13 Energetics II. Eg. Na (g) Na + (g) + e - ΔH = +550 kj mol -1

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1 13 Energetics II First ionisation energy I(1) or IE (1): the energy required to remove one mole of electrons from one mole of the gaseous atoms of an element to 1 mole of gaseous monopositive ions. Eg. Na (g) Na + (g) + e - ΔH = +550 kj mol -1 Electrón affinity EA(1): the energy change when one mole of isolated gaseous atoms gain one mole of electrons to form one mole of negatively charged ions. Eg. F(g) + e - F - (g) ΔH = kj mol -1

2 Standard enthalpy of formation: the enthalpy change when one mole of a compound is formed from its elements in their standard states (1 atm., 25 o C) Eg. C (s) + O 2 (g) CO 2(g) ΔH f 0 = kj mol -1 Standard enthalpy of combustion: the enthalpy change when one mole of a substance undergoes complete combustion. Eg. C 2 H 6 (g) + 3 1/2 O 2 (g) 2CO 2 (g) + 3H 2 O (l) ΔH c 0 = kj mol -1 Standard enthalpy of atomisation: the enthalpy change when one mole of free gaseous atoms of an element are produced from the element in its standard state. Eg. Na (s) Na (g) ΔH a 0 = kj mol -1 Hess s law: the total enthalpy change for a reaction is independent of the route taken (N.B. the 1 st law of thermodynamics).

3 Lattice Enthalpy ΔΗ latt The lattice enthalpy or energy, ΔΗ latt is defined as the exothermic energy change for the process: M + (g) + X - (g) M + X - (s) Lattice enthalpies cannot be determined directly by experiment and so we need to calculate them theoretically using a Born-Haber Cycle. The figure obtained, curiously, is called the experimental value of ΔΗ latt. Lattice enthalpies are generally large which favours the formation of an ionic solid from its constituent elements and they are a measure of the strength of the lattice i.e. the ionic bond strength.

4 A Born-Haber Cycle is an energy diagram based on Hess s Law that can be used to calculate enthalpy changes that are difficult to measure. The classic example is that of the lattice energy of Sodium Chloride The equation for the lattice enthalpy would be: ΔΗ latt =

5 This is the same information as a Hess diagram

6 BH Cycles for MgO and NaF

7 Example: Given the following 1. Standard molar enthalpy of formation of BaCl 2 (s) is kj/mol. Ba(s) + Cl 2 (g) BaCl2(s) ΔH = kj 2. Standard enthalpy of atomisation of Ba(g) is kj/mol. Ba(s) Ba(g) ΔH =175.6 kj 3. The first ionization energy of Ba is 503 kj/mol. Ba(g) Ba + (g) + 1 e- ΔH = 503 kj 4. The second ionization energy of Ba is 965 kj/mol. Ba + (g) Ba 2+ (g) + 1 e- ΔH = 965 kj 5. The electron affinity of Cl is -349 kj/mol. Cl(g) + 1 e- Cl-(g) ΔH = -349 kj 6. The standard enthalpy of atomisation of chlorine is 240 kj/mol. Cl 2 (g) 2 Cl(g) ΔH = 240 kj Determine the lattice energy of BaCl 2. That is, ΔH for the reaction: Ba 2+ (g) + 2 Cl - (g) BaCl 2 (s) Textbook pp52-55

8 Factors affecting ΔΗ latt The size of the ions and the structure of the compound the smaller the distance between the anions and cations, the larger the ΔΗ latt Charge as the charge on the ions increase, ΔΗ latt increases. If the bonding is fully ionic or not (see later). Explain any observed trends

9 Lattice Energies of Salts of the OH - and O 2- Ions (kj/mol) OH - O 2- Na Mg Al ,916 Explain the observed trends

10 Lattice Energies of Salts of F - (kj mol -1 ) Na + F Mg Al Explain the observed trend

11 Points to note and discuss from the tables: For Group I halides, the magnitude of ΔΗ latt decreases down the table. 2. The magnitude of ΔΗ latt is greater for group II fluorides than for group I. 3. The magnitude of ΔΗ latt is greater for group II oxides than for group II hydroxides. N. B. The distance between the ions depends also on the crystal structure - we do not consider this question here when identifying the trends observed in ΔΗ latt. Chemical bonding (ionic, covalent or metallic) occurs due to the fact that the bonded atoms are more stable (lower energy) that the separate atoms

12 Important implications of lattice energies Substances with high lattice enthalpies have high melting points. The solubility of an ionic compound will depend upon the relative lattice and hydration enthalpies. If we compare the theoretical lattice enthalpy (calculated using Coulombs law in 3 dimensions) with the experimental (B-H cycle) value, we achieve an indication of the degree of covalent bonding in the substance. The more covalent the bonding, the stronger the crystal structure and the magnitude of the experimental ΔΗ latt is greater. Therefore, the greater the difference between the experimental and theoretical lattice enthalpies, the greater the level of covalence. Consider the following tables:

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14 Varying the size of the cation: We observe the trend that we would expect with group II chlorides. The Mg 2+ has more polarizing power due to its high charge density (same charge, smaller volume) and so deviates more from pure ionic bonding. We note an unexpected lack of trend with group I chlorides. Really, the figures are very similar if we take into account experimental errors. This implies that these chlorides are almost completely ionic.

15 Varying the size of the anion: If the anion is larger and more polarizable, then the difference in the ΔΗ latt will be greater (more covalent character and a stronger structure). ΔΗ latt expt. (kj mol -1 ) ΔΗ latt theoretical (kj mol -1 ) NaF NaI MgF MgI Difference (kj mol -1 ) Textbook pp56-58

16 ΔΗ hyd is about polar water molecules being attracted to the ions and forming a stable situation (less energy). The size of ΔΗ hyd depends on how close the water molecules can get to the ion and the size of its charge. Small, highly charged ions have the largest ΔΗ hyd.

17 The enthalpy of hydration, ΔΗ hyd : this is the heat change per mole for the hydration of the gaseous ion with enough water for there to be no further heat change on dilution H 2 O Eg. M + (g) M + (aq) If ΔΗ hyd is about the same, or is bigger than ΔΗ latt, the energy required to break down the lattice is recuperated by the energy released by dissolving the substance and the dissolution of the solid is favoured ΔΗ soln is the sum of ΔΗ hyd and ΔΗ latt ΔΗ soln = ΔΗ hyd - ΔΗ latt So, a low solubility could be caused by a large (exothermic!) ΔΗ latt or by a small ΔΗ hyd. In either case, ΔΗ soln would be positive. Draw a Hess diagram to represent the energy changes when a salt MX (s) is completely dissolved in water. Write the general equation for ΔΗ soln. Textbook pp59-61