Thermochemistry
Dr. Ron Rusay
Energy
Joules (J) / calorie (cal) : (4.184 J = 1 cal)
 Can
be defined as the capacity
to do work.
 Chemical energy is defined as
heat.
 Name five other types of Energy.
Two Types of Energy

Potential: due to an object’s position
or material’s composition - which can be
converted to work

Kinetic: due to motion of an object
= 1/2 mv2
 (m = mass, v = velocity)
 KE
Law of Conservation
of Energy

Different forms of energy can be
inter-converted but can neither be
created nor destroyed.
 (Euniverse
 Describe
is constant)
three inter-conversions of
energy.
Temperature v. Energy

Temperature reflects random motions of
particles; i.e. the kinetic energy of a system.

Heat involves a transfer of energy
between 2 objects due to different energies
and temperature differences.
Always: HOT  cold
Heat (Energy) Loss
Energy: A State Function


Depends only on the state of the
system - not the path of how it arrived at
that state.
It is independent of pathway.
System and Surroundings
 System:
That on which we focus
attention
 Surroundings:
Everything else in the
universe
 Universe
= System + Surroundings
Defining Energy Change
Exo- and Endo- thermic
(Exergonic and Endergonic)

Two types of energy change :

Exothermic: Heat flows out of the system (to the
surroundings).…negative sign

Endothermic: Heat flows into the system (from
the surroundings).…positive sign
First of Three Laws of
Thermodynamics

First Law of Thermodynamics:
The energy of the universe is
constant or “energy is conserved”.
Heat Capacity
(Specific Heat)
heat absorbed
J
J
C =
=
or
P
increase in temperature
C
K
O
http://chemconnections.org/general/chem120/Flash/specific_heat_s.swf
Terminology

Specific heat capacity
heat capacity per gram = J/°C g or J/K g

Molar heat capacity
heat capacity per mole = J/°C mol or J/K mol
Calorimeters
http://chemconnections.org/general/chem120/Flash/calorimetry_s.html
QUESTION
Heat Capacities
Substance
H2O (l)
Speci fic Heat
J / oC . g
4.18
H2O (s)
2.03
Al (s)
0.89
Fe (s)
0.45
QUESTION
Specific Heat
H2O (l)
J / oC . g
4.18
Why can you burn the top of your mouth with
hot pizza and not the bottom?
(The top & bottom are at the same
temperature!)
http://www.dailymotion.com/video/x3hfwx_the-science-of-pizza_people
Why can you burn the top of your mouth with hot pizza
and not the bottom? (The top & bottom are at the same
temperature.)
(Cp) on body fat. In obese mice (fat content 52.76% body wt) the heat capacity was 2.65 kJ kg-1 K-1 and
in lean mice (fat content 7.55% body wt) the heat capacity was 3.66 kJ kg-1 K-1.
Specific Heat Interactive
Energy diagrams
w = - 1500 kJ
E
q = +2000 kJ
 = q + w
= +2000kJ + (-1500kJ)
= +500kJ
Can the system do 3000 kJ
of work on the
surroundings?
Process Path
If a system has 2,000 kJ put into it
and the system does work of 1500 kJ
on the surroundings, what is ?
Energy diagrams
w = - 3000 kJ
E
q = +2000 kJ
Process Path
Can the system do 3000 kJ of work on the surroundings?
QUESTION
QUESTION
Energy
The Gas Combustion Engine
http://chemconnections.org/general/movies/html-swf/workversusenergyflow.htm
Enthalpy diagrams
qp = H
H2O (l) @ 25 oC
H
H = ?
H2O (s) @- 25 oC
Process Path
If 50.0 g of ice @ -25 oC warms
to 25 oC what is H of the
process?
http://chemconnections.org/general/movies/HeatingCurves.swf
Hdeposition= (-)
H = J or kJ
cal or kcal
Hcondensation= (-)
Hsolidification= (-)
f.p.
oC
cooling
cooling
heating
heating
b.p.
oC
m.p.
oC
Hvaporization= (+)
Hfusion= (+)
Hsublimation= (+)
Enthalpies H?
slg
H2O (l) @ 25 oC
H = Cp liq x mass x T
H
H2 O (l) @ 0 oC
H2 O (s) @ 0
Substance
H2O (l)
Speci fic Heat
J / oC . g
4.18
H2O (s)
2.03
Al (s)
0.89
Fe (s)
0.45
oC
H2 O (s) @- 25 oC
Hfusion = 6.009 kJ/mol
H = Cp ice x mass x T
Process Path H = Hice +Hfusion +Hliq
If 50.0 g of ice @ -25 oC warms
H = ?
to 25 oC what is H of the
process?
QUESTION
“Heat of Reaction”
http://chemconnections.org/general/movies/hesslaw.mov
Change in Enthalpy

The heat of any reaction can be calculated
from enthalpies of formation of reactants and
products.(“Hess’s Law”)

Hrxn° =
npHf(products) 
 nrHf(reactants)
QUESTION
A) 4675 kJ B) -1545 kJ
D) -1720 kJ
C) -290 kJ
Ex. Combustion of Methane
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
Hrxn° = [1Hf(c) + 2Hf(d)] 
[1Hf(a) + 2Hf(b)]
-
Hrxn°= [-394kJ+(-572kJ)]-[-75kJ+0kJ]= -891kJ
QUESTION
Exo- and Endo- thermic
(Exergonic and Endergonic)

Heat exchange accompanies chemical
reactions.

