Q1(b). Born–Haber cycle: can NaCl₂ form?
The enthalpy of formation of an ionic compound can be calculated with accuracy by Born-Haber cycle. Predict, giving valid reasons, the possibility of formation of NaCl₂ salt. The following thermodynamic data are given for NaCl and NaCl₂: U₀ = −757 kJ mol⁻¹ (NaCl), U₀ = −2155 kJ mol⁻¹ (NaCl₂); ΔH(IE) = +495 kJ mol⁻¹; ΔH(IE₂) = +4561 kJ mol⁻¹; ΔH(EA) = −348 kJ mol⁻¹; ½ΔH(diss) = +121 kJ mol⁻¹; ΔH(sub) = +108 kJ mol⁻¹.
Answer in brief
NaCl₂ cannot form: its ΔHf is +2555 kJ mol⁻¹, against −381 kJ mol⁻¹ for NaCl; the second ionisation of sodium costs more than the lattice energy can repay.
Key points
- ΔHf(NaCl) = 108 + 121 + 495 − 348 − 757 = −381 kJ mol⁻¹.
- ΔHf(NaCl₂) = 108 + 242 + 495 + 4561 − 696 − 2155 = +2555 kJ mol⁻¹.
- The second electron must come out of the closed neon core, so the second ionisation costs 4561 kJ, 9.2 times the first.
- The lattice energy would have to reach −4710 kJ mol⁻¹ just to make ΔHf zero.
- Entropy cannot rescue NaCl₂: ΔGf has the sign of ΔHf.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversBorn–Haber cyclelattice energyionisation energyHess's lawNaCl2 stability
All parts of Question 1
- 1(a) Expectation value of position in the ground state of a box10 marks
- 1(b) Born–Haber cycle: can NaCl₂ form?10 marks
- 1(c) Rubber band: change of length on heating10 marks
- 1(d) Vapour pressure of water at 100 °C from its value at 95 °C10 marks
- 1(e) Electrochemical series and its significance10 marks