3aEnergetics

Every reaction either gives energy out or takes it in. Which one it does is decided by watching the surroundings, not the reaction itself.

Exo means out, endo means in. The temperature you measure is the surroundings — usually the water or solution the reaction happens in.

So a solution that warms up means an exothermic reaction. One that cools down means an endothermic one.

Examples worth knowing

Exothermic: combustion, neutralisation, displacement reactions. Most reactions are exothermic.

Endothermic: thermal decomposition, and dissolving certain salts such as ammonium nitrate.

Thermal decomposition is the safest endothermic example to quote, and there is a good reason: it needs continuous heating. Stop heating and the reaction stops, because it cannot sustain itself. An exothermic reaction often keeps going once started.

Measuring the change

Calorimetry experiments use a polystyrene cup with a lid, and both parts matter:

• Polystyrene is a good insulator, so less heat escapes to the room.
• The lid cuts heat loss from the surface.

Any heat exchanged with the room makes the measured temperature change too small, so insulation improves accuracy directly.

Stir before reading. Without stirring the liquid near the reaction is hotter than the rest, and the reading depends on where the thermometer happens to sit.

Record the highest (or lowest) temperature reached, not the temperature after a fixed time — the reaction may still be going.

Comparing two substances

Only the substance may change. Keep the same volume of water, the same mass of solid, and the same starting temperature. And compare the temperature change, not the final temperature.

The heat energy equation

Q = mcΔT

m is the liquid, not the reactant. Using the mass of the magnesium instead of the mass of the water is the commonest error in the whole topic. The equation calculates the energy gained or lost by the liquid.

Q comes out in joules when m is in grams. Divide by 1000 for kilojoules.

Worked example

50.0 g of water rises by 20.0 °C:

Q = 50.0 × 4.2 × 20.0 = 4200 J = 4.2 kJ

Molar enthalpy change

ΔH is the energy change per mole:

ΔH = Q in kJ ÷ amount in moles

Convert joules to kilojoules before dividing, or the answer is a thousand times too large.

Worked example

4200 J released, 0.0500 mol reacted:

4.2 kJ ÷ 0.0500 mol = 84, so ΔH = −84 kJ/mol

The sign

The negative sign is part of the answer for an exothermic reaction, and omitting it usually loses a mark.

ΔH does not change with scale

Double the amounts and you double Q, but ΔH stays the same — because it is energy per mole. Two students using 25 cm³ and 50 cm³ of the same acid should calculate the same ΔH.

Why experimental values fall short

A measured enthalpy change is always less exothermic than the accepted value. Say it that way rather than "smaller", which is ambiguous for a negative number — −980 is numerically larger than −1367.

The reasons are all forms of energy escaping:

• heat lost to the surroundings
• heat used warming the container
• incomplete combustion, releasing less energy per mole

Each makes the temperature rise smaller, and therefore the calculated value smaller in magnitude.

The assumptions

These calculations assume the solution has the same specific heat capacity and density as water, and that all the heat goes into the solution. For dilute solutions those are reasonable — and they are why experimental values differ from accepted ones.

Separate Chemistry only.

Breaking takes, making gives

Every reaction does both. Whether it is exothermic overall depends on which is larger.

• More energy released making bonds than needed breaking them → exothermic, ΔH negative.
• More needed breaking than released making → endothermic, ΔH positive.

Calculating ΔH from bond energies

ΔH = energy to break bonds − energy released making bonds

It is a subtraction because the two have opposite signs: energy in counts positive, energy out counts negative.

Worked example

H₂ + Cl₂ → 2HCl, with H–H = 436, Cl–Cl = 242, H–Cl = 431 kJ/mol.

1. Bonds broken: 436 + 242 = 678
2. Bonds made: 2 × 431 = 862
3. ΔH = 678 − 862 = −184 kJ/mol

Negative, so exothermic.

