Chapter 5 Thermochemistry

1
Chapter 5Thermochemistry
Chemistry, The Central Science, 10th
edition Theodore L. Brown H. Eugene LeMay, Jr.
and Bruce E. Bursten

• John D. Bookstaver
• St. Charles Community College
• St. Peters, MO
• ? 2006, Prentice Hall, Inc.

2
Energy

• Commonly defined as the ability to do work or
  transfer heat.
• Work Energy used to cause an object that has
  mass to move.
• Heat Energy used to cause the temperature of an
  object to rise.

See later that these are ways that energy can be
transferred
3
Potential Energy

• Energy an object possesses by virtue of its
  position or chemical composition.

Molecules possess potential energy by virtue of
their atoms positions relative to one another in
a molecular structure
PE m.g.h
same as m/s2
m mass (kg) g gravitational constant (9.8
m.s-2) h height (m)
The positions of the hydrogen atoms and oxygen
atom relative to one another influences the
degree of attraction (a P.E.) between them
units
4
Kinetic Energy

• Energy an object possesses by virtue of its
  motion.

v velocity (m.s-1)
units
5
Units of Energy

• The SI unit of energy is the joule (J).
• An older, non-SI unit is still in widespread use
  The calorie (cal).
• 1 cal 4.184 J
• The nutritional calorie, 1 Cal 1000 cal
• 1 Cal 4.184 kJ

6
System and Surroundings
system surroundings universe

• The system includes the molecules we want to
  study (here, the hydrogen and oxygen molecules).
• The surroundings are everything else (here, the
  cylinder and piston).

We are usually concerned with the energy changes
that involve the system
2H2(g) O2(g) ? 2H2O(g)
7
Transfer of Energy Work
Work is being done by surroundings on the system

• Energy used to move an object over some distance.
• w F ? d
• where w is work, F is the force (kg.m.s-2, or
  N), and d is the distance over which the force is
  exerted.
• Work can be done by a reaction as a gas expands
  against an external pressure (e.g. atmospheric
  pressure)

2NaN3(s) ? 2Na(s) 3N2(g)
Work is being done by the system on surroundings
N Newton
8
Transfer of Energy Heat

• Energy can also be transferred as heat (q)
• Heat flows from warmer objects to cooler objects.

When warm pop melts ice cubes, heat
is transferred from the drink to the ice
cube. This energy is used to warm the ice
and convert it into liquid water (work)
As heat is absorbed by the ice cubes, H2O(s) ?
H2O(l)
Heat is being transferred from the surroundings
to the system
9
?E, q, w, and their signs
E is the internal energy of a system, which is
the sum of all of its energies (potential and
kinetic) A systems internal energy is generally
impossible to measure
sign of q and w are very important
DE q w
another way of looking at it when the system
gains energy through work or heat, the sign
of w or q (respectively) is positive when the
system loses energy through work or heat, the
sign of w or q (respectively) is negative
10
Exchange of Heat between System and Surroundings
When heat flows under the conditions of constant
pressure, the heat flow (per mol of reactant) is
called enthalpy change, DH (qp DH)
H enthalpy

• When heat is absorbed by the system from the
  surroundings, the sign of the heat flow (q) is
  positive and the enthalpy change (DH) is also
  positive. The process is called endothermic

For an endothermic reaction, DH (and thus q) are
positive
Qsys,p gt 0
11
Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the
  surroundings, the process is endothermic.
• When heat is released by the system to the
  surroundings, the process is exothermic.

Heat lost by ststem q is negative, DH is
negative (exothermic)
For an exothermic reaction, DH (and thus q) are
negative
Qsys,p lt 0
12
First Law of Thermodynamics

• Energy cannot be created or destroyed during any
  physical or chemical transformation
• If the system loses energy (work or heat), this
  energy is absorbed by the surroundings.
• If the system absorbs energy in the form of work
  or heat, it came at the expense of the
  surroundings

13
State Functions
Internal energy E (sum of kinetic and potential
energies)

• The internal energy of a system is independent of
  the path by which the system achieved that state.
• In the system below, the water could have reached
  room temperature from either direction.

start
start
14
State Functions

• Therefore, internal energy is a state function.
• It depends only on the present state of the
  system, not on the path by which the system
  arrived at that state.
• And so, ?E depends only on Einitial and Efinal.

In fact, DE Efinal - Einitial
15
State Functions
DE q w

• However, q and w are not state functions.
• Whether the battery is shorted out or is
  discharged by running the fan, ?E is the same.
• But q and w are different in the two cases.

