A rising temperature does not mean the reaction has a positive enthalpy change. It means the measured solution gained thermal energy. If the reaction supplied that energy, the reaction lost it. Keeping those two accounts separate is the key to a reliable calorimetry calculation.
All measurements below are fictional supplied classroom data. They are for calculation and interpretation only, not instructions to carry out chemical reactions. Assume constant pressure, no heat exchange with the wider surroundings, negligible apparatus heat capacity unless stated otherwise, and no significant work other than pressure–volume work. Report the molar basis explicitly.
Keep the energy account separate from the mole account
For the solution, use qsolution = mcΔT, where m is the mass whose temperature changes, c is its supplied specific heat capacity and ΔT = final temperature − initial temperature. If m is in grams and c in J g⁻¹ K⁻¹, the result is in joules.
In the ideal insulated model, qreaction = −qsolution. A positive solution heat corresponds to a negative reaction heat. At constant pressure under the stated conditions, this reaction heat gives the enthalpy change for the amount that reacted. Divide by the appropriate reacting amount to obtain a molar value.
Write the two accounts on separate lines. The energy account establishes joules or kilojoules for the whole event. The mole account establishes how many moles of the stated reaction or product those joules belong to. A correct q value divided by the wrong amount still gives a wrong molar enthalpy.
Complete energy-transfer diagram: In the upper case, heat flows from the reacting system to the measured solution: solution temperature rises, q(solution) is positive and q(reaction) is negative. In the lower case, heat flows from the measured solution to an endothermic dissolving system: solution temperature falls, q(solution) is negative and q(process) is positive. Each pair has equal magnitude and opposite sign only under the stated negligible-other-heat assumptions.
Worked analysis 1: equal amounts in neutralisation
A supplied record combines 40.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid with 40.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide. Both solutions initially have temperature 24.0°C, and the final temperature is 30.5°C. Assume additive volumes, final solution density 1.00 g cm⁻³ and specific heat capacity 4.20 J g⁻¹ K⁻¹.
The equation is HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). Define the requested molar enthalpy per mole of water formed. Here the coefficients of acid, alkali and water are all one, so equal reacting amounts are needed.
- Total volume = 40.0 + 40.0 = 80.0 cm³. Solution mass = 80.0 × 1.00 = 80.0 g.
- Temperature change = 30.5 − 24.0 = +6.5 K. Temperature intervals have the same numerical size in kelvin and degrees Celsius.
- q(solution) = 80.0 × 4.20 × 6.5 = +2,184 J = +2.184 kJ.
- Each reactant amount = 1.00 × 40.0/1,000 = 0.0400 mol. Thus 0.0400 mol water forms.
- q(reaction) = −2.184 kJ, so molar enthalpy = −2.184/0.0400 = −54.6 kJ mol⁻¹ of water formed.
The negative sign agrees with heat flowing from reaction to solution. The result is an estimate from this fictional record, not a claimed standard enthalpy or a value measured by an examining body. The final rounding convention may depend on the task; the working retains sufficient digits to make the arithmetic checkable.
Using only 40 g as m would omit half the solution whose temperature changed. It would halve the calculated heat. Dividing by 0.0800 mol would make a different error: it adds acid and alkali amounts even though two reactant species together produce only 0.0400 mol water.
Worked analysis 2: temperature falls during dissolution
A fictional solute amount of 0.0500 mol dissolves completely. The final solution mass, already including solute, is 100.0 g. Use c = 4.00 J g⁻¹ K⁻¹. Temperature falls from 25.0°C to 22.0°C. Find the molar enthalpy of dissolution under the stated ideal assumptions.
ΔT = 22.0 − 25.0 = −3.0 K. Therefore q(solution) = 100.0 × 4.00 × (−3.0) = −1,200 J = −1.200 kJ. The dissolving process takes in the corresponding energy, so q(process) = +1.200 kJ.
Per mole dissolved, ΔH = +1.200/0.0500 = +24.0 kJ mol⁻¹. The positive sign means endothermic dissolution. A negative temperature change and positive process enthalpy are consistent because they describe different parts of the energy account.
Do not add the solute mass again: the supplied 100.0 g is explicitly the total solution mass. If only solvent mass were supplied in another problem, you would need to consider whether the model includes the added solute in the heated or cooled mass.
Worked analysis 3: use the supplied heat capacity
A reaction changes the temperature of 80.0 g solution by +5.0 K. The supplied solution specific heat capacity is 3.60 J g⁻¹ K⁻¹, and the defined reacting amount is 0.0250 mol. Calculate the molar enthalpy, then evaluate the error from silently using 4.20 instead.
Using the supplied value gives q(solution) = 80.0 × 3.60 × 5.0 = 1,440 J. Hence ΔH = −1.440/0.0250 = −57.6 kJ mol⁻¹. Substituting 4.20 gives 1,680 J and −67.2 kJ mol⁻¹, which is 9.6 kJ mol⁻¹ more negative.
