#### Transcript Document

```Chapter 6
Thermochemistry
Contents and Concepts
Understanding Heats of Reaction
The first part of the chapter lays the groundwork
for understanding what we mean by heats of
reaction.
1.
2.
3.
4.
5.
6.
Energy and Its Units
Heat of Reaction
Enthalpy and Enthalpy Changes
Thermochemical Equations
Applying Stoichiometry to Heats of Reaction
Measuring Heats of Reaction
6|2
Using Heats of Reaction
Now that we understand the basic properties of
heats of reaction and how to measure them, we
can explore how to use them.
7. Hess’s Law
8. Standard Enthalpies of Formation
9. Fuels—Foods, Commercial Fuels, and Rocket
Fuels
6|3
Thermodynamics
The science of the relationship between heat and
other forms of energy
Thermochemistry
An area of thermodynamics that concerns the
study of the heat absorbed or evolved by a
chemical reaction
6|4
Energy
The potential or capacity to move matter
One form of energy can be converted to another
form of energy: electromagnetic, mechanical,
electrical, or chemical.
Next, we’ll study kinetic energy, potential energy,
and internal energy.
6|5
Kinetic Energy, EK
The energy associated with an object by virtue of
its motion.
EK 
1
mv
2
2
m = mass (kg)
v = velocity (m/s)
6|6
The SI unit of energy is the joule, J, pronounced
“jewel.”
J 
kg  m
s
2
2
The calorie is a non-SI unit of energy commonly
used by chemists. It was originally defined as the
amount of energy required to raise the
temperature of one gram of water by one degree
Celsius. The exact definition is given by the
equation:
1 cal  4.184 J (exact)
6|7
?
A person weighing 75.0 kg (165 lbs)
runs a course at 1.78 m/s (4.00 mph).
What is the person’s kinetic energy?
EK = ½ mv2
m = 75.0 kg
V = 1.78 m/s
EK
m

 (75.0 kg)  1.78

2
s 

1
E K  119
kg  m
s
2
2
2
 119 J
(3 significan t figures)
6|8
Potential Energy, EP
The energy an object has by virtue of its position in
a field of force, such as gravitaitonal, electric or
magnetic field.
Gravitational potential energy is given by the
equation
E P  mgh
m = mass (kg)
g = gravitational constant (9.80 m/s2)
h = height (m)
6|9
Internal Energy, U
The sum of the kinetic and potential energies of
the particles making up a substance
Total Energy
Etot = EK + EP + U
6 | 10
Law of Conservation of Energy
Energy may be converted from one form to
another, but the total quantity of energy remains
constant.
6 | 11
Thermodynamic System
The substance under study in
which a change occurs is
called the thermodynamic
system (or just system).
Thermodynamic
Surroundings
Everything else in the vicinity
is called the thermodynamic
surroundings (or just the
surroundings).
6 | 12
Heat, q
The energy that flows into or out of a system
because of a difference in temperature between
the thermodynamic system and its surroundings
Heat flows spontaneously from a region of higher
temperature to a region of lower temperature.
• q is defined as positive if heat is absorbed by
the system (heat is added to the system)
• q is defined as negative if heat is evolved by a
system (heat is subtracted from the system)
6 | 13
Heat of Reaction
The value of q required to return a system to the
given temperature at the completion of the reaction
(at a given temperature)
6 | 14
Endothermic Process
A chemical reaction or process in which heat is
absorbed by the system (q is positive). The
reaction vessel will feel cool.
Exothermic Process
A chemical reaction or process in which heat is
evolved by the system (q is negative). The reaction
vessel will feel warm.
6 | 15
In an endothermic reaction:
The reaction vessel cools.
Heat is absorbed.
Energy is added to the system.
q is positive.
In an exothermic reaction:
The reaction vessel warms.
Heat is evolved.
Energy is subtracted from the system.
q is negative.
6 | 16
Enthalpy, H
An extensive property of a substance that can be
used to obtain the heat absorbed or evolved in a
chemical reaction
Extensive Property
A property that depends on the amount of
substance. Mass and volume are extensive
properties.
