Thermodynamics Notes
Thermodynamics can be broken down in to the following sections;
Born-Haber Cycles
Covalent Characteristics of Ionic Compounds
Enthalpy of Solution Thermocycles
Gibb’s Free Energy
Born-Haber Cycles
In order to use Born Haber cycles you must know the definitions of enthalpy changes.
Enthalpy of Formation - The enthalpy change when one mole of compound is formed from its constituent elements under standard conditions, in standard states.
Ionisation Energy - The enthalpy change when one mole of electrons is removed from one mole of gaseous atoms to produce one mole of gaseous ions with a charge of +1. Second ionisation energy is the enthalpy change when removing one mole of electrons from one more of gaseous +1 ions to produce one mole of gaseous ions with a charge of +2.
Enthalpy of Atomisation - The enthalpy change when one mole of gaseous atoms is formed from its element under standard conditions, in it’s standard state.
Bond Enthalpy - The average enthalpy change when one mole of covalent bonds are broken in the gaseous state across a range of molecules.
Electron Affinity - The enthalpy change when a mole of gaseous atoms is added to one mole of electrons to become one moles of gaseous ions with a charge of -1. The second electron affinity is the enthalpy change when one mole of electrons is added to one more of gaseous ions of -1 to become one mole of gaseous ions of -2.
Lattice Enthalpy of Formation - The enthalpy change when one mole of solid ionic compound is produced from its gaseous ions.
Lattice Enthalpy of Dissociation - The enthalpy change when one mole of solid ionic compound undergoes dissociation in to its gaseous ions.
Enthalpy of Solution - The enthalpy change when one mole of solute totally dissolves in a solvent.
Enthalpy of Hydration - The enthalpy change when one mole of gaseous ions become surrounded by water molecules. i.e dissolved.
Standard states are the states of the elements in standard conditions i.e oxygen’s standard state is gas.
To construct a Born-Haber cycle, in essence a more complex thermocycle, we are going to use Hess’s Law which states that the overall enthalpy change for a reaction is the same regardless of pathways taken.
To calculate the enthalpy of formation for an ionic compound the following route could be taken as an alternative.
Enthalpy of Atomisation for cation, Ionisation Energy for cation, Enthalpy of Atomisation for anoin, Electron affinity for anion, Lattice enthalpy of formation.
Atomisation of M
Atomisation of X
First Ionisation Energy of M
First Electron Affinity of X
Lattice Enthalpy of formation of MX
Standard Enthalpy of formation of MX
The first electron affinity is often negative, exothermic, but the second and following electron affinities will be positive, endothermic. This is because any affinities after the first a negative electron will be added to an already negative ion and these like charges will repel each other.
2x Atomisation of M
Atomisation of X
2x First Ionisation Energy of M
First Electron Affinity of X
Second Electron Affinity of X
Lattice Enthalpy of M2X
Standard Enthalpy of Formation of M2X
Covalent Characteristics of Ionic Compounds
The values for Enthalpy of lattice Dissociation/Formation can deviate significantly from experimentally recorded values. This is due to most ionic compounds having covalent characteristics, they are not “truly” ionic. The larger the difference in electronegativity the less covalent characteristics they will have, the ionic compound with the least covalent characteristic being FrF. The closer in electronegativity, and on the periodic table, the more covalent characteristic the compound has, this is why AgI will not dissolve in ammonia solution - it is too covalent to dissolve in a polar solvent. Theoterical values are developed using the perfect ionic model in which ions are perfectly spherical with no covalent character.
The larger discrepancy in the calculated Enthalpy of lattice Dissociation/ Formation and the experimental data the more covalent character the compound has. The deviation for FrF will be close to 0%, the deviation between calculated and experimental for AgI is 30%.
Answering Born-Haber Cycle Questions
To complete a Born-Haber Cycle the correct equations must be written on each line, common errors are to omit species which are not involed in the current step, for example after ionisation some students forget to add in the other atoms.
2Cl(g) + Mg(g) —> 2Cl(g) + Mg+(g) + e- [1 mark]
2Cl(g) + Mg(g) —> Mg+(g) + e- [0 marks]
In the above example the chlorine atoms have been missed from the products.
Another common error is for students not to include state symbols.
2Cl(g) + Mg2+(g) + 2e- —> 2Cl-(g) + Mg2+(g) [1 mark]
2Cl(g) + Mg2+(g) + 2e- —> 2Cl-(g) + Mg2+ [0 marks]
In the above example the gas state symbol has been omitted from the equations, note the value for the above enthalpy change will be 2x standard affinity of Cl as there is 2 moles of e- added to 2 moles of Cl.
Using the enthalpy values given and the definitions the known values need to be filled in. Be careful with lattice dissociation/formation and the direction of the arrow, and atomisation/bond enthalpy produce the correct amount of atoms.
