Enthalpy, Hess's law and calorimetry: the bookkeeping of heat, worked three ways on one screen

Enthalpy change, ΔH, is the heat a reaction gives out or takes in at constant pressure: negative when heat leaves (exothermic), positive when it enters (endothermic). You can measure it in a calorimeter from the temperature change of the water, or compute it by adding known reactions (Hess's law) or from tabulated enthalpies of formation. Below, each method is worked once with every line shown, the sign error that flips an answer is named, and a problem is left for you.

What enthalpy is, and the sign convention

Enthalpy, H, is a property of a system, the reaction mixture in your beaker. You never measure H itself. You measure the change, ΔH, and ΔH is the heat that crosses the boundary between the system and its surroundings while the pressure stays constant, which it does in an open beaker.

The sign is always the system's. Heat leaving the reaction is negative; heat entering the reaction is positive. The water in a coffee-cup calorimeter is the surroundings, not the system. So the heat the water gains is the heat the reaction lost, and the one line that carries the whole method is q_reaction = −q_water.

Exothermic means ΔH is negative: the reaction releases heat and the surroundings warm. On an energy diagram the products sit below the reactants, and the drop is ΔH. Endothermic means ΔH is positive: the reaction absorbs heat and the surroundings cool, and the products sit above the reactants.

Units: q is in joules or kilojoules, because it is a quantity of heat. ΔH is in kJ/mol, because it is heat per mole of the reaction as written. Write the equation with double the coefficients and ΔH doubles. That sentence is the whole of Hess's law in miniature.

In SCH4U this is two ideas in strand D, Enthalpy changes and calorimetry and then Hess's law and enthalpies of formation; the SCH4U course map places them. The three methods below are the three ways a test asks for ΔH, and the sign rule is the same in all three.

Method 1: calorimetry, q = mcΔT

A calorimeter measures heat by letting it change the temperature of a known mass of water. The relation is q = mcΔT: mass, specific heat capacity, temperature change. For water, c = 4.18 J/(g·°C), and the problem states that value so you are never asked to remember its last digit.

The problem. 2.00 g of sodium hydroxide (molar mass 40.00 g/mol) dissolves in 100.0 g of water in a coffee-cup calorimeter. The temperature rises from 22.0 °C to 27.3 °C. Find the enthalpy of solution per mole of NaOH.

• ΔT = 27.3 °C − 22.0 °C = 5.3 °C (subtracting; two significant figures, and that will cap the answer)
• q_water = m × c × ΔT = 100.0 g × 4.18 J/(g·°C) × 5.3 °C = 2215 J (multiplying; g and °C cancel, J remains; the water gained this heat)
• q_reaction = −2215 J (changing the sign; the dissolving released what the water gained)
• n(NaOH) = 2.00 g ÷ 40.00 g/mol = 0.0500 mol (dividing mass by molar mass)
• ΔH = −2215 J ÷ 0.0500 mol = −44,300 J/mol = −44 kJ/mol (dividing heat by moles, then converting to kJ)

Two significant figures, because ΔT has two. The thermometer's resolution is what limits the precision of a calorimetry result, and saying so in a lab write-up is worth a mark. If you carry −44.3 through later steps, fine; report −44.

The assumptions, which a report must state. The solution's heat capacity is taken as water's. The cup absorbs no heat. And the mass in q = mcΔT is the water's, 100.0 g, not the solution's. That last point is a course convention: some courses use the solution's mass, 102.0 g, which gives q = 2260 J and ΔH = −45 kJ/mol. Neither is wrong; use the one your course uses and say which. The moles in the denominator come from the stoichiometry page's mass-to-moles step, and nothing else in this calculation is new.

The common wrong line. ΔH = +44 kJ/mol, "because the temperature went up". The temperature of the water went up because the reaction gave the water heat. Heat left the reaction, so the reaction's ΔH is negative. Two more: dividing by 2.00 g instead of 0.0500 mol (which gives J/g, not J/mol), and putting 2.00 g in as m in q = mcΔT (the mass that warmed is the water's). If you have a lab to write up, the lab-report page has the sections and where the assumptions go.