Exothermic: Heat flows out of the system
(to the surroundings).…negative sign

Endothermic: Heat flows into the system
(from the surroundings).…positive sign
QUESTION
Remember from earlier slides:
http://chemconnections.org/general/chem120/Flash/specific_heat_s.swf
(Cp) of a body: In obese mice (fat content 52.76% body wt) the heat capacity was 2.65 kJ kg-1 K-1 and in
lean mice (fat content 7.55% body wt) the heat capacity was 3.66 kJ kg-1 K-1.
http://chemconnections.org/general/chem120/Flash/heat_transfer_s.swf
How does the relative (Cp) of the body of an obese person compared to that of a
lean person relate to their relative rise in body temperatures if both persons
exercise at exactly the same level?
“Heat of Reaction”
Change in Enthalpy

The heat of any reaction can be calculated
from enthalpies of formation of reactants and
products.(“Hess’s Law”)
Hrxn° =
npHf(products)

 nrHf(reactants)
http://chemconnections.org/general/chem120/Flash/hess_law_s.swf
Thermochemistry Lab
Neutralization Reactions / Hess’s law
HCl(aq) + NaOH (aq)
HNO3(aq) + NaOH (aq)
NaOH (s) + H2O(l)
HCl(aq) + NaOH (s)
NaCl (aq) + H2O(l)
NaNO3 (aq) + H2O(l)
NaOH (aq)
NaCl (aq) + H2O(l)
Calorimetry (Interactive)
Thermochemistry Lab
Neutralization Reactions / Hess’s law
Tfinal
Tinitial
QUESTION
Report Table
Vol.
Solution
(mL)
Densit Mass
Specif
y
Solution ic
Solution (g)
Heat
(g/mL)
Solution
(J/g
oC)
Tf (oC)
Ti (oC)
ΔT
(oC)
Δ Hrxn
(J)
Results/Conclusions: answer
questions #3,4,& 6 pg. 41 (DVC Lab
Manual)
Δ Hrxn
(kJ/m
ol)
QUESTION
QUESTION
Exothermic Reaction
http://www.youtube.com/watch?v=rdCsbZf1_Ng
Heats of Combustion
Combustion of octane releases 5,470 kJ per
mole of octane (Hcomb = -5,470 kJ/mol)
 How many gallons of water can be boiled by
burning 1 gallon of gasoline? (Assume the
water is at 25oC)
 How many grams of fat have the equivalent
combustion energy as 1 gallon of gasoline?
 How many pounds of CO2 are added to
atmosphere from burning 1 gallon of
gasoline? (This question relates to the Greenhouse Gas

Workshop.)
Endothermic Reaction
QUESTION
http://chemconnections.org/general/chem120/Flash/calorimetry_s.swf
Activation Energy (Ea) &
Chemical Reactions
+ 2 O2 (g)  CO2 (g) + 2 H2O (g)
N2 (g) + O2 (g)  2 NO g)
 CH4 (g)


NOTE: Ea depends on pathway.
Hess’s Law Continued
 Reactants

 Products
H = + (endothermic); H = - (exothermic)
 The
change in enthalpy is the same
whether the reaction takes place in one
step or a series of steps.
Calculations via Hess’s Law

1. If a reaction is reversed, H is also reversed.
N2(g) + O2(g)  2 NO(g)
 2 NO(g)  N2(g) + O2(g)


H = 180 kJ
H = 180 kJ
2. If the coefficients of a reaction are multiplied by
an integer, H is multiplied by that same integer.

6 NO(g)  3 N2(g) + 3 O2(g)
H = 540 kJ
QUESTION
QUESTION
Enthalpies of formation data are not always experimentally easy
to obtain. However, enthalpies of combustion data are readily
available. Calculate the enthalpy of formation of methane
from the combustion data provided.
C(graphite) + O2(g) → CO2(g)
H0comb = −393.5 kJ
H2(g) + ½ O2(g) → H2O(l)
H0comb = −285.8 kJ
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
H0comb = −890.3 kJ
A) −19.4 kJ/mol
C) −221.9 kJ/mol
B) −74.8 kJ/mol
D) −296.0 kJ/mol