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By the end of this lesson you should be able to:
All Born-Haber calculations follow the same procedure:
The critical step is sign handling - a single sign error can flip your answer from -800 to -2000 or from negative to positive. Slow down and track every sign.
Use the following data to calculate the lattice enthalpy of sodium chloride.
| Quantity | Value / kJ mol^-1 |
|---|---|
| ΔH°_f(NaCl) | -411 |
| ΔH°_at(Na) | +108 |
| ΔH°_at(Cl) | +122 |
| IE_1(Na) | +496 |
| EA_1(Cl) | -349 |
Step 1: Write the Hess-law equation.
ΔH°_f(NaCl) = ΔH°_at(Na) + ΔH°_at(Cl) + IE_1(Na) + EA_1(Cl) + ΔH°_LE
Step 2: Rearrange to isolate ΔH°_LE.
ΔH°_LE = ΔH°_f(NaCl) - ΔH°_at(Na) - ΔH°_at(Cl) - IE_1(Na) - EA_1(Cl)
Step 3: Substitute values carefully.
ΔH°_LE = (-411) - (+108) - (+122) - (+496) - (-349)
Step 4: Simplify term by term.
ΔH°_LE = -411 - 108 - 122 - 496 + 349
ΔH°_LE = -788 kJ mol^-1
Step 5: Sanity check. The answer is negative (as expected) and its magnitude is around 700-800 kJ mol^-1, typical for a 1+/1- salt. This agrees with the accepted value of -787 kJ mol^-1.
Calculate the lattice enthalpy of magnesium chloride from the following data.
| Quantity | Value / kJ mol^-1 |
|---|---|
| ΔH°_f(MgCl2) | -641 |
| ΔH°_at(Mg) | +148 |
| ΔH°_at(Cl) | +122 |
| IE_1(Mg) | +738 |
| IE_2(Mg) | +1451 |
| EA_1(Cl) | -349 |
Hess-law equation:
ΔH°_f(MgCl2) = ΔH°_at(Mg) + 2 ΔH°_at(Cl) + IE_1(Mg) + IE_2(Mg) + 2 EA_1(Cl) + ΔH°_LE
(Note the x2 on both atomisation and electron affinity of Cl.)
Rearrange:
ΔH°_LE = ΔH°_f - ΔH°_at(Mg) - 2 ΔH°_at(Cl) - IE_1(Mg) - IE_2(Mg) - 2 EA_1(Cl)
Substitute:
ΔH°_LE = (-641) - (+148) - 2(+122) - (+738) - (+1451) - 2(-349)
ΔH°_LE = -641 - 148 - 244 - 738 - 1451 + 698
ΔH°_LE = -2524 kJ mol^-1
The accepted value is around -2526 kJ mol^-1. The magnitude is very large, as expected for a 2+/1- compound with small cation.
| Quantity | Value / kJ mol^-1 |
|---|---|
| ΔH°_f(MgO) | -602 |
| ΔH°_at(Mg) | +148 |
| ΔH°_at(O) | +249 |
| IE_1(Mg) | +738 |
| IE_2(Mg) | +1451 |
| EA_1(O) | -141 |
| EA_2(O) | +798 |
Hess-law equation:
ΔH°_f(MgO) = ΔH°_at(Mg) + ΔH°_at(O) + IE_1(Mg) + IE_2(Mg) + EA_1(O) + EA_2(O) + ΔH°_LE
Rearrange and substitute:
ΔH°_LE = -602 - 148 - 249 - 738 - 1451 - (-141) - (+798)
ΔH°_LE = -602 - 148 - 249 - 738 - 1451 + 141 - 798
ΔH°_LE = -3845 kJ mol^-1
The accepted experimental value is about -3791 kJ mol^-1, and slight variations arise from different data tables. The enormous magnitude reflects the doubly charged ions and short Mg-O distance.
Born-Haber cycles can be used to calculate any unknown term if all the others are known. Suppose the lattice enthalpy and other data for KBr are given and we need to find the first electron affinity of bromine.
| Quantity | Value / kJ mol^-1 |
|---|---|
| ΔH°_f(KBr) | -394 |
| ΔH°_at(K) | +90 |
| ΔH°_at(Br) | +112 |
| IE_1(K) | +419 |
| ΔH°_LE(KBr) | -670 |
Hess-law equation:
ΔH°_f = ΔH°_at(K) + ΔH°_at(Br) + IE_1(K) + EA_1(Br) + ΔH°_LE
Rearrange for EA_1(Br):
EA_1(Br) = ΔH°_f - ΔH°_at(K) - ΔH°_at(Br) - IE_1(K) - ΔH°_LE
Substitute:
EA_1(Br) = -394 - 90 - 112 - 419 - (-670)
EA_1(Br) = -394 - 90 - 112 - 419 + 670
EA_1(Br) = -345 kJ mol^-1
This matches the accepted first electron affinity of bromine.
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