Download Chapter 10

1/7/2013
Chemistry: Atoms First
Julia Burdge & Jason Overby
10
Chapter 10
Energy Changes in
Chemical Reactions
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
10.1
Energy and Energy Changes
Thermochemistry
10.1 Energy and Energy Changes
10.2 Introduction to Thermodynamics
States and State Functions
The First Law of Thermodynamics
Work and Heat
10.3 Enthalpy
Reactions Carried Out at Constant Volume or at Constant Pressure
Enthalpy and Enthalpy Changes
Thermochemical Equations
10.4 Calorimetry
Specific Heat and Heat Capacity
Constant-Pressure Calorimetry
Constant-Volume Calorimetry
10.5 Hess’s Law
10.6 Standard Enthalpies of Formation
10.7 Bond Enthalpy and the Stability of Covalent Molecules
10.8 Lattice Energy and the Stability of Ionic Compounds
The Born-Haber Cycle
Comparison of Ionic and Covalent Compounds
Energy and Energy Changes
The system is a part of the universe that is of specific interest.
The surroundings constitute the rest of the universe outside the
system.
Thermochemistry is the study of heat (the transfer of thermal
energy) in chemical reactions.
Heat is the transfer of thermal energy.
Heat is either absorbed or released during a process.
System
heat
Surroundings
Surroundings
Universe = System + Surroundings
The system is usually defined as the substances involved in
chemical and physical changes.
Energy and Energy Changes
Energy and Energy Changes
An exothermic process occurs
when heat is transferred from the
system to the surroundings.
An endothermic process occurs
when heat is transferred from the
surroundings to the system.
“Feels hot!”
2H2(g) + O2(g)
System
2H2O(l) + energy
heat
“Feels cold”
energy + 2HgO(s)
System
2Hg(l) + O2(g)
heat
Surroundings
Surroundings
Universe = System + Surroundings
Universe = System + Surroundings
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10.2
Introduction to Thermodynamics
Introduction to Thermodynamics
Thermodynamics is the study of the interconversion of heat and
other kinds of energy.
State functions are properties that are determined by the state of the
system, regardless of how that condition was achieved.
In thermodynamics, there are three types
of systems:
The magnitude of change depends only on the initial and final states
of the system.
An open system can exchange mass and
energy with the surroundings.
 Energy
 Pressure
A closed system allows the transfer
of energy but not mass.
 Volume
An isolated system does not exchange
either mass or energy with its
surroundings.
 Temperature
The First Law of Thermodynamics
Work and Heat
The first law of thermodynamics states that energy can be converted
from one form to another, but cannot be created or destroyed.
The overall change in the system’s internal energy is given by:
ΔU = q + w
ΔUsys + ΔUsurr = 0
ΔU is the change in the internal energy.
“sys” and “surr” denote system and surroundings, respectively.
ΔU = Uf – Ui; the difference in the energies of the initial and final
states.
ΔUsys = –ΔUsurr
q is heat
 q is positive for an endothermic process (heat absorbed by the
system)
 q is negative for an exothermic process (heat released by the
system)
w is work
 w is positive for work done on the system
 w is negative for work done by the system
Work and Heat
Worked Example 10.1
Calculate the overall change in internal energy, ΔU, (in joules) for a system that
absorbs 188 J of heat and does 141 J of work on its surroundings.
ΔU = q + w
Strategy Combine the two contributions to internal energy using ΔU = q + w
and the sign conventions for q and w.
Solution The system absorbs heat, so q is positive. The system does work on
the surroundings, so w is negative.
ΔU = q + w = 188 J + (-141 J) = 47 J
Think About It Consult Table 10.1 to make sure you have used the proper sign
conventions for q and w.
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10.3
Enthalpy
Enthalpy
Sodium azide detonates to give a large quantity of nitrogen gas.
