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Energetics

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Edexcel IAL Chemistry · Energetics

Energetics

Every chemical reaction is really just a story about breaking old bonds and making new ones — and whether that story releases energy to the surroundings (exothermic) or steals energy from them (endothermic) determines everything else in this chapter.

Summary — What This Chapter Covers

  • Enthalpy level diagrams show whether reactants or products have more stored chemical energy, telling you if a reaction is exothermic (ΔH negative) or endothermic (ΔH positive).
  • Reaction profile diagrams add the missing piece — activation energy and the unstable "transition state" at the peak of the curve.
  • Standard conditions (100 kPa, 298 K, standard states) let us fairly compare enthalpy changes between different reactions.
  • Four named enthalpy changes to know cold: reaction, formation, combustion, neutralisation — each with a precise one-mole definition.
  • Calorimetry is how we measure enthalpy changes experimentally, using q = mcΔT for solutions and burning fuels to heat water for combustion.
  • Hess's Law lets us calculate enthalpy changes we can't measure directly, by building "indirect route" energy cycles through elements or combustion products.
  • Bond enthalpies give an alternative way to estimate ΔH: energy in to break bonds, energy out to make new ones.

1. Enthalpy Level Diagrams

Think of enthalpy as the total amount of chemical energy "stored" inside a substance — like the energy stored in a stretched spring or a full battery. We can never measure the absolute amount of this stored energy, but we can measure the change in it when a reaction happens. That change is called the enthalpy change, symbol ΔH (Δ = "change in", H = "enthalpy").

Here's the key mental model: picture enthalpy as a height on a vertical axis. Reactants sit at one height, products sit at another. The reaction is basically a trip from one height to the other, and ΔH is simply how far you moved and in which direction.

Exothermic Reactions (ΔH is negative)

In an exothermic reaction, the products end up with less energy than the reactants started with. Where did that "missing" energy go? It didn't disappear — it was released as heat into the surroundings. That's why exothermic reactions make the test tube feel warm: the chemical system is losing energy, and the surroundings are gaining it.

  • Energy of the system decreases
  • Temperature of the surroundings increases (you can feel/measure this)
  • ΔH is written as a negative number
ENERGY ▲ │ REACTANTS ────┐ │ │ ↓ energy released (−ΔH) │ └──── PRODUCTS └─────────────────────────────► extent of reaction
Why "thermodynamically possible" isn't the whole story
Exothermic reactions are energetically favourable (reactants are higher in energy, so rolling "downhill" to products makes sense). But if the reaction is too slow, it might not visibly happen at room temperature — it's kinetically controlled. Diamond turning into graphite is a classic example: it's exothermic and thermodynamically favourable, but so slow it never visibly happens.

Endothermic Reactions (ΔH is positive)

Endothermic is the mirror image: products end up with more energy than the reactants. That extra energy has to come from somewhere — it's absorbed from the surroundings, which is why endothermic reactions often feel cold (think of the cooling packs used for sports injuries, which rely on an endothermic dissolving process).

  • Energy of the system increases
  • Temperature of the surroundings decreases
  • ΔH is written as a positive number
ENERGY ▲ │ ┌──── PRODUCTS │ │ ↑ energy absorbed (+ΔH) │ REACTANTS ────┘ └─────────────────────────────► extent of reaction
Quick Rule
Exothermic → ΔH is negative  |  Endothermic → ΔH is positive
If the reaction gives out heat, it's "losing" enthalpy, hence the minus sign. If it takes in heat, it's "gaining" enthalpy, hence the plus sign.
Practice Question

A student mixes two solutions and notices the test tube gets noticeably colder. Is this reaction exothermic or endothermic? What is the sign of ΔH, and what has happened to the energy of the chemical system?

Reaction Profile Diagrams

Enthalpy level diagrams only show the start and end points — they don't tell you anything about how the reaction gets from reactants to products. That's where reaction profile diagrams come in. They add a curved "hill" between reactants and products, and that hill introduces two new ideas:

Activation Energy (Eₐ) The minimum energy reactant molecules need for a successful collision that starts the reaction. It's the vertical gap between the reactants' energy level and the peak of the curve.
Transition State The unstable, fleeting arrangement at the very peak of the curve, where old bonds are half-broken and new bonds are half-formed. It can never be isolated — it exists for a fraction of a second.

A crucial link: because exothermic reactions start with reactants that are already closer in energy to the transition state (higher up the diagram), they generally have a lower activation energy than endothermic reactions, where the reactants sit further below the transition state.

Don't Mix These Two Diagrams Up
Enthalpy level diagrams = flat lines only, no activation energy shown, no curve.
Reaction profile diagrams = curved hill, activation energy and transition state included.
Examiners specifically test this distinction — if asked to draw activation energy, you need the profile diagram, not the level diagram!
Practice Question

Sketch (in words) the key difference between the activation energy of an exothermic reaction and an endothermic reaction, and explain why this difference exists.