Water’s commonly used approximate heat capacity is not a universal constant for every solution. Here the assumption is unnecessary because c is given. The calculation scales directly with c: 4.20/3.60 = 7/6, so the incorrect magnitude is about 16.7% larger.
Specific heat capacity and heat capacity also have different units. A calorimeter heat capacity C in J K⁻¹ contributes CΔT without multiplying by a mass. When its value is supplied and not negligible, the energy balance must include it as well as the solution.
Worked analysis 4: find the limiting amount
A separate fictional neutralisation record uses 30.0 cm³ of 1.00 mol dm⁻³ HCl and 50.0 cm³ of 0.400 mol dm⁻³ NaOH. Use the same one-to-one equation, density 1.00 g cm⁻³, c = 4.20 J g⁻¹ K⁻¹, additive volumes and a temperature rise of 3.20 K. Both solutions start at the same temperature.
Acid amount = 1.00 × 0.0300 = 0.0300 mol. Alkali amount = 0.400 × 0.0500 = 0.0200 mol. NaOH limits the reaction, so only 0.0200 mol water forms. Some acid remains, but all 80.0 g of the resulting solution belongs in the heat calculation.
q(solution) = 80.0 × 4.20 × 3.20 = 1,075.2 J. The molar enthalpy per mole water is −1.0752/0.0200 = −53.76 kJ mol⁻¹, approximately −53.8 kJ mol⁻¹. Dividing by the initial acid amount would incorrectly spread the measured heat over acid that did not all react.
For an equation with different coefficients, compare amount divided by its stoichiometric coefficient rather than choosing the smaller raw number automatically. “Per mole” must refer to the specified reaction basis or product, not an unspecified total of every substance in the vessel.
Interpret an imperfect measurement
In a real exothermic measurement, heat escaping to the room can make the measured rise smaller than the ideal insulated rise. If all other assumptions are retained, the calculated magnitude of the negative enthalpy is then too small: the result is less negative. An ignored warming calorimeter can have a similar effect.
That is a conditional explanation, not a licence to assign every discrepancy to heat loss. An inaccurate concentration, incorrect mass, incomplete reaction or unsuitable c also changes the estimate. If the starting solutions have different temperatures, simply subtracting either starting temperature from the final one is not a sufficient energy balance.
Three practice tasks with solved answers
Practice 1: equal reacting amounts
Two 60.0 cm³ solutions, 0.500 mol dm⁻³ HCl and 0.500 mol dm⁻³ NaOH, start at the same temperature and show a 3.40 K rise. Assume additive volumes, density 1.00 g cm⁻³ and c = 4.20 J g⁻¹ K⁻¹. Find enthalpy per mole water.
Answer: m = 120.0 g; q(solution) = 120.0 × 4.20 × 3.40 = 1,713.6 J. Each amount is 0.0300 mol, so ΔH = −1.7136/0.0300 = −57.12 kJ mol⁻¹, approximately −57.1 kJ mol⁻¹.
Practice 2: endothermic sign
Dissolution of 0.0400 mol solute cools 125.0 g final solution by 2.00 K. Use c = 4.00 J g⁻¹ K⁻¹. Calculate solution heat and molar dissolution enthalpy.
Answer: q(solution) = 125.0 × 4.00 × (−2.00) = −1,000 J. The process absorbs +1.000 kJ, giving ΔH = +1.000/0.0400 = +25.0 kJ mol⁻¹.
Practice 3: excess acid
Combine supplied records for 25.0 cm³ of 1.20 mol dm⁻³ HCl and 45.0 cm³ of 0.400 mol dm⁻³ NaOH, initially at the same temperature. The rise is 3.60 K; density is 1.00 g cm⁻³ and c is 4.00 J g⁻¹ K⁻¹. Volumes are additive. Find enthalpy per mole water.
Answer: Acid = 0.0300 mol, alkali = 0.0180 mol, so water formed = 0.0180 mol. Mass = 70.0 g and q(solution) = 1,008 J. Therefore ΔH = −1.008/0.0180 = −56.0 kJ mol⁻¹. Excess acid changes the reacting amount, not the inclusion of its solution mass.
Construct an energy cycle and keep its signs
The calorimetry cases above derive enthalpy from a temperature record. Hess’s law gives another route: if reactions add to the same overall chemical change, their enthalpy changes add. The initial and final substances, amounts and physical states must match. A cycle is energy bookkeeping, not a practical route for carrying out each reaction.
Write the target first. Reverse an equation when needed and reverse its enthalpy sign. Multiply every coefficient and its enthalpy by the same factor. Add the resulting equations, then cancel identical species in identical states. A chemically different substance or a different state cannot be cancelled merely because it contains the same element.