Atomisation of Na
Atomisation of Cl or 1/2 bond enthalpy of Chlorine
First Ionisation Energy of Na
First Electron Affinity of Cl
Lattice Enthalpy of formation of NaCl
Standard Enthalpy of formation of NaCl
To calculate a missing value of the Born-Haber cycle the same approach as a thermocycle is taken, graphically a new arrow can be drawn and values taken from the adjacent enthalpy changes.
x = +349 -496-122-107-411 = -787kJ/mol
Enthalpy of Solution Thermocycles
Using Hess’s law and the Enthalpy of Solution, Enthalpy of Hydration, and Enthalpy of Lattice Dissociation/Formation we can calculate missing values and compare values of changes which are not possible in reality - just like year 1 thermocycles.
To construct a thermocycle for enthalpy of solution of NaCl we need to following equations.
NaCl(s) —> Na+(g) + Cl-(g) Enthalpy of lattice dissociation
NaCl(s) + aq —> NaCl(aq) Enthalpy of Solution
Na+(g) + aq —> Na+(aq) Enthalpy of Hydration
Cl-(g) + aq —> Cl-(aq) Enthalpy of Hydration
The thermocycle is drawn normally with Enthalpy of Solution on top, and the aqueous products at the top, multiple arrows will need to point from the gaseous ions to the dissolved ions. The thermocycle can then be used in the same way as previous thermocycles and the missing value calculated.
Enthalpy of solution of MX
Enthalpy of lattice dissociation of MX
Enthalpy of hydration of M+
Enthalpy of hydration of X-
The lattice enthalpy of dissociation could be written at the top, there will still be multiple arrows between the gaseous ions and the dissolved ions.
Enthalpy of lattice dissociation of MX
Enthalpy of solution of MX
Enthalpy of hydration of M+
Enthalpy of hydration of X-
Answering Enthalpy of Solution Questions
Enthalpy of solution questions are on par with thermocycle questions in year 1, complete the thermocycle and then find the alternative route and add the values.
In the above example, there is 2 Cl-, this is why the value for Enthalpy of Hydration of Cl- just be multiplied by 2.
The arrow can then be added and the unknown value calculated.
x = 2493 - 1920 - 2(364) = -155 kJ/mol
Gibb’s Free Energy
ΔG = ΔH - TΔS
The equation above is for calculating Gibb’s free energy for a reaction. If Gibb’s free energy is zero or negative for a given reaction it means the reaction is feasible/spontaneous, in chemistry both of these terms mean the reaction could take place, not how fast it takes place it may be so slow it effectively does not happen.
ΔG, Gibb’s free energy, is in kJmol^-1
ΔH, enthalpy change, is in kJmol^-1
T, temperature, is in Kelvin
ΔS, entropy change, is in Jmol^-1, this needs to be converted to kJmol^-1 before any calculations.
Entropy is a level of disorder in a system, solid materials generally have less disorder as the particles are tightly packed in a regular structure, liquids are closely packed but are not in a regular structure so they have a higher amount of disorder and therefor a greater value of entropy, S. If a material is a gas there is a large distance between particles and no structure, this means it has a very high level of disorder so has the largest entropy value, S, of the three states of matter.
Answering Gibb’s Free Energy Questions
[Tidy all subscripts and superscripts]
To calculate an entropy change the amount of entropy is calculated for the reactant and products and the change calculated.
H2O(l) —> H2(g) + 1/2 O2(g) ΔH = +285kJmol^-1
Entropy of H2O (l) is 70 Jmol^-1
Entropy of H2 (g) is 131 Jmol^-1, Entropy of 1/2 O2(g) is 0.5x205 Jmol^-1
There is half a mole of oxygen in the products so the entropy value of oxygen is halved.
Therefore the ΔS is (131+0.5x205)-70 = +163.5 Jmol^-1 or 0.1635kJmol^-1
ΔS = (Sum of S of products) - (Sum of Δ of reactants)
If we are to calculate the Gibb’s free energy at 150oC we need to use the equation, ΔG = ΔH - TΔS, substituting in we have ΔG= 285-(273+150)x0.1635
ΔG = 215, therefore it is not feasible. It must be 0 or negative to be feasible.
To calculate at what temperature a reaction becomes feasible we must take ΔG = 0 and use ΔG = ΔH- TΔS
substituting in from the previous calculations
0 = 285 - T * 0.1635, we rearrange so T is the subject
T = 285 / 0.1635 = 1743 K or 1470 oC
The common error in working out if a reaction is feasible or at what temperature is not converting the entropy in to kJmol^-1 and leaving it in Jmol^-1. This will give very low temperatures.
Worksheets
Identifying Enthalpy Changes
Using Born-Haber Cycles
Enthalpy of Solution Calculations
Covalent Characteristics of Ionic Compounds
Calculating Enthalpy
Calculating Gibb’s Free Energy