6 | 17
A state function is a property of a system that
depends only on its present state, which is
determined by variables such as temperature and
pressure, and is independent of any previous
history of the system.
6 | 18
The altitude of a campsite is
a state function.
It is independent of the path
taken to reach it.
6 | 19
Enthalpy of Reaction
The change in enthalpy for a reaction at a given
temperature and pressure
DH = H(products) – H(reactants)
Note: D means “change in.”
Enthalpy change is equal to the heat of reaction at
constant pressure:
DH = qP
6 | 20
The diagram illustrates the enthalpy change for the
reaction
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
The reactants are at
the top. The products
are at the bottom.
The products have less
enthalpy than the
reactants, so enthalpy
is evolved as heat.
The signs of both q and
DH are negative.
6 | 21
Enthalpy and Internal Energy
The precise definition of enthalpy, H, is
H = U + PV
Many reactions take place at constant pressure, so
the change in enthalpy can be given by
DH = DU + PDV
Rearranging:
DU = DH – PDV
The term (–PDV) is the energy needed to change
volume against the atmospheric pressure, P. It is
called pressure-volume work.
6 | 22
For the reaction
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
DV
–P
The H2 gas had to do work to raise the piston.
For the reaction as written at 1 atm, -PDV = -2.5 kJ.
In addition, 368.6 kJ of heat are evolved.
6 | 23
Thermochemical Equation
The thermochemical equation is the chemical
equation for a reaction (including phase labels) in
which the equation is given a molar interpretation,
and the enthalpy of reaction for these molar
amounts is written directly after the equation
For the reaction of sodium metal with water, the
thermochemical equation is
2Na(s) + 2H2O(l) 
2NaOH(aq) + H2(g); DH = –368.6 kJ
6 | 24
?
Sulfur, S8, burns in air to produce sulfur
dioxide. The reaction evolves 9.31 kJ of
heat per gram of sulfur at constant
pressure. Write the thermochemical
equation for this reaction.
6 | 25
We first write the balanced chemical equation:
S8(s) + 8O2(g)  8SO2(g)
Next, we convert the heat per gram to heat per mole.
ΔH  
9.31 kJ
1g S8

256.52 g S 8
1 mol S 8
Δ H   2.39 10
3
kJ
evolved; the reaction is exothermic.
Now we can write the thermochemical equation:
S8(s) + 8O2(g)  8SO2(g); DH = –2.39 × 103 kJ
6 | 26
Manipulating a Thermochemical Equation
• When the equation is multiplied by a factor, the
value of DH must be multiplied by the same
factor.
• When a chemical equation is reversed, the sign
of DH is reversed.
6 | 27
Applying Stoichiometry to Heats of Reaction
6 | 28
?
You burn 15.0 g sulfur in air. How much
heat evolves from this amount of
sulfur? The thermochemical equation is
S8(s) + 8O2(g)  8SO2(g); DH = -2.39 x 103 kJ
6 | 29
S8(s) + 8O2(g)  8SO2(g); DH = -2.39 x 103 kJ
Molar mass of S8 = 256.52 g
q  15.0 g S 8 
1 mol S 8
256.5 g S 8

 2.39 x 10
3
kJ
1 mol S 8
q = –1.40 × 102 kJ
6 | 30
Measuring Heats of Reaction
We will first look at the heat needed to raise the
temperature of a substance because this is the
basis of our measurements of heats of reaction.
6 | 31
Heat Capacity, C, of a Sample of Substance
The quantity of heat needed to raise the
temperature of the sample of substance by one
degree Celsius (or one kelvin)
Molar Heat Capacity
The heat capacity for one mole of substance
Specific Heat Capacity, s (or specific heat)
The quantity of heat needed to raise the
temperature of one gram of substance by one
degree Celsius (or one kelvin) at constant
pressure
6 | 32
The heat required can be found by using the
following equations.
Using heat capacity:
q = CDt
Using specific heat capacity:
q = s x m x Dt
6 | 33
A calorimeter is a device used to measure the
heat absorbed or evolved during a physical or
chemical change. Two examples are shown below.
6 | 34
?
