Topic 15: Energetics (HL)
15.1: Standard Enthalpy Changes of a Reaction
15.1.1: Define and apply the terms standard state, standard enthalpy change of formation, and
standard enthalpy change of combustion.
Standard State: The form an element/compound exists under standard conditions:
Temperature: 25°C (or 298K)
A. Standard Enthalpy change of Formation (∆Hf
): enthalpy change that occurs when one mole
of the substance is formed from its elements in their standards states.
B. Standard Enthalpy change of Combustion (∆Hc
): enthalpy change that occurs when one
mole of the substance burns completely under standard conditions.
15.1.2: Determine the enthalpy change of a reaction using standard enthalpy changes of formation
Example: combustion of methane
CH4 + 2O2 → 2H2O + CO2
We want to find the (Energy in/kJ – Energy out/kJ) of the reaction to determine the Enthalpy change.
We can calculate bond enthalpies using the data from the IB data booklet.
Average bond enthalpies at 298K
C-H = 413 kJ mol
However, there are four C-H bonds in methane, so the energy will be : 413kJmol
x 4 = 1652 kJmol
2O2 → O=O → 498kJmol
x 2 = 996kJmol
Total energy in = 996 + 1652 = 2648 kJmol
CO2 → 2x C=O → 746kJmol
x2 = 1492 kJmol
2H2O → 4 x H-O → 4 x464 = 1856kJmol
Total energy out = 3348 kJmol
Enthalpy Change = Energy in – Energy out = -700kJmol
15.2: Born-Haber Cycle
15.2.1: Define and apply the terms lattice enthalpy and electron affinity
lat): enthalpy change that occurs when one mole of a solid ionic compound is
separated into gaseous ions under standard conditions.We measure lattice enthalpy using kJ mol
e): enthalpy change when you add an electron to an atom in the gaseous state.
Example: X (g) + e → X – (g)
15.2.2: Explain how the relative sizes and the charges of ions affect the lattice enthalpies of
different ionic compounds.
Factors affecting the lattice enthalpy of different ionic compounds:
Size of the ions: If the ions are smaller, there will be a stronger attractive force between the
protons of one ion and the electrons of another. Conversely, if the ions are bigger, the opposite
Size of the charge: The greater the charge, the greater the attractive forces, and the higher the
15.2.3: Construct a Born–Haber cycle for group 1 and 2 oxides and chlorides, and use it to
calculate an enthalpy change.
We can construct a born haber cycle to calculate the enthalpy change
The Born Haber Cycle, just like Hess’s Law, is a method to calculate energy change that
cannot be measured directly.
The Enthalpy Change of Formation, which is what we usually want to calculate, can easily be
found by adding all the individual values of each step together.
Enthalpy of Atomization (∆H
atom):: Energy required to form one mole of gaseous atoms. Notice how
the half mole of Fluorine became 1 mole in the Enthalpy of Atomization.
Enthalpy of Ionization (∆H
i): the amount of energy required to make 1 mole of gaseous metal ions.
Born Haber Cycle: The sum of the enthalpies on any two sides must equal the enthalpy
change of the remaining side.
f = ∆H
atom + ∆H
i + ∆H
e - ∆H
15.2.4: Discuss the difference between theoretical and experimental lattice enthalpy values of
ionic compounds in terms of their covalent character.
As we discussed earlier, the Born Haber Cycle can be used to assess the Lattice Enthalpy and the
Enthalpy of Formation of an ionic compound.
When we do an experimental and extract results, and the results are very different from the theoretical
values, then this may suggest that the compound has more covalent characteristics than ionic.
The theoretical value of NaCl is 766 KJ mol-1, wheres the experimental value of NaCl is 771 KJ mol-.
There is a small difference between the two values so we can assume that there is a high degree of ionic
characteristics in NaCl.
Entropy (S) is considered to be the amount of disorder in a system.
