Chapter 5: Thermochemistry
The Study of Energy and its Transformations
The Nature of Energy
Chapter 5: Thermochemistry
(1) Kinetic Energy (Ek)
Energy of motion – a moving object has kinetic energy
• moving car
• molecules and atoms are constantly moving (thermal motion)
1
Ek  mv 2
2
mass
speed (velocity)
Chapter 5: Thermochemistry
(1) Potential Energy (Ep)
Energy an object has due to its position relative to
other objects
• potential energy due to gravity acting on the object
(diver jumping off a diving board)
• potential energy due to electrostatic attraction between
two charged objects
charge
Eel 
 Q1 Q2
d
distance between charges
• potential energy due to chemical energy stored in bonds of
molecules
Chapter 5: Thermochemistry
Different forms of energy are interconvertible:
Potential Energy
Kinetic Energy
Chemical Energy
HW: 11a,c
Chapter 5: Thermochemistry
The Units of Energy:
A person (50kg) moving at a speed of 1m/s has a kinetic energy:
2
1 2 1
kg m 2
m

Ek  mv  (50 kg )  1   25
2
2
 s
s2
 25 J
1kJ = 1000 J
A calorie is an older energy unit:
1cal = 4.184 J
1Cal = 1kcal = 1000cal
Chapter 5: Thermochemistry
The System and its Surroundings:
Energy/Heat is transferred between regions of the universe
Surroundings
Region studied: system
Molecules cannot escape:
closed system
System
Everything else: surroundings
Chapter 5: Thermochemistry
The System and its Surroundings:
2 H2
+
O2
→
2 H 2O
+ energy
Chapter 5: Thermochemistry
The System and its Surroundings:
Energy can be transferred
OR
from the system to the
surroundings
Energy can also transferred
from the surroundings to
the system
Heat is transferred from the hotter to the colder object
… until
Chapter 5: Thermochemistry
Energy can be transferred as heat (q)
Chapter 5: Thermochemistry
First law of Thermodynamics
Energy is conserved
• energy is neither created nor destroyed, it can only be
converted from one form into another
Chapter 5: Thermochemistry
Internal Energy
• The internal energy of a system (E) is the sum of all
kinetic and potential energies
• When a system undergoes change (D), the internal
energy of the system may change by DE
DE = Efinal - Einitial
Chapter 5: Thermochemistry
Internal Energy
Heat (q) transferred from
system to surroundings:
Heat (q) transferred from
surroundings to system:
Chapter 5: Thermochemistry
Internal Energy
E
E
final state
Efinal
initial state
E given off
E absorbed
initial state
Einitial
Einitial
final state
Efinal
Chapter 5: Thermochemistry
Internal Energy: sign conventions
System
Heat (q) transferred TO system
or
work (w) done TO system:
positive sign
+q, +w => DE > 0
Chapter 5: Thermochemistry
Internal Energy: sign conventions
DE = q + w
System
Heat (q) transferred TO surroundings
or work (w) done BY system:
negative sign
-q, -w => DE < 0
Chapter 5: Thermochemistry
DE = q + w
Can the sign of DE be predicted for the following experiment?
Chapter 5: Thermochemistry
What is the change in internal energy, DE, of a system that
gains 150 J of heat and does 432 J of work on the surroundings?
= -282 J
Chapter 5: Thermochemistry
Exothermic / Endothermic Processes
Endothermic process:
• heat flows into the system
Exothermic process:
• heat flows out of the system
Chapter 5: Thermochemistry
Exothermic / Endothermic Processes
H2O (l)
Water
H2O (g)
Water vapor
H2O (l)
H2O (s)
Chapter 5: Thermochemistry
Enthalpy (H)
Enthalpy measures the amount of heat exchanged
at constant pressure:
DH = qp
• Enthalpy is an EXTENSIVE property
Chapter 5: Thermochemistry
Enthalpy (H) is an extensive property
+ lots of O2
CO2
H2O
+ lots of O2
two logs
+ heat
Chapter 5: Thermochemistry
Exothermic / Endothermic Processes
H2O (l)
Water
H2O (g)
Water vapor
H2O (l)
H2O (s)
Chapter 5: Thermochemistry
Enthalpies of Reaction
DH = Hproducts - Hreactants
2 H2 (g) + O2 (g)
→
2 H2O (l)
DH = -483.6 kJ
How much heat (q) is given off by reacting 3.4 g of H2 gas?