A piece of zinc weighing 35.8 g was
heated from 20.00°C to 28.00°C. How
much heat was required? The specific
heat of zinc is 0.388 J/(g°C).
m = 35.8 g
s = 0.388 J/(g°C)
Dt = 28.00°C – 20.00°C = 8.00°C
q = m  s  Dt
 0.388 J 
 8.00  C 
q  35.8 g  
 g C 
q = 111 J
(3 significant figures)
6 | 35
?
Nitromethane, CH3NO2, an organic
solvent burns in oxygen according to
the following reaction:
CH3NO2(g) + 3/4O2(g) 
CO2(g) + 3/2H2O(l) + 1/2N2(g)
You place 1.724 g of nitromethane in a
calorimeter with oxygen and ignite it. The
temperature of the calorimeter increases
from 22.23°C to 28.81°C. The heat capacity
of the calorimeter was determined to be
3.044 kJ/°C. Write the thermochemical
equation for the reaction.
6 | 36
We first find the heat evolved for the 1.724 g of
nitromethane, CH3NO2.
q rxn   C cal Δ t
q rxn
 3.044 kJ 
 
  28.81  C  22.23  C    20.03 kJ
C


Now, covert that to the heat evolved per mole by
using the molar mass of nitromethane, 61.04 g.
q rxn 
- 20.03 kJ
1.724 g CH 3 NO

2
61.04 g CH 3 NO
1 mol CH 3 NO
2
2
DH = –709 kJ
6 | 37
We can now write the thermochemical equation:
CH3NO2(l) + ¾O2(g)  CO2(g) + 3/2H2O(l) + ½N2(g);
DH = –709 kJ
6 | 38
Hess’s Law of Heat Summation
For a chemical equation that can be written as the
sum of two or more steps, the enthalpy change for
the overall equation equals the sum of the
enthalpy changes for the individual steps.
6 | 39
Suppose we want DH for the reaction
2C(graphite) + O2(g)  2CO(g)
It is difficult to measure directly.
However, two other reactions are known:
C(graphite) + O2(g)  CO2(g); DH = -393.5 kJ
2CO2(g)  2CO(g) + O2(g); DH = – 566.0 kJ
In order for these to add to give the reaction we
want, we must multiply the first reaction by 2. Note
that we also multiply DH by 2.
6 | 40
2C(graphite) + O2(g)  2CO(g)
2C(graphite) + 2O2(g)  2CO2(g); DH = -787.0 kJ
2CO2(g)  2CO(g) + O2(g); DH = – 566.0 kJ
2 C(graphite) + O2(g)  2 CO(g); DH = –1353.0 kJ
Now we can add the reactions and the DH values.
6 | 41
DHsub = DHfus + DHvap
6 | 42
?
What is the enthalpy of reaction, DH,
for the reaction of calcium metal with
water?
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
This reaction occurs very slowly, so it is
impractical to measure DH directly. However,
the following facts are known:
H+(aq) + OH-(aq)  H2O(l); DH = –55.9 kJ
Ca(s) + 2H+(aq)
 Ca2+(aq) + H2(g); DH = –543.0 kJ
6 | 43
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
First, identify each reactant and product:
H+(aq) + OH-(aq)  H2O(l); DH = –55.9 kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
Each substance must be on the proper side.
Ca(s), Ca2+(aq), and H2(g) are fine.
H2O(l) should be a reactant.
OH-(aq) should be a product.
Reversing the first reaction and changing the sign of
its DH accomplishes this.
6 | 44
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
H2O(l)  H+(aq) + OH-(aq); DH = +55.9 kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
The coefficients must match those in the reaction
we want.
The coefficient on H2O and OH- should be 2.
We multiply the first reaction and its DH by 2 to
accomplish this.
6 | 45
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
2H2O(l)  2H+(aq) + 2OH-(aq);
DH = +111.8 kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
We can now add the equations and their DH’s.
Note that 2H+(aq) appears as both a
reactant and a product.