2NaN3(s)
2Na(s) + 3N2(g)
Under constant volume conditions, pressure increases:
Enthalpy
2NaN3(s)
2Na(s) + 3N2(g)
Under constant volume conditions, pressure increases:
Worked Example 10.2
Pressure-volume, or PV work, is done when there is a volume
change under constant pressure.
w = −PΔV
ΔV is the change in the volume of
the container.
Worked Example 10.2 (cont.)
1L
1000 mL
Determine the work done (in joules) when a sample of gas extends from 552 mL
to 891 mL at constant temperature (a) against a constant pressure of 1.25 atm,
(b) against a constant pressure of 1.00 atm, and (c) against a vacuum
(1 L∙atm = 101.3 J).
Strategy Determine change in volume (ΔV), identify the external pressure (P),
and use w = −PΔV to calculate w. The result will be in L∙atm; use the equality 1
L∙atm = 101.3 J to convert to joules.
P is the external opposing
pressure.
Solution
(c) w = -(0 atm)(339 mL)
Sodium azide detonates to give a large quantity of nitrogen gas.
101.3 J
=0J
1 L∙atm
Think About It Remember that the negative sign in the answers to part (a) and
(b) indicate that the system does work on the surroundings. When an expansion
happens against a vacuum, no work is done. This example illustrates that work is
not a state function. For an equivalent change in volume, the work varies
depending on external pressure against which the expansion must occur.
Solution ΔV = (891 – 552)mL = 339 mL. (a) P = 1.25 atm, (b) P = 1.00 atm,
(c) P = 0 atm.
(a) w = -(1.25 atm)(339 mL)
1L
1000 mL
101.3 J
= -42.9 J
1 L∙atm
(b) w = -(1.00 atm)(339 mL)
1L
1000 mL
101.3 J
= -34.3 J
1 L∙atm
Enthalpy
Pressure-volume, or PV work, is done when there is a volume
change under constant pressure.
w = −PΔV
substitute
ΔU = q + w
ΔU = q − PΔV
When a change occurs at constant volume, ΔV = 0 and no work is
done.
qV = ΔU
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Enthalpy
Enthalpy
Under conditions of constant pressure:
The thermodynamic function of a system called enthalpy (H) is
defined by the equation:
ΔU = q + w
H = U + PV
ΔU = q − PΔV
A note about SI units:
Pressure: pascal; 1Pa = 1 kg/(m . s2)
qP = ΔU + PΔV
Volume:
cubic meters; m3
PV:
1kg/(m . s2) x m3 = 1(kg . m2)/s2 = 1 J
Enthalpy: joules
U, P, V, and H are all state functions.
Enthalpy and Enthalpy Changes
Enthalpy and Enthalpy Changes
For any process, the change in enthalpy is:
ΔH = ΔU + Δ(PV)
(1)
The enthalpy of reaction (ΔH) is the difference between the
enthalpies of the products and the enthalpies of the reactants:
If pressure is constant:
ΔH = ΔU + PΔV
Rearrange to solve for ΔU:
(2)
ΔU = ΔH + PΔV
(3)
qp = ΔU + ΔV
(4)
ΔH = H(products) – H(reactants)
Assumes reactions in the lab occur at constant pressure
Remember, qp:
Substitute equation (3) into equation (4) and solve:
qp = (ΔH − PΔV) + PΔV (5)
ΔH > 0 (positive) endothermic process
ΔH < 0 (negative) exothermic process
qp = ΔH for a constant-pressure process
Thermochemical Equations
H2O(s)
Thermochemical Equations
H2O(l)
ΔH = +6.01 kJ/mol
Concepts to consider:
Is this a constant pressure process?
What is the system?
What are the surroundings?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
ΔH = −890.4 kJ/mol
Concepts to consider:
Is this a constant pressure process?
What is the system?
What are the surroundings?
ΔH < 0 exothermic
ΔH > 0 endothermic
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Thermochemical Equations
Thermochemical Equations
Enthalpy is an extensive property.
The following guidelines are useful when considering
thermochemical equations:
Extensive properties are dependent on the amount of matter involved.