Formula: ΔS = Sfinal – Sinitial
15.3.1: State and explain the factors that increase the entropy in a system
Changes of state: When the product changes from state from solid to liquid and lastly, to gas,
the state of disorder increases.
Mixing and Dissolving: If you mix two substances together, you will be increasing the disorder
in the system. Therefore, the disorder of aqueous solutions will be larger than that of a liquid.
Formation of gaseous products: In a reaction, if one of the products is a gas, then the disorder
in the system will increase, as the disorder of a gas is typically larger than that of a solid or a
Increased movement of particles: If you heat a liquid or a gas, you are providing the system
with more kinetic energy, hence increasing the disorder in the system.
Increased number of particles: This increases the disorder in the system.
15.3.2: Predict whether the entropy change (ΔS) for a given reaction or process is positive or
solid liquid increase (+)
solid gas increase (+)
liquid gas increase (+)
liquid solid decrease (-)
gas solid decrease (-)
gas liquid decrease (-)
Example of Entropy Predicting:
1) X (g) + Y (g) –> A (g) + B (g)
2) X (g) + Y (g) –> A (s) + A (l)
3) X (s) + Y( s) –> A (s) + A (g)
4) X (s) + Y( s) –> A (s) + A (s)
1) There is no change in state so the entropy change is zero, or very close to zero.
2) There is a change of state from gas to liquid/solid, so the change in entropy is negative, as the
disorder in the system decreases.
3) Notice that although one of the substances is still a solid, substance B changed state to a gas, so
the change in entropy is positive, as the disorder in the system increases
4) There is no change in state, so the change in entropy is zero.
15.3.3: Calculate the standard entropy change for a reaction (S°) using standard entropy values
Enthalpy is very difficult to measure directly.
However, the absolute entropy can be measured directly. We give the absolute entropy the letter
products – ΣS
Let’s do an example
1.Calculate entropy change in the following reaction:
3H2(g) +N2 (g) –> 2NH3 (g)
However, we need to know the values of some absolute entropies
Hydrogen: 131 KJ
Nitrogen: 192 JK
Reactants = (3×131) + 192= 585
2.Next, the Absolute Entropy is simply Entropy of products – Entropy of reactants.
So, Products – Reactants= 585-192 =393 KJ mol
A spontaneous reaction is often defined as “a reaction that causes a system to move from a less stable
to a more stable state”.
Spontaneous processes are those that can occur naturally without the need to do work.
15.4.1: Predict whether a reaction or process will be spontaneous by using the sign of ∆G
Spontaneous reactions produce substantial amounts of products at equilibrium and release free
o Free energy is energy that is available to do work.
The Gibbs free energy change, G is the maximum amount of free energy that can be extracted to do
If ΔG < 0, then the reaction is spontaneous
If ΔG >0, then the reaction is non-spontaneous.
15.4.2: Calculate ∆G for a reaction using the equation ∆G= ∆H- T∆S and by using values of the
standard free energy change of formation, ∆G
Whether a reaction is spontaneous or not depends on three factors:
Enthalpy Change (ΔH)
Gibbs Free Energy (ΔG)
An equation linking these three values is:
ΔG = ΔH – T ΔS
Gibbs Free Energy = Enthalpy Change – Temperature x Entropy
Note: Temperature is measured in Kelvin, which can be calculated by 273 to the °C value you are given.
176 298 284
G H T S
100 176 284
G H T S
15.4.3: Predict the effect of a change in temperature on the spontaneity of a reaction using
standard entropy and enthalpy changes and the equation ∆G= ∆H -T∆S
ΔH ΔS TΔS ΔH – T ΔS Gibbs Free energy (ΔG) Spontaneous?
+ + + (+) – (+) Depends on “T” If ΔH < TΔS
+ - - (+) – (-) + Never spontaneous
- + + (-) – (+) - Always spontaneous
- - - (-)-(-) Depends on “T” If ΔH > TΔS
0 + + 0 – (+) - Always spontaneous
0 - - 0-(-) + Never spontaneous