Energy Changes¶
Syllabus mapping
CH-11-03 — Chemical Reactions. Covers: the energy changes in endothermic and exothermic reactions (including a practical investigation); relating heat energy changes to bond energy changes and the law of conservation of energy; energy profile diagrams in terms of chemical potential energy and bond breaking and forming; a laboratory experiment comparing temperature changes in substances with different heat capacities; problems using \(q = mc\Delta T\); enthalpy change problems using \(\Delta H = -\dfrac{q}{n}\); a calorimetry experiment to calculate \(\Delta H\), evaluating its accuracy against secondary-source data; how catalysts lower activation energy by providing an alternative pathway; a secondary-source investigation into the economic advantages of catalysts in an industrial process.
Wording above is paraphrased for teaching use — check the official syllabus before treating any line as verbatim NESA text.
You need to know: the difference between exothermic and endothermic reactions, why breaking bonds takes energy and forming bonds releases it, how to read and draw an energy profile diagram, and how to calculate heat energy (\(q = mc\Delta T\)) and enthalpy change (\(\Delta H = -\dfrac{q}{n}\)) from experimental data.
Assumed knowledge check¶
Before this page makes sense, you should be comfortable with:
- Moles and \(n = \dfrac{m}{MM}\) and \(n = CV\)
- Collision theory and activation energy
- Conservation of energy, and the difference between heat and temperature
Core¶
Exothermic and endothermic reactions¶
When chemists talk about energy changes, they split the world into two parts:
- the system — the reacting chemicals
- the surroundings — everything else (the water they're dissolved in, the beaker, the air, your hand)
| Exothermic | Endothermic | |
|---|---|---|
| Energy goes… | Out of the system into the surroundings | Into the system from the surroundings |
| The surroundings get… | Hotter — the temperature rises | Colder — the temperature falls |
| Sign of \(\Delta H\) | Negative (\(\Delta H < 0\)) — the system lost energy | Positive (\(\Delta H > 0\)) — the system gained energy |
| Examples | Combustion, neutralisation, respiration, metal + acid, hand warmers | Photosynthesis, thermal decomposition (e.g. heating \(\ce{CaCO3}\)), citric acid + sodium hydrogen carbonate, instant cold packs |
\(\Delta H\) ("delta H") is the enthalpy change — the heat energy absorbed or released by a reaction, usually in kilojoules per mole (\(\text{kJ mol}^{-1}\)). Memory hook: exo = exit (heat leaves); endo = enter (heat goes in).
Where the energy comes from: bonds¶
- Breaking bonds takes in energy (endothermic). You have to pull the atoms apart.
- Making bonds releases energy (exothermic). The atoms fall together into a more stable arrangement.
Every reaction breaks some bonds and makes new ones. Which one wins decides the overall energy change:
- If more energy is released making bonds than was taken in breaking them → exothermic.
- If more energy is taken in breaking bonds than is released making them → endothermic.
This is the law of conservation of energy at work: energy isn't created or destroyed — it's moved between chemical potential energy (stored in the bonds) and heat in the surroundings.
Worked example — using bond energies
Use bond energies to estimate \(\Delta H\) for \(\ce{H2 + Cl2 -> 2HCl}\).
Bond energies: H–H \(436\), Cl–Cl \(242\), H–Cl \(431\ \text{kJ mol}^{-1}\).
Bonds broken (energy in): one H–H + one Cl–Cl \(= 436 + 242 = 678\ \text{kJ}\)
Bonds made (energy out): two H–Cl \(= 2 \times 431 = 862\ \text{kJ}\)
Negative → exothermic: more energy was released making the new bonds than was used breaking the old ones.
Energy profile diagrams¶
An energy profile diagram shows how the chemical potential energy changes as a reaction happens.
- The vertical axis is chemical potential energy; the horizontal axis is the progress of the reaction.
- Reactants are on the left, products on the right.
- The hump in the middle is the activation energy (\(E_a\)) — the energy needed to start breaking bonds, measured from the reactants up to the top of the hump.
