Lattice enthalpies and Born-Haber cycles
In a nutshell
The lattice enthalpy change of formation is the enthalpy change when gaseous ions form one mole of a solid ionic compound under standard conditions. Born-Haber cycles can be drawn to calculate the lattice enthalpy change of formation. The theoretical values of lattice enthalpy may not match the experimental values.
Equations
There are general chemical equations for different enthalpy changes you should be able to use with any given reaction.
element in its standard state→one mole of gaseous atomelement\>in\>its\>standard\>state→one\>mole\>of\>gaseous \>atomelementinitsstandardstate→onemoleofgaseousatom
one mole of compound in its standard state→ gaseous atomsone\>mole\>of\>compound \>in\>its\>standard\>state→\>gaseous\>atomsonemoleofcompoundinitsstandardstate→gaseousatoms
gaseous compound → gaseous atomsgaseous\>compound\>→\>gaseous\>atomsgaseouscompound→gaseousatoms
one mole of gaseous atoms →one mole of gaseous ions1++electronone\>mole\>of\>gaseous\>atoms\>→one\>mole\>of\>gaseous\>ions^{1+}+electrononemoleofgaseousatoms→onemoleofgaseousions1++electron
one mole of gaseous ions1+ →one mole of gaseous ions2++electronone\>mole\>of\>gaseous\>ions^{1+}\>→one\>mole\>of\>gaseous\>ions^{2+}+electrononemoleofgaseousions1+→onemoleofgaseousions2++electron
one mole of gaseous atoms + electron →one mole of gaseous ions1−one\>mole\>of\>gaseous\>atoms\>+\>electron\>→one\>mole\>of\>gaseous\>ions^{1-}onemoleofgaseousatoms+electron→onemoleofgaseousions1−
one mole of gaseous ions1− + electron →one mole of gaseous ions2−one\>mole\>of\>gaseous\>ions^{1-}\>+\>electron\>→one\>mole\>of\>gaseous\>ions^{2-}onemoleofgaseousions1−+electron→onemoleofgaseousions2−
one mole of gaseous ions → one mole of aqueous ionsone\>mole\>of\>gaseous\>ions\>→\>one\>mole\>of\>aqueous\>ionsonemoleofgaseousions→onemoleofaqueousions
one mole of solute → aqueous soluteone\>mole\>of\>solute\>→\>aqueous\>soluteonemoleofsolute→aqueoussolute
elements in standard states → one mole of compoundelements\>in\>standard\>states\>→\>one\>mole\>of\>compoundelementsinstandardstates→onemoleofcompound
Lattice enthalpy
Formation and dissociation
Positive and negative ions are held together via electrostatic attractions to form giant ionic lattices. The energy released when the solid lattice is formed from gaseous ions is called lattice energy. The lattice enthalpy of formation, △fH\triangle_{f}H△fH, is the enthalpy change when gaseous ions form one mole of a solid ionic compound under standard conditions.
The lattice enthalpy measures how strong an ionic bond is. The more negative it is (more exothermic), the more stronger the ionic bond is.
Example
The following lattice formation enthalpies are given below. Which species has the strongest ionic bonds?
K+(g)+Cl−(g)→KCl(s)K^+(g)+Cl^-(g)→KCl(s)K+(g)+Cl−(g)→KCl(s)
△latticeH⦵=−698 kJ mol−1\triangle_{lattice}H^{⦵}=-698\>kJ\>mol^{-1}△latticeH∘−=−698kJmol−1
Ca2+(g)+2Cl−(g)→CaCl2(s)Ca^{2+}(g)+2Cl^-(g)→CaCl_2(s)Ca2+(g)+2Cl−(g)→CaCl2(s)
△latticeH⦵=−1145 kJ mol−1\triangle_{lattice}H^{⦵}=-1145\>kJ\>mol^{-1}△latticeH∘−=−1145kJmol−1
CaCl2CaCl_2CaCl2 has stronger ionic bonds as the standard lattice energy is more exothermic.
The lattice enthalpy of dissociation is the enthalpy change under standard conditions when one mole of the solid ionic compound dissociates into its gaseous ions completely. It is endothermic, opposite to the enthalpy of formation.
Example
The lattice dissociation enthalpy of potassium chloride is given below.
