Chapter 11

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Chapter 11
Chapter 5
• Thermochemistry - the study of how
energy in the form of heat is consumed
and produced by chemical reactions.
• Energy is the capacity to do work or to
transfer heat.
• Work is done when a force moves an
object through a distance or a charge is
moved through a potential.
• Heat is energy transferred between
objects because of a difference in their
• Thermodynamics is the study of
relationship between chemical reactions
and changes in heat energy.
• Heat transfer is the process of heat
energy flowing from one object into
Work and Energy
Two Types (at least) of
• Potential: due to position or composition can be converted to work
PE = m x g x h
(m = mass, g = force of gravity, and h =
vertical distance)
• Kinetic: due to motion of the object
KE = 1/2 mv 2 (m = mass, v = velocity)
Potential Energy: A State Function
• Depends only on the present state of
the system - not how it arrived there.
 It is independent of pathway.
The Nature of Energy
Energy is conserved: Law of
Conservation of Energy states that
energy can be converted from one
form to another but can be neither
created nor destroyed.
Energy at the Molecular Level
• Kinetic energy at the molecular level depends
on the mass and velocity of the particle but
because its velocity depends on temperature
KE does too.
 As temperature increases then the KE of the same
will increase.
• One of the most important forms of potential
energy at the atomic-molecular level arises
from electrostatic interactions.
Electrostatic Potential Energy
E = k(q1q2/d)
• Coulombic attraction, not gravitational
force, determines the potential energy
of matter at the atomic level.
Q1 x Q2
Eel 
Eel is the electrostatic potential energy
Terms Describing Energy Transfer
• System: the part of the universe that is
the focus of a thermodynamic study.
• Surroundings: everything in the
universe that is not part of the system.
• Universe = System + Surroundings
• An isolated system exchanges neither
energy nor matter with the surroundings.
Heat Flow
• In an exothermic process, heat flows from a
system into its surroundings.
• In an endothermic process, heat flows from
the surroundings into the system
Heat Flow and Phase Changes
Internal Energy
• The internal energy of a system is the sum of
all the KE (translational, rotational, and
vibrational) as well as the PE of all of the
components of the system.
First Law of Thermodynamics
• The first law of thermodynamics states
that the energy gained or lost by a
system must equal the energy lost or
gained by surroundings.
• The calorie (cal) is the amount of heat
necessary to raise the temperature of 1
g of water 1oC.
• The joule (J) is the SI unit of energy;
4.184 J = 1 cal.
Energy Flow Diagrams
Change in Internal Energy
• E = q + w
 E = change in system’s internal
 q = heat
 w = work
• Work done by the system = PV
(P is the atmospheric pressure (a
constant) and V is the change
volume of the system).
Calculation of Work
Calculate the work in L•atm and joules
associated with the expansion of a gas in a
cylinder from 54 L to 72 L at a constant
external pressure of 18 atm.
1.00 L۰atm = 101.32 J
The following reactions take place in a cylinder
equipped with a movable piston at atmospheric
pressure . Which reactions will result in work being
done on the surroundings? Assume the system returns
to an initial temperature of 110°C. Hint: The volume
of a gas is proportional to number of moles (n) at
constant temperature and pressure.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
N2(g) + 2 O2(g) → 2 NO2(g)
Enthalpy and Change in Enthalpy
Enthalpy (H) = E + PV
Change in Enthalpy (H) = E + PV
At constant pressure qP = E + PV, therefore
qP = H
H = change in enthalpy: an energy flow as heat
(at constant pressure)
H > 0, Endothermic; H < 0, Exothermic
Since most heating changes occur at constant pressure, we can
define the heat transfer at constant pressure, qp, as a new
variable, H, called the enthalpy
H = E + PV
H = E + PV
E = q – PV = q + w
H = qp
How would the heat change at constant pressure be different?
Heating Curves
Changes in temperature (T) depend on four factors:
1. T depends on q, the amount of heat transferred. Double
the heat transferred and T doubles.
2. T depends on the direction in which heat is transferred. If
the system gains heat T is positive, and if the system loses
heat T is negative. This assumes the temperature changes.
In some processes, for example phase changes, heat is
transferred but T = 0.
