Enthalpies of Formation

1
Enthalpies of Formation

• An enthalpy of formation, ?Hf, is defined as the
  enthalpy change for the reaction in which a
  compound is made from its constituent elements in
  their elemental forms.
• Standard enthalpies of formation, ?Hf, are
  measured under standard conditions (25C and 1.00
  atm pressure).
• Elemental source of oxygen is O2 and not O
  because O2 is the stable form of oxygen at 25
  and 1 atm, likewise with H2
• Elemental source of carbon is specified as
  graphite (and not, for example, diamond) because
  graphite is the lowest energy form of carbon at
  room temp and 1 atm
• Why is the O2 stoichiometry left at "1/2"? The
  stoichiometry of formation reactions always
  indicates the formation of 1 mol of product.
  Thus, ?Hf values are reported as kJ / mole of the
  substance produced
• If C(graphite) is the lowest energy form of
  carbon under standard conditions, then what is
  the ?Hf for C(graphite)?
• By definition, the standard enthalpy of formation
  of the most stable form of any element is zero
  because there is no formation reaction needed
  when the element is already in its standard state
• ?Hf C(graphite), H2 (g) and O2 (g) 0

2
Enthalpies of Formation
Which reaction represents the DHf reaction for
NaNO3?
3
Calculation of ?H

• We can use Hesss law in this way
• ?H ??n??Hf(products) - ??m??Hf(reactants)
• where n and m are the stoichiometric
  coefficients.
• C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)
• ??????H 3(-393.5 kJ) 4(-285.8 kJ) -
  1(-103.85 kJ) 5(0 kJ)
• (-1180.5 kJ) (-1143.2 kJ) - (-103.85
  kJ) (0 kJ)
• (-2323.7 kJ) - (-103.85 kJ)
• -2219.9 kJ

4
Chemical Bonds

• Three basic types of bonds
• Ionic
• Electrostatic attraction between ions
• Covalent
• Sharing of electrons
• Metallic
• Metal atoms bonded to several other atoms

5
Ionic BondingEnergetics of Ionic Bonding

• As we saw in the last chapter, it takes 495
  kJ/mol to remove electrons from sodium.
• We get 349 kJ/mol back by giving electrons to
  chlorine.
• But these numbers dont explain why the reaction
  of sodium metal and chlorine gas to form sodium
  chloride is so exothermic!

6
Energetics of Ionic Bonding

• There must be a third piece to the puzzle.
• What is as yet unaccounted for is the
  electrostatic attraction between the newly formed
  sodium cation and chloride anion.
• Lattice EnergyThis third piece of the puzzle is
  the lattice energyThe energy required to
  completely separate a mole of a solid ionic
  compound into its gaseous ions.The energy
  associated with electrostatic interactions is
  governed by Coulombs law

7
Lattice Energy

• Lattice energy, then, increases with the charge
  on the ions.
• It also increases with decreasing size of ions.

Which one of the following has the largest
lattice energy? LiF, NaF, CaF2, AlF3 Which one
of the following has the largest lattice
energy? LiCl, NaCl, CaCl2, Al2O3
8
Energetics of Ionic Bonding

• By accounting for all three energies (ionization
  energy, electron affinity, and lattice energy),
  we can get a good idea of the energetics involved
  in such a process.

Na(s) 1/2 Cl2(g) --- gt NaCl(s) Na(g) --- gt
Na(s) Na(g) e --- gt Na(g)
Cl(g) --- gt 1/2Cl2 (g) Cl-(g) --- gt Cl(g) 2
e Add all the above equations leading
to Na(g) Cl-(g) --- gt NaCl(s)
9
Energetics of Ionic Bonding

• The lattice energy of calcium chloride
• The ionization energy of potassium
• The second ionization energy of strontium
• The electron affinity of bromine atom
• The sublimation of lithium
• The lattice energy of silver iodide

