Exothermic, Endothermic Reactions & Hess's Law

Enthalpy and Enthalpy Change

Enthalpy (H) is the total heat content of a system. It accounts for both the internal energy of the system and the work done by the system when it expands or contracts against a constant external pressure. Since it depends only on the current state of the system and not on how that state was reached, enthalpy is a state function.
Enthalpy equals internal energy plus the product of pressure and volume
=Enthalpy of the system(J or kJ)
=Internal energy of the system(J or kJ)
=Pressure of the system(Pa or atm)
=Volume of the system(m³ or L)
Absolute enthalpy cannot be measured: only the change in enthalpy () between two states is experimentally accessible
At constant pressure: , meaning the enthalpy change equals the heat exchanged at constant pressure, making the most practical way to track heat in chemical reactions
For solids and liquids: volume changes are negligible (), so
The change in enthalpy relates to the change in internal energy through the pressure-volume work term. Starting from the definition of enthalpy and expanding the change, we arrive at the key relationship between and .
Enthalpy change equals the change in internal energy plus the pressure-volume work done by the system
=Change in enthalpy at constant pressure(kJ mol⁻¹)
=Change in internal energy(kJ mol⁻¹)
=Constant external pressure(atm)
=Change in volume of the system(L)
(solids/liquids)
→
moles of gas are produced
→
From first law: where , so
Substituting into $\Delta H$:
Gas-phase shortcut: where (moles of gas only)

Exothermic and Endothermic Reactions

In an exothermic reaction, the enthalpy of the products is less than the enthalpy of the reactants. Heat is released to the surroundings, and the value is negative. The energy profile shows reactants at a higher energy level and products at a lower energy level.
For exothermic reactions, products have lower enthalpy than reactants, so is negative
=Enthalpy change of the reaction(kJ mol⁻¹)
=Total enthalpy of products(kJ mol⁻¹)
=Total enthalpy of reactants(kJ mol⁻¹)
Heat flows out: the reaction vessel and surroundings feel warmer
Energy diagram: reactants start high, products end low — the drop represents energy released as heat
Examples: combustion of fuels, neutralization of strong acids and bases, formation of water from and
Neutralization example: with kJ mol⁻¹

Exothermic Reaction Examples

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— kJ mol⁻¹
•
— kJ mol⁻¹
•
— kJ mol⁻¹
•
— kJ mol⁻¹
In an endothermic reaction, the enthalpy of the products is greater than the enthalpy of the reactants. Heat is absorbed from the surroundings, and the value is positive. The energy profile shows reactants at a lower energy level and products at a higher energy level.
For endothermic reactions, products have higher enthalpy than reactants, so is positive
=Enthalpy change of the reaction(kJ mol⁻¹)
=Total enthalpy of products(kJ mol⁻¹)
=Total enthalpy of reactants(kJ mol⁻¹)
Heat flows in: the reaction vessel and surroundings feel cooler
Energy diagram: reactants start low, products end high — the rise represents energy absorbed from surroundings
Examples: thermal decomposition, dissolution of in water, photosynthesis
Dissolution example: with kJ mol⁻¹

Endothermic Reaction Examples

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Dissolution of : kJ mol⁻¹ (solution cools)
•
Decomposition of : — requires continuous heating
•
Photosynthesis: — absorbs solar energy
Several specific types of enthalpy change are important for classifying and comparing reactions. Each type describes a particular process under standard conditions (25°C, 1 atm), and the sign convention ( or ) tells you whether the process is endothermic or exothermic.
Standard enthalpy of formation ($\Delta H^\circ_f$): heat absorbed or evolved when one mole of a compound forms from its elements in their standard states — e.g., of kJ mol⁻¹
Standard enthalpy of combustion ($\Delta H^\circ_c$): heat evolved when one mole of a substance burns completely in excess oxygen — e.g., of ethanol = −1368 kJ mol⁻¹
Standard enthalpy of neutralization ($\Delta H^\circ_n$): heat evolved when one mole of reacts with one mole of — always ≈ −57.4 kJ mol⁻¹ for strong acid–strong base pairs
Standard enthalpy of solution ($\Delta H^\circ_{\text{sol}}$): heat absorbed or evolved when one mole of solute dissolves in a large excess of solvent — can be positive (endothermic, e.g., : +16.2 kJ mol⁻¹) or negative (exothermic, e.g., : −25.0 kJ mol⁻¹)

Sign Convention Summary

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: Exothermic — heat released, surroundings gain energy
•
: Endothermic — heat absorbed, surroundings lose energy