KCl(s)→K+(g)+Cl−(g)KCl(s)→K^+(g)+Cl^-(g)KCl(s)→K+(g)+Cl−(g)
△latticeH⦵=698 kJ mol−1\triangle_{lattice}H^{⦵}=698\>kJ\>mol^{-1}△latticeH∘−=698kJmol−1
Factors affecting lattice energy
The lattice energy is affected by ionic charge and ion size. The larger the ionic charge, the larger the lattice energy. This is because more energy is released upon formation of the ionic lattice. The smaller the radius of the ion, the lattice energy will be more exothermic. This is because the ions are held closer together in the lattice.
Example
Predict which lattice energy is larger, KClKClKCl or CaCl2CaCl_2CaCl2.
A potassium ion is singly charged and has a larger ionic radius. A calcium ion is doubly charged and has a smaller ionic radius. The charge density for calcium chloride is larger. Therefore, calcium chloride has a larger lattice energy.
Enthalpy change
The standard enthalpy change, △H⦵\triangle H^{⦵}△H∘−, is the transfer of heat energy in a reaction under standard conditions. A negative enthalpy change indicates an exothermic reaction, heat energy is released. A positive enthalpy change indicates an endothermic reaction, heat energy is absorbed.
There are many types of enthalpy changes. You must know all of these definitions and be able to identify them.
Enthalpy change of atomisation of an element
Enthalpy change of atomisation of an element, △atH\triangle_{at}H△atH, is the enthalpy change when an element in its standard state forms one mole of its gaseous atoms.
element in its standard state→one mole of gaseous atomelement\>in\>its\>standard\>state→one\>mole\>of\>gaseous \>atomelementinitsstandardstate→onemoleofgaseousatom
Example
one gaseous bromine element →one mole of gaseous bromine atomone\>gaseous\>bromine\>element\>→one\>mole\>of\>gaseous\>bromine\>atomonegaseousbromineelement→onemoleofgaseousbromineatom
12Br2(g)→Br(g)\dfrac{1}{2}Br_2(g)→Br(g)21Br2(g)→Br(g)
Enthalpy change of atomisation of a compound
Enthalpy change of atomisation of a compound, △atH\triangle_{at}H△atH, is the enthalpy change when gaseous atoms are formed from one mole of a compound in its standard state.
one mole of compound in its standard state→ gaseous atomsone\>mole\>of\>compound \>in\>its\>standard\>state→\>gaseous\>atomsonemoleofcompoundinitsstandardstate→gaseousatoms
Example
solid potassium chloride→ gaseous potassium atom + gaseous chlorine atomsolid \>potassium\>chloride→\>gaseous\>potassium\>atom\>+\>gaseous\>chlorine\>atomsolidpotassiumchloride→gaseouspotassiumatom+gaseouschlorineatom
KCl(s)→K(g)+Cl(g)KCl(s)→K(g)+Cl(g)KCl(s)→K(g)+Cl(g)
Bond dissociation enthalpy
Bond dissociation enthalpy, △dissH\triangle_{diss}H△dissH, is the enthalpy change when one mole of gaseous bonds of all the same type are broken.
gaseous compound → gaseous atomsgaseous\>compound\>→\>gaseous\>atomsgaseouscompound→gaseousatoms
Example
gaseous bromine → gaseous bromine atomsgaseous\>bromine\>→\>gaseous\>bromine\>atomsgaseousbromine→gaseousbromineatoms
Br2(g)→2Br(g)Br_2(g)→2Br(g)Br2(g)→2Br(g)
First ionisation energy
First ionisation energy, △ie1H\triangle_{ie1}H△ie1H, is the enthalpy change when one mole of gaseous atoms forms one mole of gaseous 1+1+1+ ions.
one mole of gaseous atoms →one mole of gaseous ions1++electronone\>mole\>of\>gaseous\>atoms\>→one\>mole\>of\>gaseous\>ions^{1+}+electrononemoleofgaseousatoms→onemoleofgaseousions1++electron
Example
one mole of gaseous calcium →one mole of gaseous calcium ion1++electronone\>mole\>of\>gaseous\>calcium\>→one\>mole\>of\>gaseous\>calcium\>ion^{1+}+electrononemoleofgaseouscalcium→onemoleofgaseouscalciumion1++electron
Ca(g)→Ca1+(g)+e−Ca(g)→Ca^{1+}(g)+e^-Ca(g)→Ca1+(g)+e−
Second ionisation energy
Second ionisation energy, △ie2H\triangle_{ie2}H△ie2H, is the enthalpy change when one mole of gaseous 1+1+1+ ions forms one mole of gaseous 2+2+2+ ions.
