Every chemical and physical process involves energy changes. The system is the reaction or process being studied; the surroundings are everything else (including the container, the solvent, and the air).
Feature
Exothermic
Endothermic
Energy flow
System releases energy TO surroundings
System absorbs energy FROM surroundings
ΔH
Negative (ΔH < 0)
Positive (ΔH > 0)
Temperature of surroundings
Increases (feels warm/hot)
Decreases (feels cold)
Bond perspective
Energy released forming bonds > energy absorbed breaking bonds
Energy absorbed breaking bonds > energy released forming bonds
Examples
Combustion, neutralization, formation of ionic compounds
Photosynthesis, melting ice, dissolving NH₄NO₃
Energy perspective on bonds:
• Breaking bonds always requires energy (endothermic process).
• Forming bonds always releases energy (exothermic process).
• The overall ΔH depends on the balance: if more energy is released forming new bonds than is absorbed breaking old bonds, the reaction is exothermic.
6.2 — Energy Diagrams
Energy diagrams (enthalpy diagrams) visually represent the energy changes during a reaction. They show the relative enthalpy of reactants and products.
Energy diagram showing activation energy, transition state, and ΔH. Products are lower than reactants for an exothermic reaction. (Wikimedia Commons)
Exothermic reaction: Products are at a LOWER energy level than reactants. Energy is released. ΔH < 0.
Endothermic reaction: Products are at a HIGHER energy level than reactants. Energy is absorbed. ΔH > 0.
Reading the diagram:
• The vertical axis = enthalpy (H) or potential energy
• ΔH = Hproducts − Hreactants
• Ea (forward) = height of the peak above the reactants
• Ea (reverse) = height of the peak above the products
• Ea(reverse) = Ea(forward) − ΔH (for exothermic) or Ea(forward) + |ΔH| (for endothermic)
6.3 — Heat Transfer and Thermal Equilibrium
Heat (q) is the transfer of thermal energy between objects at different temperatures. Heat always flows spontaneously from hot to cold until thermal equilibrium is reached (both objects at the same temperature).
Conservation of energy in heat transfer:
qlost = −qgained
The heat lost by the hotter object equals the heat gained by the cooler object (assuming no heat loss to the environment).
Important Distinctions
Temperature (T): A measure of the average kinetic energy of particles. Intensive property (does not depend on amount).
Heat (q): The total energy transferred. Extensive property (depends on amount of substance).
Internal energy (U): The total kinetic + potential energy of all particles in a system.
Example: A 50.0 g piece of copper (c = 0.385 J/g·°C) at 100.0°C is placed in 200.0 g of water (c = 4.184 J/g·°C) at 25.0°C. Find the final temperature.
Calorimetry is the experimental measurement of heat changes in chemical or physical processes.
q = mcΔT
• q = heat (J or kJ)
• m = mass (g)
• c = specific heat capacity (J/g·°C) — amount of heat needed to raise 1 g by 1°C
• ΔT = Tfinal − Tinitial
Specific heat of water: 4.184 J/g·°C (one of the highest of any common substance — water is an excellent heat sink)
Coffee Cup Calorimeter (Constant Pressure)
Used for reactions in aqueous solution (acid-base, dissolution, etc.).
Open to the atmosphere → pressure is constant → qrxn = ΔH.
Assumes all heat is absorbed by the water: qwater = mcΔT, and qrxn = −qwater.
To find ΔH per mole: divide qrxn by the moles of limiting reagent.
Bomb Calorimeter (Constant Volume)
Used for combustion reactions.
Sealed, rigid container → constant volume → measures ΔU (internal energy change), which is approximately equal to ΔH for most reactions.
q = Ccal × ΔT where Ccal = heat capacity of the entire calorimeter (J/°C).
Example: 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH (both at 25.0°C) are mixed in a coffee cup calorimeter. Tf = 31.9°C. Find ΔH for the neutralization (per mole).
qwater = (100.0 g)(4.184 J/g·°C)(31.9 − 25.0) = 2887 J = 2.887 kJ
qrxn = −2.887 kJ (exothermic)
Moles = 0.0500 mol HCl (limiting)
ΔH = −2.887 / 0.0500 = −57.7 kJ/mol
6.5 — Energy of Phase Changes
Phase changes involve energy changes even though the temperature remains constant during the transition. The energy goes into overcoming (or forming) intermolecular forces, not into changing kinetic energy.
