CP 2 Enthalpy Change via Hess’s Law
Edexcel International A Level Chemistry revision notes for Core Practical 2: finding the enthalpy change of decomposition of potassium hydrogencarbonate by measuring the reactions of K₂CO₃ and KHCO₃ with hydrochloric acid in a polystyrene cup, then combining them with Hess’s law.
Method with reasons, q = mcΔT, moles, the excess check, the cycle, uncertainties and the direction of every error.
Calorimetry Bench: K₂CO₃ and KHCO₃ with HCl
Measure the temperature change when each potassium salt reacts with hydrochloric acid, then combine the two results in a Hess cycle to find the enthalpy change for the decomposition of potassium hydrogencarbonate.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
GCSE Recap: Energy Changes
Four quick questions on exothermic and endothermic changes, the polystyrene cup and the units of energy from GCSE.
Method: Step by Step
The apparatus, and weighing by difference: the mass actually added is the difference between the two balance readings.
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.
Why a polystyrene cup?: Polystyrene is a better insulator than glass, so less heat is transferred to or from the surroundings, and its own heat capacity is so small that the heat it absorbs can be ignored.
Standing the cup in a beaker stops it tipping over and is worth stating.
Check: Order the Method
Put the steps of the calorimetry method for a different carbonate into the correct order.
Measuring the Temperature Change Accurately
Why the cooling line is extrapolated back to the time of addition, and how ΔT is read as the vertical gap between the two lines at that time. Model data, not the results below.
In this practical the reactions are fast, so the highest or lowest temperature reached is used as the final temperature.
Even so, heat exchange with the surroundings starts the moment the solid is added, so the maximum recorded is slightly below the true maximum, and the minimum recorded is slightly above the true minimum.
The magnitude of ΔT, and therefore of ΔH, is always a little too small.
The correction is a temperature-time graph. Readings are taken every 30 s for 3 minutes before the solid is added, and for 8 to 10 minutes afterwards.
A straight line is drawn through the readings after the peak (never through the rising part of the curve) and extrapolated back to the time of addition; the pre-addition readings are extrapolated forward to the same time.
ΔT is the vertical gap between the two lines at the time of addition. For an endothermic reaction the extrapolated minimum is lower than the recorded minimum, so ΔT is again larger in magnitude than the uncorrected value.
Describe the maximum or minimum method when asked for this practical, and offer extrapolation as the improvement when asked how to reduce the effect of heat loss.
Common mistake: Students lose marks by mixing the two: saying they extrapolated when the method records the maximum, or drawing a cooling line through the rising points.
Exam wording: “Plot temperature against time, draw a line of best fit through the readings after the maximum and extrapolate it back to the time of addition; ΔT is the difference between this value and the start temperature at that time.”
Name the line, the direction and where ΔT is read.
Sample Results
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.
The same test tube (empty mass 23.04 g) was used for both solids. Temperatures are the start temperature and the highest or lowest temperature reached; masses are read from a 2 d.p. balance.
| Measurement | K₂CO₃ run (reaction 1) | KHCO₃ run (reaction 2) |
|---|---|---|
| Mass of test tube with solid / g | 25.12 | 26.37 |
| Mass of test tube after emptying / g | 23.04 | 23.04 |
| Mass of solid added / g | 2.08 | 3.33 |
| Volume of 2.00 mol dm⁻³ HCl / cm³ | 30.0 | 30.0 |
| Start temperature / °C | 23.2 | 23.1 |
| Maximum temperature / °C | 27.2 | |
| Minimum temperature / °C | 19.2 | |
| Temperature change ΔT / °C | +4.0 | −3.9 |
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.
Exam focus: Record the mass by difference, not the nominal mass, and record both the start and the extreme temperature so that ΔT can be checked. A table with headings, units and consistent decimal places earns marks on its own.
Check: Completing a Results Table
Fill in the mass by difference, ΔT and the sign of ΔH for a fresh set of balance and thermometer readings.
The Core Equation: q = mcΔT
The equation, the meaning of each symbol, the insulated-cup method and its limitations in one figure. Its diagram shows the version in which two solutions are mixed; the method on this page adds a solid to the acid.
The energy transferred to or from the solution is calculated with q = mcΔT.
Here q is the energy change in joules, m the mass of solution in grams, c the specific heat capacity in J g⁻¹ K⁻¹ and ΔT the temperature change in K or °C.
A change of 1 °C is a change of 1 K, so the two are interchangeable here.
Three assumptions are built in and are asked for by name: the density of the solution is 1.00 g cm⁻³; the specific heat capacity of the solution equals that of water; and the heat capacity of the cup and thermometer is negligible.
