THERMOCHEMISTRY OR
THERMODYNAMICS
Chapter 6
Chemical Reactivity
What drives chemical reactions? How do they occur?
The first is answered by THERMODYNAMICS and
the second by KINETICS.
We have already seen a number of “driving forces” for
reactions that are PRODUCT-FAVORED.
• formation of a precipitate
• gas formation
• H2O formation (acid-base reaction)
• electron transfer in a battery
Energy and Chemistry
ENERGY is the capacity to do work or
transfer heat.
HEAT is the form of energy that flows
between 2 samples because of their
difference in temperature.
Other forms of energy —
light
kinetic
electrical
potential
nuclear
Law of Conservation
of Energy
Energy can be converted from one form to
another but can neither be created nor
destroyed.
(Euniverse is constant)
The Two Types of Energy
Potential: due to position or composition - can
be converted to work
PE = mgh
(m = mass, g = acceleration of gravity, and h =
height)
Kinetic: due to motion of the object
KE = 1/2 mv2
(m = mass, v = velocity)
Kinetic and Potential Energy
Potential
energy —
energy a
motionless
body has by
virtue of its
position.
Kinetic and Potential Energy
Kinetic energy
— energy of
motion.
• Translation
• Rotation
• Vibration
Units of Energy
1 calorie = heat required to raise temp.
of 1.00 g of H2O by 1.0 oC.
1000 cal = 1 kilocalorie = 1 kcal
1 kcal = 1 Calorie (a food “calorie”)
But we use the unit called the JOULE
1 cal = 4.184 joules
James Joule
1818-1889
Temperature v. Heat
Temperature reflects random motions of
particles, therefore related to kinetic
energy of the system.
Heat involves a transfer of energy between
2 objects due to a temperature difference
Extensive & Intensive
Properties
Extensive properties
depends directly on the
amount of substance
present.
•mass
•volume
•heat
•heat capacity (C)
Intensive properties is
not related to the amount
of substance.
•temperature
•concentration
•pressure
•specific heat (s)
State Function
Depends only on the present state of the
system - not how it arrived there.
It is independent of pathway.
Energy change is independent of the pathway
(and, therefore, a state function), while
work and heat are dependent on the
pathway.
System and Surroundings
System: That on which we focus attention
Surroundings: Everything else in the universe
Universe = System + Surroundings
Exo and Endothermic
Heat exchange accompanies chemical
reactions.
Exothermic: Heat flows out of the system
(to the surroundings).
Endothermic: Heat flows into the system
(from the surroundings).
ENERGY DIAGRAMS
Exothermic
Endothermic
Endo- and Exothermic
Surroundings
System
heat
qsystem > 0
w>0
E goes up
ENDOTHERMIC
Endo- and Exothermic
Surroundings
System
heat
Surroundings
System
heat
qsystem > 0
w>0
qsystem < 0
w<0
E(system) goes up
E(system) goes down
ENDOTHERMIC
EXOTHERMIC
First Law
First Law of Thermodynamics:
The energy of the universe is
constant.
Enthalpy
DH = Hfinal - Hinitial
If Hfinal > Hinitial then DH is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then DH is negative
Process is EXOTHERMIC
First Law
DE = q + w
DE = change in system’s internal energy
q = heat
w = work
06_73
P=
P =
F
A
F
A
Area = A
A
Dh
(a) Initial
state
Dh
DV
(b) Final
state
Piston moving a distance against
a pressure does work.
Work
P = F/A
F = PA
w = F h
w = PA h
w = P V
w & V must have opposite
signs, since work is being
done by the system to
expand the gas.
