Chapter 6
*******************************************************************************************************************************************************************
Goals of the Chapter:
1.
To be familiar with the concept of energy, energy units, type
of energy, and type of systems
2.
To understand the concept of enthalpy, know the definition
of endothermic and exothermic processes
3.
To use calorimetry to find
a.
b.
c.
4.
the heat of a reaction, qrxn,
the heat capacity, C
the specific heat, s
To calculate the enthalpy of a reaction DHrxn
a.
b.
c.
from calorimetry data
using the Hess’s Law
using enthalpies of formation
Some Definitions
• Thermodynamics is the study of energy
and its transformations.
• Thermochemistry is the study of the
relationships between chemical reactions
and energy changes involving heat.
• Energy is the ability to do work or transfer
heat.
Work
• Work is energy used to cause an object with
mass to move.
w=Fxd
• A force is any kind of push or pull exerted on an
object.
• The most familiar force is the pull of gravity.
Heat
• Heat is the energy used to cause the temperature
of an object to increase.
• Heat flows from warmer objects to cooler objects.
Potential and Kinetic Energy
Energy an object possesses by
virtue of its position or chemical
composition.
Ep = mgh
where m is the mass, g is the gravity constant
and h is the height
Energy an object possesses
by virtue of its motion.
1
KE = ¾ mv2
2
Types of Energy
• Potential energy “Stored Energy” the energy that an object
possesses in virtue of its composition or its position with
respect of another object
• Chemical energy a type of potential energy is the energy
stored within the bonds of chemical substances
•
Kinetic energy “ Energy of motion” is the energy that an
object possesses in virtue of its motion and depends on
the magnitude of the object and its velocity
•
Thermal energy the energy that an object possesses
because of its temperature and is associated with the
random motion of atoms and molecules (i.e. their kinetic
energy)
Units of Energy
• The SI unit of energy is the joule (J).
kg m2
1 J = 1 ¾¾
s2
• An older, non-SI unit is still in
widespread use: The calorie (cal).
1 cal = 4.184 J
Understanding Energy
1. A moving racquetball has __________.
2. A motionless racquetball has __________.
3. A racquetball player perspires during the game,
giving off __________.
4. The sum of all the kinetic and potential energies of
a system’s components is known as its __________.
a.
b.
c.
d.
Kinetic energy
Potential energy
Work
Heat
e. work energy.
f. internal energy.
g. dynamic energy.
h. integral energy.
Calculating the Kinetic Energy
Problem 1: Calculate the kinetic energy in
joules of an automobile weighing 2135 lb
and traveling at 55 mph.
(1 mile = 1.6093 km, 1lb = 453.59 g).
Answer: 2.9 x 105 J
Study of Energy Changes
Divide the universe in 2 parts:
1. System: part of the universe that is of interest in the study.
2. Surroundings: includes everything else
Consider reactants, products, and flask as our system
Consider the room, or anything else as surroundings
open
Exchange:
closed
isolated
mass & energy energy
nothing
First Law of Thermodynamics
• Energy is neither created nor destroyed.
• In other words, the total energy of the
system and the surrounding is constant.
So,
if the system loses energy, it must be
gained by the surroundings, and vice
versa.
Internal Energy
The internal energy of a system is the sum of all kinetic and
potential energies of all components of the system; we call it E.
E = Ep + KE
By definition, the change in internal energy, DE, is the final
energy of the system minus the initial energy of the system:
DE = Efinal − Einitial
DE < 0, Efinal < Einitial
DE is exergonic
DE > 0, Efinal > Einitial
DE is endergonic
Changes in Internal Energy
• When energy is
exchanged between
the system and the
surroundings, it is
exchanged as either
heat (q) or work (w)
or both.
DE = q + w
Calculating DE
Problem 2: The value of ΔE for a system that performs
213 kJ of work on its surroundings and loses 79 kJ of
heat is __________ kJ.
A) +292
B) -292
C) +134
D) -134
E) -213
State Functions
The internal energy of a system is independent
of the path by which the system achieved that
state.
DE depends only on Einitial and Efinal
State Functions
• q and w are not state
functions.
