Going over basics:
Going over basics:
Enthalpy symbol H, is stored energy, its not possible to measure enthalpy but enthalpy changes ((Η), by measuring temperature changes of reactions at constant pressure. ( Pronounced ‘delta’ means ‘change of’
Enthalpy change, (Η is the measure of the transfer of energy into or out of a reacting system at constant pressure
Exothermic reaction – Energy given out by reactants as they form products
Energy exits in exothermic reactions (Η is negative, enthalpy(stored energy) decreases, temperature increases
|Bond making, amount of heat released is greater than the amount of heat used to start the reaction |
| |
Endothermic reaction – Energy is taken in by reactants to form products
Energy enters in endothermic reactions (Η is positive, enthalpy(stored energy) increases, temperature decreases
|Bond breaking, energy continues to be absorbed as long as the reaction continues |
| |
Definitions to remember:
|Lattice Enthalpy ΔHlatt Energy released per mole for exothermic process M+(g) + X–(g) ( M+X-(s) |
|(Enthalpy/heat energy released when gaseous ions come together to form 1 mole of solid) |
|Standard enthalpy of atomisation ΔHa θ Enthalpy change for production of one mole of gaseous atoms from the element in its standard state |
|Enthalpy of hydration ΔHfhyd θ Enthalpy change per mole for dissolving the gaseous ions, with enough water to form an infinite dilute solution. |
|(When water is used as the solvent, the dissolving process is called hydration) |
|(For a unipositive cation ΔHhyd is exothermic: M+(g) + aq (M+(aq)) |
Understand that:
|Enthalpy of formation ΔHf Enthalpy change when one mole of a compound is formed from its elements |
|Na(s) + (1/2)Cl2(g) → NaCl(s) |
|1st Ionisation Energy M(g) ( M+(g) + e– 2nd Ionisation Energy M+(g) ( M2+(g) + e– |
|1st Electron affinity Enthalpy change per mole for the process, X(g) + e– ( X–(g) |
|- Negative(exothermic), since the electron is attracted by the positive charge on the atoms nucleus |
|2nd Electron affinity Enthalpy change per mole for the process, X–(g) + e– ( X2–(g) O–(g) + e →O2−(g) |
|- Positive(endothermic), since energy needed to overcome repulsion between the electron and negative ions |
Questions
|Write the equation, with state symbols, for the enthalpy of atomisation of chlorine |
|½Cl2(g) ( Cl(g) |
|Write an equation which represents the change when the second electron affinity of oxygen is measured |
|O–(g) + e →O2−(g) |
|[pic] |Explain the trend in IE within the group 1 elements |
| | |
| |electron further away from nucleus |
| |inceased shielding effect inner shells |
| |less energy needed to remove outer electron |
Construct a Born-Haber cycle and carry out associated calculations
|[pic] |A Born-Haber cycle calculates the lattice enthalpy by comparing |
| |the ΔHf of the ionic compound (from the elements) to the enthalpy|
| |required to make gaseous ions from the elements |
| | |
| |The Born-Haber cycle involves the formation of an ionic compound |
| |from the reaction of a metal(often a Group1/2 element) with a |
| |non-metal |
| | |
| |All endothermic reactions shown by arrow pointing upwards (vice |
| |versa) |
| |Atomisation enthalpy of metal (in this case lithium) |
| |Ionisation enthalpy of metal |
| |Atomisation enthalpy of non-metal (in this case fluorine) |
| |Electron affinity of non-metal |
| |Lattice enthalpy |
Questions
|Presenting an ionic solid, NaH |Construct a Born-Haber cycle and to obtain ΔHlatt of SrCl2(s) |
|Draw a Born-Haber cycle which could be used to determine the electron affinity|ΔHf of SrCl2(s) |
|of hydrogen |-829 kJmol–1 |
|[pic] | |
| |ΔHa of strontium |
| |+164 kJmol–1 |
| | |
| |ΔHa of chlorine |
| |+122 kJmol–1 |
| | |
| |1st IE of strontium |
| |+550 kJmol–1 |
| | |
| |2nd IE of strontium |
| |104 kJmol–1 |
| | |
| |Ε“ of chlorine |
|Construct a Born-Haber cycle |+736 |
|And find 2nd electron affinity of oxygen | |
| |2nd ionisation energy of magnesium |
|ΔH / kJ mol–1 |+1450 |
| | |
|Enthalpy of atomisation of magnesium |Ist electron affinity of oxygen |
|+150 |–142 |
| | |
|Bond energy of O == O in oxygen |Lattice enthalpy of magnesium oxide |
|+496 |–3889 |
| | |
|1st ionisation energy of magnesium |Enthalpy of formation of magnesium oxide |
| |–602 |
| | |
|Use the data below to calculate the first electron affinity of chlorine. |-642 = 150 +736 +1450 +2(121) +2x +(-2493) |
|Enthalpy change |2x = 727 |
|Enthalpy change |x = –363 ± 1 |
| | |
|ΔHat of magnesium | |
|1st IE of magnesium | |
|2nd IE of magnesium | |
|ΔHf of MgCl2 | |
|ΔHat of chlorine | |
|ΔHlatt of MgCl2 | |
|+150 kJ mol–1 | |
|+736 kJ mol–1 | |
|+1450 kJ mol–1 | |
|–642 kJ mol–1 | |
|+121 kJ mol–1 | |
|–2493 kJ mol–1 | |
| | |
Understand the factors that influence the value of the lattice energies
Factors affecting theoretical value of lattice enthalpy/energy •Radius/size of ions •Charges on ions