- The difference in height between reactants and products is \(\Delta H\).
| Exothermic | Endothermic | |
|---|---|---|
| Products are… | Lower than reactants | Higher than reactants |
| \(\Delta H\) arrow points… | Down | Up |
🖼️ Diagram needed — two energy profile diagrams side by side in one consistent style (this is the "free-body diagram" of Chemistry — used again on Rates of Reactions and in Year 12). (1) Exothermic: reactants high, products low, hump in between, \(E_a\) arrow from reactants to peak, \(\Delta H\) arrow pointing down from reactant level to product level. (2) Endothermic: products higher than reactants, \(\Delta H\) arrow pointing up. Label "bonds breaking" on the way up the hump and "bonds forming" on the way down. TODO — placeholder until sourced/drawn.
Catalysts on an energy profile diagram: a catalyst gives a lower hump (lower \(E_a\)) but the reactants and products stay at the same heights — so \(\Delta H\) doesn't change. A catalyst makes a reaction faster; it doesn't change how much energy it releases or absorbs.
Specific heat capacity¶
Different substances need different amounts of energy to warm up. Specific heat capacity (\(c\)) is the energy needed to raise the temperature of 1 g of a substance by 1 °C (or 1 K — same size step).
| Substance | \(c\) (\(\text{J g}^{-1}\,\text{K}^{-1}\)) |
|---|---|
| Water | 4.18 |
| Vegetable oil | about 2.0 |
| Aluminium | 0.90 |
| Copper | 0.39 |
Water has a high specific heat capacity — it takes a lot of energy to heat up. (Your data sheet gives it as \(4.18 \times 10^3\ \text{J kg}^{-1}\,\text{K}^{-1}\) — the same value per kilogram.)
The heat energy absorbed or released:
where \(q\) = heat energy (J), \(m\) = mass (g), \(c\) = specific heat capacity (\(\text{J g}^{-1}\,\text{K}^{-1}\)), \(\Delta T\) = temperature change (°C or K) \(= T_{\text{final}} - T_{\text{initial}}\).
Worked example
How much heat energy is needed to warm \(200\ \text{g}\) of water from \(20.0\ ^\circ\text{C}\) to \(35.0\ ^\circ\text{C}\)?
\(\Delta T = 35.0 - 20.0 = 15.0\ ^\circ\text{C}\)
Enthalpy change from calorimetry¶
In a calorimetry experiment, a reaction heats (or cools) a known mass of water. You measure the temperature change, calculate \(q\), then divide by the moles that reacted to get the energy per mole:
The minus sign is there because \(q\) is the heat gained by the water (surroundings), and \(\Delta H\) is for the reaction (system). If the water gains heat, the reaction lost it — so \(\Delta H\) is negative.
Units: \(q\) is usually in joules, but \(\Delta H\) is given in kJ mol⁻¹ — divide \(q\) by 1000 first.
Worked example — neutralisation
\(50.0\ \text{mL}\) of \(1.00\ \text{mol L}^{-1}\) \(\ce{HCl}\) is mixed with \(50.0\ \text{mL}\) of \(1.00\ \text{mol L}^{-1}\) \(\ce{NaOH}\) in a polystyrene cup. The temperature rises by \(6.8\ ^\circ\text{C}\). Calculate \(\Delta H\) for the neutralisation.
Step 1 — heat absorbed by the solution. Treat the solution as water: \(100\ \text{mL}\) → \(m = 100\ \text{g}\), \(c = 4.18\).
Step 2 — moles that reacted. \(n(\ce{HCl}) = 1.00 \times 0.0500 = 0.0500\ \text{mol}\) (and the same for \(\ce{NaOH}\)), so \(0.0500\ \text{mol}\) of water formed.
Step 3 — \(\Delta H\):
Negative → exothermic, matching the temperature rise. (The accepted value is about \(-57\ \text{kJ mol}^{-1}\).)