• whether the battery is
discharged by
running the fan or by
a coil, its DE is the
same.
– But q and w are
different in the two
cases.
Exchange of Heat: System
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.
Examples of endothermic reactions
energy + 2HgO (s)
2Hg (l) + O2 (g)
energy + H2O (s)
H2O (l)
Examples of exothermic reactions
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Enthalpy
Definition: is the measurement of the
energy that is either absorbed or
released by a system at P = const.
Enthalpy (H) is a state function
Pressure-Volume Work
Pressure-Volume Work
Zn (s) + HCl (aq)
We can measure the work done by the gas if the reaction is
done in a vessel that has been fitted with a piston.
w = −PDV
where DV= Vf-Vi
Enthalpy
If a process takes place at:
1. constant pressure and
2. the only work done is this pressure-volume work,
we can account for energy flow during the process
by measuring the enthalpy of the system.
Enthalpy is the measurement of the heat flow into
or out of a system in a constant pressure process.
H = E + PV
Enthalpy
• When the system changes at constant pressure, the
change in enthalpy, DH, is
DH = D(E + PV)
• This can be written as
DH = DE + PDV
• Since DE = q + w and w = −PDV,
we can substitute these into the enthalpy expression:
DH = DE + PDV
DH = (q+w) − w
DH = qp
At P = const., the DH is the heat gained or lost in a reaction
Endothermicity and Exothermicity
• A process is endothermic,
then, when DH is positive
• A process is exothermic
when DH is negative
Problem 3: Which of the following statements is true for the diagram?
1.
2.
3.
4.
5.
Internal energy > 0
Internal energy < 0
Enthalpy > 0
Enthalpy < 0
Internal energy = enthalpy
System
Surroundings
heat
work
Problem 4: An endothermic reaction causes the surroundings to
1. warm up.
2. become acidic.
3. condense.
4. decrease in temperature.
5. release CO2.
Problem 5: How much work, !, must be done on a system to decrease
its volume from 14.0 L to 7.0 L by exerting a constant pressure
of 4.0 atm? (101.325 J = 1 L*atm)
Answer: 2.8 kJ
Enthalpy of Reaction
1. Enthalpy is an extensive property
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) ΔH= -890.4 kJ/mol
•890.4 kJ are released for every 1 mole of
methane that is burned at 250C and 1 atm
•If you multiply both sides of the equation by a
factor n, then DH must be n x (890.4 kJ)
Enthalpy of Reaction
2. DH for a reaction in the forward direction is
equal in size, but opposite in sign, to DH for
the reverse reaction
Enthalpy of Reaction
3. DH of a reaction depends on the state
of the products and the state of the
reactants.
H2O (s)
H2O (l) DH = 6.01 kJ
H2O (l)
H2O (g) DH = 44.0 kJ
Calculating Enthalpy of Reaction
Problem 6A: How much heat is evolved when 266 g of
white phosphorus (P4) burns in air ?
P4 (s) + 5O2 (g)
P4O10 (s)
DH = -3013 kJ
mol
Answer: -6.47 x 103 kJ
Problem 6B E.C.: If the heat of combustion for a
specific compound is −1340.0 kJ/mol and its molar
mass is 43.65 g/mol, how many grams of this
compound must you burn to release 317.50 kJ of heat?
Answer: 10.34 g
Calorimetry: Measuring Heat
•
Definition: is a technique used for the
measurement of the heat flow in a system.
•
A calorimeter is an apparatus that measures
heat flow.
We use calorimetry to find the heat of a
reaction, qrxn,
Using
a. the specific heat, s
b. the heat capacity, C
Heat Capacity
The Heat capacity (C) of a substance is the amount of heat (q)
required to raise the temperature of a substance of mass = m
by 1 K (or 1°C)
C = m•s
Specific Heat
Specific heat capacity (or simply specific heat) is the amount
of energy required to raise the temperature of 1 g of a substance
by 1 K (or 1°C)
s=
q
m ´ DT
Heat Capacity, C and Specific Heats, s
C = m•s
q
s=
m ´ DT
q = msDT
q = CDT
where DT = tfinal - tinitial
Heat (q) Absorbed or Released
Problem 7: How much heat is given off when an 869 g iron
bar cools from 94.0 °C to 5.00 °C? (sFe = 0.444 J/g • °C)
Answer: - 3.43 x 104 J
Problem 8: Large beds of rocks are used in some solarheated homes to store heat. Assume that the specific
heat of the rocks is 0.0820 J/g-K.
