Chemistry note on thermodynamics
A born haber cycle= a thermochemical cycle that includes all the enthalpy changes involved in the formation of an ionic compound.
e.g the born haber cycle for sodium chloride if we know any five we can calculate the other: starting from the elements in their standard states.
Na s -> Na g +108kjmol-1
½ CL2-> Cl g +122kjmol-1
Na g -> Na+ + e- +496kjmol-1
Cl g + e+ -> Cl- g -349kjmol-1
Na s + ½ Cl2 -> NaCl s -411kjmol-1
When drawing the born haber cycles: * Make up a rough scale 1 line of paper to 100kjmol-1 * Plan out roughly first to avoid going off the top or bottom of the paper. The zero line representing elements in their standard state will need to be in the middle of the paper. * Remember to put in the sign of each enthalpy change and an arrow to show its direction. Possitive enthalpy changes go up, negative enthalpy changes go down.

Using born haber cycle we are able to see the formation of an ionic compound from its elements is an exothermic process. This is mainly due to the large amount of energy given out when the lattice forms.

1. Elements in their standard states. This is the energy zero of the diagram 2. Add in the atomisation of sodium. This is positive, drawn uphill. 3. Add in the atomisation of chlorine. This is positive, drawn uphill. 4. Add in the ionisation of sodium, also posstive, drawn uphill. 5. Add in the electron affinity of chlorine. This is negative drawn downhill 6. Add in the enthalpy of formation of sodium chloride negative= downhill. 7. The final unknown quantity is the lattice formation enthalpy of sodium chloride. The size of this is 788kjmol-1 from the diagram. Lattice energy is the change from separate ions to solid lattice and we must therefore go downhill so LE (Na+ + Cl) = 788kjmol-1.

Trends in lattice enthalpies.
The lattice formation enthalpies of some simple ionic compounds of formula M+ X-
Larger anions
Larger anions
Larger possitive ions
Larger possitive ions

| F- | Cl- | Br- | I- | Li+ | -1031 | -848 | -803 | -759 | Na+ | -918 | -788 | -742 | -705 | K+ | -817 | -711 | -679 | -651 | Rb+ | -783 | -685 | -656 | -628 | Cs+ | -747 | -661 | -635 | -613 |

The trend that the larger ions lead to smaller lattice enthalpies. This is because the opposite charges do not approach each other as closely when the ions are larger. The lattice enthalpies increase with the size of the charge. This is because ions with double the charge give out roughly twice as much energy when they come together.
Predicting enthalpies of formation of theoretical compounds
Born haber cycle can be used to investigate the enthalpy of formation of theoreitical compounds to see if they might be expected to exist. They use lattice enthalpies they have been calculated using sensible assumption about the crystal structures of these compounds and the sizes of the Ca+ and Ca3+ ions.
Look at the enthalpies of formation. A large amount of energy would have to be put in to form CaF3, CaF formation would given out energy but not as much as CaF2. This explains why only CaF2 has been prepared a stable compound. CaF has indeed been made but readily turns into CaF + Ca. we say it is unstable with respect to CaF2 + Ca.
Below the relative enthalpy levels of CaF and CaF2, shows CaF2 is 1185-287 = 898kjmol-1 below CaF so we can work out the ΔH for the reaction
ΔH= +574 -1185=-611 kjmol-1

[ space for diagram here]

More enthalpy changes * Ionic solids dissolve well in polar solvents * To dissolve ionic solvents lattice must be broken up * Lattice enthalpy is required * The separate ions are solvated by the solvent molecules These cluster round the ions so that the positive ions are surrounded by the negative ends of the dipole of the solvent molecules and the negative ions are surrounded by the positive ends of the dipoles of the solvent molecules. This is called hydration when the solvent is water.
The enthalpy change of hydration shows the same trends as lattice enthalpy: it is more negative for more highly charged ions and less negative for bigger ions.
We can think of dissolving an ionic compound in water as the sum of three processes: 1. Breaking the ionic lattice to give separate gaseous ions- the lattice dissociation enthalpy has to be put in. 2. Hydrating the positive ions- the enthalpy of hydration is given out 3. Hydrating the negative ions- the enthalpy of hydration is given out
For ionic compounds the enthalpy change of hydration has rather a small value and may be positive or negative. For example the enthalpy of hydration ΔH hyd for sodium chloride is given by the equation.
NaCl s + aq -> Na+ + Cl-
Here are the steps: 1. Nacl s -> Na+ + Cl- +788kjmol-1 2. Na+ + aq + Cl- -> Na+ + Cl- -406kjmol-1 3. Na+ + Cl- + aq -> Na+ + Cl- -363 kjmol-1 4. So ΔH NaCl= ΔH L (NaCl) + ΔH hyd (Na+) + ΔH hyd (Cl-) +788 -406 -363= +19kjmol-1 The process of dissolving can also be represented on an enthalpy diagram or calculated like above. Either method is equally acceptable.

