Born-Haber Calculations
Thermodynamics Worksheet
Reference Data
Enthalpy of Atomisation (kJ mol⁻¹)
| Li | O | Na | Mg | Cl | K | Ca | Cu | Br |
|---|---|---|---|---|---|---|---|---|
| 159.4 | 249.2 | 107.3 | 147.7 | 121.7 | 89.2 | 178.2 | 338.3 | 111.9 |
Ionisation Enthalpy (kJ mol⁻¹)
| Element | 1st IE | 2nd IE |
|---|---|---|
| Li | 520.0 | 7298 |
| Na | 520.0 | 4563 |
| K | 419.0 | 3052 |
| Mg | 738.0 | 1451 |
| Ca | 590 | 1145 |
| Cu | 746.0 | 1958 |
Electron Affinity (kJ mol⁻¹)
| Element | 1st EA | 2nd EA |
|---|---|---|
| Br | -324.6 | – |
| O | -141.1 | 798 |
Lattice Enthalpy of Formation (kJ mol⁻¹)
| Na₂O | -2478.0 |
| MgCl₂ | -2526.0 |
| KBr | -679.0 |
| CaBr₂ | -2178.0 |
| CuO | -4104.0 |
| Cu₂O | -3244.7 |
Enthalpy of Formation (kJ mol⁻¹)
| LiBr | -351.2 |
| MgCl₂ | -641.3 |
| KBr | -393.8 |
| CaBr₂ | -682.8 |
Calculations
1. Construct a Born Haber Cycle for LiBr and use the information in the table to calculate the lattice enthalpy of formation of LiBr.
Working Out:
ΔHf = ΔHat(Li) + ΔHIE1(Li) + ΔHat(Br) + ΔHEA1(Br) + LE
-351.2 = 159.4 + 520.0 + 111.9 + (-324.6) + LE
-351.2 = 466.7 + LE
LE = -351.2 – 466.7
LE = -817.9 kJ mol⁻¹
2. Construct a Born Haber Cycle for Na2O and use the information in the table to calculate the enthalpy of formation of Na2O.
Working Out:
ΔHf = 2ΔHat(Na) + 2ΔHIE1(Na) + ΔHat(O) + ΔHEA1(O) + ΔHEA2(O) + LE
ΔHf = 2(107.3) + 2(520.0) + 249.2 + (-141.1) + 798 + (-2478.0)
ΔHf = 214.6 + 1040.0 + 249.2 – 141.1 + 798 – 2478.0
ΔHf = -317.3 kJ mol⁻¹
3. Construct a Born Haber Cycle for MgCl2 and use the information in the table to calculate the first electron affinity of chlorine.
Working Out:
ΔHf = ΔHat(Mg) + ΔHIE1(Mg) + ΔHIE2(Mg) + 2ΔHat(Cl) + 2EA + LE
-641.3 = 147.7 + 738.0 + 1451 + 2(121.7) + 2EA + (-2526.0)
-641.3 = 2336.7 + 243.4 – 2526.0 + 2EA
-641.3 = 54.1 + 2EA
2EA = -695.4
EA = -347.7 kJ mol⁻¹
4. Construct Born Haber cycles for CuO and Cu2O and calculate the enthalpy of formation of each compound. Explain why the results of these Born Haber cycles do not support the known stability of the two compounds and suggest why this discrepancy does not matter.
Working Out (Cu₂O):
ΔHf = 2ΔHat(Cu) + 2IE1 + ΔHat(O) + EA1 + EA2 + LE
ΔHf = 2(338.3) + 2(746.0) + 249.2 – 141.1 + 798 – 3244.7
ΔHf = 676.6 + 1492.0 + 249.2 – 141.1 + 798 – 3244.7
ΔHf = -170.0 kJ mol⁻¹
Working Out (CuO):
ΔHf = ΔHat(Cu) + IE1 + IE2 + ΔHat(O) + EA1 + EA2 + LE
ΔHf = 338.3 + 746.0 + 1958 + 249.2 – 141.1 + 798 – 4104.0
ΔHf = -155.6 kJ mol⁻¹
5. Construct Born Haber cycles for CaBr2 and KBr and use them to calculate the first electron affinity of bromine according to each cycle. Suggest a reason why they do not give the same value.
Working Out (CaBr₂):
ΔHf = ΔHat(Ca) + IE1 + IE2 + 2ΔHat(Br) + 2EA + LE
-682.8 = 178.2 + 590 + 1145 + 2(111.9) + 2EA – 2178.0
-682.8 = 1913.2 + 223.8 – 2178.0 + 2EA
-682.8 = -41.0 + 2EA
2EA = -641.8
EA = -320.9 kJ mol⁻¹
Working Out (KBr):
ΔHf = ΔHat(K) + IE1 + ΔHat(Br) + EA + LE
-393.8 = 89.2 + 419.0 + 111.9 + EA – 679.0
-393.8 = -58.9 + EA
EA = -334.9 kJ mol⁻¹
The formation data is determined from an experiment, so there is a small amount of error.