( Strong attraction - Small ionic radius and high charge
( Weak attraction – Large ionic radius and small charge
Lattice energy is the measure of the strength of bonds in that ionic compound. It is the equivalent to the amount of energy required to separate a solid ionic compound into gaseous ions (always negative, exothermic)
- When lattice energy increases it becomes more negative
- Lattice energies increase when ions are smaller with high charge
- (( (( Strong attraction between ions because their ionic radii are small
(( (( Less attraction between ions because their ionic radii are larger
Understand that values of lattice energies calculated from the Born-Haber cycle may differ from those calculated from a purely ionic model - limited to the radius and charge of the ions
- Experimental lattice energies are from Born-Haber cycles
- Theoretical lattice energies are from equations, assumes ionic lattice is totally ionic, when actually it has covalent character(electron sharing)
( If the metal cation is small and/or highly charged, it will distort the electron cloud of the anion , more polarising
( If the non metal anion is larger it is more polarisable
( This polarisation of the negative ion leads to partial covalency
Questions:
|Theoretically ΔHlatt MgCl2 is –2326 kJ mol–1 | Theoretical ΔHlatt MgI2, is –1944 kJ mol–1 |
|Experimental ΔHlatt MgCl2 is –2526 kJ mol–1 |Experimental ΔHlatt MgI2, is –2327 kJ mol–1 |
|Explain why this difference occurs |Explain why this difference occurs |
|• MgCl2 has (a degree of ) covalent character |• magnesium ion is small and highly charged |
|• due to polarisation of the anion |• leading to polarisation of the (large) iodide ion |
| |• and (causing) covalency (into the lattice) |
|The theoretical and actual values of the lattice enthalpy of magnesium fluoride are very similar because magnesium fluoride is almost completely ionic. |
|Explain why magnesium fluoride is almost completely ionic |
|• F− ion is small• Mg2+ ion does not have a high enough charge density to polarise F− |
|Magnesium iodide compound. Radius of magnesium ion is 0.072 nm, iodide ion is 0.215 nm. |
|(i) Describe the effect that the magnesium ion has on an iodide ion next to it in the magnesium iodide lattice - The electrons around the iodide ion are drawn|
|towards the magnesium ion |
|(ii) What TWO quantities must be known about the ions in a compound in order to calculate a theoretical lattice energy? •Radius/size of ions •Charges on ions|
|(iii) Suggest how the value of the theoretical lattice energy would compare with the experimental value from a Born-Haber Cycle for magnesium iodide. - Less |
|(exothermic) - covalent character (strengthens lattice) |
|Why is the lattice energy of magnesium hydroxide more exothermic than that of barium hydroxide? |
|• as magnesium has a much smaller ion (than barium ion) |
|• and has same charge |
|• so stronger attraction between ions “charge density” scores 1 (out of first 2 marks) |
|Explain why the lattice enthalpy of magnesium fluoride, MgF2, is more exothermic than that of calcium chloride. |
|• smaller size of cation • smaller size of anion • greater attraction between (oppositely charged) ions |
|Lattice energies |Lattice energies |
|NaCl(s) is –771 kJ mol–1 |MgCl2 –2526 kJ mol–1 |
|MgO(s) is –3889 kJ mol–1 |CaCl2 –2237 kJ mol–1 |
|Explain the difference in lattice energies |SrCl2 –2112 kJ mol–1 |
| |BaCl2 –2018 kJ mol–1 |
|• Lattice enthalpy depends on charges and the ionic radii |Explain why lattice energies become less exothermic |
|• Comparison of Na+/Mg2+ size and charge | |
|• Comparison of Cl– / O2– size and charge |• As group descended, radius of M 2+ (ion) increases |
|• (High LE results from) higher interaction |• Charge on ions remains the same |
| |• (down group) weaker forces of attraction between ions |
|[pic] |Find enthalpy of solution of NaCl: |
| |Lattice enthalpy of sodium chloride = -771 |
| |hydration enthalpy of Na+ = -406 |
| |hydration enthalpy of Cl- = -364 |
| | |
| |Hsol = Hhyd (Na+) + Hhyd (Cl-) – Hlatt(NaCl) |
| |= (-406) + (-364) – (-711) = +1kJmol-1 |
Understand how ΔHlatt and ΔHhyd vary the solubilities of the hydroxides and sulphates of Group 2
Enthalpy change of solution - Enthalpy change when one mole of a substance is dissolved completely in a large volume of a solvent at constant pressure. Remember: ΔHsoln = − ΔHlatt (Μ+X–) + ΔHhyd(Μ+) + ΔHhyd(X–)
(Always small because they almost cancel out)
To dissolve, ΔHhyd ≥ ΔHlattice so enough hydration energy needed to overcome breaking the lattice
When an ionic substance dissolves enthalpy change depends on • ΔHlatt of the solid • ΔHhyd of the ions
Energy has to be supplied to break up the lattice of ions
Energy is released when these ions form bonds with water molecules
Trends in solubility depend on how fast both enthalpy terms fall relative to each other.