(a) Calculate the quantity of heat absorbed by 50.0 kg of
rocks if their temperature increases by 12.00°C at
constant pressure.
(b) What temperature change would these rocks undergo
if they emitted 450. kJ of heat.
Answer: (a) 4.92 x 104 J; (b) DT= 110. K (or oC)
Understanding Specific Heat
Problem 9: Which metal will undergo the greatest temperature change
if an equal amount of heat is added to each? Explain.
1. Fe, s = 0.45 J/g K
2. Al, s = 0.90 J/g K
3. Cu, s = 0.38 J/g K
4. Pb, s = 0.13 J/g K
5. Sn, s = 0.22 J/g K
Problem 10 EC: If a piece of metal at 85°C is added to water at 25°C, the
final temperature of the system is 30°C. Which of the following is true?
Assume that the metal is the only system and the water is the only
surrounding.
1.Heat lost by the metal > heat gained by water.
2.Heat gained by water > heat lost by the metal.
3.Heat lost by metal > heat lost by the water.
4.Heat lost by the metal = heat gained by water.
5.More information is required.
Constant-Pressure Calorimetry
• For reactions in dilute solution (except for
combustions reactions), DH=qp is
measured using a constant-pressure
calorimeter
qrxn
qsol
qcal
qsys = react. + prod. + aq. sol. + cup
qsurr = air
qsurr + qsys = 0
Assume: No heat enters or leaves!
qsurr = 0
qsys = 0
Constant-Pressure Calorimetry
qsys = qrxn + qsol + qcal
qsys = 0
qrxn = - (qsol + qcal)
qsol = ms*DT
qcal = CcalDT = 0 (assume Ccal~ 0)
Reaction at Constant P
DH = qrxn = - qsol
*s of the solution is assumed to be equal to
the specific heat for water sH O = 4.184 J/g-K
2
Measuring DH Using a Coffee-Cup Calorimeter
Problem 11: When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100
M HCl are mixed in a constant-pressure calorimeter, the
temperature of the mixture increases from 22.20°C to 23.11°C.
Calculate DH for this reaction in kJ/mol of AgNO3 assuming that the
combination of the two solutions has a mass of 100.0 g and a
specific heat of the combined solutions is s = 4.184 J/g-°C.
Answer: - 76 kJ/mol
Problem 12 EC: Together, a pure gold ring and a pure titanium ring have a
mass of 15.98 g. Both rings are heated to 79.2 °C and dropped
into 17.0 mL of water at 15.4 °C. The water and the rings reach thermal
equilibrium at a temperature of 20.2 °C.
The density of water is 0.998 g/mL and swater is 4.18 Jg·°C, the sAu
is 0.129 Jg·°C, and the sTi is 0.544 Jg·°C. Calculate the mass of
each ring.
Answer: massAu= 6.99 g; massTi = 8.99 g
Constant-Volume Calorimetry
For combustions reactions, DE is
measured using a constant-volume
calorimeter (or bomb calorimeter)
qsys = qwater + qbomb + qrxn
qsys = 0
qrxn = - (qwater + qbomb)
qwater = msDt
(incorporated in in qbomb)
qrxn = - qbomb = - CbombDT
Reaction at Constant V
DE = qrxn – PDV
DV = 0 è DE qrxn = - qbomb
Assume: No heat enters or leaves!
DE = - CbombDT
DE ~DH
Measuring qrxn Using a Bomb Calorimeter
Problem 13: A 0.5865-g sample of lactic acid (HC3H5O3) is burned
in a calorimeter whose heat capacity is 4.812 kJ/°C. The
temperature increases from 23.10°C to 24.95°C. (a) Write the
equation of the combustion reaction of the lactic acid. Calculate
the heat of combustion of lactic acid (b) per gram and (c) per mole.