H2O(l) → H2O(g)
Double the amount of matter
ΔH = +44 kJ/mol
Double the enthalpy
2H2O(l) → 2H2O(g)
1) Always specify the physical states of reactants and products
because they help determine the actual enthapy changes.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
ΔH = +88 kJ/mol
Units refer to
mole of reaction as written
Thermochemical Equations
different states
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
ΔH = −802.4 kJ/mol
different enthalpies
ΔH = +890.4 kJ/mol
Worked Example 10.3
The following guidelines are useful when considering
thermochemical equations:
2) When multiplying an equation by a factor (n), multiply the ΔH
value by same factor.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
ΔH = − 802.4 kJ/mol
2CH4(g) + 4O2(g)
2CO2(g) + 4H2O(g)
ΔH = − 1604.8 kJ/mol
Given the thermochemical equation for photosynthesis,
6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g)
ΔH = +2803 kJ/mol
calculate the solar energy required to produce 75.0 g of C6H12O6.
Strategy The thermochemical equation shows that for every mole of C6H12O6
produced, 2803 kJ is absorbed. We need to find out how much energy is absorbed
for the production of 75.0 g of C6H12O6. We must first find out how many moles
there are in 75.0 g of C6H12O6.
The molar mass of C6H12O6 is 180.2 g/mol, so 75.0 g of C6H12O6 is
3) Reversing an equation changes the sign but not the magnitude of
ΔH.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
ΔH = − 802.4 kJ/mol
CO2(g) + 2H2O(g)
CH4(g) + 2O2(g)
ΔH = +802.4 kJ/mol
1 mol C6H12O6
= 0.416 mol C6H12O6
180.2 g C6H12O6
We will multiply the thermochemical equation, including the enthalpy change, by
0.416, in order to write the equation in terms of the appropriate amount of
C6H12O6.
10.4 Calorimetry
Worked Example 10.3 (cont.)
Solution
(0.416 mol)[6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g)]
and
(0.416 mol)(ΔH) = (0.416 mol)(2803 kJ/mol) gives
2.50H2O(l) + 2.50CO2(g) → 0.416C6H12O6(s) + 2.50O2(g)
75.0 g C6H12O6 ×
ΔH = +1.17×103 kJ
1.17×103
Therefore,
kJ of energy in the form of sunlight is consumed in the
production of 75.0 g of C6H12O6. Note that the “per mole” units in ΔH are
canceled when we multiply the thermochemical equation by the number of moles
of C6H12O6.
Calorimetry is the measurement of heat changes.
Heat changes are measured in a device called a calorimeter.
The specific heat (s) of a substance is the amount of heat required to
raise the temperature of 1 g of the substance by 1°C.
Think About It The specified amount of C6H12O6 is less than half a mole.
Therefore, we should expect the associated enthalpy change to be less than half
that specified in the thermochemical equation for the production of 1 mole of
C6H12O6.
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Specific Heat and Heat Capacity
Specific Heat and Heat Capacity
The heat capacity (C) is the amount of heat required to raise the
temperature of an object by 1°C.
The heat associated with a temperature change may be calculated:
q = msΔT
The “object” may be a given quantity of a particular substance.
heat capacity of 1 kg of water =
4.184 J
 1000 g = 4184 J/C
1 g  C
Specific heat capacity of water
heat capacity of
1 kg water
q = CΔT
m is the mass.
s is the specific heat.
Specific heat capacity has units of J/(g • °C)
ΔT is the change in temperature (ΔT = Tfinal – Tinitial).
Heat capacity has units of J/°C
C is the heat capacity.
Worked Example 10.4
Calculate the amount of heat (in kJ) required to heat 255 g of water from 25.2°C
to 90.5°C.
Strategy Use q = msΔT to calculate q. s = 4.184 J/g∙°C, m = 255 g,
ΔT = 90.5°C – 25.2°C = 65.3°C.
Solution
4.184 J
4
q=
g∙°C × 255 g × 65.3°C = 6.97×10 J or 69.7 kJ
Think About It Look carefully at the cancellation of units and make sure that
the number of kilojoules is smaller than the number of joules. It is a common
error to multiply by 1000 instead of dividing in conversions of this kind.