16
Enthalpies of Reaction
H and DH are also state functions

• The enthalpy change that accompanies a chemical
  reaction is called the enthalpy of reaction,
  ?Hrxn (or sometimes the heat of reaction).

2H2(g) O2(g) ? 2H2O(g) DHrxn -436.6 kJ
DH
this reaction is one that gives heat energy off
to the surroundings (exothermic)
17
The Truth about Enthalpy
Guidelines for Using DH Values
intensive property value doesnt depend on the
amount of material present
a property whose value depends on the amount of
material involved

• Enthalpy and enthalpy change (DH) are extensive
  properties. The size of DH will depend on the
  amount of matter involved in the chemical
  reaction
• For
• 2H2(g) O2(g) ? 2H2O(g)
• If 2 mol H2(g) are combusted, DHrxn -483.6 kJ
• If 4 mol H2(g) are combusted, DHrxn -967.2 kJ

enthalpy of reaction
i.e. twice the amount involved in the
first reaction
18
Problem 5.41b
This info tells you that 2 mol Mg produces 1204
kJ of heat (gives off to surroundings) when it
reacts with O2(g) to produce MgO(s)

• 2Mg(s) O2(g) ? 2MgO(s) DH -1204 kJ
• How much heat is evolved when 2.4 g of Mg(s)
  reacts with O2(g) under constant pressure?

Notice this is not 2 mol Mg(s)
from the periodic table
from the information given above
19
The Truth about Enthalpy
Guidelines for Using DH Values

• The enthalpy change for a reaction is equal in
  magnitude, but opposite in sign, to DH for the
  reverse reaction
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) DH -890
  kJ
• CO2(g) 2H2O(l) ? CH4(g) 2O2(g) DH 890 kJ

20
Problem 5.41d
You know that 2 mol of MgO(s) decomposing into 2
mol Mg(s) and 1 mol O2(g) absorbs 1204
kJ 2MgO(s) ? 2Mg(s) O2(g) DH 1204 kJ

• 2Mg(s) O2(g) ? 2MgO(s) DH -1204 kJ
• How many kilojoules of heat are absorbed when
  7.50 g of MgO(s) are decomposed into Mg(s) and
  O2(g) at constant pressure?

21
The Truth about Enthalpy
Guidelines for Using DH Values

• The enthalpy change for a reaction depends on the
  state of the reactants and products
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) DH -890
  kJ
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g) DH -802
  kJ
• (because 2H2O(l) ? 2H2O(g) DH 88 kJ)

22
Enthalpy of Reaction

• So an enthalpy can be determined for a reaction
  by measuring heat flow under constant pressure
  conditions (qP)
• This can be done using a device called a
  calorimeter (two types coffee cup and bomb)
• qP DH
• DH can be determined for many reactions in this
  way, but it is not necessary to do a calorimetry
  measurement get this information
• Use Hesss Law

23
Hesss Law
CH4(g) 2O2(g) ? CO(g) 2H2O(l)
1/2O2(g) DH2 CO(g) 2H2O(l) 1/2O2(g) ? CO2(g)
2H2O(l) DH3 ------------------------------------
---------------------- CH4(g) 2O2(g) ? CO2(g)
2H2O(l) DH1

• Hesss law states that If a reaction is carried
  out in a series of steps, ?H for the overall
  reaction will be equal to the sum of the enthalpy
  changes for the individual steps.

CH4(g) 2O2(g) ? CO2(g) 2H2O(l)
two steps
one step
this works because enthalpy is a state function
24
Hesss Law

• Overall enthalpy change is independent of the
  number of steps or the particular nature of the
  path by which the reaction is carried out
• Useful for calculating enthalpy changes for
  reactions without actually carrying out the
  experiment

25
Hesss Law

• For example, lets say we want to determine DH
  for the reaction
• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• And we are given the following information
• C2H4(g) ? 2C(s) 2H2(g) DH1 -52.3 kJ
• 2H2(g) 2F2(g) ? 4HF(g) DH2 -1074 kJ
• 2C(s) 4F2(g) ? 2CF4(g) DH3 -1360 kJ
• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)

DH DH1 DH2 DH3 -2486.3 kJ -2.49 x 103 kJ
26
Hesss Law

• Usually, this information is not given so nicely.
  For the same problem
• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• And we know the following information
• H2(g) F2(g) ? 2HF(g) DH -537 kJ
• C(s) 2F2(g) ? CF4(g) DH -680 kJ
• 2C(s) 2H2(g) ? C2H4(g) DH 52.3 kJ