2. Enthalpy Change — Definitions

To compare enthalpy changes fairly between different experiments (maybe done in different labs, on different days, at different pressures), chemists agree on a fixed set of standard conditions. Think of it like agreeing to always measure people's height while they're standing on flat ground, not on a hill — otherwise the comparison is meaningless.

Standard Conditions
Pressure = 100 kPa  |  Temperature = 298 K (25 °C)  |  Standard physical states
The symbol (or ≡) on ΔH shows a value was measured under these standard conditions, e.g. ΔHf⦵.
Physical states matter — a lot
Changing physical state can massively change ΔH. Compare:
NaCl (s) → Na⁺(aq) + Cl⁻(aq)   ΔH = +4 kJ/mol
NaCl (g) → Na⁺(g) + Cl⁻(g)   ΔH = +500 kJ/mol
Same substance, wildly different values — always include state symbols in your equations!

The Four Named Enthalpy Changes

These four definitions are the backbone of the whole chapter — they show up constantly in Hess's Law questions, and examiners love asking you to identify which one an equation represents. Learn the exact wording, especially the phrase "one mole", because that's what makes each definition precise.

TypeDefinitionSymbolExo/Endo
ReactionThe enthalpy change when the reactants in the stoichiometric equation react to give the products, under standard conditionsΔHrEither
FormationThe enthalpy change when one mole of a compound is formed from its elements, under standard conditionsΔHfEither
CombustionThe enthalpy change when one mole of a substance is burnt completely in excess oxygen, under standard conditionsΔHcAlways exothermic
NeutralisationThe enthalpy change when one mole of water is formed by reacting an acid and an alkali, under standard conditionsΔHneutAlways exothermic

A fifth definition, used less at AS level but still worth knowing: Standard enthalpy of atomisation — the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state, under standard conditions.

Golden rule about formation
The ΔHf⦵ of any element in its standard state is always zero. This makes sense — you're not really "forming" an element from itself, so there's no enthalpy change to speak of. E.g. ΔHf⦵ of O₂(g) = 0 kJ/mol. This fact is the secret weapon behind almost every Hess cycle in this chapter!
Practice Question

Identify which type of enthalpy change (reaction, formation, combustion, or neutralisation) each of these represents:
(a) C (graphite) + O₂(g) → CO₂(g)
(b) HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

3. Using Calorimetry

Calorimetry is simply the practical technique of measuring enthalpy changes by tracking temperature. The core idea across every version of this experiment: heat energy released or absorbed by a reaction changes the temperature of a known mass of liquid (usually water), and since we know water's specific heat capacity, we can work backwards to calculate exactly how much energy was transferred.

The Core Calorimetry Equation
q = m × c × ΔT
q = heat transferred (J)  |  m = mass of water/solution (g)  |  c = specific heat capacity (4.18 J g⁻¹ K⁻¹ for water)  |  ΔT = temperature change (K or °C — the size of a degree is the same in both)

Reactions in Solution (e.g. displacement, neutralisation)

You carry out the reaction directly inside an insulated cup (polystyrene cup, vacuum flask, or metal can), stir continuously, and record temperature over time. One reagent is used in excess so the other is fully consumed — this makes the "number of moles reacted" calculation straightforward.

┌─────┐ │ 🌡 │ ← thermometer (reads to 0.2°C) ___│_____│___ │ reaction │ ← polystyrene cup │ mixture │ (acts as calorimeter) └─────────────┘

To actually calculate an enthalpy change per mole, we divide the total energy transferred by the number of moles of the limiting reagent:

Enthalpy Per Mole
ΔH = q ÷ n
n = moles of the limiting reagent. Remember: if temperature rose, the reaction released heat, so ΔH must come out negative — don't forget the sign!

Temperature Correction Graphs

Real reactions aren't always instant — sometimes it takes a few seconds for the maximum temperature to be reached, and during that delay heat is already leaking to the surroundings. This means the "true" maximum temperature is never actually recorded directly. The fix: plot temperature against time, then extrapolate the cooling section of the graph backwards to the exact moment the second reactant was added.

Temp T₂ ┤ ╲ │ extrapolated╲___cooling │ line ╲ ╲___section │ ↕ ΔT ╲ T₁ ┤━━━●●●━ ← steady before mixing └──────┬──────────────► Time reactant added here

You use both extrapolated lines to read off ΔT at the moment of mixing, which gives a far more accurate (and larger) value than just reading the actual peak temperature from the raw data.

Enthalpy of Combustion

Here the logic flips slightly: instead of the reaction happening in the water, we burn a fuel underneath a container of water and measure how much the water heats up. The fuel is weighed before and after burning so we know exactly how much mass was burnt.