  411.1 kJ
Chapter 5: Thermochemistry
2 H2 (g) + O2 (g)
→
2 H2O (l)
DH = -483.6 kJ
H
2 H2 (g) + O2 (g)
DH > 0
DH < 0
q absorbed
q given off
2 H2O (l)
2 H2O (l)
→
2 H2 (g) + O2 (g)
DH = +483.6 kJ
Chapter 5: Thermochemistry
2 H2O (l)
DH = -483.6 kJ
2 H2 (g) + O2 (g)
DH = +483.6 kJ
2 H2 (g) + O2 (g)
2 H2O (l)
→
→
DH is equal in magnitude, but opposite in sign, to DH for
the reverse reaction
Chapter 5: Thermochemistry
Calorimetry
… is a way to measure the amount of heat given off or
absorbed in the course of a reaction
+ NaOH
dissolves
heat (q)
Chapter 5: Thermochemistry
Calorimetry
… when the reaction occurs in a container that is
thermally insulated …
NaOH dissolves
Chapter 5: Thermochemistry
Calorimetry
The temperature change that a substance undergoes when it
absorbs heat depends on its specific heat (s)
1 g of H2O
+ 4.184 J of heat
1 g of H2O
11oC
+ 1 cal of heat
12oC
The specific heat (s) is the amount of heat required to heat
1 g of a substance by 1K (or 1oC)
Chapter 5: Thermochemistry
Calorimetry
The temperature change that a substance undergoes when it
absorbs heat depends on its specific heat (s)
mass
q
s
m  DT
q
DT 
ms
q  DT  m  s
temperature change
Chapter 5: Thermochemistry
Calorimetry
“Coffee-cup Calorimeter”
Chapter 5: Thermochemistry
The temperature of the aqueous NaOH solution (250g) was found
to have risen from room temperature (21oC) to 30oC. How much
heat was transferred? the specific heat of water is 4.184 J/g-K.
q
s
m  DT
DT= 30oC – 21oC
= 9 oC
= 9K
q  DT  m  s
m = 250 g
J
s = 4.184
gK
DT ( K )  (T (o C ) final  273)  (T (o C )initial  273)
 T (o C ) final  273  T (o C )initial  273
 T (o C ) final  T (o C )initial
J
q  9 K  250 g  4.184
gK
 9414 J  9.41 kJ
Chapter 5: Thermochemistry
The temperature of the aqueous NaOH solution (250g) was found
to have risen from room temperature (21oC) to 30oC. How much
heat was transferred? The specific heat of water is 4.184 J/g-K.
9.41 kJ were transferred – was the reaction exo- or endothermic?
NaOH (s)
→
Na+(aq)
+
OH- (aq)
+ heat
The reaction gives off heat => exothermic
Chapter 5: Thermochemistry
NaOH (s)
→
Na+(aq)
+
OH- (aq)
+ heat
The reaction gives off heat => exothermic
qsolution in calorimter = -qreaction
DH = qreaction = -9.41 kJ
Chapter 5: Thermochemistry
Calorimetry
Temperature increase in calorimeter:
→ reaction is exothermic
→ DH = negative
Temperature decrease in calorimeter:
→ reaction is endothermic
→ DH = positive
Chapter 5: Thermochemistry
Calorimetry
Where are the system and the surroundings in a thermally
insulated calorimeter?
- both are part of the solution inside the calorimeter!
Chapter 5: Thermochemistry
Which of the following substances requires the smallest amount
of energy to increase the temperature of 50.0 g of that
substance by 10K?