10
Energetics of Ionic Bonding

• What is the lattice energy of LiBr? The
  ionization energy (IE) of lithium is 520 kJ/mol,
  the vaporization energy (VE) of lithium is 134.7
  kJ/mol, the VE for Br2(l) is 15.46 kJ/mol, the
  energy of atomization (AE) of gaseous bromine
  (Br22 Br) is 111.7 kJ/mol, the electron affinity
  (EA) of bromine is 324 kJ/mol, the formation
  energy (FE) (Li(s) 1/2 Br2(l) LiBr(s)) is
  351.2 kJ/mol.
• IE Li (g) Li(g) e
• VE Li(s) Li(g)
• VE Br2(l) 2 Br2(g)
• AE Br2(g) 2 Br(g)
• EA Br(g) e Br(g)
• FE Li(s) 1/2 Br2(l) LiBr(s)

Li(g) Br(g) --- gt LiBr(s) IE Li(g) e
--gt Li (g) E 520 kJ EA Br(g) --- gt
Br(g) e E 324 kJ FE Li(s) 1/2 Br2(l)
--- gt LiBr(s) E 351.2 kJ VE Li(g) --- gt
Li(s) E 134.7 kJ VE 1/2 Br2(g) ---
gt1/2Br2(l) E 7.73 kJ
11
Covalent Bonding

• In these bonds atoms share electrons.
• There are several electrostatic interactions in
  these bonds
• Attractions between electrons and nuclei
• Repulsions between electrons
• Repulsions between nuclei

12
Electronegativity

• The ability of atoms in a molecule to attract
  electrons to itself.
• On the periodic chart, electronegativity
  increases as you go
• from left to right across a row.
• from the bottom to the top of a column.

13
Polar Covalent Bonds

• Although atoms often form compounds by sharing
  electrons, the electrons are not always shared
  equally.
• Fluorine pulls harder on the electrons it shares
  with hydrogen than hydrogen does.
• Therefore, the fluorine end of the molecule has
  more electron density than the hydrogen end.

14
Polar Covalent Bonds

• When two atoms share electrons unequally, a bond
  dipole results.
• The dipole moment, ?, produced by two equal but
  opposite charges separated by a distance, r, is
  calculated
• Qr
• It is measured in debyes (D).
• The greater the difference in electronegativity,
  the more polar is the bond.

15
Lewis Structures

• Lewis structures represent molecules showing all
  electrons, bonding and nonbonding.
• Allows for two dimensional representation of
  molecular bonding.
• Shows the order of connectivity
• Shows the types of bonds
• Shows all electrons bonding and non-bonding

16
Anatomy of Writing Lewis Structures

• PCl3

• Find the sum of valence electrons of all atoms in
  the polyatomic ion or molecule.
• If it is an anion, add one electron for each
  negative charge.
• If it is a cation, subtract one electron for each
  positive charge.
• The central atom is the least electronegative
  element that isnt hydrogen. Connect the outer
  atoms to it by single bonds.
• Keep track of the electrons26 ? 6 20
• Fill the octets of the outer atoms.
• Keep track of the electrons26 ? 6 20 ? 18 2
• Fill the octet of the central atom.
• Keep track of the electrons26 ? 6 20 ? 18 2
  ? 2 0

5 3(7) 26
17
Writing Lewis Structures

• If you run out of electrons before the central
  atom has an octet
• form multiple bonds until it does.

18
Writing Lewis Structures

• Then assign formal charges.
• For each atom, count the electrons in lone pairs
  and half the electrons it shares with other
  atoms.
• Subtract that from the number of valence
  electrons for that atom The difference is its
  formal charge.

• The best Lewis structure
• is the one with the fewest charges.
• puts a negative charge on the most
  electronegative atom.

19
Resonance

• This is the Lewis structure we would draw for
  ozone, O3.

• But this is at odds with the true, observed
  structure of ozone, in which
• both OO bonds are the same length.
• both outer oxygens have a charge of ?1/2.