Calorimetry
Calculate the amount of heat required to heat 1.01 kg of water from
0.05°C to 35.81°C.
Solution:
Step 1:Use the equation q = msΔT to calculate q.
q  1.01 kg 
1000 g 4.184 J

 [35.81C  0.05C] = 151000 J = 151 kJ
1 kg
g  C
Calorimetry
Calorimetry
A coffee-cup calorimeter may be used to measure the heat
exchange for a variety of reactions at constant pressure:
Concepts to consider for coffee-cup calorimetry:
Heat of neutralization:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Heat of ionization:
H2O(l) → H+(aq) + OH‒(aq)
Heat of fusion:
H2O(s) → H2O(l)
qP = ΔH
System:
reactants and products (the reaction)
Surroundings: water in the calorimeter
For an exothermic reaction:

the system loses heat

the surroundings gain (absorb) heat
qsys = −msΔT
qsurr = msΔT
qsys = −qsurr
Heat of vaporization:
H2O(l) → H2O(g)
The minus sign is used to keep
sign conventions consistent.
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Worked Example 10.5
Worked Example 10.5 (cont.)
A metal pellet with a mass of 100.0 g, originally at 88.4°C, is dropped into 125
g of water originally at 25.1°C. The final temperature of both pellet and the
water is 31.3°C. Calculate the heat capacity C (in J/°C) of the pellet.
Strategy Water constitutes the surroundings; the pellet is the system. Use qsurr =
msΔT to determine the heat absorbed by the water; then use q = CΔT to determine
the heat capacity of the metal pellet.
mwater = 125 g, swater = 4.184 J/g∙°C, and ΔTwater = 31.3°C – 25.1°C = 6.2°C.
The heat absorbed by the water must be released by the pellet: qwater = -qpellet,
mpellet = 100.0 g, and ΔTpellet = 31.3°C – 88.4°C = -57.1°C.
Solution
4.184 J
qwater =
g∙°C × 125 g × 6.2°C = 3242.6 J
Thus,
qpellet = -3242.6 J
From q = CΔT we have
-3242.6 J = Cpellet × (-57.1°C)
Thus,
Cpellet = 57 J/°C
Think About It The units cancel properly to give appropriate units for heat
capacity. Moreover, ΔTpellet is a negative number because the temperature of the
pellet decreases.
Constant-Volume Calorimetry
Constant-Volume Calorimetry
Constant volume calorimetry is carried out in a device known as a
constant-volume bomb.
To calculate qcal, the heat capacity of the calorimeter must be
known.
A constant-volume
calorimeter is an isolated
system.
qcal = CcalΔT
Bomb calorimeters are
typically used to determine
heats of combustion.
qrxn = −qcal
qrxn = −CcalΔT
qcal = −qrxn
10.5 Hess’s Law
A Famous Amos bite-sized chocolate chip cookie weighing 7.25 g is burned in a
bomb calorimeter to determine its energy content. The heat capacity of the
calorimeter is 39.97 kJ/°C. During the combustion, the temperature of the water
in the calorimeter increases by 3.90°C. Calculate the energy content (in kJ/g) of
the cookie.
Think About It According to the label on the cookie package, a
service
size
or 29 g, the
andheat
eachreleased
serving by
contains
150
Strategy
Use
qrxnis =four
-Ccalcookies,
ΔT to calculate
the combustion
of
Cal. Convert
the heat
energy
per gram
to Calories
per cookie
servingtotodetermine
verify its
the cookie.
Divide the
released
by the
mass of the
the
result.
energy content per gram Ccal = 39.97 kJ/°C and ΔT = 3.90°C.
21.5 kJ
1 Cal
29 g
×
= 1.5×102 Cal/serving
×
g
4.184 kJ
serving
Solution
qrxn = -CcalΔT = -(39.97 kJ/°C)(3.90°C) = -1.559×102 kJ
Because energy content is a positive quantity, we write
energy content per gram =
1.559×102
7.25 g
kJ
= 21.5 kJ/g
Hess’s law states that the change in enthalpy for a stepwise process is
the sum of the enthalpy changes for each of the steps.