Lattice enthalpies and bonding
It is possible to work out a theoretical value for the lattice formation enthalpy of an ionic compound if we know the charge on the ions, their distance apart and the geometry of their structure.
For many ionic compounds, the lattice formation enthalpy determined from experimental values via a born-haber cycle agrees very well with the calculated theoretically, and this confirms that we have the correcy model of ionic bonding in that compoiund. However, there are some compounds when there is a large discrepancy between the two values for lattice formation enthalpy because the bond in question has some covalent character.
e.g. ZnSe has ΔHL -3611kjmol, where as the theoretical value is -3305kjmol-1 which is 10% lower than the experimental value, shows some extra bonding is present.
The Zn+ is very small and has a high positive charge , while Se2- is relatively large and has a high negative charge. The small Zn2+ can approach closely to the electron clouds of the Se2- and distort them by attracting them towards it. The Se2- is fairly easy to distort, because its large size means the electrons are far from the nucleaus and its double charge means there Is plenty of negative charge to distort. This distoration means there are more electrons than expected concentrated between the Zn and Se nuclei, and represents a degree of electron sharing or covalency which accounts for the lattice enthalpy discrepancy. The Se2- ion is said to be polarised.
The factors which increase polarisation are: * Positive ion; small size, high charge * Negative ion: larger size, high charge
All ionic and covalent bonds can be seen as part of a continuum from ionic to purely covalent. E.g. caesium fluoride Cs+F- which has a large, singly charged positive ion and a small singly charged negative ion, is hardly polarised at all and is almost completely ionic, whereas a bond between two identical atoms must be 100% covalent.

Calculating enthalpy changes using mean bond enthalpies
The more negative the lattice formation enthalpy the more energy is required to break the structure apart, so the stronger the ionic bonding. Both enthalpies give a measure of the strength of covalent bonding.
The standard enthalpy of bond dissociation is the enthalpy change when a mole of gaseous molecules each breaks a covalent bond to form two free radicals.
Bond enthalpies refer to a specific bond in a molecule and have slightly different values depending on the environment of the bond. E.g. the C-F bonds enthalpy is 452kjmol-1 in CH3F, but 485kjmol-1 in CF4. 1. Comparing the strengths of bonds
e.g. bromoethane C2H5Br
C-H 413KJMOL-1
C-C 437kjmol-1
C-Br 290kjmol-1
So we might predict that the C-Br bond is most likely to break and this is in fact the case in most of the reaction of bromoethane. 2. Hess’s law cycles:
These can be used to calculate approximately enthanlipes changes for reactions where this cannot be measured experimentally, or when enthalpies of formation or combustion are not available, they can therefore be used to help predict whether a compound can be prepared.
Hess law states that the enthalpy change of a reaction is the same, whatever route taken.
We can work out the approximate enthalpy changes by using mean bond enthalpies, by adding the mean bond enthalpies of every bond in the reactants. We then add up all the mean bond enthalpies of every bond in the products
More energy out than in= exothermic = -ve
Less energy out than in= endothermic = +ve
e.g.
CH2=CH2 + H2O -> CH3CH2OH
Breaking bonds into separate atom=enthothermic
4X C-H= 4 X 413 = 1652
1 X C=C = 612
2 X 0-H = 2 X 464 = 928
Total energy put in= 3192kjmol-1
To form bonds to make products = exothermic
5 X C-H = 5 X 413 = 2065
1 X C-C = 347
1 X C-O = 358
1 X O-H = 464
Total energy given out = 3234kjmol-1
Difference = 42kjmol-1 and more energy is given out than put in so the ΔH is negative
We can see this through the thermochemical cycle:

[ space for diagram]

These are only mean values because we’ve used mean bond enthalpies which are normally 10% within the actual value range.
Bond enthalpy calculations give an indication of the sign and magnitude of the ΔH which may indicate whethee the reaction is likely to take place or not. Exothermic reactions tend to take place where as endothermic ones tend not to.

Why do chemical reactions take place?
Fesible/spontaneous= happen on their own accord- no significance to its rate though.
Most spontaneous reactions give out heat so are exothermic.
Randomness or entropy * Mixing and spreading out e.g. liquids evaporating, solids dissolving * Gas= more random than liquids due to their arrangement of particles * Endothermic reactions may be spontaneous if they involve spreading out, randomising or disordering. This is true of the two reactions above the arrangement of the particles in the products is more random than in the reactants.
The randomness of a system, expressed mathematically= entropy of the system and is given symbol S.
Entropies have been determinded for a vast range of substances and can be looked up. With standard conditions 298k and 100kpa. * Gases have larger values than liquids, which have larger values than solids. * Entropy units= JK-1Mol-1 * Entropy increase when water turns to steam * Entropies increase with temperature * Higher temperatures, particles spread out and randomness increases. Calculating entropy changes
Can be calculated by adding all entropies of products and subtracting sum of entropies of the reactants.
e.g. CaCO3 -> CaO + CO2 entropy of products = 40+ 214 = 254 jkmol-1 entropy of reactant = 93Jk-1mol-1
ΔS = 254 – 93 = + 161Jk-1mol-1
Large positive value so gas has formed from a solid