one mole of gaseous ions1+ →one mole of gaseous ions2++electronone\>mole\>of\>gaseous\>ions^{1+}\>→one\>mole\>of\>gaseous\>ions^{2+}+electrononemoleofgaseousions1+→onemoleofgaseousions2++electron
Example
one mole of gaseous calcium ion1+→one mole of gaseous calcium ion2++electronone\>mole\>of\>gaseous\>calcium\>ion^{1+}→one\>mole\>of\>gaseous\>calcium\>ion^{2+}+electrononemoleofgaseouscalciumion1+→onemoleofgaseouscalciumion2++electron
Ca1+(g)→Ca2+(g)+e−Ca^{1+}(g)→Ca^{2+}(g)+e^-Ca1+(g)→Ca2+(g)+e−
First electron affinity
First electron affinity, △ea1H\triangle_{ea1}H△ea1H, is the enthalpy change when one mole of gaseous atoms form one mole of gaseous 1−1-1− ions.
one mole of gaseous atoms + electron →one mole of gaseous ions1−one\>mole\>of\>gaseous\>atoms\>+\>electron\>→one\>mole\>of\>gaseous\>ions^{1-}onemoleofgaseousatoms+electron→onemoleofgaseousions1−
Example
one mole of gaseous sulfur + electron→one mole of gaseous sulfur ion1−one\>mole\>of\>gaseous\>sulfur\>+\>electron→one\>mole\>of\>gaseous\>sulfur\>ion^{1-}onemoleofgaseoussulfur+electron→onemoleofgaseoussulfurion1−
S(g)+e−→S1−(g)S(g)+e^-→S^{1-}(g)S(g)+e−→S1−(g)
Second electron affinity
Second electron affinity, △ea2H\triangle_{ea2}H△ea2H, is the enthalpy change when one mole of gaseous 1−1-1− ions form one mole of gaseous 2−2-2− ions.
one mole of gaseous ions1− + electron →one mole of gaseous ions2−one\>mole\>of\>gaseous\>ions^{1-}\>+\>electron\>→one\>mole\>of\>gaseous\>ions^{2-}onemoleofgaseousions1−+electron→onemoleofgaseousions2−
Example
one mole of gaseous sulfur1− + electron→one mole of gaseous sulfur ion2−one\>mole\>of\>gaseous\>sulfur^{1-}\>+\>electron→one\>mole\>of\>gaseous\>sulfur\>ion^{2-}onemoleofgaseoussulfur1−+electron→onemoleofgaseoussulfurion2−
S1−(g)+e−→S2−(g)S^{1-}(g)+e^-→S^{2-}(g)S1−(g)+e−→S2−(g)
Enthalpy change of hydration
Enthalpy change of hydration, △hydH\triangle_{hyd}H△hydH, is the enthalpy change when one mole of gaseous ions form one mole of aqueous ions.
one mole of gaseous ions → one mole of aqueous ionsone\>mole\>of\>gaseous\>ions\>→\>one\>mole\>of\>aqueous\>ionsonemoleofgaseousions→onemoleofaqueousions
Example
one mole of gaseous potassium ion → one mole of aqueous potassium ionone\>mole\>of\>gaseous\>\>potassium\>ion\>→\>one\>mole\>of\>aqueous\>potassium\>iononemoleofgaseouspotassiumion→onemoleofaqueouspotassiumion
K+(g)→K+(aq)K^+(g)→K^+(aq)K+(g)→K+(aq)
Enthalpy change of solution
Enthalpy change of solution, △solutionH\triangle_{solution}H△solutionH, is the enthalpy change when sufficient solvent dissolves one mole of solute where there is not any further enthalpy change upon further dilution.
one mole of solute → aqueous soluteone\>mole\>of\>solute\>→\>aqueous\>soluteonemoleofsolute→aqueoussolute
Example
solid potassium chloride → aqueous potassium chloridesolid\>potassium\>chloride\>→\>aqueous\>potassium\>chloridesolidpotassiumchloride→aqueouspotassiumchloride
KCl(s)→KCl(aq)KCl(s)→KCl(aq)KCl(s)→KCl(aq)
Enthalpy change of formation
Enthalpy change of formation, △fH\triangle_{f}H△fH, is the enthalpy change when elements in their standard state forms one mole of a compound under standard conditions.
