement fails for two reasons that both earn marks. First, the solid must be heated continuously to decompose it, so the energy supplied by the Bunsen cannot be separated from the energy absorbed by the reaction.

Second, the reaction is solid to solid and gas: there is no solution into which a thermometer can be placed to record a temperature change.

Instead, two related reactions that do take place in solution are measured with a thermometer in a polystyrene cup, and Hess’s law is used to combine them.

Water is written as H₂O(l) because both measured reactions produce liquid water at room temperature, so the value obtained is for the equation with H₂O(l).

The written papers (at least 15% of the marks test practical skills) and the practical endorsement.

For this practical that means: the method with reasons, q = mcΔT with the assumptions stated, moles and ΔH with the correct sign, the excess check, the Hess cycle, percentage uncertainties and the direction of the error caused by heat loss.

Every step has a reason, and the reason is what the mark scheme rewards. The method below is for reaction 1; reaction 2 repeats it with about 3.5 g of KHCO₃ and a fresh 30.0 cm³ of the same acid.

The apparatus, and weighing by difference: the mass actually added is the difference between the two balance readings.

The same test tube (empty mass 23.04 g) was used for both solids. Temperatures are the start temperature and the extrapolated maximum or minimum; masses are read from a 2 d.p. balance.

Masses are to 4 significant figures and temperatures to 3, so ΔT is known to 2 significant figures and a final ΔH quoted to 3 significant figures is the most that can be justified.

The sign of ΔT tells you whether the reaction is exothermic or endothermic; q = mcΔT is then worked with the magnitude of ΔT and the sign is put on ΔH by inspection: rise → negative, fall → positive.

Where the table values come from: both runs on one temperature-time axis, the K₂CO₃ run rising to 27.2 °C and the KHCO₃ run falling to 19.2 °C after the solid is added.

Three steps, each with units, for each reaction. Convert joules to kilojoules before dividing by moles so that ΔH comes out in kJ mol⁻¹.

Reaction 1: K₂CO₃ (exothermic)

q = m × c × ΔT = 30.0 g × 4.18 J g⁻¹ K⁻¹ × 4.0 K = 501.6 J = 0.5016 kJ (energy given out by 2.08 g of K₂CO₃)

Mr(K₂CO₃) = (2 × 39.1) + 12.0 + (3 × 16.0) = 138.2, so n = 2.08 g ÷ 138.2 g mol⁻¹ = 0.01505 mol

ΔH₁ = −0.5016 kJ ÷ 0.01505 mol = −33.3 kJ mol⁻¹ (negative because the temperature rose)

Reaction 2: KHCO₃ (endothermic)

q = 30.0 g × 4.18 J g⁻¹ K⁻¹ × 3.9 K = 489.1 J = 0.4891 kJ (energy taken in by 3.33 g of KHCO₃)

Mr(KHCO₃) = 39.1 + 1.0 + 12.0 + (3 × 16.0) = 100.1, so n = 3.33 g ÷ 100.1 g mol⁻¹ = 0.03327 mol

ΔH₂ = +0.4891 kJ ÷ 0.03327 mol = +14.7 kJ mol⁻¹ (positive because the temperature fell)

Keep the moles to 4 significant figures inside the calculation and round only the final ΔH. Rounding n to 0.0151 mol first and then writing −33.3 gives working that does not check (0.5016 ÷ 0.0151 = 33.2), and examiners do check.

Hess’s law: the enthalpy change of a reaction is independent of the route taken. Draw the cycle rather than listing equations: examiners ask for it.

Across the top is the target, 2KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l), with 2HCl(aq) added to both sides so that each side can react with the same acid.

At the bottom are the common products, 2KCl(aq) + 2CO₂(g) + 2H₂O(l): one CO₂ and one H₂O come from reaction 1 and the other pair from the target itself.

The Hess cycle. The left arrow is two lots of reaction 2 (2ΔH₂), the right arrow is reaction 1 (ΔH₁); going round the cycle, ΔH₃ + ΔH₁ = 2ΔH₂.

Route 1 (left arrow): 2KHCO₃(s) + 2HCl(aq) → 2KCl(aq) + 2CO₂(g) + 2H₂O(l), enthalpy change 2ΔH₂ because two moles of KHCO₃ react.