wsystem = -P V
1 Latm = 101.3 J
1 J = kgm2/s2
Enthalpy
Enthalpy = H = E + PV
DE = DH  PDV
DH = DE + PDV
At constant pressure,
qP = DE + PDV,
where qP = DH at constant pressure
DH = energy flow as heat (at constant pressure)
Some Heat Exchange Terms
specific heat capacity (s)
heat capacity per gram = J/°C g or J/K g
molar heat capacity (s)
heat capacity per mole = J/°C mol or J/K mol
Specific Heat Capacity
Substance
H2O
Al
glass
Spec. Heat (J/g•K)
4.184
0.902
0.84
Aluminum
06_74
Thermometer
Styrofoam
cover
Ho
= - qp
qp = mst
Styrofoam
cups
Stirrer
q = heat (J)
Simple Calorimeter
m = mass (g)
s = specific heat (j/gCo)
t = “change” in temperature (Co)
Ho = “change in” enthalpy (kJ)
Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC,
how many joules of heat energy are lost
by the Al?
heat gain/lost = q = m s DT
where DT = Tfinal - Tinitial
Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how
many joules of heat energy are lost by the Al?
heat gain/lost = q = m s DT
where DT = Tfinal - Tinitial
q = (0.902 J/g•K)(25.0 g)(37 - 310)K
q = - 6160 J
Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC,
how many joules of heat energy are lost
by the Al?
q = - 6160 J
Notice that the negative sign on q signals
heat “lost by” or transferred out of Al.
06_75
Stirrer
Ignition
wires
Thermometer
Insulating
container
Steel
bomb
Reactants
in sample
cup
Water
Bomb Calorimeter
Heat Capacity
heat absorbed
J
J
C =
=
or
increase in temperature
C K
E = qv & qv = -(C t + ms t)
E = “change in” internal energy (J)
qv = heat at constant volume (J)
C = heat capacity (J/Co)
t = “change”in temperature (Co)
REMEMBER!!!
In regular calorimetry pressure is constant, but the
volume will change so:
qp = -H
qp = E + p  V
In bomb calorimetry, volume is constant so:
qv = E
since p
 V = zero.
Measuring Heats of Reaction
CALORIMETRY
Calculate heat of combustion of octane.
C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O
• Burn 1.00 g of octane
• Temp rises from 25.00 to 33.20 oC
• Calorimeter contains 1200 g water
• Heat capacity of bomb = 837 J/K
Hcm = -(Ct + mst) where Hc is heat of
combustion.
Measuring Heats of Reaction
CALORIMETRY
Step 1Calc. heat transferred from reaction to water.
q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J
Step 2Calc. heat transferred from reaction to bomb.
q = C t
= (837 J/K)(8.20 K) = 6860 J
Step 3Total heat evolved
41,170 J + 6860 J = 48,030 J
Heat of combustion of 1.00 g of octane = - 48.0 kJ
Hess’s Law
Reactants  Products
The change in enthalpy is the same
whether the reaction takes place in one
step or a series of steps.
Standard States
Compound
- For a gas, pressure is exactly 1 atmosphere.
- For a solution, concentration is exactly 1 molar.
- Pure substance (liquid or solid), it is the pure
liquid or solid.
Element
- The form [N2(g), K(s)] in which it exists at
1 atm and 25°C.
Calculations via Hess’s Law
1. If a reaction is reversed, DH is also reversed.
N2(g) + O2(g)  2NO(g) DH = 180 kJ
2NO(g)  N2(g) + O2(g) DH = 180 kJ
2. If the coefficients of a reaction are multiplied
by an integer, DH is multiplied by that same
integer.
6NO(g)  3N2(g) + 3O2(g)
DH = 540 kJ
Using Enthalpy
Consider the decomposition of water
H2O(g) + 243 kJ ---> H2(g) + 1/2 O2(g)
Endothermic reaction — heat is a “reactant”
DH = + 243 kJ
Using Enthalpy
Making H2 from H2O involves two steps.
H2O(l) + 44 kJ ---> H2O(g)
H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g)
----------------------------------------------------------------H2O(l) + 286 kJ --> H2(g) + 1/2 O2(g)
Example of HESS’S LAW—
If a rxn. is the sum of 2 or more others, the net DH is
the sum of the DH’s of the other rxns.