Why combustion calorimetry is less accurate
Not all the heat produced actually reaches the water — some escapes to the surrounding air, and some is absorbed by the calorimeter (e.g. the copper can) itself. This is why experimental values for enthalpy of combustion are usually less exothermic (a smaller negative number) than the accepted data book values. Main sources of error: heat loss and incomplete combustion (producing soot/CO instead of only CO₂).
Practice Question

In a calorimetry experiment, 1.50 g of an organic liquid (Mᵣ = 58.0) is completely combusted, and the heat raises the temperature of 100 g of water from 20°C to 75°C. Calculate the enthalpy of combustion of the liquid.

4. Using Hess Cycles

Some enthalpy changes simply can't be measured directly in a lab. For example, you can't easily react carbon and hydrogen together to directly form propane under standard conditions. So how do chemists find ΔHf⦵ for propane? They use Hess's Law.

The Mountain Analogy
Imagine hiking up a mountain. Whether you take the steep direct path or a long winding trail with several stops, your total change in altitude from base to summit is exactly the same. Hess's Law says enthalpy works the same way: it doesn't matter which "route" a reaction takes — direct or via several indirect steps — the total enthalpy change is identical, as long as start and end points match.
Hess's Law (Official Wording)
"The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place, as long as the initial and final conditions are the same."

Building a Cycle Using ΔHf (Formation Data)

When you're given enthalpy of formation data, you build a cycle where elements sit at the bottom, with arrows pointing up to both reactants and products (because formation always goes from elements upward to a compound).

REACTANTS ────ΔHr────► PRODUCTS ▲ ▲ │ ΔH1 │ ΔH2 │ │ └────── ELEMENTS ───┘ Rule: ΔH2 = ΔH1 + ΔHr Rearranged: ΔHr = ΔH2 − ΔH1
  1. Write the balanced equation for the reaction
  2. Write the elements (with correct moles + state symbols) underneath
  3. Draw upward arrows from elements to both reactants and products
  4. Label each arrow with the ΔHf value × number of moles
  5. Go around the cycle: reverse the sign of any value if you travel against the arrow's direction
Practice Question

Given ΔHf⦵ values: H₂O(g) = −242, CO₂(g) = −394, NH₄NO₃(s) = −365, C(s) and N₂(g) = 0 kJ/mol, calculate ΔHr⦵ for:
NH₄NO₃(s) + ½C(s) → N₂(g) + 2H₂O(g) + ½CO₂(g)

Building a Cycle Using ΔHc (Combustion Data)

Combustion data is often easier to measure experimentally than formation data, so it's frequently used instead. Here the logic flips: since burning things always converts them down to combustion products (CO₂ and H₂O), the arrows point downward to the combustion products, which sit at the bottom of the cycle this time.

REACTANTS ────ΔHr────► PRODUCTS │ │ │ ΔHc(react) │ ΔHc(prod) ▼ ▼ └── COMBUSTION PRODUCTS ──┘ (e.g. CO₂ + H₂O) Rule: ΔHr = ΔHc(reactants) − ΔHc(products)
  1. Write the equation for formation of the compound
  2. Write the combustion products (usually CO₂ + H₂O) below the equation
  3. Draw downward arrows from every substance to its combustion products
  4. Label arrows with ΔHc × moles
  5. Travel around the cycle, flipping signs against arrow direction
Practice Question

Using ΔHc⦵ values: C(s) = −394, H₂(g) = −286, CH₃COCH₃(l) = −1821 kJ/mol, calculate ΔHf⦵ of propanone from:
3C(s) + 3H₂(g) + ½O₂(g) → CH₃COCH₃(l)

Practical tip that saves marks
Keep every ΔH value inside its own bracket while doing the arithmetic — it's incredibly easy to accidentally drop a minus sign when combining several negative numbers in a row. Also always double-check your equation is balanced before building the cycle — an unbalanced starting equation guarantees a wrong final answer.

5. Bond Enthalpies

Zoom right into a reaction at the molecular level, and every reaction is really just a rearrangement of atoms — old bonds break, new bonds form. Since bonds are forces of attraction, pulling atoms apart always costs energy (endothermic), while atoms coming together to form new bonds releases energy (exothermic).

BOND BREAKING (needs energy in) = ENDOTHERMIC ●─●─● +energy→ ● ● ● BOND MAKING (releases energy) = EXOTHERMIC ● ● ● → ●─●─● +energy released
Bond Dissociation Enthalpy The exact energy needed to break one specific bond in one specific molecule. Also just called "bond enthalpy" or "bond energy".
Average Bond Enthalpy The energy needed to break one mole of a type of bond in a gaseous molecule, averaged across many different (similar) compounds — since the exact same bond type behaves slightly differently depending on its surroundings.