If the temperature falls (an endothermic reaction), \(\Delta T\) is negative, so \(q\) is negative, and \(\Delta H = -\dfrac{q}{n}\) comes out positive. The formula takes care of the sign for you — as long as you use \(\Delta T = T_{\text{final}} - T_{\text{initial}}\).
Calorimetry and accuracy¶
Calorimetry results are almost always less than the accepted value (smaller in size), because some heat is lost to the surroundings instead of going into the water. Common sources of error:
- Heat lost to the air and the container (the cup, beaker or can also warms up)
- For combustion: incomplete combustion (soot on the can), and heat escaping around the sides of the can
- Evaporation of the fuel or water
- Assuming the solution has the same mass and specific heat capacity as pure water
Worked example — combustion of ethanol, and evaluating accuracy
Burning \(0.50\ \text{g}\) of ethanol (\(\ce{C2H5OH}\), \(MM = 46.07\)) heats \(200\ \text{g}\) of water in a metal can by \(12.0\ ^\circ\text{C}\). Calculate \(\Delta H_{\text{combustion}}\) and compare it with the accepted value of \(-1367\ \text{kJ mol}^{-1}\).
Step 1: \(q = 200 \times 4.18 \times 12.0 = 10\,032\ \text{J} = 10.0\ \text{kJ}\)
Step 2: \(n = \dfrac{0.50}{46.07} = 0.0109\ \text{mol}\)
Step 3: \(\Delta H = -\dfrac{10.0}{0.0109} = -925\ \text{kJ mol}^{-1}\)
Evaluate: the experimental value is much smaller in size than the accepted \(-1367\ \text{kJ mol}^{-1}\):
Most of the "missing" energy heated the can and the air instead of the water. Improvements: put a draught shield around the flame, insulate and lid the can, keep the flame close to the can. See Trends and Relationships for percentage error.
Secondary-source investigation: catalysts in industry¶
Your task: research the economic advantages of using a catalyst in one industrial process. Possible choices:
- The Haber process (making ammonia for fertilisers) — an iron catalyst
- The Contact process (making sulfuric acid) — a vanadium(V) oxide catalyst
- Catalytic converters in cars — platinum-group metals
Questions to answer: what does the catalyst do to the activation energy? How does that let the process run at a lower temperature or faster? What does that save — energy, fuel, equipment, time? Does the catalyst have any costs or downsides (expensive metals, being "poisoned" by impurities, needing replacement)?
Practical¶
The syllabus lists three pracs here:
1. Measuring exothermic and endothermic changes. In polystyrene cups, measure the temperature change for, e.g., magnesium ribbon in dilute acid (exothermic) and citric acid with sodium hydrogen carbonate (endothermic). Record starting and highest (or lowest) temperatures.
2. Comparing heat capacities. Heat equal masses of water and vegetable oil (or blocks of different metals) with the same energy input, and compare temperature rises. The substance with the lower \(c\) heats up more. (Oil: heat gently in a water bath, never directly over a flame.)
3. Calorimetry. Either a neutralisation in a polystyrene cup (like the worked example above), or a spirit burner heating water in a metal can (like the ethanol example). Weigh the burner before and after to find the mass of fuel burned. Calculate \(\Delta H\), then evaluate accuracy against a data book value.
Advanced
Bond energies are averages. Using bond energies to estimate the combustion of methane (\(\ce{CH4 + 2O2 -> CO2 + 2H2O}\)): broken \(4(\text{C–H } 414) + 2(\text{O=O } 498) = 2652\ \text{kJ}\); made \(2(\text{C=O } 803) + 4(\text{O–H } 463) = 3458\ \text{kJ}\); \(\Delta H \approx -806\ \text{kJ mol}^{-1}\). The accepted value is \(-890\ \text{kJ mol}^{-1}\). The difference is because bond energies are averages across many molecules, and the calculation assumes everything is a gas (the accepted value makes liquid water, which releases extra energy as it condenses).