3. T depends inversely on the amount of material. Double
the amount of material and T changes by one-half.
4. T depends on the identity of the material.
Heat Capacities
• Molar heat capacity (cp) is the heat required
to raise the temperature of 1 mole of a
substance by 1oC at constant pressure.
• q = ncpT (where q is the heat transferred)
• Specific heat (cs) is the heat required to
raise the temperature of 1 gram of a
substance by 1oC at constant pressure.
• Heat capacity (Cp) is the quantity of heat
needed to raise the temperature of some
specific object by 1oC at constant pressure.
The same quantity of energy is added to 10.00 g pieces of gold,
magnesium, and platinum, all initially at 25°C.
Which metal will have the highest final temperature?
Cn (J/mol۰C°)
Calculating Energy Through a
Change in State
• Molar heat of fusion (Hfus) - the heat
required to convert 1 mole of a solid
substance at its melting point to 1 mole of
• q = nHfus
• Molar heat of vaporization (Hvap) - the heat
required to convert 1 mole of a substance at
its boiling point to 1 mole of vapor.
• q = nHvap
Heating Curves
The heat required to change the temperature of a substance
can be calculated at constant pressure by the equation.
q (amount of heat) = nCpT
The heat required to change the phase of a substance requires
either the heat of fusion or heat of vaporization.
q = nH(fusion or vaporization)
What is the enthalpy change associated with the following
process (at 1.00 atm)?
H2O(s) (10.0 g, -12.0C)  H2O(g) (10.0g, 150.0C)
H(fusion, H2O) = 6.01 kJ/mol
H(vaporization, H2O) = 40.67 kJ/mol
Cp(ice) = 37.1 J/molC
Cp(water) = 75.37 J/molC
Cp(steam) = 43.1 J/molC
How much heat is transferred when 100. g of Cu at 90.0 °C is
added to 200.g of H2O at 25.0C? What is the final
temperature of the mixture? (assume that no heat is lost to
the surroundings)
Specific Heat: Cu = 0.385 J/gK ; H2O = 4.184 J/gK
During a strenuous workout, a student
generates 2000 kJ of heat energy. What
mass of water would have to evaporate from
the student’s skin to dissipate this much
Hvap(H2O) = 40.67kJ/mol x 1 mol/18.015g
= 2.257 kJ/g
Exactly 10. mL of water at 25oC was added
to a hot iron skillet. All of the water was
converted into steam at 100oC. If the mass of
the pan was 1.20 kg and the molar heat
capacity of iron is 25.19 J/mol•oC, what was
the temperature change of the skillet?
Heat of Reaction
• Heat of reaction is also known as
enthalpy of reaction (Hrxn) is the heat
absorbed or released by a chemical
A Specific Enthalpy
• The standard enthalpy of formation (Hfo) is
also called the standard heat of formation and
is the enthalpy change of the a formation
• A formation reaction is the process of forming
1 mole of a substance in its standard state
from its component elements in their standard
• H2(g) + 1/2 O2(g) ---> H2O(l) Hfo for water
• The standard state of a substance is its most
stable form under 1 bar pressure and 25oC.
The standard enthalpy of formation (Hf) is the enthalpy change
that takes place when 1 mole of a substance in its standard state is
formed from its pure elements in their standard states (p. 537; also
see Appendix Table A4.3).
ΔH (reaction)  n  Hf (products)  n  Hf (reactants
Standard enthalpies of formation can be used to calculate reaction
enthalpies. Enthalpies of formation of elements under standard
conditions are defined as equal to zero.
Methods of Determining Hrxn
1. from calorimetry experiments
2. from enthalpies of formation
3. using Hess’s Law
• Calorimetry is the measurement of the
change in heat that occurs during a
physical change or chemical process.
• A calorimeter is the device used to
measure the absorption or release of
heat by a physical or chemical process.
Measuring Heat Capacity
-qaluminum = qwater; qwater = ncpT and -qaluminum = ncsT
Calorimetry: Bomb Calorimeter
• H = -qcal = -CcalT
• A bomb calorimeter
is a constant-volume
device used to
measure the heat of a
combustion reaction.
Write the standard enthalpy of formation
reaction for nitric acid.