-
20
Resonance

• One Lewis structure cannot accurately depict a
  molecule such as ozone.
• We use multiple structures, resonance structures,
  to describe the molecule.

• Just as green is a synthesis of blue and
  yellowozone is a synthesis of these two
  resonance structures.

21
Resonance

• In truth, the electrons that form the second CO
  bond in the double bonds below do not always sit
  between that C and that O, but rather can move
  among the two oxygens and the carbon.
• They are not localized, but rather are
  delocalized.

• The organic compound benzene, C6H6, has two
  resonance structures.
• It is commonly depicted as a hexagon with a
  circle inside to signify the delocalized
  electrons in the ring.

22
Exceptions to the Octet Rule

• There are three types of ions or molecules that
  do not follow the octet rule
• Ions or molecules with an odd number of
  electrons.
• Ions or molecules with less than an octet.
• Ions or molecules with more than eight valence
  electrons (an expanded octet).

Odd Number of Electrons
Though relatively rare and usually quite
unstable and reactive, there are ions and
molecules with an odd number of electrons.
23
Fewer Than Eight Electrons

• Consider BF3
• Giving boron a filled octet places a negative
  charge on the boron and a positive charge on
  fluorine.
• This would not be an accurate picture of the
  distribution of electrons in BF3.

Therefore, structures that put a double bond
between boron and fluorine are much less
important than the one that leaves boron with
only 6 valence electrons.
The lesson is If filling the octet of the
central atom results in a negative charge on the
central atom and a positive charge on the more
electronegative outer atom, dont fill the octet
of the central atom.
24
More Than Eight Electrons

• The only way PCl5 can exist is if phosphorus has
  10 electrons around it.
• It is allowed to expand the octet of atoms on the
  3rd row or below.
• Presumably d orbitals in these atoms participate
  in bonding.

25
More Than Eight Electrons

• Even though we can draw a Lewis structure for the
  phosphate ion that has only 8 electrons around
  the central phosphorus, the better structure puts
  a double bond between the phosphorus and one of
  the oxygens.

• This eliminates the charge on the phosphorus and
  the charge on one of the oxygens.
• The lesson is When the central atom is on the
  3rd row or below and expanding its octet
  eliminates some formal charges, do so.

26
Covalent Bond Strength

• Most simply, the strength of a bond is measured
  by determining how much energy is required to
  break the bond.
• This is the bond enthalpy.
• The bond enthalpy for a ClCl bond,
• D(ClCl), is measured to be 242 kJ/mol.

27
Average Bond Enthalpies
28
Average Bond Enthalpies

• This table lists the average bond enthalpies for
  many different types of bonds.
• Average bond enthalpies are positive, because
  bond breaking is an endothermic process.

• NOTE These are average bond enthalpies, not
  absolute bond enthalpies the CH bonds in
  methane, CH4, will be a bit different than theCH
  bond in chloroform, CHCl3.

29
Enthalpies of Reaction

• Yet another way to estimate ?H for a reaction is
  to compare the bond enthalpies of bonds broken to
  the bond enthalpies of the new bonds formed.

?Hrxn ?(bond enthalpies of bonds broken) ?
?(bond enthalpies of bonds formed)
30
Enthalpies of Reaction

• CH4(g) Cl2(g) ???
• CH3Cl(g) HCl(g)
• In this example, one
• CH bond and one
• ClCl bond are broken one CCl and one HCl bond
  are formed.

So, ?Hrxn D(CH) D(ClCl) ? D(CCl)
D(HCl) (413 kJ) (242 kJ) ? (328 kJ)
(431 kJ) (655 kJ) ? (759 kJ) ?104 kJ
31
Bond Enthalpy and Bond Length

• We can also measure an average bond length for
  different bond types.
• As the number of bonds between two atoms
  increases, the bond length decreases.