Chapter 05: Chemical Energetics
Chapter 5
Chemical Energetics
Chapter 5
Chemical Energetics
At the crux of it all, this chapter deals with the energy changes during reactions. As much as this chapter appears wholly abstract and theoretical in nature, there are many real life applications from understanding chemical energetics, such as burning fuel for energy or predicting the solubility of a solute in a solvent.
There are many new terms in this chapter which would take a while to memorise and practice before you may feel comfortable with them. However, the work to memorize the definitions would pay off in helping you answer not just definition questions but calculation questions too.
Here is the list of definitions you should memorise
- Standard enthalpy change of Reaction
- Standard enthalpy change of Combustion
- Standard enthalpy change of Neutralisation
- Standard enthalpy change of Formation
- Standard enthalpy change of Atomisation
- Standard enthalpy change of Hydration
- Standard enthalpy change of Solution
- 1st Ionisation energy, 1ˢᵗ E.
- 1st Electron Affinity, 1ˢᵗ E.A.
- Bond dissociation energy, B.E.
- Standard Lattice Energy, L.E.
- Hess’ Law
- Entropy
A common mistake made in calculation questions involving standard conditions is to assume its values are the values of standard temperature and pressure.
In actual fact, standard conditions are different from standard temperature and pressure, where the latter is 0°C and 1 bar.
What are standard conditions?
- Temperature = 25℃ or 298 K
- Pressure = 1 bar or 10⁵ Pa
- Substance is in most stable physical/allotropic form
E.g. For carbon which can exist as either graphite or diamond, graphite is used as it is the most stable form of carbon at standard conditions.
Hess’ law
ΔHfθ = ∑ BE (bonds broken) - ∑BE (bonds formed)
All products and reactants must be in gaseous state.
Must remember to multiply the bond energies with the number of bonds broken or formed.
Worked Example 1:
What is the value of the ΔH for the reaction NH3(g) + 3F2(g) → NF3(g) + 3HF(g)
Consider the bonds broken and formed.
(Bond energies/ kJ mol-1: N-H = 390; F-F = 158; H-F = 562; N-F =272)
Step 1
Draw out the bonds of every reactant and product. This is to make sure you do not miss out any bonds.
Step 2
Count the number of each bond.
| N-H | F-F | N-F | H-F |
|---|---|---|---|
| 3(+390) | 3(+158) | 3(+272) | 3(+562) |
Step 3
Apply the formula:
ΔHθ = ∑ BE (bonds broken) - ∑BE (bonds formed)
= [ 3(+390) + 3(+158) ] - [ 3(+272) + 3(+562) ]
= -858 kJ mol-1
Note: Even if the answer is endothermic, the positive ‘+’ sign must still be written.
Bond cycles
Worked Example 2:
Given the following enthalpy changes
I₂(g) + 3Cl₂(g) → 2ICl₃(s)
ΔHθ = -214 kJ mol⁻¹
I₂(s) → l₂(g)
ΔHθ = +38 kJ mol⁻¹
By drawing a suitable energy cycle, what is the enthalpy change of formation of ICI₃?
- Always start by writing an equation for the reaction that the question is asking for. Hence in this question, we construct an equation for the enthalpy change of formation of ICI₃.
¹/² l₂(s) + ³/² Cl₂(g) → ICI₃(s) - Assess why the equation l₂(g) + 3Cl₂(g) → 2ICl₃(s) does not represent the enthalpy change of formation of ICI₃(s).
It does not since l₂ is solid at standard conditions. Enthalpy change of formation is the energy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions. - Hence connect the equations at hand to form a cycle.
Answer: ΔHfӨ = (38-214) ÷ 2
= -88 kJ mol-1
Born-Haber cycles
Checklist for Born-Haber cycles
❏ Equation are balanced with state symbol included
❏ All Arrows are labelled with enthalpy changes
❏ Label y-axis “ energy/kJ mol-¹ ” and “zero” for elements at standard states.
❏ ↓ arrow for exothermic reaction; ↑ arrow for endothermic reaction
Worked Example 3:
Construct a Born-Haber cycle for the formation of CaF₂ from its elements.