If you added these equations together, youd get
this equation (which is nice, but not really what
you want)
3H2(g) 3F2(g) 3C(s) ? 2HF(g) CF4(g)
C2H4(g)
Using the guidelines for DH values, we should be
able to rearrange the Three equations given so
that we can add them together to create the top
(blue) equation (and thus get DH for this
reaction)
27
Hesss Law

• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• H2(g) F2(g) ? 2HF(g) DH -537 kJ
• C(s) 2F2(g) ? CF4(g) DH -680 kJ
• 2C(s) 2H2(g) ? C2H4(g) DH 52.3 kJ

multiply by 2
flip
Each of the three lower equations involves at
least one of the molecules in the top
equation Rearrange the three red equations
(multiplying by coefficients when necessary
so that they can be added together to yield the
blue equation
28
Hesss Law

• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• 2H2(g) 2F2(g) ? 4HF(g) DH 2(-537 kJ)
• 2C(s) 4F2(g) ? 2CF4(g) DH 2(-680 kJ)
• C2H4(g) ? 2C(s) 2H2(g) DH -52.3 kJ

29
Hesss Law

• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• 2H2(g) 2F2(g) ? 4HF(g) DH 2(-537 kJ)
• 2C(s) 4F2(g) ? 2CF4(g) DH 2(-680 kJ)
• C2H4(g) ? 2C(s) 2H2(g) DH -52.3 kJ
• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)

Adding, get DH -2.49x103 kJ
30
Enthalpies of Formation

• An enthalpy of formation, ?Hf, is defined as the
  enthalpy change for the reaction in which one
  mole of compound is made from its constituent
  elements in their elemental forms.
• C(graphite) O2(g) ? CO2(g)
• 2C(graphite) 2H2(g) O2(g) ? HC2H3O2(l)
• Mg(s) ½O2(g) ? MgO(s)

these are the elemental (natural) forms of carbon
and oxygen
qP DHf
31
Standard Enthalpies of Formation

• Standard enthalpies of formation, ?Hf, are
  enthalpies of formation measured when all
  reactants and products are preent in their
  standard states at (25C (298 K) and 1.00 atm
  pressure (for any gases present).
• The symbol o indicates standard conditions

32
Enthalpies of formation

• Thus, the enthalpy change that accompanies the
  formation of one mole of a compound from its
  constituent elements, with all substances in
  their standard states, is DHof
• e.g.
• 2C(graphite) 3H2(g) ½O2(g) ? C2H5OH(l)
  DHof -277.7 kJ
• This reaction represents DHofC2H5OH(l)
• Notice that all the reactants are pure elements
  (this reaction shows the formation of 1 mol of
  ethanol from the elements that make up an ethanol
  molecule, C, H, and O. These elements are
  written as they appear in nature)

33
Elements in their standard states

• Some examples
• Carbon C(s) (graphite)
• Oxygen O2(g)
• Hydrogen H2(g)
• Sodium Na(s)
• Chlorine Cl2(g)
• Bromine Br2(l)
• Iodine I2(s)
• Phosphorus P4(s)
• Iron Fe(s)
• Trés important Enthalpy of formation of any
  element in its standard state is zero

34
Enthalpies of formation

• Which of the following represent standard
  enthalpy of formation reactions?
• 2K(l) Cl2(g) ? 2KCl(s)
• C6H12O6(s) ? 6C(diamond) 6H2(g) 3O2(g)
• 2Na(s) ½O2(g) ? Na2O(s)

35
Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• The sum of these equations is

C3H8 (g) ?? 3 C(graphite) 4 H2 (g) 3
C(graphite) 3 O2 (g) ?? 3 CO2 (g) 4 H2 (g) 2
O2 (g) ?? 4 H2O (l)
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)
36
Calculation of ?H

• We can use Hesss law in this way
• ?H ??n??Hf(products) - ??m??Hf(reactants)
• where n and m are the stoichiometric
  coefficients.

?
?
37
Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• ??????H 3(-393.5 kJ) 4(-285.8 kJ) -
  1(-103.85 kJ) 5(0 kJ)
• (-1180.5 kJ) (-1143.2 kJ) - (-103.85
  kJ) (0 kJ)
• (-2323.7 kJ) - (-103.85 kJ)
• -2219.9 kJ