Presentation on theme: "Title: Lesson 6 Born-Haber Cycles and Lattice Enthalpies Learning Objectives: – Understand the term lattice enthalpy – Use Born-Haber cycles to calculate."— Presentation transcript:
Title: Lesson 6 Born-Haber Cycles and Lattice Enthalpies Learning Objectives: – Understand the term lattice enthalpy – Use Born-Haber cycles to calculate lattice enthalpy – Identify and explain trends in lattice enthalpy
Main Menu Refresh The standard enthalpy change of three combustion reactions is given below in kJ. 2C 2 H 6 (g) + 7O 2 (g) → 4CO 2 (g) + 6H 2 O(l)∆H o = –3120 2H 2 (g) + O 2 (g) → 2H 2 O(l)∆H o = –572 C 2 H 4 (g) + 3O 2 (g) → 2CO 2 (g) + 2H 2 O(l) Δ H o = –1411 Based on the above information, calculate the standard change in enthalpy, ∆H o, for the following reaction. C 2 H 6 (g) → C 2 H 4 (g) + H 2 (g)
Main Menu Bond Enthalpies Look at the covalent bond enthalpies on Table 10 in the Data Booklet. Why do you think there are no ionic bond enthalpies?
Main Menu Recap of First Ionisation Energies and Electron Affinities We know that metals loses electrons and non-metals gain electrons. We can use Ionisation Energies and Electron Affinities to work out the enthalpy changes within an ionic compound. The first ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms: Sodium (on the left of the periodic table) has a relatively low ionisation energy. The first electron affinity is the enthalpy change when one mole of gaseous atoms attracts one mole of electrons. Values can be found in section 7 of the IB data booklet. As the electron is attracted to the positively charged nucleus of the Cl atom, the process is EXOTHERMIC
Main Menu Lattice Enthalpies Add the equations for the first ionisation energy and first electron affinity. The process is ENDOTHERMIC overall. This is energetically UNFAVOURABLE (despite the fact it leads to the formation of ions with stable noble gas configurations). Oppositely charged ions come together to form an ionic lattice. The strong attraction between the oppositely charged ions means its very EXOTHERMIC. Lattice Enthalpy H θ lat expresses this enthalpy change in terms of the reverse ENDOTHERMIC process. The lattice enthalpy relates to the ‘formation of gaseous ions from one mole of a solid crystal breaking into gaseous ions’. (As seen above)
Main Menu The lattice enthalpy relates to the enthalpy change of ‘formation of gaseous ions from one mole of a solid crystal breaking into gaseous ions’. Think of lattice enthalpy as ‘lattice disassociation enthalpy’...
Main Menu Lattice Enthalpy, H lat This is the equivalent of ‘bond strength’ for ionic compounds It is the enthalpy change when one mole of an ionic compound is converted to gaseous ions. This is an endothermic process, requiring energy to be put in. MX(s) M + (g) + X - (g) Compound Lattice Enthalpy kJ mol-1 LiF1049 LiBr820 KF829 CaF Note: Most places define lattice enthalpy the opposite way round, i.e: M + (g) + X - (g) MX(s) The values would be the same magnitude, but with a negative sign to show they are exothermic. It is just a strange quirk of the IB that they do it this way round….I think so that it fits with average bond enthalpies, which are also represent bond breaking
H θ EA ClH θ 1st IE Na H θ at Cl H θ f NaCl Born-Haber Cycle eg for sodium chloride: NaCl (s) Cl - (g) H θ LAT NaCl Na (s) + ½ Cl 2 (g) Na (g) Cl (g) + Na + (g) + e - + NB: H θ f = formation H θ at = atomisation H θ IE = ionisation H θ EA = electron affinity H θ LAT = lattice enthalpy Using the ‘FAIL’ technique F = formation A = atomisation I = ionisation L = lattice enthalpy FORMATION IONISATION Construction of a H θ at Na ATOMISATION LATTICE ENTHALPY
Born-Haber Cycle There are two routes fromelements to ionic compound H at Na + H at Cl + H 1st IE Na + H EA Cl - H LAT NaCl = H f NaCl Apply Hess’s Law: : Applying Hess’s Law Clockwise Anti-clockwise = The Indirect route and the Direct route Clockwise arrows must equal the Anti-clockwise arrows H θ EA Cl H θ 1st IE Na H θ at Cl H θ f NaCl NaCl (s) Cl - (g) H θ LAT NaCl Na (s) + ½ Cl 2 (g) Na (g) Cl (g) + Na + (g) + e - + H θ at Na
Born-Haber Cycle: Rearrange to find the lattice energy: So Born-Haber cycles can be used to calculate a measure of ionic bond strength based on experimental data. H at Na + H at Cl + H 1st IE Na + H EA Cl - H LAT NaCl = H f NaCl If you want to calculate H LAT NaCl, it is as follows:- H LAT NaCl = [H at Na + H at Cl + H 1st IE Na + H EA Cl] - H f NaCl Calculation Values: H at Na = 107 H at Cl = 121 H 1st IE Na = 496 (kJmol -1 ) H EA Cl = -349 H f NaCl = -411 H LAT NaCl = [(107) + (121) + (496) + (-349)] - (-411) H LAT NaCl = 786 kJmol -1 It’s important to keep all numbers in ( ), whether +ve or –ve, when entering information into your calculator.
2xH θ EA Cl H θ 1st +2nd IE Mg 2xH θ at Cl H θ f MgCl 2 Born-Haber Cycle eg for magnesium chloride: MgCl 2 (s) 2Cl - (g) H θ LAT MgCl 2 Mg (s) + Cl 2 (g) Mg (g) 2Cl (g) + Mg 2+ (g) + 2e - + NB: H θ f = formation H θ at = atomisation H θ IE = ionisation H θ EA = electron affinity H θ LAT = lattice enthalpy Using the ‘FAIL’ technique F = formation A = atomisation I = ionisation L = lattice enthalpy FORMATION IONISATION Construction of a H θ at Mg ATOMISATION LATTICE ENTHALPY
Main Menu Born-Haber Cycles Lattice enthalpies are difficult to measure directly, so we use Born-Haber cycles These are just a specialised type of Hess cycle To calculate H lat : Start at the bottom and work round clockwise, adding and subtracting according to the arrows. solid ionic compound elements in their standard states metal and atomised non-metal atomised metal and atomised non-metal ionised metal and atomised non-metal ionised metal and ionised non-metal Enthalpy of atomisation (metal) Ionisation energy/s (Metal) Enthalpy of atomisation (non-metal) Enthalpy of formation H lat = ? Electron affinity/s (Non-metal)
Main Menu Building Born-Haber Cycles Work through the activity herehere Cut out the equations first and arrange them sensibly on your desk. We will work through LiF together. Hint: Don’t forget to include both ionisations for Mg/Ca and both electron affinities for O