Using Enthalpy
Calc. DH for S(s) + 3/2 O2(g) --> SO3(g)
S(s) + O2(g) --> SO2(g)
-320.5 kJ
SO2(g) + 1/2 O2(g) --> SO3(g) -75.2 kJ
_______________________________________
S(s) + 3/2 O2(g) --> SO3(g)
-395.7 kJ
energy
S solid
direct path
+ 3/2 O2
DH =
-395.7 kJ
SO3 gas
+O2 DH 1 =
-320.5 kJ
SO2 gas
+ 1/2 O2
DH 2 = -75.2 kJ
 DH along one path =
 DH along another path
DH along one path =
DH along another path
This equation is valid because
DH is a STATE FUNCTION
These depend only on the state
of the system and not how it
got there.
Unlike V, T, and P, one cannot
measure absolute H. Can only
measure DH.
Standard Enthalpy Values
NIST (Nat’l Institute for Standards and Technology)
gives values of
DHof = standard molar enthalpy of formation
This is the enthalpy change when 1 mol of compound is
formed from elements under standard conditions.
DHof is always stated in terms of moles of product
formed.
See Appendix A21-A24.
DHof, standard molar enthalpy of
formation
H2(g) + 1/2 O2(g) --> H2O(g)
DHof = -241.8 kJ/mol
By definition, DHof = 0 for
elements in their standard
states.
Using Standard Enthalpy Values
Use DH’s to calculate enthalpy change for
H2O(g) + C(graphite) --> H2(g) + CO(g)
(product is called “water gas”)
Using Standard Enthalpy Values
H2O(g) + C(graphite) --> H2(g) + CO(g)
From reference books we find
H2(g) + 1/2 O2(g) --> H2O(g)
DHf of H2O vapor = - 242 kJ/mol
C(s) + 1/2 O2(g) --> CO(g)
DH f of CO = - 111 kJ/mol
Using Standard Enthalpy Values
H2O(g) --> H2(g) + 1/2 O2(g) DHo = +242 kJ
C(s) + 1/2 O2(g) --> CO(g) DHo = -111 kJ
----------------------------------------------------------------H2O(g) + C(graphite) --> H2(g) + CO(g)
DHonet = +131 kJ
To convert 1 mol of water to 1 mol each of H2 and CO
requires 131 kJ of energy.
The “water gas” reaction is ENDOthermic.
Change in Enthalpy
Can be calculated from enthalpies of
formation of reactants and products.
DHrxn° = npDHf(products)  nrDHf(reactants)
H is an extensive property--kJ/mol
For the reaction: 2H2 (g) + O2 (g) ---> 2H2O(g)
Enthalpy would be twice the H value for the
combustion of hydrogen.
Using Standard Enthalpy Values
Calculate the heat of combustion of
methanol, i.e., DHorxn for
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2
H2O(g)
DHorxn =  DHof (prod) -  DHof (react)
Using Standard Enthalpy Values
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
DHorxn =  DHof (prod) -  DHof (react)
DHorxn = DHof (CO2) + 2 DHof (H2O)
- {3/2 DHof (O2) + DHof (CH3OH)}
= (-393.5 kJ) + 2 (-241.8 kJ)
- {0 + (-201.5 kJ)}
DHorxn = -675.6 kJ per mol of methanol
DHorxn is always in terms of moles of reactant.
06_77
C(s)
CH4(g)
(a)
(c)
CO2(g)
2H2(g)
(d)
2O2(g)
(b)
Reactants
2H2O(l)
2O2(g)
Elements
Products
Pathway for the Combustion of Methane
06_1551
Reactants
Elements
0 kJ
C(s)
2O2(g)
2O2(g)
75 kJ
2H2(g)
- 394 kJ
- 572 kJ
Energy
CH4(g)
= Reactants
= Elements
= Products
CO2(g)
2H2O(l)
Products
Schematic diagram of the energy changes for the
combustion of methane.
Greenhouse Effect
-- a warming effect exerted by the earth’s atmosphere due to
thermal energy retained by absorption of infrared radiation.
Infrared
radiated by
the earth
06_80
Greenhouse Gases:
CO2
Visible light
from the sun
CO2
and H2O
molecules
H2O
Earth
CH4
N2O
Earth’s
atmosphere