Measuring Enthalpy Changes

A glass calorimeter is an insulated container fitted with a thermometer and stirrer, used to measure enthalpy changes of reactions in solution. Reactants in stoichiometric amounts are mixed inside, and the temperature change is recorded before and after the reaction. The heat exchanged is calculated from the mass, specific heat, and temperature change of the reaction mixture.
Heat absorbed or released by the reaction mixture equals its mass times its specific heat times the temperature change
=Heat exchanged (negative for exothermic)(J or kJ)
=Mass of the reaction mixture (solution)(g)
=Specific heat capacity of the solution (~4.18 J g⁻¹ K⁻¹ for water)(J g⁻¹ K⁻¹)
=Temperature change ($T_{\text{final}} - T_{\text{initial}}$)(K or °C)
Density assumption: for dilute aqueous solutions, density ≈ 1 g cm⁻³, so volume in cm³ ≈ mass in grams
Moles from molarity: moles = where M is molarity
Per-mole enthalpy: divide total heat by the number of moles of the limiting reagent to get in kJ mol⁻¹
A bomb calorimeter is used for accurate determination of enthalpy of combustion. The sample is placed in a sealed steel vessel (the bomb) pressurized with oxygen to about 20 atm, then immersed in an insulated water jacket. The sample is ignited electrically, and the temperature rise of the water is measured. Since the bomb has a fixed volume, the heat measured equals the internal energy change, though for most purposes this approximates .
Heat released by combustion equals the heat capacity of the calorimeter system times the temperature rise
=Heat evolved by combustion (negative)(kJ)
=Heat capacity of the entire calorimeter system (bomb + water + hardware)(kJ K⁻¹)
=Temperature rise recorded(K)
Heat capacity: given directly for the calorimeter as a whole — no need to find separate mass and specific heat
Convert to per mole: divide by the number of moles of the sample burnt (mass / molar mass)
Always negative: combustion always releases heat, so is always negative

Hess's Law of Constant Heat Summation

Hess's law states that if a chemical change takes place by several different routes, the overall enthalpy change is the same regardless of the route taken, provided the initial and final conditions are identical. This is a direct consequence of the first law of thermodynamics (conservation of energy) because enthalpy is a state function. Mathematically, the sum of all enthalpy changes around a complete cycle equals zero.
Around any closed thermodynamic cycle, the total enthalpy change is zero
=Sum of all enthalpy changes in the cycle (going one way is positive, returning is negative)(kJ mol⁻¹)
directly
→
via intermediates B and C
Why it matters: many values cannot be measured directly — e.g., of CO cannot be measured because burning carbon always produces a mixture of CO and
Practical approach: combine known enthalpy values for other reactions algebraically (adding, subtracting, reversing) to find the unknown enthalpy
Reversing rule: if a reaction is reversed, the sign of changes; if coefficients are multiplied by a factor, is multiplied by the same factor
Hess's law can be verified experimentally. Consider the formation of sodium carbonate: it can occur in a single step or in two steps via sodium hydrogen carbonate. The total enthalpy change is identical in both routes, confirming that enthalpy is path-independent.
Single step: with kJ
Step 1: with kJ
Step 2: with kJ
Verification: kJ

Applying Hess's Law — Energy Cycles

A classic application of Hess's law is finding the enthalpy of formation of carbon monoxide. Since burning carbon in limited oxygen always produces a mixture of CO and , of CO cannot be measured directly. Instead, we use the known combustion enthalpies of graphite and CO in an energy cycle.
The enthalpy change for the direct route (graphite to ) equals the sum of the two-step route (graphite to CO, then CO to )
=Enthalpy of combustion of graphite to $CO_2$(kJ mol⁻¹)
=Enthalpy of formation of CO from graphite (unknown)(kJ mol⁻¹)
=Enthalpy of combustion of CO to $CO_2$(kJ mol⁻¹)
Direct route: with kJ mol⁻¹
Indirect step 1: with
Indirect step 2: with kJ mol⁻¹
Solving: kJ mol⁻¹

Energy Cycle for CO Formation

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— starting point
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Route A (direct): — kJ mol⁻¹
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Route B (via CO): — ; then — kJ mol⁻¹
The Born-Haber cycle is a specific application of Hess's law used to calculate lattice energy — the enthalpy change when one mole of an ionic solid forms from its gaseous ions under standard conditions. Lattice energy cannot be measured directly, so the Born-Haber cycle combines measurable quantities to find it indirectly.
The standard enthalpy of formation equals the total energy to form gaseous ions plus the lattice energy
=Standard enthalpy of formation of the ionic compound(kJ mol⁻¹)
=Sum of all steps to convert elements to gaseous ions (atomization + ionization + electron affinity)(kJ mol⁻¹)
=Lattice energy of the ionic compound(kJ mol⁻¹)
Step 1 — Atomization of metal: — kJ mol⁻¹
Step 2 — Ionization of metal: — kJ mol⁻¹
Step 3 — Atomization of non-metal: — kJ mol⁻¹
Step 4 — Electron affinity: — kJ mol⁻¹
$\Delta H^\circ_x$ for NaCl: kJ mol⁻¹
Lattice energy: kJ mol⁻¹

Born-Haber Cycle Steps for NaCl

1
Start:
2
Atomize Na: — kJ mol⁻¹
3
Ionize Na: — kJ mol⁻¹
4
Atomize Cl: — kJ mol⁻¹
5
Electron affinity: — kJ mol⁻¹
6
Lattice formation: — (unknown)
7
Direct: — kJ mol⁻¹