Count the bonds carefully

The coefficients matter. In N₂ + 3H₂ → 2NH₃:

• Broken: one N≡N plus three H–H
• Made: 2NH₃ contains six N–H bonds, not three

Miscounting there is the largest single source of error in these calculations.

Set the working out in three lines — bonds broken, bonds made, subtract — because each line earns its own mark even if the final number goes wrong.

Energy level diagrams

Energy runs up the vertical axis, progress of reaction across the horizontal.

The vertical gap between the two levels is ΔH.

Activation energy

The curve rises to a peak before falling. The height of that peak above the reactants is the activation energy — the minimum energy a collision needs for a reaction to happen.

Every reaction has one, including exothermic ones, because bonds must be broken before any energy is released. That is why a fuel does not burn until it is lit.

Examiners report the same handful of errors on this material every year.

1. Bond breaking and bond making the wrong way round

Examiners note that candidates often fail to distinguish clearly between bond breaking being endothermic and bond making being exothermic, and that answers such as "more energy was needed for bond breaking than bond making" appear when the reverse was true.

Fix the phrase in your head: breaking takes, making gives. Then read the calculation to check which is larger.

2. Using the wrong mass in Q = mcΔT

m is the mass of the liquid being heated. Not the reactant, not the total, not the mass of the container.

3. Forgetting to convert joules to kilojoules

Q comes out in joules. ΔH is quoted in kJ/mol. Divide by 1000 before dividing by the moles, not after.

4. Dropping the negative sign

An exothermic ΔH is negative. Writing 84 instead of −84 loses the mark that the sign carries, because the sign is what says which direction the energy went.

5. Saying "smaller" about a negative value

If an experimental value is −980 and the accepted one is −1367, the experimental value is less exothermic or smaller in magnitude. Calling it simply "smaller" is ambiguous, since −980 is the larger number.

6. Miscounting bonds in 2NH₃

Two ammonia molecules hold six N–H bonds. Write out how many of each bond type appear on each side before multiplying by the bond energies.

7. Confusing activation energy with particle energy

Examiners report this specifically about catalysts: candidates say a catalyst gives the particles more energy. It does not. It lowers the activation energy they need.

The distinction matters because it is the difference between changing the particles and changing the barrier.

The same simple apparatus measures the temperature change of four different kinds of reaction.

Apparatus

Polystyrene cup with a lid, standing in a beaker for stability, thermometer, measuring cylinder.

Method

1. Measure a fixed volume of the first solution into the cup.
2. Record the initial temperature.
3. Add the second reagent, replace the lid and stir.
4. Record the highest temperature reached — or the lowest, if the mixture cools.

ΔT = final temperature − initial temperature

The four reaction types

Dissolving is the useful one to remember, because it can go either way — ammonium salts cool the water noticeably, while sodium hydroxide warms it strongly.

Why polystyrene, and why a lid

Polystyrene is a good insulator, so less heat escapes through the walls than with a glass beaker.

The lid matters just as much: without one, heat is lost to the air above and by evaporation. Both losses make an exothermic rise look smaller than it really is.

These are the two answers to "how would you improve this experiment", and the reason to name is always reducing heat loss to the surroundings.

Reading the maximum, not the last value

Once the reaction stops, the mixture begins cooling back towards room temperature. Waiting and then reading gives a value that is too low.

The temperature to record is the highest the thermometer reaches, so watch it continuously rather than reading after a fixed time.

Calculating the energy change

Q = m × c × ΔT

where m is the mass of the solution in grams (1 cm³ of dilute solution ≈ 1 g) and c is 4.2 J/g/°C.

Dividing by the number of moles of the limiting reactant gives the energy change per mole. An exothermic reaction takes a negative sign, because energy has left the chemicals.

Where marks are lost

• Using the mass of the solid rather than the mass of the solution in Q = mcΔT.
• Recording a final temperature instead of the maximum.
• Giving an exothermic ΔH as positive.