elements in standard states → one mole of compoundelements\>in\>standard\>states\>→\>one\>mole\>of\>compoundelementsinstandardstates→onemoleofcompound
Example
hydrogen + oxygen → one mole of waterhydrogen\>+\>oxygen\>→\>one\>mole\>of\>waterhydrogen+oxygen→onemoleofwater
H2(g)+12O2(g)→H2O(l)H_2(g)+\dfrac{1}{2}O_2(g)→H_2O(l)H2(g)+21O2(g)→H2O(l)
Born-Haber cycles
Born-Haber cycles are used to calculate lattice enthalpies.
procedure
1. | Draw the Born-Haber cycle. |
2. | Find a route to calculate the lattice enthalpy. |
3. | Calculate lattice enthalpy by adding all arrows facing in the correct direction and subtracting all arrows not facing in the correct direction. |
Example
The Born-Haber cycle for calcium chloride is given below. Calculate the lattice enthalpy of formation for calcium chloride.
Label 1=△ie2HLabel 2=△ie1HLabel 3=△atH (Ca)Label 4=2×△atH (Cl)Label 5=△fHLabel 6=2×△eaHLabel 7=△LEHLabel\>1=\triangle _{ie2}H \newline Label\>2=\triangle _{ie1}H \newline Label\>3=\triangle _{at}H \>(Ca) \newline Label\>4=2 \times \triangle _{at}H\>(Cl) \newline Label\>5=\triangle _{f}H \newline Label\>6=2 \times \triangle _{ea}H \newline Label\>7=\triangle _{LE}H \newlineLabel1=△ie2HLabel2=△ie1HLabel3=△atH(Ca)Label4=2×△atH(Cl)Label5=△fHLabel6=2×△eaHLabel7=△LEH
The enthalpies values are given below.
Type of enthalpy | Symbol | Value |
Enthalpy of formation | △fH\triangle_{f}H△fH | −795.8 kJ mol−1-795.8\>kJ\>mol^{-1}−795.8kJmol−1 |
Enthalpy of atomisation for chlorine | △atH(Cl)\triangle_{at}H(Cl)△atH(Cl) | 121 kJ mol−1121\>kJ\>mol^{-1}121kJmol−1 |
Enthalpy of atomisation for calcium | △atH(Ca)\triangle_{at}H(Ca)△atH(Ca) | 178 kJ mol−1178\>kJ\>mol^{-1}178kJmol−1 |
Enthalpy of first ionisation energy for calcium | △ie1H\triangle_{ie1}H△ie1H | 589.8 kJ mol−1589.8\>kJ\>mol^{-1}589.8kJmol−1 |
Enthalpy of second ionisation energy for calcium ion | △ie2H\triangle_{ie2}H△ie2H | 1145.5 kJ mol−11145.5\>kJ\>mol^{-1}1145.5kJmol−1 |
Enthalpy of electron affinity for chlorine | △eaH\triangle_{ea}H△eaH | −346 kJ mol−1-346\>kJ\>mol^{-1}−346kJmol−1 |
Find a route with known values to calculate the lattice enthalpy. Subtract all arrows facing in the opposite direction and add all arrows facing in the right direction:
△LEH=−2(△eaH)−△ie2H−△ie1H−△atH(Ca)−2(△atH(Cl))+△fH\triangle_{LE}H=-2(\triangle_{ea}H)-\triangle_{ie2}H-\triangle_{ie1}H-\triangle_{at}H(Ca)-2(\triangle_{at}H(Cl))+\triangle_{f}H△LEH=−2(△eaH)−△ie2H−△ie1H−△atH(Ca)−2(△atH(Cl))+△fH
The atomisation and electron affinity of chlorine are multiplied by two as there are two chlorine atoms.
Calculate the lattice enthalpy by inserting known the values:
△LEH=−2(−346)−1145.5−589.8−178−2(121)+(−795.8)=−2259.1 kJ mol−1\triangle_{LE}H=-2(-346)-1145.5-589.8-178-2(121)+(-795.8)=-2259.1\>kJ\>mol^{-1}△LEH=−2(−346)−1145.5−589.8−178−2(121)+(−795.8)=−2259.1kJmol−1
The lattice enthalpy of calcium chloride is −2259.1 kJ mol−1‾\underline{-2259.1\>kJ\>mol^{-1}}−2259.1kJmol−1.
Theoretical and experimental
The purely ionic model of a lattice can be used to calculate the the theoretical lattice enthalpy. This model uses the assumption that all ions are spherical because they have an evenly distributed charge. However, this isn't true in real life.
Some ionic compounds have covalent character. Some ions have high polarising character that arises covalent character. This distorts the ideal spherical ionic model. This results in the experimental lattice enthalpy value being more exothermic.