Method 2: Hess's law, adding equations

Hess's law says that ΔH depends only on where a reaction starts and where it ends, not on the route between them. So if you can write the reaction you want as a sum of reactions whose ΔH values you know, its ΔH is the same sum. Two moves are allowed when you rearrange: reverse an equation and flip the sign of its ΔH; multiply an equation by a number and multiply its ΔH by the same number.

The problem. Find ΔH for C(s) + ½ O₂(g) → CO(g), given:

• (1) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ

• (2) CO(g) + ½ O₂(g) → CO₂(g), ΔH = −283.0 kJ

• Keep (1) as written: it has C(s) on the left, where the target wants it. ΔH = −393.5 kJ.

• Reverse (2) to put CO(g) on the right: CO₂(g) → CO(g) + ½ O₂(g), ΔH = +283.0 kJ (the sign flips with the reversal)
• Add the two: C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½ O₂(g)
• Cancel CO₂(g) from both sides, and ½ O₂(g) from the O₂(g) on the left: C(s) + ½ O₂(g) → CO(g) (the target, which is the check that the rearrangement was right)
• ΔH = −393.5 kJ + 283.0 kJ = −110.5 kJ (adding the two ΔH values the same way the equations were added)

Four significant figures, because both given values carry four. The target is the formation of one mole of CO from its elements, so this number is also the standard enthalpy of formation of CO(g), and it agrees with the tabulated value. That agreement is a good sign that the method is sound, not a coincidence.

The common wrong line. Adding (1) and (2) as given: −393.5 + (−283.0) = −676.5 kJ. That is the enthalpy of a reaction nobody asked about, C + CO + 1½ O₂ → 2 CO₂. The clue is on the page: CO appears on the left of the sum and the target has it on the right, so nothing cancels into the target. Always write the summed equation out and cancel before you touch the numbers. If the species do not cancel into the target, a rearrangement is wrong.

Method 3: enthalpies of formation

A standard enthalpy of formation, ΔH°_f, is the ΔH for making one mole of a substance from its elements in their standard states. An element in its standard state has ΔH°_f = 0 by definition: O₂(g), C(s) as graphite, H₂(g). For any reaction:

ΔH°_rxn = Σ n ΔH°_f(products) − Σ n ΔH°_f(reactants)

where n is each coefficient. Products minus reactants, every term multiplied by its coefficient.

The problem. Methane burns: CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l). The values are given in the problem, in kJ/mol (they are the usual tabulated values to one decimal place; published tables differ by a few tenths for some species): ΔH°_f CH₄(g) = −74.6; CO₂(g) = −393.5; H₂O(l) = −285.8; O₂(g) = 0.

• Products: 1 × (−393.5) + 2 × (−285.8) = −393.5 − 571.6 = −965.1 kJ (each product's value times its coefficient, summed)
• Reactants: 1 × (−74.6) + 2 × 0 = −74.6 kJ (the oxygen contributes nothing)
• ΔH° = −965.1 − (−74.6) = −965.1 + 74.6 = −890.5 kJ (products minus reactants; subtracting a negative)

So burning one mole of methane releases 890.5 kJ. Four significant figures, from the data; the equation as written is per mole of CH₄, so the answer is also −890.5 kJ/mol of methane.

The common wrong line. Reactants minus products, which gives +890.5 kJ and says that burning methane absorbs heat. Methane is a fuel; the sign is a sense check you can always run. The second wrong line is forgetting the coefficient 2 on water: (−393.5) + (−285.8) − (−74.6) = −604.7 kJ, which is wrong by one mole of water's formation. Write the coefficient beside each value before you multiply.