An endothermic calorimetry calculation. \(4.00\ \text{g}\) of ammonium nitrate (\(0.0500\ \text{mol}\)) dissolves in \(50.0\ \text{g}\) of water and the temperature falls by \(6.0\ ^\circ\text{C}\). \(q = 50.0 \times 4.18 \times (-6.0) = -1254\ \text{J} = -1.25\ \text{kJ}\). \(\Delta H = -\dfrac{-1.25}{0.0500} = +25.1\ \text{kJ mol}^{-1}\) — positive, endothermic. (This is why ammonium nitrate is used in instant cold packs.)
"Per mole of what?" \(\Delta H\) always refers to the equation as written. For \(\ce{2H2 + O2 -> 2H2O}\), \(\Delta H\) is for 2 mol of water; for \(\ce{H2 + 1/2O2 -> H2O}\), it's half as much. In calorimetry, divide by the moles of whichever substance the question asks about (usually the limiting reagent, or the fuel).
Extension
Beyond the syllabus for this course — won't appear in the HSC, included for interest / depth study inspiration.
Hess's law. Because energy is conserved, the total enthalpy change for a reaction is the same no matter which route you take from reactants to products. That means you can calculate \(\Delta H\) for reactions that are impossible to measure directly (such as carbon turning into carbon monoxide without making any \(\ce{CO2}\)) by adding up the \(\Delta H\) values of other reactions that get you to the same place.
📎 Depth study idea
Compare the energy released per gram by different fuels (ethanol, methanol, propan-1-ol, butan-1-ol, or even different foods like nuts and corn chips) using a spirit burner or a burning-food calorimeter. Look for a pattern with the size of the molecule, evaluate your heat losses, and design an improved calorimeter that reduces them.
Video/visual resources¶
- 🖥️ PhET Simulation — Reactions & Rates — shows the energy profile (potential energy vs reaction progress) alongside the colliding particles (older Java-based sim — runs in the browser but can be slow to load)
- 🎥 TODO: source a calorimetry technique video (spirit burner and can, weighing the burner, stirring and reading the thermometer).
- 🎥 Khan Academy — TODO: source an exothermic/endothermic and bond energy explainer.
Check yourself¶
-
A reaction makes the test tube feel cold. Is it exothermic or endothermic, and what is the sign of \(\Delta H\)?
Answer
Endothermic — it takes heat in from the surroundings (including your hand), so they get colder. \(\Delta H\) is positive.
-
Explain, in terms of bonds, why combustion reactions are exothermic.
Answer
More energy is released when the new bonds in the products (\(\ce{CO2}\) and \(\ce{H2O}\)) form than is taken in to break the bonds in the fuel and oxygen. The extra energy is released to the surroundings as heat.
-
On an energy profile diagram for an exothermic reaction, where are the products compared with the reactants? What does adding a catalyst change?
Answer
The products are lower than the reactants. A catalyst lowers the hump (the activation energy) but doesn't change the heights of the reactants or products, so \(\Delta H\) stays the same.
-
How much energy is needed to heat \(500\ \text{g}\) of water from \(18.0\ ^\circ\text{C}\) to \(100.0\ ^\circ\text{C}\)?
Answer
\(q = 500 \times 4.18 \times 82.0 = 171\,380\ \text{J} = 171\ \text{kJ}\)
-
When \(0.0250\ \text{mol}\) of a substance dissolves in water, the water absorbs \(1.50\ \text{kJ}\) of heat. Calculate \(\Delta H\) for dissolving.
Answer
\(\Delta H = -\dfrac{q}{n} = -\dfrac{1.50}{0.0250} = -60.0\ \text{kJ mol}^{-1}\) (exothermic).
-
A student's calorimetry experiment gives \(\Delta H = -720\ \text{kJ mol}^{-1}\) for a fuel, but the accepted value is \(-1300\ \text{kJ mol}^{-1}\). Give two likely reasons and one improvement.
Answer
Reasons: heat lost to the air and the can instead of the water; incomplete combustion (soot). Improvement: use a draught shield and an insulated, lidded can placed close to the flame (or any other change that reduces heat loss).