Phase Change
Direction
Energy
Equation
Melting (fusion)
Solid → Liquid
Endothermic
q = n × ΔHfus
Vaporization
Liquid → Gas
Endothermic
q = n × ΔHvap
Sublimation
Solid → Gas
Endothermic
q = n × (ΔHfus + ΔHvap)
Freezing
Liquid → Solid
Exothermic
q = −n × ΔHfus
Condensation
Gas → Liquid
Exothermic
q = −n × ΔHvap
ΔHvap > ΔHfus because going from liquid to gas requires breaking ALL remaining IMFs, while going from solid to liquid only requires partially overcoming them (particles gain enough freedom to flow but remain close together).
To calculate the total energy needed to go from one state to another (e.g., ice at −10°C to steam at 120°C), add up the energy for each segment:
Heat the solid: q₁ = mcsolidΔT
Melt the solid: q₂ = nΔHfus
Heat the liquid: q₃ = mcliquidΔT
Boil the liquid: q₄ = nΔHvap
Heat the gas: q₅ = mcgasΔT
Total q = q₁ + q₂ + q₃ + q₄ + q₅
6.6 — Introduction to Enthalpy of Reaction
The enthalpy of reaction (ΔHrxn) is the heat change associated with a chemical reaction at constant pressure. It is one of the most important quantities in thermochemistry.
Properties of ΔH:
1. Extensive: ΔH is proportional to the amount of reactants. If you double the reaction, ΔH doubles.
2. Sign reversal: Reversing the reaction reverses the sign of ΔH.
3. State-dependent: ΔH depends on the states of reactants and products (s, l, g, aq).
4. ΔHrxn = Hproducts − Hreactants
Thermochemical equations:
A balanced equation with ΔH specified:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890.4 kJ
This means 890.4 kJ of heat is released when 1 mol CH₄ reacts with 2 mol O₂ to form CO₂ and liquid water.
If 2 mol CH₄ react: ΔH = 2(−890.4) = −1780.8 kJ
The reverse reaction: CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890.4 kJ
6.7 — Bond Enthalpies
Bond enthalpy (bond energy) is the energy required to break one mole of a particular bond in the gas phase. It can be used to estimate ΔHrxn when standard enthalpies of formation are not available.
ΔHrxn ≈ Σ(bonds broken) − Σ(bonds formed)
• Breaking bonds is endothermic (+)
• Forming bonds is exothermic (−)
• If more energy is released forming bonds than absorbed breaking bonds → exothermic (ΔH < 0)
Bond enthalpies are averages over many different molecules. The actual O—H bond energy in water is slightly different from the O—H in ethanol.
This method gives estimates, not exact values.
Only applies to gas-phase reactions. If reactants or products are liquids or solids, additional energy terms (e.g., ΔHvap) are needed.
Works best for simple molecules; less accurate for complex ones or resonance structures.
6.8 — Enthalpy of Formation
The standard enthalpy of formation (ΔH°f) is the enthalpy change when one mole of a compound is formed from its elements in their standard states (most stable form at 1 atm and 25°C).
Key rules:
• ΔH°f of any element in its standard state = 0 (by definition).
Examples: O₂(g), N₂(g), C(graphite), Fe(s), H₂(g) all have ΔH°f = 0
• ΔH°f of compounds can be positive (endothermic formation) or negative (exothermic formation).
• Compounds with very negative ΔH°f are very stable (e.g., CO₂: −393.5 kJ/mol; H₂O(l): −285.8 kJ/mol).
Using ΔH°f to calculate ΔH°rxn:
ΔH°rxn = Σ nΔH°f(products) − Σ nΔH°f(reactants)
where n = stoichiometric coefficients from the balanced equation.
Example: Calculate ΔH° for the combustion of methane:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Hess’s Law states that the total enthalpy change for a reaction is the same regardless of whether it occurs in one step or multiple steps. This works because enthalpy is a state function — it depends only on the initial and final states, not the path taken.
Rules for manipulating thermochemical equations:
1. If you reverse an equation, change the sign of ΔH.
2. If you multiply an equation by a factor, multiply ΔH by the same factor.
3. If you add equations, add their ΔH values.
4. Species that appear on both sides of the summed equation cancel out.
Worked Example: Find ΔH for: C(s) + 2H₂(g) → CH₄(g)
Strategy: We need C and H₂ as reactants and CH₄ as the product.
