THERMOCHEMISTRY OR
THERMODYNAMICS
Chapter 6
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
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
Law of Conservation
of Energy
Energy can be converted from one form to
another but can neither be created nor
destroyed.
(Euniverse is constant)
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.
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
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)
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).
Endo- and Exothermic
Surroundings
System
heat
Surroundings
System
heat
E(system) goes up
E(system) goes down
ENDOTHERMIC
EXOTHERMIC
Enthalpy
DH = Hfinal - Hinitial
If Hfinal > Hinitial then DH is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then DH is negative
Process is EXOTHERMIC
Upon adding potassium hydroxide to water
the following reaction takes place
NaOH(S)
NaOH(aq)
for this reaction at constant
pressure, ∆H= -43 kj/mol
1 .Is the reaction exo- or
endothermic
2. Does the water get warmer?
3. What is the enthalpy change for
the solution if 14 g of NaOH is
Exercise
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
ΔH = –2221 kJ
Assume that all of the heat comes from the
combustion of propane. Calculate ΔH in
which 5.00 g of propane is burned in excess
oxygen at constant pressure.
–252 kJ
Consider the reaction
H2(g) + O2(g)  H2O(l) DH° = –286 kJ
Which of the following is true?
a) The reaction is exothermic.
b) The reaction is endothermic.
c) The enthalpy of the products is less than that of
the reactants.
d) Heat is absorbed by the system.
e)
Both A and C are true. AC
Consider the reaction:
C2 H5OH(l ) + 3O2 (g )  2CO2 (g ) + 3H 2O(l ), DH  1.37 × 103 kJ
When a 24.8-g sample of ethyl alcohol
(molar mass = 46.07 g/mol) is burned,
how much energy is released as heat?
c
The total volume of hydrogen gas needed to fill
the Hindenburg was 2.01x108 L at 1.00 atm
and 24.7°C.
How much energy was evolved when it
burned?
2H2(g) + O2(g)  2H2O(l) DH° = –286 kJ
2.35 ´ 109 kJ
7.37 ´ 102 kJ
06_74
Thermometer
Styrofoam
cover
Styrofoam
cups
q = msDt
q
m
s
Dt
Stirrer
= heat (J)
Simple Calorimeter
= mass (g)
= specific heat (j/gCo)
= “change” in temperature (Co)
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
Heat Capacity
heat absorbed
J
J
C =
=
or
increase in temperature
C K
Specific Heat Capacity
Substance
H2O
Al
glass
Spec. Heat (J/g•K)
4.184
0.902
0.84
Aluminum
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?
Spec of Al=0.902
where DT = Tfinal - Tinitial
heat gain/lost = q = m s DT
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)
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.bv
233hhbn
Copper has a specific hear of
.382j/goC. If 2.51 g of cooper
absorbs 2.75 j of heat , what is
the change in temp ?
Cooper has a specific heat of .382j/g/
oC. the temperature of an
unknown mass of cooper
increases by 4.50 oC when it
absorbs 3.97J of heat. What is the
mass of the copper?
Heating curves
REMEMBER!!!
In regular calorimetry pressure is constant, but
the volume will change.
In bomb calorimetry, volume is constant.
Calorimetry
Constant volume calorimeter is called a bomb
calorimeter.
Material is put in a container with pure
oxygen. Wires are used to start the
combustion. The container is put into a
container of water.
The heat capacity of the calorimeter is known
and tested.
Bomb Calorimeter
thermometer
stirrer
full of water
ignition wire
Steel bomb
sample
Suppose we wish to measure the energy of
combustion of octane (C8H18), a component
of gasoline. A 0.5269-g sample of octane is
placed in a bomb calorimeter known to have
a heat capacity of 11.3 kJ/ºC.
This means that 11.3 kJ of energy is required to
raise the temperature of the water and other
parts of the calorimeter by 1ºC. The octane
is ignited in the presence of excess oxygen,
and the temperature increase of the
calorimeter is 2.25ºC.