Answer: (b) -15.2 kJ/g; (c) -1.37 x 103 kJ/mol
Problem 14 EC: The combustion of 1.769 g of carbon disulfide, CS2(l),
in a bomb calorimeter with a heat capacity of 4.60 kJ/°C results in an
increase in the temperature of the calorimeter and its contents
from 22.61 °C to 32.41 °C. What is the internal energy change, ΔE, for
the combustion of carbon disulfide?
Answer: -45.09 kJ
Hess’s Law
If a reaction is carried out in a series of
steps, DH for the overall reaction will be
equal to the sum of the enthalpy changes
for the individual steps
DHrxn = DH1 + DH2+ ….DHn
Hess’ Law and Enthalpies of Reactions
Hess’s Law can be used to determine DH for
different types of physical and chemical
changes such as:
1. Enthalpies of vaporization
2. Enthalpies of fusion
3. Enthalpies of combustion
Calculation of DHrxn using Hess’s Law
C3H8 (g) + 5 O2 (g) ¾® 3 CO2 (g) + 4 H2O (l)
The sum of these equations is:
C3H8 (g) ¾® 3 C(graphite) + 4 H2 (g)
DH1
3 C(graphite) + 3 O2 (g) ¾® 3 CO2 (g)
DH2
4 H2 (g) + 2 O2 (g) ¾® 4 H2O (l)
DH3
C3H8 (g) + 5 O2 (g) ¾® 3 CO2 (g) + 4 H2O (l)
DHrxn = DH1 + DH2 + DH3
DHrxn = 103.85 kJ + (-1181 kJ) + (-1143 kJ ) = -2220 kJ
Another Example of Hess’s Law
If a reaction is carried out in a series of
steps, DH for the overall reaction will be
equal to the sum of the enthalpy changes
for the individual steps
D H1 = D H2 + D H3
DH1 = (-607 kJ) + (-283 kJ ) = - 890 kJ
Using Hess’s Law
Problem 16: What is the value of the unknown DH in the diagram?
1.
2.
3.
4.
5.
+329.5 kJ
–329.5 kJ
+285.8 kJ
–241.8 kJ
+241.8 kJ
H2 (g) + 1/2 O2 (g)
DH = –285.8 kJ
DH = ?
H2O (g)
DH = –44.0 kJ
Answer: –241.8 kJ kJ/mol
H2O (l)
Problem 17: Carbon occurs in two forms, graphite and diamond. The
enthalpy of the combustion of graphite is –393.5 kJ/mol and that of
diamond is –395.4 kJ/mol. Calculate DH for the conversion of graphite to
diamond.
Answer: 1.9 kJ/mol
Another Problem
Problem 14: Barium and oxygen react together to form
barium oxide. The value of DHo for the reaction is -558.1 kJ.
a) How many kJ of heat are released when 5.75 g of barium
reacts completely with oxygen to form barium oxide?
b) How many kJ of heat are released when 5.75 g of barium
oxide is produced?
c) How many kJ of heat are released when 15.75 g of
barium reacts completely with oxygen to form barium
oxide?
Answer: a) -23.4 kJ b) -20.8 kJ c) -64.00 kJ
More Problems on Hess’s Law
Problem 18: The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood
Rowland, and Mario Molina for their work concerning the formation and decomposition of
ozone in the stratosphere. Rowland and Molina hypothesized that chlorofluorocarbons
(CFCs) in the stratosphere break down upon exposure to UV radiation, producing chlorine
atoms. Chlorine was previously identified as a catalyst in the breakdown of ozone into
oxygen gas. Using the enthalpy of reaction for two reactions with ozone, determine the
enthalpy of reaction for the reaction of chlorine with ozone. Answer: -162.5 kJ
(1)ClO(g)+O3(g)⟶Cl(g)+2O2(g)
(2)2O3(g)⟶3O2(g)
(3)O3(g)+Cl(g)⟶ClO(g)+O2(g)
Δ"∘rxn=−122.8 kJ
Δ"∘rxn=−285.3 kJ
Δ"∘rxn= ?
