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
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)
|Ca(g) → Ca²⁺(g) = 2e⁻|
|2F(g) + 2e⁻ → 2F⁻(g)|
|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.
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
- 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.