Specific Heat
CH4 (g)
2.20 J/g-K
Hg (l)
0.14 J/g-K
H2O (l)
4.18 J/g-K
Chapter 5: Thermochemistry
Hess’s law
You want to convert water vapor into ice at a constant
temperature. You can do this in two steps: first, convert H2O
into liquid water, then, in a second step, into ice:
add:
H2O (g)
H2O (l)
DH1 = -44 kJ
H2O (l)
H2O (s)
DH1 = -6.0 kJ
H2O (g) + H2O (l)
net:
H2O (g)
H2O (l) + H2O (s)
H2O (s)
DH = DH1 + DH2
= -44kJ + (-6.0kJ)
= -50 kJ
Chapter 5: Thermochemistry
Hess’s law
If a reaction is carried out in a series of steps, DH for the overall
reaction will equal the sum of the enthalpy changes for the
individual steps
Chapter 5: Thermochemistry
Hess’s law
H
H2O (g)
-44kJ
-50kJ
H2O (l)
-6 kJ
H2O (s)
Chapter 5: Thermochemistry
From the enthalpies of reaction
a) 2 H2 (g)
+
b)
O2 (g)
→
2 H2O (g)
DH = -483.6 kJ
3 O2 (g)
→
2 O3 (g)
DH = +284.6 kJ
calculate the heat of the reaction
3 H2 (g)
+
O3 (g)
→
3 H2O (g)
Strategy:
Find two equations, the addition of which will give you
the final reaction
Chapter 5: Thermochemistry
a) 2 H2 (g)
+
O2 (g)
→
2 H2O (g)
3 H2 (g)
+
O3 (g)
→
3 H2O (g)
DH = -483.6 kJ
(1) start by considering the first reactant: find an equation where
H2 appears - reaction a). Then, multiply reaction a) so that
the coefficients match the final equation
O2 (g)
→
2 H2O (g)
DH = -483.6 kJ ) 
+ 3/2 O2 (g)
→
3 H2O (g)
DH = -725.4 kJ
( 2 H2 (g)
3 H2 (g)
+
3
2
Chapter 5: Thermochemistry
b)
3 O2 (g)
→
2 O3 (g)
DH = +284.6 kJ
b) inverted:
2 O3 (g)
→
3 O2 (g)
DH = -284.6 kJ
O3 (g)
→
3 H2O (g)
3 H2 (g)
+
(2) Consider the second reactant: we need to introduce O3 to get
to the final equation. Find equation where O3 appears – rxn. b).
O3 appears on the product side – invert it to get O3 on the
reactant side
multiply “b) inverted” so that the coefficients match:
(2 O3 (g)
→
3 O2 (g)
DH = -284.6 kJ) x 1/2
O3 (g)
→
3/2 O2 (g)
DH = -142.3 kJ
Chapter 5: Thermochemistry
3) Now add the two partial reactions - and their DH’s - and check
whether you get the overall reaction:
3 H2 (g)
add:
+ 3/2 O2 (g)
→
3 H2O (g)
DH = -725.4 kJ
O3 (g)
→
3/2 O2 (g)
DH = -142.3 kJ
3 H2 (g) + O3 (g)
→
3 H2O (g)
DH = -867.7 kJ
Chapter 5: Thermochemistry
Calculate the enthalpy change for the reaction
A
+
2B
→
D3
DH = ?
Given the following enthalpies of reaction:
a) A + B
b) D3
→
→
C
C + B
DH = -125 kJ
DH = -41 kJ
Chapter 5: Thermochemistry
Calculate the enthalpy change for the reaction
A
+
2B
→
D3
DH = ?
Chapter 5: Thermochemistry
Calculate the enthalpy change for the reaction
A
+
2B
→
D3
DH = ?
Chapter 5: Thermochemistry
Calculate the enthalpy change for the reaction
A
+
2B
→
D3
DH = ?
Chapter 5: Thermochemistry
Calculate the enthalpy change for the reaction
P4O6 (s) + 2 O2(g)
→
P4O10 (s)
DH = ?
Given the following enthalpies of reaction:
a) P4 (s) + 3 O2(g)
→
P4O6 (s)
DH = -1640.1 kJ
b) P4 (s) + 5 O2(g)
→
P4O10 (s)
DH = -2940.1 kJ
Chapter 5: Thermochemistry
P4O6 (s) + 2 O2(g)
→
P4O10 (s)
DH = ?
Where in the following reactions does P4O6 appear?
a) P4 (s) + 3 O2(g)
→
P4O6 (s)
DH = -1640.1 kJ
b) P4 (s) + 5 O2(g)
→
P4O10 (s)
DH = -2940.1 kJ
Chapter 5: Thermochemistry
P4O6 (s) + 2 O2(g)
→
P4O10 (s)
DH = ?
Where in the following reactions does P4O6 appear?
a) P4 (s) + 3 O2(g)
→
P4O6 (s)
DH = -1640.1 kJ
b) P4 (s) + 5 O2(g)
→
P4O10 (s)
DH = -2940.1 kJ
- it appears on the product side, so a) needs to be inverted
a) inv.