Calculating Hrxn°
• Hrxn° = npHf(products)  nrHf(reactants)
Use Table 5.2 to calculate an
approximate enthalpy of reaction for
CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l)
One step in the production of nitric acid is the
combustion of ammonia. Using the data in the
appendix to calculate the enthalpy of this
4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g)
Enthalpy changes for reactions can be estimated from
average bond enthalpies.
ΔH(reaction)  ( bond enthalpyreactants) ( bond enthalpyproducts)
Fuel Values
Fuel Value
* Based on the formation of H2O (g)
Hess’s Law
• Hess’s law states that the enthalpy change of
a reaction that is the sum of two or more
reactions is equal to the sum of the enthalpy
changes of the constituent reactions.
1. The equation for the standard enthalpy of formation of
hydrazine, N2H4, is
2 N2H4 (g)  2 NH3(g) + H2(g)
2 NH3 + H2(g)  N2H4 (g)
N2(g) + 2 H2O (g)  N2H4 (g) + O2 (g)
N2(g) + 2 H2(g)  N2H4(g)
2 NO2 (g) + 2 H2 (g)  N2H4(g) + 4 H2O (g)
Calculations via Hess’s Law
1. If a reaction is reversed, H sign changes.
N2(g) + O2(g)  2NO(g) H = 180 kJ
2NO(g)  N2(g) + O2(g) H = 180 kJ
2. If the coefficients of a reaction are multiplied by
an integer, H is multiplied by that same integer.
6NO(g)  3N2(g) + 3O2(g)
H = 540 kJ
Calculate the enthalpy change for
C2H4(g) + H2(g) --> C2H6(g) using the following
H2(g) + 1/2O2(g) ---> H2O(l)
-285.8 kJ
C2H4(g) + 3O2(g) --> 2H2O(l) + 2CO2(g) -1411 kJ
C2H6(g) + 7/2O2(g) --> 3H2O(l) + 2CO2(g) -1560 kJ
ChemTour: State Functions and
Path Functions
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This ChemTour defines and explores the difference
between state and path functions using a travel analogy that
leads into a discussion of energy, enthalpy, heat, and work.
ChemTour: Internal Energy
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This ChemTour explores how energy is exchanged between
a system and its surroundings as heat and/or work, and
how this transfer in turn affects the internal energy (E) of a
ChemTour: Pressure-Volume
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An animated ChemTour of an internal combustion engine
shows how a system undergoing an exothermic reaction
can do work on its surroundings; students can explore the
relationship among pressure, volume, and work.
ChemTour: Heating Curves
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In this ChemTour, students use interactive heating curve
diagrams to explore phase changes, heat of fusion, and
heat of vaporization. Macroscopic views of ice melting and
water boiling are shown in sync with the appropriate
sections of the heating curve.
ChemTour: Calorimetry
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This ChemTour demonstrates how a bomb calorimeter
works, and walks students through the equations used to
solve calorimetry problems. Includes an interactive
ChemTour: Hess’s Law
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This ChemTour explains Hess’s law of constant heat of
summation using animated sample problems and step-bystep descriptions.
An ideal gas in a sealed piston is
allowed to expand isothermally (at a
constant temperature) against a
pressure of 1 atm. In what direction, if
at all, does heat flow for this process?
A) into the system
B) out of the system
Isothermal Expansion of Ideal Gas
C) heat does not flow
Consider the following arguments for each answer and
vote again:
A. When the gas expands isothermally, it does work
without a decrease in its energy, so heat must flow into
the system.
B. During the expansion, the gas pressure decreases,
thereby releasing heat to the surroundings.
C. The fact that the process is isothermal means that heat
does not flow.
Isothermal Expansion of Ideal Gas
An ideal gas in an insulated piston is
compressed adiabatically (q = 0) by its
surroundings. What can be said of the
change in the temperature (ΔT) of the
gas for this process?
A) ΔT > 0
Adiabatic Compression of an Ideal Gas
B) ΔT = 0 C) ΔT < 0
Consider the following arguments for each answer
and vote again:
A. The surroundings are doing work on the system, and
no heat is flowing. Therefore, ΔE > 0 and so ΔT > 0.
