1. Chapter 6: Energy and Chemical Reactions
Chem 103: Chapter 6
Chapter 6: Energy and Chemical Reactions
Copyright 2006, David R. Anderson

2. Energy
Types of energy
Energy units
Specific heat and energy transfer
Changes of state
Enthalpy: Thermochemical Equations
Definitions
Enthalpies of reaction
Hess's law
Enthalpy of formation
standard state; standard molar enthalpies of formation
enthalpies of reaction from enthalpies of formation
Calorimetry
Coffee-cup calorimetry
Bomb calorimetry

3. = capacity to do work or transfer heat
Energy
= capacity to do work or transfer heat
chemical energy in
gasoline & O2
move car & generate heat
(w = F•d) 
A. Types of energy
Kinetic energy: energy of motion
mechanical: motion of objects (KE = ½mv2)
thermal: motion of atoms and molecules
electrical: motion of electrons
Potential energy: stored energy
gravitational energy (PE = mgh)
chemical energy: energy stored in chemical bonds
electrostatic energy: attraction or repulsion of charged particles
 Total energy = KE + PE
conserved! (law of conservation of energy)

4. SI: define 1 joule (J) = KE of a 2-kg object traveling at 1 m/s
Energy
Energy units
SI: define 1 joule (J) = KE of a 2-kg object traveling at 1 m/s
KE = ½mv2 = ½(2 kg)(1 m/s)2 = 1 kg·m2/s2 = 1 J
1 kJ = 1000 J
English: 1 cal = heat required to raise the temperature of 1 g of water by 1ºC
1 kcal = 1000 cal
1 cal = J
1 kcal = kJ
1 Calorie (Cal) = 1000 cal = 1 kcal
(food calorie)

5. heat capacity: heat required to raise the temperature of something
Energy
Specific heat
heat capacity: heat required to raise the temperature of something
(extensive) by 1ºC (1 K)
specific heat: heat required to raise the temperature of 1 g of something
(intensive) by 1ºC
for water,
specific heat, C = = or C = 1 cal/g ºC
or J/g ºC
or rearranging for heat: q = C  m  DT
How many kJ of heat are required to raise the temperature of 5.00 x 102 g of water by 24.0 ºC? q
m•T
cal
g ·ºC J
g·K

6. qfusion = heat required to melt a substance vaporization (boiling):
Energy
Changes of state
fusion (melting):
solid + heat  liquid
qfusion = heat required to melt a substance
vaporization (boiling):
liquid + heat  gas
qvaporization = heat required to vaporize a substance

7. Heating/cooling curves:
Energy
Changes of state
Heating/cooling curves:
gas
(e)
liquid + gas
b.p.
(d) T
liquid
(c)
solid + liquid
m.p.
(b)
solid
(a)
heat added  (heating)
 heat removed (cooling)
e.g. H2O (a) C(s) = 2.1 J/g·ºC (d) qvap = 2256 J/g or 40.7 kJ/mol
(b) qfus = J/g or 6.01 kJ/mol (e) C(g) = 2.0 J /g·ºC
(c) C(l) = 4.2 J /g·ºC

8. Energy
Changes of state
How much heat is required to convert 10.0 g of solid water at -10ºC to gas at 100 ºC?

9. Internal energy, heat, and work: For a system: DEsys = qsys + wsys
Enthalpy: Thermochemical Equations
Definitions
universe:
surroundings
system
(energy can be
transferred)
Internal energy, heat, and work:
For a system: DEsys = qsys wsys
change in energy
for a system
heat absorbed
or released by
the system
work done on or
by the system

10. Esys = energy of a system = sum of all energies (kinetic, bond, etc.)
Enthalpy: Thermochemical Equations
Definitions
DEsys = qsys + wsys
Esys = energy of a system
= sum of all energies (kinetic, bond, etc.)
DEsys = Efinal - Einitial
system gains energy: DEsys > 0 (DEsurr < 0)
system loses energy: DEsys < 0 (DEsurr > 0)
qsys = heat
heat added to system: qsys > 0 endothermic (qsurr < 0)
heat lost from system: qsys < 0 exothermic (qsurr > 0)
wsys = work
work done on system: wsys > 0 (wsurr < 0)
work done by system: wsys < 0 (wsurr > 0)
can’t
measure
can
measure

11. A balloon is heated by adding 240 J of heat. It expands, doing 135 J
Enthalpy: Thermochemical Equations
Definitions
A balloon is heated by adding 240 J of heat. It expands, doing 135 J
of work on the atmosphere. What is DE for the system?

