Lecture 19 Enthalpy

1
Lecture 19 - Enthalpy
2
There are 3 ways to find DH

• 1. Measure it in the lab
• 2. Use Hess law
• 3. Use enthalpies of formation

3
Enthalpies of Formation

• Standard Enthalpy of Formation
• DHfo

standard state
formation
4
Enthalpies of Formation

• DHfo is the hypothetical DH for formation of one
  mole of a compound in its standard state from the
  elements in their standard states

5
Standard States

• defined at 25oC, 1 atm pressure
• e.g. C graphite
• O2 O2(g)
• CO2 CO2(g)
• H2O H2O(l)
• Zn2(aq) 1 M Zn2(aq)

6
by definition,

• DHfo for an element in its standard state 0

7
for example,

• DHfo (CO2(g)) is DH for the reaction
• C(s) O2(g) CO2(g)

8
for example,

• DHfo (CO2(g)) is DH for the reaction
• C(s) O2(g) CO2(g)

graphite
gases at 1 atm pressure, 25oC
9

• DHfo (CO(g)) is DH for the reaction
• C(s) ½ O2(g) CO(g)
• DHfo (CH3OH(l)) is DH for the reaction
• C(s) ½ O2(g) 2 H2(g) CH3OH(l)
• DHfo (Al2O3(s)) is DH for the reaction
• 2 Al(s) 3/2 O2(g) Al2O3(s)

10
some examples of DHfo (kJ/mol)

• Ca(s) 0
• CaO(s) -635.1
• CaCO3(s) -1206.9
• N2(g) 0
• NH3(g) -45.9
• NO(g) 90.2

11
e.g. The Thermite Reaction

• Fe2O3(s) 2 Al(s) 2 Fe(s) Al2O3(s)
• 2 Al(s) 3/2 O2(g) Al2O3(s) (rxn 1)
• DH DHfo (Al2O3(s))
• 2 Fe(s) 3/2 O2(g) Fe2O3(s) (rxn 2)
• DH DHfo (Fe2O3(s))

12
According to Hess

• 2 Al(s) 3/2 O2(g) Al2O3(s) (rxn 1)
• Fe2O3(s) 2 Fe(s) 3/2 O2(g) (- rxn 2)
• Fe2O3(s) 2 Al(s) 2 Fe(s) Al2O3(s)
• DH DHfo (Al2O3(s)) -DHfo (Fe2O3(s))

13
DH DHfo (Al2O3(s)) -DHfo (Fe2O3(s))

• -1675.7 - (-821.4)
• -854.3 kJ mol-1

14
In general,

• DHreaction
• S(np DHfo,p) - S(nr DHfo,r)

stoichiometric coefficient
sum of
15
for example,

• find DH for the reaction
• 2 H2S(g) 3 O2(g) 2 SO2(g) 2 H2O(g)
• DH 2 DHfo (SO2(g)) 2 DHfo (H2O(g))
• -
• 2 DHfo (H2S(g)) 3 DHfo (O2(g))

16

• (2 x (-296.8) 2 x (-241.8))
• -
• (2 x (-20.2) 3 x (0))
• -1036.8 kJ/mol

17
another example...

• NaOH(s) Na(aq) OH-(aq)
• DH DHfo (Na(aq)) DHfo (OH-(aq))
• -
• DHfo (NaOH(s))

18

• - 239.4 (- 229.7) - (- 426.3)
• - 42.8 kJ/mol
• (exothermic, as observed)

19
another example...

• NH4Cl(s) NH4(aq) Cl-(aq)
• DH DHfo (NH4(aq)) DHfo (Cl-(aq))
• -
• DHfo (NH4Cl(s))

20

• -132.7 (-167.3) - (-315.1)
• 15.1 kJ/mol
• (endothermic, as observed)

21
Another endothermic reaction...

• Ba(OH)2 8 H2O(s) 2 NH4NO3(s)
• ?
• Ba(NO3)2(s) 2 NH3(aq) 10 H2O(l)

water of hydration
22
Where do the DHfo values come from?

• e.g. find DHfo CH3OH(l)
• CH3OH(l) 3/2 O2(g) CO2(g) 2 H2O(g)
• 1. DHorxn from calorimetry -638.5 kJ/mol
• 2. DHorxn DHfo (CO2(g)) 2 DHfo (H2O(g))
• - DHfo (CH3OH(l))

23

• 3. Thus, DHfo (CH3OH(l))
• - DHorxn DHfo(CO2(g)) 2 DHfo(H2O(g))
• 638.5 (-393.5) 2(-241.8)
• -238.6 kJ/mol

24
another example

• How much energy can the body obtain from 1 g
  glucose (a carbohydrate)?
• food glucose (during digestion)
• glucose energy (via metabolism)

a very complicated mechanism!
25

• glucose energy has the overall reaction
• C6H12O6(s) 6 O2(g) 6 CO2(g) 6 H2O
• DHo -2803 kJ/mol
• -15.6 kJ/g

26
Eating too much fat?

• A typical fat is glycerol trilaurate
• C39H74O6(s) 54.5 O2(g) 39 CO2(g) 37 H2O
• DHo -23900 kJ/mol
• -37.4 kJ/g

27
The next time you go to class in Dunton Tower...

• Use the stairs!
• 70 kg x 100 m x 9.8 m s-2
• 68,600 J
• (68.6 kJ)
• (youll burn off less than 2 g of fat)

28
Why IS heat released or absorbed?

• Breaking bonds requires energy.
• Making bonds gives energy back.
• DH DH bonds broken DH bonds formed

DH gt 0
DH lt 0
29

• Bond Dissociation Energies
• e.g. H-H H H
• DH DH-H 436 kJ/mol

30

• e.g. find DH for the reaction
• ½ N2(g) 3/2 H2(g) NH3(g)

31

• ½ N2(g) N(g) (rxn 1)
• DH ½ (DN-N)
• 3/2 H2(g) 3 H(g) (rxn 2)
• DH 3/2 (DH-H)
• N(g) 3 H(g) NH3(g) (rxn 3)
• DH - 3 (DN-H)

32

• rxn 1 rxn 2 rxn 3
• ½ N2(g) 3/2 H2(g) NH3(g)
• DH - 3(DN-H) ½ (DN-N) 3/2 (DH-H)
• - 3(389) ½ (946) 3/2 (436)
• - 40 kJ/mol
• (actual is -46.1 kJ/mol)