Route 2 (across then down): the target ΔH₃ followed by reaction 1, K₂CO₃(s) + 2HCl(aq) → 2KCl(aq) + CO₂(g) + H₂O(l), enthalpy change ΔH₁.

Equating the routes: ΔH₃ + ΔH₁ = 2ΔH₂, so

ΔH₃ = 2ΔH₂ − ΔH₁ = 2(+14.7) − (−33.3) = +29.4 + 33.3 = +62.7 kJ mol⁻¹

The decomposition is endothermic, as expected for a reaction that needs heating.

Say what “per mole” means here: +62.7 kJ mol⁻¹ is per mole of the equation as written, that is per 2 mol of KHCO₃ or per mole of K₂CO₃ formed.

Comparison with the accepted value. From standard enthalpies of formation the value for the equation with H₂O(l) is about +96 kJ mol⁻¹, with ΔH₁ about −33 kJ mol⁻¹ and ΔH₂ about +32 kJ mol⁻¹.

The measured ΔH₁ (−33.3) agrees within its uncertainty, but the measured ΔH₂ (+14.7) is much less endothermic than the true value because the cold solution gained heat from the surroundings, so its temperature fall was too small.

Since ΔH₂ is doubled in the cycle, that shortfall is doubled in ΔH₃, which is why the experimental value is well below +96 kJ mol⁻¹.

The same three-step pattern (q = mcΔT, moles of the reactant not in excess, q ÷ n with the sign) covers every calorimetry question. What changes is the mass in q = mcΔT and what “per mole” refers to.

Displacement: 25.0 cm³ of 0.20 mol dm⁻³ copper(II) sulfate with excess zinc; the mass is the 25.0 g of solution and q is divided by the 0.0050 mol of Cu²⁺, the reactant not in excess.

Neutralisation: the mass is the total 50.0 g of both solutions and the result is per mole of water formed, here the same as per mole of HCl.

Combustion: the mass heated is the 150 g of water, the moles are those of propan-1-ol burned, and the answer is much less negative than the accepted −2021 kJ mol⁻¹ because of heat loss and incomplete combustion.

• Wrong sign: calculating q correctly and then giving an exothermic reaction a positive ΔH. Rise → negative; fall → positive.
• Forgetting the 2: writing ΔH₃ = ΔH₂ − ΔH₁ when the target contains two moles of KHCO₃.
• Wrong “per mole”: quoting +62.7 kJ mol⁻¹ as the value per mole of KHCO₃ (it is +31.4).
• Dividing joules by moles and forgetting to convert to kJ, giving an answer 1000 times too large.
• Using the mass of solid as m in q = mcΔT, or adding it to the 30.0 g. The mass is the solution being heated.
• Nominal mass: using “3.00 g” instead of the mass found by difference.
• Cooling line through the rising points on a temperature-time graph, or reading ΔT anywhere other than the time of addition.
• “Human error” or “use better equipment” as an evaluation. Name the error, its direction and a specific improvement.
• One reading, one uncertainty: forgetting that ΔT and a mass by difference each come from two readings.

Say

“The acid is in excess (0.0600 mol available, 0.0333 mol needed) so the solid limits the reaction.”

“q = 30.0 × 4.18 × 4.0 = 501.6 J, assuming a density of 1.00 g cm⁻³ and the specific heat capacity of water.”

“ΔH₃ = 2ΔH₂ − ΔH₁ because the target contains 2 mol of KHCO₃ and reaction 1 is reversed.”

“Heat gained from the surroundings makes ΔH₂ less positive, and the error is doubled in the cycle.”

“Percentage uncertainty in ΔT = (0.2 ÷ 4.0) × 100 = 5.0 %.”

Do not say

“The reaction is exothermic so ΔH is positive.” “The mass is 2.08 g” (in q = mcΔT). “Heat loss makes the value more negative.”

“Use a more accurate specific heat capacity” or “calibrate the calorimeter” as improvements.

“Extrapolate” when describing a method that records the highest temperature, or “highest temperature” when the question asked for the graphical method.

Watch for

State symbols on every equation. Units on every line of working: g, J g⁻¹ K⁻¹, K, J, kJ, mol, kJ mol⁻¹. Significant figures: 3 in the final ΔH.

Whether a question wants ΔH per mole of KHCO₃ or per mole of equation. Whether the cooling line has been drawn only through the readings after the peak. Whether both readings have been counted in each percentage uncertainty.