chemical energetics

1. r , △Hθhyd 2. r , △Hθlatt 3. 4.
(SO42->>M2+, sum of radius depend more on SO42-) (△Hθlatt decreases less than △Hθhyd) △Hθhyd decreases more than △Hθlatt so, △Hθsol gets more endothermic down the group,
Mg2+(g) +2Cl-(g) → MgCl2(s) △Hθlatt = -2526 kJ mol the more exothermic △Hθlatt, the stronger the ionic bond in lattice.
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Standard enthalpy change of atomisation:△Hθat is the enthalpy change when 1 mole of gaseous atoms formed from its element under standard conditions. 1/2I2(s)→ I(g) △Hθat = +188 kJ mol
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The greater the polarizing power of the cation, the greater the polarization of anions, the more distorted the anion becomes, The distorted CO32- breaks up into CO2 and O2more readily. • As a result, the MgCO3 lattice is less strong than the BaCO3 lattice and decomposes at a lower temperature. Lattice enthalpy: MgCO3 ＜ CaCO3 ＜ SrCO3 ＜ BaCO3
• According to Hess’s law,
θ H H ( Mg ( s )) H H 2 Δ H (Cl (g)) 2 H H f at i1 i2 at 2 ea1 latt
• This can be rearranged,
2.Charge of the ions:
As the charge on the ion increases,△Hθlatt becomes more exothermic. Ions with silimar radius: the charge increases, the charge density increases, the attraction between ions increase, so△Hθlatt is more exothermic.
MgCl2(s) + aq →Mg2+(aq) +2Cl-(aq) △Hθsol = -55kJ mol
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Standard enthalpy change of hydration:△Hθhyd
is the enthalpy change when 1 mole of the gaseous ions dissolves in in sufficient water to form a very dilute solution.
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△Hθi1 = +1260kJ mol
Electron affinity △Hθea(first) is the enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms, to form 1 mole of gaseous 1- ions. Cl(g) + e-(g) → Cl-(g) △Hθea1 = - 348kJ mol
)
The small Mg2+ has a large charge density and so it has the greatest ability to polarize the carbonate ion.
The larger barium cation has a lower charge density and so it polarizes the carbonate ion to a lesser extent:
1 H H ( Li ( s )) H ( F ( g )) H H H f at at 2 2 i1 ea1 latt
• This can be rearranged,
1 Hlatt H H ( Li ( s )) H ( F ( g )) H H f at at 2 formula is:
lattice enthalpy = heat of formation – heat of atomisation – ionisation energy – electron affinity
Draw a Born-Haber cycle for magnesium chloride, naming each step.
calculating enthalpy changes in solution
△Hθlatt + △Hθsol = △Hθhyd
△Hθsol = △Hθhyd - △Hθlatt ( + or - ) ( - ) ( - )
The solubility (△Hθsol)of Group II sulfates (depend on△Hθhyd and △Hθlatt)
• To form O2-, you will use two electron affinities:
O( g ) e O ( g )

Hea 1

O ( g) e O2 (g)


Hea 1
Trends in the lattice enthalpy
1.Size of the ions:
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Ionisation energy △Hθi (first) is the energy needed to remove one electron from each atom in one mole of gaseous atoms of an element.
Cl(g) - e-(g) → Cl+(g)
As the size of the ion increases,△Hθlatt becomes less exothermic. (This applies to both cations and anions. ) Ions with the same charge: the radius increases, the charge density decreases, the attraction between ions decrease, so△Hθlatt is less exothermic.
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Standard enthalpy change of solution:△Hθsol
is the energy absorbed or released when 1 mole of an ionic solid dissolves in sufficient water to form a very dilute solution.
solubility decreases, less soluble.
1 Hlatt H H ( Mg ( s )) H H 2 H ( Cl ( g )) 2 H f at i1 i2 at 2 2 ea1
1st key point
Both 2Hat and 2H are required in the Born-Haber cycle for MgCl2.
This is caused by the charge density of the cation polarizing ion polarization the anion.
Polarization
Factors affecting polarization: the polarizing abilty of cation(r ,charge density , the polarizability of anion(r , charge density , )

ea1
2nd key point
• To get to Al3+,you will use three ionisation energies:
Al ( g) Al ( g) e Al ( g) e Al 3 ( g) e
Hi 1 Hi2 2 Hi3
Mg2+(g)+ aq →Mg2+(aq) Cl-(g)+ aq →Cl-(aq)
△Hθhyd = - 1660kJ mol △Hθhyd = - 364kJ mol
Draw a Born-Haber cycle for lithium fluoride, naming each step.
• According to Hess’s law,
The thermal stability of Group II carbonates and nitrates
The digrame suggest that the lattice structure of MgCO3 is weaker than BaCO3,MgCO3 should therefore have the lower decomposition temperature.