Enthalpy on one screen

Method	What you need	The lines
Calorimetry	mass of water, c, ΔT, moles reacted	q_water = mcΔT; q_rxn = −q_water; ΔH = q_rxn ÷ n
Hess's law	known equations and their ΔH	reverse → flip the sign; multiply → multiply ΔH; add and cancel
Formation	ΔH°_f of each species	Σ n ΔH°_f(products) − Σ n ΔH°_f(reactants); elements are 0

• The sign rule. Negative is exothermic, heat leaving the reaction, surroundings warming. The water warming means the reaction's q is negative.
• The two Hess moves. Reverse and flip; multiply and multiply. Cancel species before adding numbers.
• Significant figures. A calorimetry answer is limited by ΔT, usually two figures. Table-based answers carry the tables' precision.

Attempting the problems below before reading their answers is chapter 3 of the free playbook: the try comes first, and the check comes after.

Try one cold

Cover the page above. Three short problems, one per method.

(a) Hess's law. Find ΔH for 2 H₂(g) + O₂(g) → 2 H₂O(l), given H₂(g) + ½ O₂(g) → H₂O(l), ΔH = −285.8 kJ.

(b) Enthalpies of formation. Ethanol burns: C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l). Given ΔH°_f: ethanol(l) −277.6, CO₂(g) −393.5, H₂O(l) −285.8 kJ/mol, O₂(g) 0.

(c) Calorimetry. A reaction in a coffee-cup calorimeter warms 50.0 g of water by 4.0 °C, with c = 4.18 J/(g·°C). How much heat did the reaction release?

Check

• (a) The target is the given equation multiplied by 2, so ΔH = 2 × (−285.8) = −571.6 kJ. One move, no reversal.
• (b) Products: 2 × (−393.5) + 3 × (−285.8) = −787.0 − 857.4 = −1644.4 kJ. Reactants: −277.6 kJ. ΔH° = −1644.4 − (−277.6) = −1366.8 kJ per mole of ethanol.
• (c) q_water = 50.0 g × 4.18 J/(g·°C) × 4.0 °C = 836 J, so the reaction released 836 J, reported as 0.84 kJ (two significant figures, from the 4.0 °C). If you were asked for ΔH you would still need the moles that reacted; the question did not give them, so q is where this one stops.

If (a) went wrong, you reversed when you should have multiplied. If (b) came out positive, you subtracted the wrong way round. If (c) came out positive, reread the sign rule once more.

Where this sits in the course

These are ideas 11 and 12 of 24 in SCH4U, Enthalpy changes and calorimetry and Hess's law and enthalpies of formation, the first two ideas of strand D. In the course order they come after strand C's Intermolecular forces and physical properties, and they rest on Gas laws and stoichiometry (prerequisite review): the moles in every denominator, and the coefficients that multiply every ΔH°_f.

The ideas after them are 13, 14 and 15: Reaction rates and factors affecting them, Rate laws and reaction order, and Activation energy, catalysts and reaction mechanisms. The energy diagram there has two floors, reactants and products, and the gap between the floors is this page's ΔH; the hill between them is a different quantity, activation energy, which the rate laws and activation energy page works. Bond enthalpies, a fourth way to estimate ΔH, are a note in this course rather than a method, and are not worked here.

The habit of combining equations returns in the equilibrium strand that follows, where adding two equations multiplies their equilibrium constants instead of adding them. Every idea of the three science courses has a page like this one.

How to practise it

One problem of each method a night, mixed, for a week. Rohrer and Taylor (2007) found that students who practised problem types shuffled together did worse during practice and much better on the test a week later than students who practised each type in its own block, because the shuffled set makes you decide which method the problem wants, and that decision is what a test asks. Sweller and Cooper (1985) found that studying a worked example and then solving a matching problem teaches a procedure faster than solving problems alone, which is why each method above is one example and one sibling.

Keep an error log in two columns: the sign, and the step. Most calorimetry errors are the sign; most Hess errors are a species that did not cancel; most formation errors are a coefficient. Mixed practice has the schedule, and how to study chemistry treats this strand as what it is: three procedures, each a short list of lines.