Going over basics:
Enthalpy symbol H, is stored energy, its not possible to measure enthalpy but enthalpy changes (, by measuring temperature
changes of reactions at constant pressure. Pronounced ‘delta’ means ‘change of’
Enthalpy change,  is the measure of the transfer of energy into or out of a reacting system at constant pressure
Exothermic reaction – Energy given out by reactants as they form products
Energy exits in exothermic reactions is negative, enthalpy(stored energy) decreases, temperature increases
Bond making, amount of heat released is greater than the amount of heat used to start the reaction
Endothermic reaction – Energy is taken in by reactants to form products
Energy enters in endothermic reactions  is positive, enthalpy(stored energy) increases, temperature decreases
Bond breaking, energy continues to be absorbed as long as the reaction continues
Definitions to remember:
Lattice Enthalpy Hlatt Energy released per mole for exothermic process M+(g) + X–(g)  M+X-(s)
(Enthalpy/heat energy released when gaseous ions come together to form 1 mole of solid)
Standard enthalpy of atomisation Ha θ Enthalpy change for production of one mole of gaseous atoms from the element in its
standard state
Enthalpy of hydration Hfhyd θ Enthalpy change per mole for dissolving the gaseous ions, with enough water to form an
infinite dilute solution.
(When water is used as the solvent, the dissolving process is called hydration)
(For a unipositive cation Hhyd is exothermic: M+(g) + aq M+(aq))
Understand that:
Enthalpy of formation Hf Enthalpy change when one mole of a compound is formed from its elements
Na(s) + (1/2)Cl2(g) → NaCl(s)
1st Ionisation Energy M(g)  M+(g) + e– 2nd Ionisation Energy M+(g)  M2+(g) + e–
1st Electron affinity Enthalpy change per mole for the process, X(g) + e–  X–(g)
- Negative(exothermic), since the electron is attracted by the positive charge on the atoms nucleus
2nd Electron affinity Enthalpy change per mole for the process, X–(g) + e–  X2–(g)
O–(g) + e O2(g)
- Positive(endothermic), since energy needed to overcome repulsion between the electron and negative ions
Questions
Write the equation, with state symbols, for the enthalpy of atomisation of chlorine
½Cl2(g)  Cl(g)
Write an equation which represents the change when the second electron affinity of oxygen is measured
O–(g) + e O2(g)
Explain the trend in IE within the group 1 elements
Ionisation Energy/kJ mol –1
Element
Li
+519
Na
+494
K
+418
Rb
+402
Cs
+376
electron further away from nucleus
inceased shielding effect inner shells
less energy needed to remove outer electron
Construct a Born-Haber cycle and carry out associated calculations
A Born-Haber cycle calculates the lattice enthalpy
by comparing the Hf of the ionic compound (from
the elements) to the enthalpy required to make
gaseous ions from the elements
The Born-Haber cycle involves the formation of an
ionic compound from the reaction of a metal(often a
Group1/2 element) with a non-metal
All endothermic reactions shown by arrow pointing
upwards (vice versa)
1.
2.
3.
4.
5.
Questions
Presenting an ionic solid, NaH
Draw a Born-Haber cycle which could be used to determine the
electron affinity of hydrogen
Na + (g) + H(g) + e –
Na + (g) + H – (g)
Na(g) + H(g)
Atomisation enthalpy of metal (in this case
lithium)
Ionisation enthalpy of metal
Atomisation enthalpy of non-metal (in this
case fluorine)
Electron affinity of non-metal
Lattice enthalpy
Construct a Born-Haber cycle and to obtain Hlatt of SrCl2(s)
Hf of SrCl2(s)
-829 kJmol–1
Ha of strontium
+164 kJmol–1
Ha of chlorine
+122 kJmol–1
st
1 IE of strontium
+550 kJmol–1
2nd IE of strontium 104 kJmol–1
of chlorine
-349 kJmol–1
Remember:
2x EA of chlorine used
2x Ha of chlorine used
Na(g) + 1/2 H 2 (g)
Na(s) + 1/2 H 2 (g)
NaH(s)
Construct a Born-Haber cycle
And find 2nd electron affinity of oxygen
H / kJ
mol–1
Enthalpy of atomisation of
+150
magnesium
Bond energy of O == O in
+496
oxygen
1st ionisation energy of
+736
magnesium
2nd ionisation energy of
+1450
magnesium
Ist electron affinity of
–142
oxygen
Lattice enthalpy of
–3889
magnesium oxide
Enthalpy of formation of
–602
magnesium oxide
150 + ½(496) + 736 + 1450 + 602 + × = 142 + 3889 x = +845 k J mol– 1
½ × (496) is key point
Use the data below to calculate the first electron affinity of chlorine.