State all three when asked why the calculated value is approximate.
Exam wording: c = 4.18 J g⁻¹ K⁻¹ is the value given in the exam. If a question quotes 4.2, use 4.2 and round to 2 significant figures; never mix the two in one calculation.
Using Hess’s Law to Find the Target ΔH
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.
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.
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.
Exam tip: If a question asks for the enthalpy change per mole of KHCO₃ decomposed, the answer is +62.7 ÷ 2 = +31.4 kJ mol⁻¹. This is a frequent trap.
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.
Exam wording: “ΔH₃ = 2ΔH₂ − ΔH₁: reaction 2 is doubled because the target contains 2 mol of KHCO₃, and reaction 1 is reversed because K₂CO₃ is a product in the target.” Give the reason for the 2 and the reason for the minus.
Check: Building the Hess Cycle
Build the cycle for the sodium salts from two new measured values and avoid the per-mole trap.
Other Calorimetry Calculations You Must Recognise
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.
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.
Exam strategy: Identify the target enthalpy change first, then decide which measured equation is reversed, which is multiplied and what is being heated. Only then pick up the calculator.
What This Practical Is Testing
The aim of Core 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 Unit 3 practical skills paper (WCH13), which asks for methods, apparatus choices, data processing, uncertainty and evaluation.
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.
Key idea: Measure two reactions you can do in a cup; calculate the one you cannot. Hess’s law says the enthalpy change is the same whichever route is taken from reactants to products, provided the conditions are the same.
The Two Reactions Being Measured
Both solids react with the same acid: 30.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid in a polystyrene cup.
Using the same volume and concentration of acid in both runs is deliberate: the acid appears on both sides of the Hess cycle and cancels out. The temperature change is measured after adding the solid.
| Reaction | Equation | Observation and sign |
|---|---|---|
| Reaction 1: potassium carbonate, ΔH₁ | K₂CO₃(s) + 2HCl(aq) → 2KCl(aq) + CO₂(g) + H₂O(l) | The temperature rises: exothermic, ΔH₁ negative. Effervescence as CO₂ is given off. |
| Reaction 2: potassium hydrogencarbonate, ΔH₂ | KHCO₃(s) + HCl(aq) → KCl(aq) + CO₂(g) + H₂O(l) | The temperature falls: endothermic, ΔH₂ positive. Effervescence as CO₂ is given off. |
Notice the mole ratios. One mole of K₂CO₃ needs two moles of HCl; one mole of KHCO₃ needs one mole of HCl. The target equation contains two moles of KHCO₃, so ΔH₂ will have to be doubled when the cycle is built.
Sign trap: A temperature rise gives a negative ΔH; a temperature fall gives a positive ΔH. Students calculate the size of q correctly and then lose the mark by giving the wrong sign.
Check: Exothermic or Endothermic?
Decide the sign of ΔH for temperature changes in reactions that are not the ones on this page.
Safety and Apparatus
Hazards and precautions: 2.00 mol dm⁻³ hydrochloric acid is an irritant: wear eye protection and wipe up spills.
Potassium carbonate is an irritant to the eyes and skin: avoid raising dust and wash hands after use.
Both reactions give off CO₂ rapidly, so add the solid steadily enough that the mixture does not froth over the rim of the cup, and keep the cup in the beaker for stability.
Worked Calculation: From ΔT to ΔH
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.
Important: ΔH₁ is for one mole of K₂CO₃ reacting and ΔH₂ for one mole of KHCO₃ reacting, exactly as the equations are written. The Hess cycle has to respect that ratio.
Check: Full Calculation
Work through q, moles, ΔH with its sign and the excess check for sodium hydrogencarbonate, then flip each card to compare your working.
Checking That the Acid Is in Excess
The acid must be in excess so that the solid is the limiting reagent: q is then the energy change for the moles of solid added, and dividing by those moles gives ΔH per mole of solid.
The concentration and volume of acid are chosen so that even the largest mass of solid that might be added reacts completely.
n(HCl) available = 2.00 mol dm⁻³ × 30.0 cm³ ÷ 1000 = 0.0600 mol
| Solid | Moles of solid | HCl needed | HCl available | Excess? |
|---|---|---|---|---|
| K₂CO₃ (1 : 2) | 2.08 ÷ 138.2 = 0.01505 mol | 2 × 0.01505 = 0.0301 mol | 0.0600 mol | Yes: 0.0600 > 0.0301 |
| KHCO₃ (1 : 1) | 3.33 ÷ 100.1 = 0.03327 mol | 1 × 0.03327 = 0.0333 mol | 0.0600 mol | Yes: 0.0600 > 0.0333 (about 1.8 times) |
The KHCO₃ run is the closer of the two, which is why the method says “about 3.5 g” and not more: with 6 g of KHCO₃ the acid would run out and the moles of solid would no longer limit the reaction.