12. taken to get from one to the other
Enthalpy: Thermochemical Equations
Definitions
DE is a state function
-depends only on the initial and final states of a system, not the path
taken to get from one to the other
1 g H2O, 10 ºC
1 g H2O, 65 ºC
1 g H2O, 25 ºC
1 g H2O, 50 ºC DE = 25 cal
First Law of Thermodynamics: energy is neither created nor destroyed;
it can only change forms (conservation of energy).
For a system: DEsys = qsys wsys

13. Enthalpy change, DH = heat gained or lost at constant pressure
Enthalpy: Thermochemical Equations
Enthalpies of reaction
Enthalpy change, DH = heat gained or lost at constant pressure
DH = Hfinal - Hinitial = qP (does not include work)
-state function
Enthalpy of reaction
Reactants  Products
DH = Hproducts - Hreactants
DH > 0: endothermic (products higher energy than reactants)
DH < 0: exothermic (products lower energy than reactants)

14. 1. DH is an extensive property What is DH per mole of Fe?
Enthalpy: Thermochemical Equations
Enthalpies of reaction
e.g., 2Al + Fe2O3  2Fe + Al2O3 DH = kJ (thermochemical equation)
1. DH is an extensive property
What is DH per mole of Fe?
2. reverse equation: change sign of DH
2Fe + Al2O3  2Al + Fe2O3 DH = ?
3. enthalpy diagrams:
2Al + Fe2O3
DH =
–851.5 kJ
DH = kJ H
2Fe + Al2O3

15. DHoverall process = SDHindividual steps (by any path; 1st law)
Enthalpy: Thermochemical Equations
Hess’s law
DHoverall process = SDHindividual steps (by any path; 1st law)
3Mg + N2  Mg3N DH = ?
given: Mg3N2 + 3H2  3Mg + 2NH3 DH = +371 kJ
and: 1/2N2 + 3/2H2  NH3 DH = –46 kJ

16. S + O2  SO2 DH = ? given: 2SO2 + O2  2SO3 DH = –196 kJ
Enthalpy: Thermochemical Equations
Hess’s law
S + O2  SO DH = ?
given: 2SO2 + O2  2SO3 DH = –196 kJ
and: 2S + 3O2  2SO3 DH = –790 kJ

17. 1. standard state = most stable form of a substance at 1 bar pressure
Enthalpy: Thermochemical Equations
Enthalpy of formation
1. standard state = most stable form of a substance at 1 bar pressure
and a given temperature, usually 25ºC (298 K).
noble gases: monatomic gases
diatomic gases: brinclhof
metals: Mº
carbon: graphite
standard molar enthalpy of formation, DHfº
= enthalpy of reaction for formation of one mole of a substance
from its elements, all in their standard states (DHfº(element) = 0)
(Table 6.2, Appendix L)
e.g., DHfº(CO(g)) = –110.5 kJ/mol
means: C(graphite) + ½O2(g)  CO(g) DHrxnº = –110.5 kJ/mol
= DHfº

18. DHrxnº =  nDHfº(products) –  nDHfº(reactants)
Enthalpy: Thermochemical Equations
Enthalpy of formation
enthalpies of reaction from enthalpies of formation
DHrxnº =  nDHfº(products) –  nDHfº(reactants)
C2H2(g) + 2H2(g)  C2H6(g) DHrxnº = ?
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l) DHrxnº = ?
DHfº (kJ/mol)
C2H2(g) 226.7
C2H6(g)
C2H5OH(l)
CO2(g)
H2O(l)

19. Enthalpy: Thermochemical Equations
Enthalpy of formation
enthalpies of reaction from enthalpies of formation
The heat of combustion of glucose is – kJ. What is DHfº for glucose?
C6H12O6(s) +

20. two styrofoam cups (insulation)
Calorimetry
Coffee-cup calorimetry
thermometer
stopper or lid
two styrofoam cups (insulation)
-enclose the universe: system (reaction)
and surroundings (solution)
reaction in solution generates or
consumes heat, raising or lowering
temperature
qsolution = C  m  T
Hrxn = qrxn = –qsolution
i.e., if T increases (T>0), rxn is exothermic (H<0)
if T decreases (T<0), rxn is endothermic (H>0)

21. Calorimetry
Coffee-cup calorimetry
When 4.25 g of NH4NO3(s) dissolves in 60.0 g of H2O in a coffee cup calorimeter, the temperatures drops from 22.0ºC to 16.9ºC. Assuming the
specific heat of the solution is J/g·ºC, what is H (in kJ/mol) for the dissolution of NH4NO3? (fw NH4NO )

22. thermometer heating element stirrer insulation heat capacity of
Calorimetry
Bomb calorimetry
thermometer
heating element
stirrer
insulation
heat capacity of
calorimeter = C
qrxn = -qcal = –C  T
bomb

23. Calorimetry
Bomb calorimetry
A 1.80-g sample of octane, C8H18, was combusted in a bomb calorimeter with a heat capacity of kJ/ºC. The temperature increased from 21.36ºC to 28.78ºC. What is H of combustion per mole of octane? (mw C8H )