Worked cycle: find a missing direct change
This original paper problem supplies rounded teaching data at one common reference condition. They are not new experimental measurements. Step A: C(s, graphite) + O₂(g) → CO₂(g), ΔHA = −394 kJ per equation as written. Step B: CO(g) + ½O₂(g) → CO₂(g), ΔHB = −283 kJ per equation as written. Find ΔH for C(s, graphite) + ½O₂(g) → CO(g).
| Route | Inventory changes | Enthalpy |
|---|---|---|
| Direct | C(s, graphite) + O₂(g) → CO₂(g) | −394 kJ |
| Via carbon monoxide | C(s, graphite) + O₂(g) → CO(g) + ½O₂(g) → CO₂(g) | x + (−283 kJ) |
The unused ½O₂ carried through the first part of the indirect route is explicit, so both routes have exactly the same endpoints. Therefore x − 283 = −394 and x = −111 kJ for formation of one mole of CO in the written target. A Hess cycle does not imply that a pure CO preparation follows from combining carbon and oxygen in a particular way.
Check by equation addition instead of relying on a diagram: reverse B to CO₂(g) → CO(g) + ½O₂(g), ΔH = +283 kJ. Add it to A. Cancel CO₂, then cancel ½O₂ from the oxygen appearing on both sides. The target remains, and −394 + 283 = −111 kJ. Adding −394 and −283 without reversing B would correspond to the wrong net reaction.
Worked cycle: a state change changes the answer
A separate original exercise supplies H₂(g) + ½O₂(g) → H₂O(l), ΔH = −286 kJ per equation, and H₂O(l) → H₂O(g), ΔH = +44 kJ per equation. Both refer to matching reference conditions and one mole of water. Find the enthalpy for forming water vapour, then for the equation with whole-number coefficients.
Add the equations: the liquid water cancels, leaving H₂(g) + ½O₂(g) → H₂O(g). ΔH = −286 + 44 = −242 kJ for one mole of water vapour. Double everything to get 2H₂(g) + O₂(g) → 2H₂O(g), ΔH = −484 kJ per equation as now written. The molar value per mole of water remains −242 kJ mol⁻¹.
Formation of the vapour is less exothermic than formation of the liquid in this supplied account because vaporisation requires energy. “Both are water” is not enough to ignore their states. Nor does doubling the equation double the value per mole of water; it doubles the amount of water described.
Independent cycle calculations
- New supplied rounded data: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890 kJ; C(s, graphite) + O₂(g) → CO₂(g), ΔH = −394 kJ; H₂(g) + ½O₂(g) → H₂O(l), ΔH = −286 kJ. Construct and calculate the change for C(s, graphite) + 2H₂(g) → CH₄(g).
- Using the state-change exercise above, find ΔH for 2H₂O(g) → 2H₂(g) + O₂(g), and then the energy change for decomposition of 0.250 mol water vapour on that basis.
Checked cycle answers
1. Add the graphite equation and twice the liquid-water equation: the products are CO₂ + 2H₂O and ΔH = −394 + 2(−286) = −966 kJ. Add the reversed methane-combustion equation, ΔH = +890 kJ. CO₂, 2H₂O and 2O₂ cancel, leaving C + 2H₂ → CH₄. ΔH = −966 + 890 = −76 kJ per equation, or −76 kJ mol⁻¹ of CH₄ formed. The target contains C = 1 and H = 4 on both sides. 2. Reverse the two-water equation: ΔH = +484 kJ per equation. For 0.250 mol H₂O(g), +484 × (0.250/2) = +60.5 kJ. This is the supplied thermodynamic change, not a statement about the mechanism or actual energy efficiency of a device.
A final audit should show the uncancelled target, its physical states, its amount basis and its sign. The existing phase-change lesson distinguishes energy for a temperature rise from energy for a change of state; neither should silently replace the reaction enthalpy in a cycle.
Background: OpenStax, Hess’s law and thermochemical equations. All task wording, cycle layout and worked arithmetic above are original. The supplied rounded data are teaching values and are kept consistent within each exercise.
Frequently asked questions
Why does a temperature rise give negative reaction enthalpy?
The solution gains heat while the reaction releases it. Under the stated ideal balance, the two heat values have equal magnitude and opposite sign.
Should I add 273 to the temperature change?
No. A temperature difference of one degree Celsius equals a difference of one kelvin. Calculate final minus initial; adding 273.15 converts an individual Celsius temperature to kelvin, not its change.
Which mass belongs in mcΔT?
Use the mass whose temperature change and specific heat capacity are supplied. For the mixed-solution cases here, that includes both solution volumes under the stated density and additive-volume assumptions.
Which number of moles should I divide by?
Use the reacting amount for the defined molar basis. In the one-to-one neutralisations here, that is the moles of water formed, fixed by the limiting reactant.
Can I always use 4.20 J g⁻¹ K⁻¹?
No. Use the given specific heat capacity or an explicitly authorised approximation. A solution need not have exactly the same value as water.
What if the calorimeter absorbs heat too?
Include its heat, CΔT, when its heat capacity C is supplied and significant. Then reaction heat balances both solution and calorimeter heat, with any other stated terms.
Are these standard enthalpy values?
No. They are results calculated from fictional supplied records under stated assumptions. A standard value requires specified reference conditions and appropriate evidence.
Sources and further reading
The references below support the scientific background. The cases, calculations and illustrations in this lesson are original. External teaching guidance does not establish current KCSE marking rules or grade placement.
Continue your revision
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