Why "average"? Take methane, CH₄. You might assume all four C–H bonds are identical, but breaking the first C–H bond is actually easier than breaking the second, because the remaining hydrogens get pulled in more strongly toward the carbon once one hydrogen leaves. Since we can't practically measure each individual bond's energy, we take the total dissociation energy for the whole molecule and divide by the number of bonds — then compare that average against similar compounds to settle on an accepted value.

Overall Enthalpy Change from Bond Enthalpies
ΔrH⦵ = (energy for bonds broken) + (energy for bonds formed)
Bonds broken values are positive (endothermic — costs energy). Bonds formed values are negative (exothermic — releases energy). Add them together to get the overall ΔH.

If more energy is released forming new bonds than was spent breaking old ones, the overall reaction is exothermic and the products are more stable than the reactants. If the opposite is true, the reaction is endothermic and the products are less stable.

Worked Example — Haber Process

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), using average bond energies: N≡N = 945, H–H = 436, N–H = 391 kJ/mol

Bonds BrokenBonds Formed
1 × N≡N = 945
3 × H–H = 3 × 436 = 1308
6 × N–H = 6 × 391 = 2346
Total = +2253Total = −2346

ΔrH⦵ = (+2253) + (−2346) = −93 kJ/mol

Practice Question

Using average bond enthalpies: C–H = 414, C≡C = 839, O=O = 498, C=O = 804, O–H = 463 (all kJ/mol), calculate the enthalpy of combustion of ethyne:
2C₂H₂(g) + 5O₂(g) → 2H₂O(g) + 4CO₂(g)
(Hint: simplify to one mole of ethyne first — combustion is defined per mole of fuel.)

Bond enthalpies predict reactivity too
Bonds with high bond enthalpy values are strong and need lots of energy to break — reactions involving them often need heating or a catalyst. Bonds with low bond enthalpy values are weaker and break more easily — reactions involving them can often happen at room temperature. This is why the weakest bond in a molecule is usually the first one to break in a reaction.

What to Memorise

q = mcΔTHeat transferred = mass × specific heat capacity × temperature change
ΔH = q ÷ nEnthalpy change per mole = energy transferred ÷ moles of limiting reagent
Specific heat capacity of water4.18 J g⁻¹ K⁻¹
Standard conditions100 kPa, 298 K, standard physical states
ΔHf⦵ of any elementAlways exactly 0 kJ/mol
Bond enthalpy equationΔrH⦵ = bonds broken (+) + bonds formed (−)
Exothermic signΔH is negative; temperature of surroundings rises
Endothermic signΔH is positive; temperature of surroundings falls
Hess's LawTotal ΔH is independent of route, as long as start/end points match
Combustion definitionOne mole of substance burnt completely in excess oxygen, standard conditions

Concepts Checklist

Exam Tips & Common Mistakes

Forgetting the sign
The single most common mistake in this chapter: students calculate a correct magnitude but forget whether ΔH should be positive or negative. Always ask yourself: "did temperature rise (exothermic, negative) or fall (endothermic, positive)?" before writing your final answer.
Mixing up enthalpy level and reaction profile diagrams
If a question asks you to show activation energy or a transition state, you MUST use a reaction profile diagram (with the curved hill), not a flat enthalpy level diagram.
Missing state symbols
Enthalpy values can change dramatically with physical state. Always include state symbols in equations — a question about NaCl(s) dissolving is completely different from NaCl(g) dissociating.
Not multiplying by coefficients in Hess cycles
If 2 moles of a product form, you must multiply that substance's ΔHf value by 2 before using it in your cycle. Forgetting this is one of the most common ways marks are lost in Hess's Law questions.
Losing sign when reversing a cycle arrow
Whenever you travel against the direction of an arrow in a Hess cycle, you must flip the sign of that enthalpy value. Keep your working laid out clearly with brackets around each value — bond enthalpy and Hess's Law questions often carry 3 marks, with 2 available purely for showing correct working even if your final number is wrong.
Not drawing full structures for bond enthalpy questions
It's very easy to miss a bond (especially double or triple bonds counted as "one" bond by accident) if you don't draw the full displayed structure first. Always sketch every atom and every bond before starting the calculation.

What examiners are actually looking for:

  • Precise definitions using the exact phrasing (e.g. "one mole", "standard conditions", "excess oxygen") — vague wording loses marks even if the general idea is right.
  • Clear, methodical working in multi-step calculations — don't skip straight to an answer.
  • Correct sign conventions throughout, especially in final answers.
  • Recognising which named enthalpy change type an equation represents (sometimes more than one applies at once).
  • Awareness of experimental limitations — heat loss, assumptions about specific heat capacity of solutions being the same as water, incomplete combustion.
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