-642 = 150 +736 +1450 +2(121) +2x +(-2493)
2x = 727
Enthalpy change
Enthalpy change
–1
x = –363 ± 1
+150
kJ
mol
Hat of magnesium
–1
+736
kJ
mol
1st IE of magnesium
+1450 kJ mol–1
2nd IE of magnesium
–642 kJ mol–1
Hf of MgCl2
+121 kJ mol–1
Hat of chlorine
–2493 kJ mol–1
Hlatt of MgCl2
Understand the factors that influence the value of the lattice energies
Factors affecting theoretical value of lattice enthalpy/energy •Radius/size of ions •Charges on ions
 Strong attraction - Small ionic radius and high charge
 Weak attraction – Large ionic radius and small charge
Lattice energy is the measure of the strength of bonds in that ionic compound. It is the equivalent to the amount of energy required
to separate a solid ionic compound into gaseous ions (always negative, exothermic)
- When lattice energy increases it becomes more negative
- Lattice energies increase when ions are smaller with high charge
-   Strong attraction between ions because their ionic radii are small
  Less attraction between ions because their ionic radii are larger
Understand that values of lattice energies calculated from the Born-Haber cycle may differ from those calculated from a purely
ionic model - limited to the radius and charge of the ions
- Experimental lattice energies are from Born-Haber cycles
- Theoretical lattice energies are from equations, assumes ionic lattice is totally ionic, when actually it has covalent
character(electron sharing)
 If the metal cation is small and/or highly charged, it will distort the electron cloud of the anion , more polarising
 If the non metal anion is larger it is more polarisable
 This polarisation of the negative ion leads to partial covalency
Questions:
Theoretically Hlatt MgCl2 is –2326 kJ mol–1
Experimental Hlatt MgCl2 is –2526 kJ mol–1
Explain why this difference occurs
• MgCl2 has (a degree of ) covalent character
• due to polarisation of the anion
Theoretical Hlatt MgI2, is –1944 kJ mol–1
Experimental Hlatt MgI2, is –2327 kJ mol–1
Explain why this difference occurs
• magnesium ion is small and highly charged
• leading to polarisation of the (large) iodide ion
• and (causing) covalency (into the lattice)
The theoretical and actual values of the lattice enthalpy of magnesium fluoride are very similar because magnesium fluoride is
almost completely ionic. Explain why magnesium fluoride is almost completely ionic
• Fion is small• Mg2+ ion does not have a high enough charge density to polarise F
Magnesium iodide compound. Radius of magnesium ion is 0.072 nm, iodide ion is 0.215 nm.
(i) Describe the effect that the magnesium ion has on an iodide ion next to it in the magnesium iodide lattice - The electrons
around the iodide ion are drawn towards the magnesium ion
(ii) What TWO quantities must be known about the ions in a compound in order to calculate a theoretical lattice energy?
•Radius/size of ions •Charges on ions
(iii) Suggest how the value of the theoretical lattice energy would compare with the experimental value from a Born-Haber Cycle
for magnesium iodide. - Less (exothermic) - covalent character (strengthens lattice)
Why is the lattice energy of magnesium hydroxide more exothermic than that of barium hydroxide?
• as magnesium has a much smaller ion (than barium ion)
• and has same charge
• so stronger attraction between ions
“charge density” scores 1 (out of first 2 marks)
Explain why the lattice enthalpy of magnesium fluoride, MgF2, is more exothermic than that of calcium chloride.
• smaller size of cation • smaller size of anion • greater attraction between (oppositely charged) ions
Lattice energies
Lattice energies
MgCl2 –2526 kJ mol–1
NaCl(s) is –771 kJ mol–1
CaCl2 –2237 kJ mol–1
MgO(s) is –3889 kJ mol–1
SrCl2 –2112 kJ mol–1
Explain the difference in lattice energies
BaCl2 –2018 kJ mol–1
Explain why lattice energies become less exothermic
• Lattice enthalpy depends on charges and the ionic radii
• Comparison of Na+/Mg2+ size and charge
• As group descended, radius of M 2+ (ion) increases
• Comparison of Cl– / O2– size and charge
• Charge on ions remains the same
• (High LE results from) higher interaction
• (down group) weaker forces of attraction between ions
M+ (g)
H Lattice
+
X– (g)
H Hydration
H Hydration
Find enthalpy of solution of NaCl:
Lattice enthalpy of sodium chloride = -771
hydration enthalpy of Na+
= -406
hydration enthalpy of Cl= -364
Hsol = Hhyd (Na+) + Hhyd (Cl-) – Hlatt(NaCl)
= (-406) + (-364) – (-711) = +1kJmol-1
M + X – (s)
H Solution
M+ (aq) +
X – (aq)
Understand how Hlatt and Hhyd vary the solubilities of the hydroxides and sulphates of Group 2
Enthalpy change of solution - Enthalpy change when one mole of a substance is dissolved completely in a large volume of a
solvent at constant pressure. Remember: Hsoln = Hlatt +X– + Hhyd+HhydX–
(Always small because they almost cancel out)
To dissolve, Hhyd ≥ Hlattice so enough hydration energy needed to overcome breaking the lattice
When an ionic substance dissolves enthalpy change depends on • Hlatt of the solid • Hhyd of the ions
Energy has to be supplied to break up the lattice of ions
Energy is released when these ions form bonds with water molecules
Trends in solubility depend on how fast both enthalpy terms fall relative to each other.