6 | 46
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g)
2H2O(l)  2H+(aq) + 2OH-(aq); DH = +111.8 kJ
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g); DH = –543.0 kJ
Ca(s) + 2H2O(l)  Ca2+(aq) + 2OH-(aq) + H2(g);
DH = –431.2 kJ
6 | 47
Standard Enthalpies of Formation
The term standard state refers to the standard
thermodynamic conditions chosen for substances
when listing or comparing thermodynamic data:
1 atm pressure and the specified temperature
(usually 25°C). These standard conditions are
indicated with a degree sign (°).
When reactants in their standard states yield
products in their standard states, the enthalpy of
reaction is called the standard enthalpy of
reaction, DH°. (DH° is read “delta H zero.”)
6 | 48
Elements can exist in more than one physical
state, and some elements exist in more than one
distinct form in the same physical state. For
example, carbon can exist as graphite or as
diamond; oxygen can exist as O2 or as O3 (ozone).
These different forms of an element in the same
physical state are called allotropes.
The reference form is the most stable form of the
element (both physical state and allotrope).
6 | 49
The standard enthalpy of formation, DHf°, is the
enthalpy change for the formation of one mole of
the substance from its elements in their reference
forms and in their standard states. DHf° for an
element in its reference and standard state is zero.
For example, the standard enthalpy of formation
for liquid water is the enthalpy change for the
reaction
H2(g) + 1/2O2(g)  H2O(l)
DHf° = –285.8 kJ
Other DHf° values are given in Table 6.2 and
Appendix C.
6 | 50
Standard enthalpies of formation can be used to
calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
1/
2
H2(g) + 1/2 Cl2(g)  HCl(g); DHf° = -92.3 kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = +74.9 kJ
We first identify each reactant and product from
the reaction we want.
6 | 51
Standard enthalpies of formation can be used to
calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
1/
2
H2(g) + 1/2 Cl2(g)  HCl(g); DHf° = -92.3 kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = +74.9 kJ
Each needs to be on the correct side of the arrow
and is. Next, we’ll check coefficients.
6 | 52
Standard enthalpies of formation can be used to
calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
1/
2
H2(g) + 1/2 Cl2(g)  HCl(g); DHf° = -92.3 kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = +74.9 kJ
Cl2 and HCl need a coefficient of 4. Multiplying the
second equation and its DH by 4 does this.
6 | 53
Standard enthalpies of formation can be used to
calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
2H2(g) + 2Cl2(g)  4HCl(g); DHf° = -369.2 kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = +74.9 kJ
Now, we can add the equations.
6 | 54
Standard enthalpies of formation can be used to
calculate the standard enthalpy for a reaction, DH°.
CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g); DH° = ?
Table 6.2 shows the DHf° values:
C(graphite) + 2Cl2(g)  CCl4(l); DHf° = –135.4 kJ
2H2(g) + 2Cl2(g)  4HCl(g); DHf° = -369.2 kJ
CH4(g)  C(graphite) + 2H2(g); DHf° = +74.9 kJ
CH4(g) + 4Cl2(g)  CCl4(g) + 4HCl(g); DH° = –429.7 kJ
6 | 55
Methyl alcohol, CH3OH, is toxic
because liver enzymes oxidize it to
formaldehyde, HCHO, which can
coagulate protein. Calculate DHo for the
following reaction:
2CH3OH(aq) + O2(g)  2HCHO(aq) + 2H2O(l)
?
Standard enthalpies of formation, Δ H fo :
CH3OH(aq): -245.9 kJ/mol
HCHO(aq):
-150.2 kJ/mol
H2O(l):
-285.8 kJ/mol
6 | 56
We want DH° for the reaction:
2CH3OH(aq) + O2(aq)  2HCHO(aq) + 2H2O(l)


Δ H reaction 
products
ΔH
o
reacton

nΔH f 


nΔH f
reactants

kJ 
kJ  


  2 mol   150.2
  2 mol  - 285.8

mol 
mol  




kJ 
kJ  


  2 mol   245.9
  1 mol  0

mol 

 mol  

   300.4 kJ     571.6 kJ    491.8
o
Δ H reaction    872.0 kJ    491.8 kJ 
Δ H reaction 
o
kJ 
Δ H reaction   380.2 kJ
o