Use equation (1) as is: C(s) + O₂(g) → CO₂(g) ΔH = −393.5
Multiply equation (2) by 2: 2H₂(g) + O₂(g) → 2H₂O(l) ΔH = 2(−285.8) = −571.6
Reverse equation (3): CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890.3
Look at each given equation and decide if you need to use it as is, reversed, or multiplied.
Focus on species that appear in the target equation — ensure they end up on the correct side.
Add equations and ΔH values. Verify that unwanted species cancel.
📝 Practice Multiple-Choice Questions
Test your knowledge of Unit 6. Click “Show Answer” to reveal the correct choice and explanation.
1. In an exothermic reaction:
(A) The system absorbs energy from the surroundings
(B) ΔH is positive
(C) The temperature of the surroundings increases
(D) Energy is required to form products
Answer: (C)
In an exothermic reaction, the system releases heat to the surroundings, causing the temperature of the surroundings to increase. ΔH is negative.
2. On an energy diagram, the activation energy of the forward reaction is the difference between:
(A) Products and reactants
(B) Transition state and products
(C) Transition state and reactants
(D) Reactants and 0
Answer: (C)
Ea(forward) = energy of the transition state minus the energy of the reactants. This is the energy barrier molecules must overcome for the reaction to proceed.
3. A 100.0 g sample of water absorbs 4184 J of heat. By how much does its temperature increase? (cwater = 4.184 J/g·°C)
(A) A bomb calorimeter operates at constant pressure
(B) A coffee cup calorimeter operates at constant volume
(C) qrxn = −qsolution in a coffee cup calorimeter
(D) Heat capacity and specific heat are the same thing
Answer: (C)
In a coffee cup calorimeter (constant pressure), the heat released by the reaction is absorbed by the solution: qrxn = −qsolution. A bomb calorimeter operates at constant volume, not pressure.
5. Why is ΔHvap greater than ΔHfus for the same substance?
(A) Vaporization requires breaking all remaining intermolecular forces
(B) Fusion produces a gas with more kinetic energy
(C) Fusion requires more energy because solids are more ordered
(D) Vaporization involves breaking covalent bonds
Answer: (A) Vaporization requires breaking essentially ALL remaining intermolecular forces to completely separate molecules into the gas phase. Fusion only requires partially overcoming IMFs (enough for particles to flow but not completely separate).
6. If a reaction has ΔH = −200 kJ, what is ΔH for the reverse reaction?
(A) −200 kJ
(B) +200 kJ
(C) −400 kJ
(D) 0 kJ
Answer: (B)
Reversing a reaction changes the sign of ΔH. If the forward reaction releases 200 kJ (exothermic), the reverse reaction absorbs 200 kJ (endothermic): +200 kJ.
7. Using bond enthalpies: H—H = 436 kJ/mol, Cl—Cl = 242 kJ/mol, H—Cl = 431 kJ/mol. Estimate ΔH for H₂ + Cl₂ → 2HCl.
8. The standard enthalpy of formation of O₂(g) is:
(A) −249 kJ/mol
(B) +249 kJ/mol
(C) 0 kJ/mol
(D) Cannot be determined
Answer: (C)
O₂(g) is oxygen in its standard state (most stable form at 1 atm, 25°C). By definition, ΔH°f of any element in its standard state = 0 kJ/mol.
9. Hess’s Law works because:
(A) All reactions are reversible
(B) Enthalpy is a state function
(C) Energy is always conserved as heat
(D) Bond enthalpies are exact values
Answer: (B)
Hess’s Law works because enthalpy is a state function — it depends only on the initial and final states, not the pathway. So the total ΔH is the same whether the reaction occurs in one step or many.
10. Given: (1) A → B, ΔH = −100 kJ; (2) B → C, ΔH = +50 kJ. What is ΔH for A → C?
(A) −150 kJ
(B) −50 kJ
(C) +50 kJ
(D) +150 kJ
Answer: (B)
By Hess’s Law: ΔH(A → C) = ΔH(A → B) + ΔH(B → C) = (−100) + (+50) = −50 kJ.
💡 Study Tip for Unit 6
Thermodynamics connects to almost every other unit. Master the q = mcΔT and q = nΔH calculations — they appear in many FRQ problems. Practice Hess’s Law manipulations (reversing, multiplying, adding equations) until they feel automatic. Know the ΔH°f formula (products minus reactants) and remember that elements in standard states = 0. For bond enthalpies, remember the formula is Σbroken − Σformed. Always check your signs — the most common error is mixing up endothermic and exothermic.