The amount of energy released is
calculated as follows:
Energy released by the reaction
= ΔT x heat capacity of calorimeter
A bomb calorimeter has a heat capacity of
9.47 kJ/K. When a 2.01-g sample of
(C3H6) was burned in this calorimeter, the
temperature increased by 4.26 K.
Calculate the energy of combustion for the
sample.
A bomb calorimeter has a heat capacity of
9.47 kJ/K. When a 2.01-g sample of
(C3H6) was burned in this calorimeter, the
temperature increased by 4.26 K.
Calculate the energy of combustion for the
sample.
A 2.200-g sample of quinine (C6H4O2) is burned in
a bomb calorimeter whose total heat capacity is
7.854kj/ºC. the temperature of the calorimeter
plus the contents increased from 23.44ºC to
30.57ºC.
What is the heat of combustion?
per gram of quinine?
Per mole of Quinine?
.5865g sample of lactic acid HC3H5O3 is
burned in a calorimeter whose heat
capacity is 4.812kj/ºC. The temperature
increases from 23.10ºC to 24.95ºC .
Calculate the heat of combustion of lactic
acid per gram ?
The heat of combustion of pentane, is -131.64
kJ/g. Combustion of 4.50 g of pentane
causes a temperature rise of 2.00°C in a
certain bomb calorimeter. What is the heat
capacity of this bomb calorimeter?
The combustion of 0.1584g benzoic acid
increase the Temperature of a bomb
calorimeter by 2.54°C. The energy
released by the combustion is
26.42kj/g.Calculate the heat capacity
of the bomb Calorimeter .
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.
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.
Using Enthalpy
Consider the decomposition of water
H2O(g) + 286 kJ ---> H2(g) + 1/2 O2(g)
Endothermic reaction — heat is a “reactant”
DH = + 286 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 all rxns.
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
Calc. DH for
S(s) + 3/2 O2(g) --> SO3(g)
S(s) + O2(g) --> SO2(g)
-320.5 kJ
SO3(g) --> SO2(g) + 1/2 O2(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
Determine the heat of reaction for the decomposition of one
mole of benzene to acetylene
C6H6(l)  3C2H2(g)
given the following thermo chemical equations:
2C6H6(l) + 15O2(g)  12CO2(g) + 6H2O(g) DH = -6271 kJ
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(g)
DH = -2511 kJ
See smart Hess Law
Using Standard Enthalpy Values
H2O(g) + C(graphite) --> H2(g) + CO(g)
(product is called “water gas”)
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.
Change in Enthalpy
Can be calculated from enthalpies of
formation of reactants and products.
DHrxn° = npDHf(products)  nrDHf(reactants)
D H is an extensive property--kJ/mol
For the reaction: 2H2 (g) + O2 (g) ---> 2H2O(g)
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.
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f’s to calculate enthalpy change DH for
H2O(g) + C(graphite) --> H2(g) + CO(g)
From reference books we find
DHf of H2O vapor = - 242 kJ/mol
H2(g) + 1/2 O2(g) --> H2O(g)
DH f of CO = - 111 kJ/mol
C(s) + 1/2 O2(g) --> CO(g)
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)
DHof (-201.5kj)
(-393.5kj)
(-241.8kj)
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 = (-393.5 kJ) + 2 (-241.8 kJ)
- {0 + (-201.5 kJ)}
DHorxn = -675.6 kJ per mol of methanol
h
2Al(s) + Fe2O3(s)
Al2O3 DHof = 646kj/mol
Fe2O3(s) DHof =-826kj/mol
Al2O3(s)+Fe(s)
2KIO3 + 12HCl
KIO3= -501
6H20= -286
HCl= -92
ICl= -24
KCL= -435
2ICl +KCL+6H20+4Cl2
The standard enthalpy of combustion of ethane gas
C2H4 is –1411.1 kj/mol a 298k
Given the following enthalpy of formation calculate
the enthalpy of formation of ethane gas
CO2DHof =-393.5kj/mol
o
DH
H20
f =-285.8kj/mol
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