Exam wording: Compare the moles: “0.0600 mol of HCl is available and only 0.0333 mol is needed, so the acid is in excess and the moles of solid limit the reaction.” Show both numbers.
Errors, Uncertainty and Improvements
Questions ask for the source of an error, its direction (is the result too large or too small?) and a specific improvement. Vague answers (“human error”, “use better equipment”) score nothing.
Percentage uncertainty in the K₂CO₃ run
Every quantity found from two readings carries twice the reading uncertainty.
Temperature: ±0.1 °C on each of two readings gives ±0.2 °C on ΔT, so (0.2 ÷ 4.0) × 100 = 5.0 %.
Mass: ±0.01 g on each of two readings gives ±0.02 g on 2.08 g, so (0.02 ÷ 2.08) × 100 = 1.0 %.
Volume: ±0.05 cm³ on each burette reading gives ±0.1 cm³ on 30.0 cm³, so (0.1 ÷ 30.0) × 100 = 0.3 %.
Total about 6.3 %, which on ΔH₁ = −33.3 kJ mol⁻¹ is ±2.1 kJ mol⁻¹. For the KHCO₃ run the temperature term is (0.2 ÷ 3.9) × 100 = 5.1 % and the mass term 0.6 %.
The temperature change is by far the largest contributor, so the worthwhile improvements attack ΔT.
Use a larger mass of solid or a smaller volume of acid to give a bigger temperature change (keeping the acid in excess).
Or use a thermometer reading to 0.1 °C if a 1 °C thermometer was used (±0.5 °C per reading would make the ΔT term 25 %). Improving the balance changes almost nothing.
Exam wording: “Percentage uncertainty = (2 × 0.1 ÷ 4.0) × 100 = 5.0 %, the largest of the three, so the thermometer limits the accuracy; a larger ΔT would reduce it.” Two readings, the arithmetic, the comparison and the improvement.
Check: Uncertainty and Error Direction
Percentage uncertainties on a new data set, which instrument to improve first, and which way heat loss pushes each value.
Common Mistakes
- 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.
Check: Assumptions and Final Checks
Pick the accurate statement about each assumption behind the calculation.
Common Exam Points
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.
FAQs
The questions students ask most about calorimetry, enthalpy calculations and Hess’s law for this practical.
Why can the decomposition of potassium hydrogencarbonate not be measured directly?
The solid has to be heated continuously to decompose, so the energy supplied by the Bunsen cannot be separated from the energy absorbed by the reaction.
There is no solution into which a thermometer can be placed to measure a temperature change. Two reactions that do happen in solution are measured instead and combined with Hess’s law.
Why is a polystyrene cup used, and why is it stood in a beaker?
Polystyrene is a poor conductor with a very small heat capacity, so almost all of the energy change stays in the solution and very little is absorbed by the cup or lost to the surroundings.
The beaker simply supports the light cup so it cannot tip over.
Why must the hydrochloric acid be in excess?
So that the solid is the limiting reagent. The energy change q then belongs to the moles of solid added, and dividing q by those moles gives ΔH per mole of solid.
With 30.0 cm³ of 2.00 mol dm⁻³ acid there is 0.0600 mol of HCl against a maximum of about 0.033 mol needed.
Why is reaction 2 doubled and reaction 1 reversed in the Hess calculation?
The target equation contains 2 mol of KHCO₃ but reaction 2 is written for 1 mol, so ΔH₂ is multiplied by 2.
K₂CO₃ is a reactant in reaction 1 but a product in the target, so reaction 1 is reversed and the sign of ΔH₁ changes. That gives ΔH₃ = 2ΔH₂ − ΔH₁.
Why is the experimental value so much smaller than the data-book value?
Mainly because the endothermic KHCO₃ reaction gains heat from the surroundings while the solution is cold, so its temperature fall and ΔH₂ are too small; ΔH₂ is doubled in the cycle, so the shortfall is doubled in ΔH₃.
Heat exchange with the surroundings always makes the magnitude of a measured enthalpy change too small.
What is the single most common calculation mistake?
The sign. A temperature rise means the reaction is exothermic and ΔH is negative; a temperature fall means endothermic and ΔH is positive. Work q with the magnitude of ΔT, then put the sign on by inspection.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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