B. The volume of the gas decreases, but the pressure
increases to keep the product of the pressure and
volume constant. Therefore, the temperature is also
C. The gas is being compressed to a more ordered state,
which corresponds to a lower temperature.
Adiabatic Compression of an Ideal Gas
Consider a stretched rubber band that is
suddenly released. What can be said of
the change in the temperature (ΔT) of
the rubber band for this process?
A) ΔT > 0
ΔT of a Released Rubber Band
B) ΔT = 0 C) ΔT < 0
Consider the following arguments for each answer
and vote again:
A. The stretched rubber band is at a higher energy state
than the unstretched rubber band. Releasing the
stretched rubber band causes the energy to be
B. Because the recoil of the rubber band is rapid, this
process is essentially adiabatic.
Therefore, the
temperature of the rubber band will not change.
C. As the rubber band contracts, it does work and its
energy decreases, resulting in a decrease in its
ΔT of a Released Rubber Band
A 1.0 gram block of Al (cs = 0.9 J·°C1·g-1) at 100 °C and a 1.0 gram block
of Fe (cs = 0.4 J·°C-1·g-1) at 0 °C are
added to 10 mL of water (cs = 4.2 J·°C1·g-1) at 50 °C. What will be the final
temperature of the water?
A) < 50 °C
Specific Heat Capacity of Al and Fe
B) 50 °C C) > 50 °C
Consider the following arguments for each answer and
vote again:
A. The specific heat capacity of Fe(s) is smaller than that
of Al(s), so heat from both the Al(s) and the water will
be required to warm the Fe(s).
B. The average initial temperature of the three
components is 50 °C. Therefore, the final temperature
of the water will be 50 °C.
C. The specific heat capacity of Al(s) is greater than that
of Fe(s), so the Al block at 100 °C will heat the water
more than the Fe block will cool it.
Specific Heat Capacity of Al and Fe
Cyclooctatetraene, C8H8, can undergo a
transformation between two possible states, A and B,
by rearranging its 4 double bonds.
Which of the following graphs depicts the
dependence of the equilibrium constant (K) on
temperature for the conversion from state A to state
Equilibium Constant of Cyclooctatetraene
Consider the following arguments for each answer
and vote again:
A. The intermediate is at a higher energy state than
either states A or B, so at high temperatures, the
reaction will favor the intermediate and K will
B. The enthalpies of formation for states A and B are
equal, so ΔH° = 0 and K is not temperature
C. At high temperatures, the conversion from state A to
state B will occur at a much faster rate, thus
increasing the value of K.
Equilibium Constant of Cyclooctatetraene
The reaction of N2(g) and O2(g) to form
N2O(g) is an endothermic process. The
reaction of N2(g) and H2(g) to form
NH3(g) is an exothermic process.
Given this information, which of the
following species has the lowest
enthalpy of formation, ?
A) N2(g)
Enthalpies of N , NH , and N O
B) NH3(g)
C) N2O(g)
Consider the following arguments for each answer
and vote again:
A. N2(g) is the elemental form of nitrogen, which by
definition will have H f= 0, the lowest possible
enthalpy of formation.
B. N2O(g) has a higher H f than N2(g), O2(g), and
H2(g), whereas NH3(g) has a lower H f than N2(g),
O2(g), and H2(g).
C. The formation of an N−N double bond and a N−O
double bond, as found in N2O, releases more energy
than does the creation of 3 N−H bonds to form
Enthalpies of N , NH , and N 0
Which of the following is true of ΔH° for the
polymerization of ethylene to form polyethylene?
Note: the C-C single bond enthalpy is ~350 kJ/mole
and the C-C double bond enthalpy is ~600 kJ/mole.
A) ΔH° > 0
Polymerization of Ethylene
B) ΔH° = 0
C) ΔH° < 0
Consider the following arguments for each answer
and vote again:
A. The energy required to break double bonds is more
than the energy released by forming new single
B. The total number of C-C bonds (if we count double
bonds twice) does not change with polymerization.
Therefore, there can be no change in ΔH°.
C. For each C-C double bond that breaks (~600
kJ/mole), two single bonds form, (2×~350 = ~700
Polymerization of Ethylene
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