- As you go down a Group:
• Energy needed to break up the lattice falls because, bigger ions, larger distance between ions, less attraction between + and - ions
• Hydration enthalpies falls - bigger ions, less charge density, reduces the attraction of water, the less exothermic the hydration enthalpy.
- Hydroxides become more soluble
The lattice enthalpy falls faster than the hydration enthalpy, ΔHsoln becomes more exothermic(-) (more soluble)
- Sulphates become less soluble
The hydration enthalpy falls faster the ΔHsoln becomes more endothermic(+) (less soluble)
Because sulphate ion is bigger, change in ionic radius of Group 2 cations doesn’t have as much affect on ΔHlatt
The greater the charge density the easier it is for the Group 2 cation to hydrate and hence dissolve in water due to greater attraction with the polar water molecules.
|Salt |Relative solubility |Explain the reasons for this trend in solubility in terms of changes of lattice energies and enthalpies of |
| | |hydration. |
| | |• salt likely to be more soluble if ΔHsol exothermic |
| | |• both lattice energy and hydration enthalpies become less exothermic |
| | |• as cations increase in size |
| | |• but lattice energy changes less so enthalpy of solution less exothermic |
|MgSO4 |1 | |
|CaSO4 |10–2 | |
|SrSO4 |10–4 | |
|BaSO4 |l0–6 | |
| |
|ΔH[pic]/kJ mol–1 |
| |
|ΔHhydration of Sr2+ |
|–1480 |
| |
|ΔHhydration of Ba2+ |
|–1360 |
| |
|ΔHhydration of OH– |
|–460 |
| |
|Lattice enthalpy of Sr(OH)2 |
|–1894 |
| |
|Lattice enthalpy of Ba(OH)2 |
|–1768 |
| |
|(i) Explain why the lattice enthalpy of strontium hydroxide is different from that of barium hydroxide. |
|charge density of Sr2+ < Ba2+ ∴ stronger force of attraction between ions |
|(ii) Explain why the hydration enthalpy of a cation is exothermic. |
|• Negative part of water attracted to (+ ion) and forms bond • bond formation releases energy |
|(iii) Use the lattice enthalpy and hydration enthalpy values to explain why barium hydroxide is more soluble in water than strontium hydroxide. |
|• ΔHsol = ΔHhyd – ΔHlattice |
|• ΔHlatt and ΔHhydr decrease down Group 2 (Barium lower than Strontium) |
|• ΔHlatt decreases more than the ΔHhydr |
|• ΔHsoln Ba(OH)2 more exothermic (than for Sr(OH)2,(so more soluble)) |
|ΔHsoln = –ΔHlatt + ΔHhydrM2+ion + 2 × ΔHhydr OH– ion or = –ΔHlatt + ∑ΔHhydr ions |
|ΔHsoln Sr(OH)2 = – (– 1894) + (– 1480) + 2 ×(– 460) = – 506 kJ mol–1 |
|ΔHsoln Ba(OH)2 = – (– 1768) + (– 1360) + 2 ×(– 460) = – 512 kJ mol–1 |
|Enthalpy of hydration of Mg2+ –1890 kJ mol–1 |Calculate ΔHsoln of Mg(OH)2 |
|Enthalpy of hydration of Ba2+ –1275 kJ mol–1 |ΔHsoln = -1890 -550 -550 +2995 = 5kJ mol–1 |
|Enthalpy of hydration of OH– –550 kJ mol–1 | |
|Lattice energy of Mg(OH)2 –2995 kJ mol–1 |Use the data to explain how the solubility of Ba(OH)2 compares with Mg(OH)2|
|Lattice energy of Ba(OH)2 –2320 kJ mol–1 |• ΔHlatt down but ΔHhyd down by less |
| |• ∴ ΔHsolution is more exothermic |
|Draw a labelled Hess’s law cycle for Mg(OH)2(s) |• ∴ solubility is greater |
|[pic] | |
|Explain why magnesium oxide is insoluble in water. |
|• not enough energy generated by hydration to overcome breaking the lattice |
|• solubility due to balance between – ΔHlattice and ΔHhydration of the ions ΔHsol = ΔHhyd – ΔHlattice |
Further:
|• Formation of MgCl2 is energetically favoured because of its higher lattice enthalpy than MgCl, which is almost never formed |
|• High lattice enthalpy more than compensates for the additional energy that has to be supplied for the 2nd ionisation of magnesium |
|[pic] |
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