- As you go down a Group:
 Energy needed to break up the lattice falls because, bigger ions, larger distance between ions, less attraction between +
and - ions
 Hydration enthalpies falls - bigger ions, less charge density, reduces the attraction of water, the less exothermic the
hydration enthalpy.
- Hydroxides become more soluble
The lattice enthalpy falls faster than the hydration enthalpy, Hsoln becomes more exothermic(-) (more soluble)
- Sulphates become less soluble
The hydration enthalpy falls faster the Hsoln becomes more endothermic(+) (less soluble)
Because sulphate ion is bigger, change in ionic radius of Group 2 cations doesn’t have as much affect on Hlatt
The greater the charge density the easier it is for the Group 2 cation to hydrate and hence dissolve in water due to greater
attraction with the polar water molecules.
Salt
MgSO4
CaSO4
SrSO4
BaSO4
Relative solubility
1
10–2
10–4
l0–6
Explain the reasons for this trend in solubility in terms of changes of lattice energies and
enthalpies of hydration.
• salt likely to be more soluble if Hsol exothermic
• both lattice energy and hydration enthalpies become less exothermic
• as cations increase in size
• but lattice energy changes less so enthalpy of solution less exothermic
H /kJ mol–1
–1480
Hhydration of
–1360
Hhydration of Ba2+
–460
Hhydration of OH–
Lattice enthalpy of Sr(OH)2
–1894
Lattice enthalpy of Ba(OH)2
–1768
(i) Explain why the lattice enthalpy of strontium hydroxide is different from that of barium hydroxide.
charge density of Sr2+ < Ba2+  stronger force of attraction between ions
(ii) Explain why the hydration enthalpy of a cation is exothermic.
• Negative part of water attracted to (+ ion) and forms bond • bond formation releases energy
(iii) Use the lattice enthalpy and hydration enthalpy values to explain why barium hydroxide is more soluble in water than
strontium hydroxide.
• Hsol = Hhyd – Hlattice
• Hlatt and Hhydr decrease down Group 2 (Barium lower than Strontium)
• Hlatt decreases more than the Hhydr
• Hsoln Ba(OH)2 more exothermic (than for Sr(OH)2,(so more soluble))
Hsoln = –Hlatt + HhydrM2+ion + 2 × Hhydr OH– ion or = –Hlatt + Hhydr ions
Hsoln Sr(OH)2 = – (– 1894) + (– 1480) + 2 ×(– 460)
= – 506 kJ mol–1
Hsoln Ba(OH)2 = – (– 1768) + (– 1360) + 2 ×(– 460)
= – 512 kJ mol–1
Enthalpy of hydration of Mg2+ –1890 kJ mol–1
Calculate Hsoln of Mg(OH)2
Enthalpy of hydration of Ba2+ –1275 kJ mol–1
Hsoln = -1890 -550 -550 +2995 = 5kJ mol–1
Enthalpy of hydration of OH– –550 kJ mol–1
Lattice energy of Mg(OH)2 –2995 kJ mol–1
Use the data to explain how the solubility of Ba(OH) 2
Lattice energy of Ba(OH)2 –2320 kJ mol–1
compares with Mg(OH)2
Sr2+
Draw a labelled Hess’s law cycle for Mg(OH)2(s)
• Hlatt down but Hhyd down by less
• ∴ ΔHsolution is more exothermic
• ∴ solubility is greater
Explain why magnesium oxide is insoluble in water.
• not enough energy generated by hydration to overcome breaking the lattice
• solubility due to balance between – Hlattice and Hhydration of the ions
Hsol = Hhyd – Hlattice
Further:
• Formation of MgCl2 is energetically favoured because of its higher lattice enthalpy than MgCl, which is almost never formed
• High lattice enthalpy more than compensates for the additional energy that has to be supplied for the 2 nd ionisation of magnesium