For ionic compounds like CaF₂ above, we can use the acronym FAIEL to check if we have included the necessary equations. Hopefully by using this acronym, you will “fail” to get such questions wrong again.
| F | Standard enthalpy change of Formation |
|---|---|
| Ca(s) + F₂(g) → CaF₂(s) | |
| A | Standard enthalpy change of Atomisation |
| Ca(s) → Ca(g)
F₂(g) → 2F(g) |
|
| I | Ionisation energy |
| Ca(g) → Ca²⁺(g) = 2e⁻ | |
| E | Electron affinity |
| 2F(g) + 2e⁻ → 2F⁻(g) | |
| L | Lattice energy |
| Ca²⁺(g) + 2F⁻(g) → CaF₂(s) |
| F | Standard enthalpy change of Formation | Ca(s) + F₂(g) → CaF₂(s) |
|---|---|---|
| A | Standard enthalpy change of Atomisation | Ca(s) → Ca(g)
F₂(g) → 2F(g) |
| I | Ionisation energy | Ca(g) → Ca²⁺(g) + 2e⁻ |
| E | Electron affinity | 2F(g) + 2e⁻ → 2F⁻(g) |
| L | Lattice energy | Ca²⁺(g) + 2F⁻(g) → CaF₂(s) |
Do note that this acronym is only applicable to ionic compounds.
Solubility
Ever wondered why salt dissolves in water? The short answer is that although the ΔHsolθ of salt dissolving in water is endothermic, it is still relatively small enough for salt to be soluble in water. Required heat from the surroundings can be absorbed such that salt dissolves into water.
Generally, the more exothermic the ΔHsolθ , the more soluble a solute is in a solvent. So, how do we determine the ΔHsolθ and hence solubility of an ionic compound in water?
Two stages are involved in the dissolution of ionic compounds,
Stage 1 : The ions are pulled far apart to form separate gaseous ions.
The heat change would be the reverse of lattice energy -ΔHlattθ since ionic bonds are breaking to form the constituent gaseous ions of the ionic compound. Since energy is taken in to break bonds, this stage is always endothermic.
Stage 2: Solvent molecules, H₂O, form bonds to the gaseous ions to form a solution of aqueous ions. The heat change would be the enthalpy change of hydration, ΔHhydθ. Since energy is released to form ion-dipole interactions between H₂O molecules and ions, this stage is always exothermic.
Thus, we can conclude that ΔHsolθ = ΔHhydθ - ΔHlattθ
Generally, entropy, ΔS, increases when an ionic compound dissolves in water.
In stage 1, As the number of particles increases, the number of ways to arrange the gaseous particles increases too, thus disorder of the system increases and entropy increases.
In stage 2, as water molecules form bonds around the ions, their positions become more fixed and the number of ways to rearrange the particles decreases. Thus, disorder of the system decreases and entropy decreases.
Since overall entropy depends on both stages, and usually the disordering process in stage 1 is dominant, thus ΔS increases overall.
Gibbs Free Energy and Entropy
ΔGӨ = ΔHӨ - TΔSӨ
You can remember this equation using a mnemonic aid, which is the acronym Go Home To Sleep.
Do note that SI units for ΔGӨ and ΔHӨ are in kJ mol-1 while units for ΔSӨ are in JK-1mol-1 and T is in K.
| ΔGӨ < 0 | Forward reaction is spontaneous |
|---|---|
| ΔGӨ > 0 | Reaction is not spontaneous |
| ΔGӨ > 0 | Reaction is in equilibrium |
Examples where reaction is in equilibrium and ΔGӨ = 0 would be when a phase change is taking place.
There are limits to ΔGӨ, however.
The spontaneity of a reaction using ΔGӨ is only possible under standard conditions. Also, a reaction might be spontaneous as ΔGӨ < 0, but kinetically too slow for an observable reaction.
Checklist for answering entropy change questions
❏ Describe the change
❏ State if there are more or less ways of arranging the particles/ distribute energy “packets” among particles
❏ Hence state if there is greater or less disorder of the system
❏ Thus entropy increases or decreases.
Using the above checklist,
Worked Example 4:
Describe and explain how the entropy of the following system will change during the stated process. Assume the pressure remains constant throughout.
1 mol of Cl2(g) at 298 K is heated to 373 K
Solution:
- As temperature increases,
- The system possesses more kinetic energy, hence there are more ways to arrange the particles.
- This leads to greater disorder of the system,
- Thus, entropy increases.
Continue Reading on Chapter 06: Reaction Kinetics