Problem 19:
???? Is there
something wrong
about these
equations????
Answer: -59.0 kJ
Using Enthalpies of
Formation to Determine
Enthalpies of Reaction
Enthalpies of Formation
Definition: An enthalpy of formation,
DHf, is the enthalpy change for the
reaction in which a compound is made
from its constituent elements in their
elemental forms.
Standard Enthalpy of Formation
Definition: Standard Enthalpy of Formation (DHf0) is
the heat change that results when 1 mole of a
compound is formed from its elements at P = 1 atm
An arbitrary scale with the DHf0 as a reference point
for all enthalpy expressions has been established
For any element in its most stable form allotrope
DHf0 is zero
DH0f (O2) = 0
DH0f (C, graphite) = 0
DH0f (O3) = 142 kJ/mol
DH0f (C, diamond) = 1.90 kJ/mol
Standard Enthalpies of Formation
Standard enthalpies of formation, DHof, are
measured under normal conditions (25°C
and 1.00 atm pressure).
Standard Enthalpy of Formation
Problem 20: The standard heat of formation of carbon in its diamond
form is +1.90 kJ/mole. This means that diamond is _______ graphite.
Explain.
a.
b.
c.
more stable than
less stable than
an isotope of
Problem 21: Which reaction represents the DHf reaction for NaNO3?
Explain.
1.
2.
3.
4.
5.
Na+ (aq) + NO3- (aq)
NaNO3 (aq)
Na+ (g) + NO3- (g)
NaNO3 (s)
Na (s) + NO3 (s)
NaNO3 (s)
2 Na (s) + N2 (g) + 3 O2 (g)
2 NaNO3 (s)
Na (s) + 1/2 N2 (g) + 3/2 O2 (g)
NaNO3 (s)
Calculation of DHrxn using DHof
We can also use Hess’s law in this way:
°
°
DH = S n DHf(products)
- S m DHf(reactants)
where n and m are the stoichiometric
coefficients.
Calculation of DHrxn using DHof
C3H8 (g) + 5 O2 (g) ¾® 3 CO2 (g) + 4 H2O (l)
DH = [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
Calculating an DHorxn from DHof and vice versa
Problem 22: Benzene (C6H6) burns in air to produce carbon
dioxide and liquid water. How much heat is released per mole of
C6H6 burned? The standard enthalpy of formation of C6H6 is 49.04
kJ/mol.
Answer: -3268 kJ/mol
Problem 23: Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn= -393.5 kJ
S(rhombic) + O2 (g)
SO2 (g) DH0rxn= -296.1 kJ
CS2(l) + 3O2 (g)
CO2 (g) + 2SO2 (g) DH0rxn= -1072 kJ
Answer: 86.3 kJ/mol
Important Equations in Thermochemistry
1. Internal Energy
2. Enthalpy
E = Ep + KE
DE = q + w; DE = qv
DE = Efinal − Einitial
DH = DE + P DV
DH = q p
DH = Hproducts − Hreactants
3. Heat Capacity & Specific Heat
C = m•s ;
s = ____q___
m ´ DT
4. Heat using s or C
q = msDT; q = CDT
5. Calorimetry P = const
6. Calorimetry V = const
DH = qrxn = - qsol
DE= qrxn= - qcal DH ~ - qcal
7. Hess’s Law
8. Enthalpies of Formation
DHrxn = DH1 + DH2 + …
DH = S n DHf (products) - S m DHf (reactants)
Summary Chapter 6
1.
Review concept of energy, energy units, type of energy, and type
of systems
2.
Review definition of internal energy, enthalpy, endothermic and
exothermic processes
3.
Review calorimetry definition and know what to use to calculate
a.
b.
c.
4.
the heat of a reaction, qrxn, (q = msDt ; q = CDt)
the heat capacity, C
the specific heat, s
To calculate the enthalpy of a reaction DHrxn
a.
b.
c.
from calorimetry data (P = const; V = const)
using the Hess’s Law (DHrxn = DH1 + DH2 + DH3 )
using enthalpies of formation DH = S n DHf(products) - S m DHf(reactants)