The gibbs free energy change ΔG
Two factors govern spontaneity of a reaction 1. The enthalpy change 2. The entropy change
Combine they are the gibbs free energy G, if the ΔG of a reaction is negative the reaction is fesible, if its positive it is not fesible.
ΔG= ΔH – TΔS
ΔG depends on temeperature, due to TΔS, so some reactions are fesible at one temperature and not another.
So an endothermic reaction csn be fesible when the temp is increased if there is a large enough positive entropy change. ( positive entropy make ΔG more negative as the negative sign in the TΔS term)
e.g.
caCO3 -> CaO + CO2 ΔH= +178kjmol-1 the ΔS = +161Jk-1mol-1 so = 0.161 kJk-1mol-1 so at room temp (298k)
ΔG = ΔH – TΔS
ΔG = 178 – (298 X 0.161) = + 130kjmol-1
So not fesible, but the reverse reaction will have ΔG = -130 kjmol-1, so will be fesible.
But if we do it at 1500k
The ΔG = -64kjmol-1
And so this is fesible at this temperature. This reaction occurs to make lime calcium oxide from limestone.

What happens when ΔG = 0
At this point the reaction is just fesible.
ΔG = ΔH – TΔS
0 = ΔH – TΔS
ΔH = TΔS where ΔH= + 178kjmol-1 and ΔS= 0.16kjk-1mol-1
So T= 178/ 0.161 = 1105.6 K
It is in equilibrium where both products and reactants are present.

Calculating entropy change
Use ΔG= 0
e.g. a solid at its melting point = exists as a solid or liquid- an equilibrium exists between solid and liquids so the ΔG for the melting process = 0
0=ΔH – TΔS
Temp= 273 K
Enthalpy change for melting = 6.0kjmol-1
0= 6.0 – 273 X ΔS
ΔS= 6.0/273 = 0.022kjk-1mol-1 = + 22kjmol-1
This is the entropy change when ice changes to water. It is positive, we would expect as the molecules in water= more disordered than those in ice.

Extracting metals * Heat with carbon -> CO2 and leaves metal * Carbon is cheap * Gaseous carbon dioxide diffuses away, sot here isn’t a problem separating it from the metal * We can us ΔG to investigate under what conditions the reaction might be fesible for different metals * One of the most important metals is iron and its ore = irom oxide fe2o3 so we can calculate ΔG from a thermochemical cycle.

[space for diagram]

2Fe2O3 + 3C -> 4fe + 3CO2 ΔG= +302kjmol-1
Not fesible under standard conditions (298k)
Will the reaction take place at a higher temperature?
We can work out the temperature from the following thermochemical cycle- when ΔG = ΔH – TΔS = 0

[ space for diagram ]

Calculating ΔS
We can calculate the entropy change of the reaction by finding the difference between the sum of the entropies of all the products and the sum of the entropies of all the reactants:
2FeO3 + 3C -> 4Fe + 3CO2
2 X 87.4 + 3 X 5.7 - > 4 X 27.3 + 3 X 213.6
191.9 750.0
ΔS = +558 JK-1mol-1 turning from solid to gas
ΔG = ΔH – TΔS
0= 467.9-T X 558.1/1000
= 838.4 K NOT FESIBLE
Kinetic factors
Neither enthalpy nor entropy tell us anything about how quickly or slowly a reaction is likely to occur. So we may predict that a certain reaction should occur spontaneously as of enthalpy and entropy changes, but the reaction may take place so slowly that for practical purposes it doesn’t occur at all. In other words there is a large activation energy barrier for the reaction.
e.g. C + O2 -> CO2 ΔH= -393.5KJMOL-1 the reaction is exothermic and entropy increases as we go from an ordered solid to a disordered gas, so we would expect the reaction to go on both counts. We can calculate the actual value of ΔS and hence find ΔG

calculating ΔS the ΔS for a reaction is the sum of the entropies of the product minus the sum of the entropies of the reactants C + O2 -> CO2
S/JK-1MOL-1 5.7 205.0 213.6
SO ΔS= 213.6 – 5.7+205.0
ΔS = +2.9jk-1mol-1 possitive as predicted.

Calculating ΔG
ΔG =ΔH – TΔS
So USSC : ΔG= -394 –(298 X 2.9/1000)
ΔG= -395-0.86
ΔG= -394.86KJMOL-1 negative so reaction = fesible
However , experience with graphite tells us that the reaction doesn’t take place at room temp. although it will take place at higher temperatures. At room temperature , the reaction is so slow that in practice it doesn’t take place at all.
Since the branch of chemistry dealing with enthalpy and entropy changes is called thermodynamics, and that dealing with rates is called kinetics, it is said to be thermodynamically unstable by kinetically stable.

http://web.alfretongrange.derbyshire.sch.uk/web/docs/sci-resources/Ks5/yr13/Chemistry/Thermodynamics.pdf