P4O6 (s)
→
P4 (s) + 3 O2(g)
DH = +1640.1 kJ
Chapter 5: Thermochemistry
P4O6 (s)
→
P4 (s) + 3 O2(g)
P4O6 (s) + 2 O2(g)
→
P4O10 (s)
DH = +1640.1 kJ
DH = ?
II) O2 needs to be introduced on the reactant side, P4O10
on the product side, and P4 needs to be eliminated equation b)
b) P4 (s) + 5 O2(g)
→
P4O10 (s)
DH = -2940.1 kJ
Chapter 5: Thermochemistry
P4O6 (s)
→
add: P4 (s) + 5 O2(g)
P4 (s) + 3 O2(g)
→
P4O6 (s) + P4(s) + 5 O2(g)
P4O10 (s)
DH = +1640.1 kJ
DH = -2940.1 kJ
→ P4(s) + 3 O2(g) + P4O10 (s)
Chapter 5: Thermochemistry
P4O6 (s)
→
add: P4 (s) + 5 O2(g)
P4 (s) + 3 O2(g)
→
P4O6 (s) + P4(s) + 2 O2(g)
P4O6 (s) + 2 O2(g)
P4O10 (s)
DH = +1640.1 kJ
DH = -2940.1 kJ
→ P4(s) + 3 O2(g) + P4O10 (s)
→ P4O10 (s)
DH = -1300.0 kJ
Chapter 5: Thermochemistry
Enthalpy of Formation, DHof, …
... is the DH for the reaction that forms one mole of that
compound from its elements, with all substances in their
standard states [25oC, 1atm]
this is what the “ o “ stands for
H2 (g)
½ N2 (g)
2 C (s)
+
½ O2 (g)
→
H2O (l)
+ ½ O2 (g)
→
NO (g)
+ 3 H2 (g) + ½ O2 (g) → C2H5OH (l)
DHof = -285.8 kJ
DHof = 90.4 kJ
DHof = -277.7 kJ
Chapter 5: Thermochemistry
Enthalpy of Formation, DHof
DHof of the most stable form of any element is zero
(it does not need to be formed)
½ O2 (g) + ½ O2 (g)
→
O2 (g)
DHof = 0
DHof for O2 (g) , N2 (g) , H2 (g), Br2 (l) etc. = 0
Chapter 5: Thermochemistry
For which of the following reactions would DH represent a
standard enthalpy of formation?
2 Na (s)
K (l)
+
→
½ O2 (g)
+ ½ Cl2 (g)
CO (g)
+ ½ O2 (g)
2 Na (s)
+ Cl2 (g)
→
→
→
Na2O (s)
KCl (s)
CO2 (g)
2 NaCl (s)
Chapter 5: Thermochemistry
Enthalpies of formations can be used to calculate
enthalpies of reactions (under standard conditions)
DHorxn = sum of all DHof(products) - sum of all DHof(reactants)
DHorxn = Σ n DHof(products) – Σ m DHof(reactants)
“sum”
moles of reactant
moles of product
Chapter 5: Thermochemistry
Chapter 5: Thermochemistry
Enthalpies of formations (DHof) can be used to calculate
enthalpies of reactions (DHorxn)
C3H8 (g)
+ 5 O2 (g)
→
3 CO2 (g) + 4 H2O (l)
DHorxn = [3 x DHof [CO2 (g)] + 4 x DHof [H2O (l)]) - (DHof [C3H8] (g) + 0)
= [(3 x -394 kJ) + (4 x -286 kJ)] - [-104 kJ]
= -1182 kJ - 1144 kJ + 104 kJ = -2222 kJ
Chapter 5: Thermochemistry
What is DHo for the reaction below?
Mg(OH)2 (s) →
MgO (s) + H2O (l)
DHorxn = [ DHof [MgO (s)] + DHof [H2O] (l)] - [DHof [Mg(OH)2 (s)]
= [-601.8 kJ + (-285.8 kJ)] - [-924.7 kJ]
= 37.1 kJ
DHof [MgO (s)] = -601.8 kJ
DHof [H2O (l)] = -285.8 kJ
DHof [Mg(OH)2 (s)] = -924.7 kJ
DHof values are in Table 5.3 and
Appendix C