CH4(g) + 2O2(g)
2H2O(l)
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
ΔH = −890.4 kJ/mol
2H2O(g)
ΔH = +88.0 kJ/mol
CO2(g) + 2H2O(g)
ΔH = −802.4 kJ/mol
CH4(g) + 2O2(g)
Enthalpy
Worked Example 10.6
ΔH = −802.4 kJ
ΔH = −890.4 kJ
CO2(g) + 2H2O(g)
CO2(g) + 2H2O(l)
ΔH = +88.0 kJ
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Hess’s Law
Worked Example 10.7
When applying Hess’s Law:
1) Manipulate thermochemical equations in a manner that gives the
overall desired equation.
2) Remember the rules for manipulating thermochemical equations:
Always specify the physical states of reactants and products because
they help determine the actual enthalpy changes.
When multiplying an equation by a factor (n), multiply the ΔH value by
same factor.
Reversing an equation changes the sign but not the magnitude of ΔH.
Given the following thermochemical equations,
NO(g) + O3(g) → NO2(g) + O2(g)
3
O3(g) → O2(g)
2
O2(g) → 2O(g)
ΔH = –198.9 kJ/mol
ΔH = –142.3 kJ/mol
ΔH = +495 kJ/mol
determine the enthalpy change for the reaction
NO(g) + O(g) → NO2(g)
Strategy Arrange the given thermochemical equations so that they sum to the
desired equation. Make the corresponding changes to the enthalpy changes, and
add them to get the desired enthalpy change.
3) Add the ΔH for each step after proper manipulation.
4) Process is useful for calculating enthalpies that cannot be found
directly.
Worked Example 10.7 (cont.)
Worked Example 10.7 (cont.)
Solution The first equation has NO as a reactant with the correct coefficients, so
we will use it as is.
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = –198.9 kJ/mol
The second equation must be reversed so that the O3 introduced by the first
equation will cancel (O3 is not part of the overall chemical equation). We also
must change the sign on the corresponding ΔH value.
3
ΔH = +142.3 kJ/mol
O (g) → O3(g)
2 2
These two steps sum to give:
NO(g) + O3(g) → NO2(g) + O2(g)
1
3
+ O2(g) O2(g) → O3(g)
2
2
1
NO(g) + O2(g) → NO2(g)
2
10.6
ΔH = –198.9 kJ/mol
ΔH = +142.3 kJ/mol
ΔH = –56.6 kJ/mol
Standard Enthalpies of Formation
The standard enthalpy of formation (ΔH °)
f is defined as the heat
change that results when 1 mole of a compound is formed from its
constituent elements in their standard states.
C(graphite) + O2(g)
Elements in standard
states
CO2(g)
1 mole of product
ΔH f° = −393.5 kJ/mol
1
Solution We then replace 2 O2 on the left with O by incorporating the last
equation. To do so, we divide the third equation by 2 and reverse its direction. As
a result, we must also divide ΔH value by 2 and change its sign.
1
O(g) → O2(g)
ΔH = –247.5 kJ/mol
2
Finally, we sum all the steps and add their enthalpy changes.
NO(g) + O3(g) → NO2(g) + O2(g)
3
O (g) → O3(g)
2 2
1
+ O(g) → O2(g)
2
NO(g) + O(g) → NO2(g)
ΔH = –198.9 kJ/mol
ΔH = +142.3 kJ/mol
ΔH = –247.5 kJ/mol
ΔH = –304 kJ/mol
Think About It Double-check the cancellation of identical items–especially
where fractions are involved.
Standard Enthalpies of Formation
The standard enthalpy of formation (ΔH f°) is defined as the heat
change that results when 1 mole of a compound is formed from its
constituent elements in their standard states.
The superscripted degree sign denotes standard conditions.
1 atm pressure for gases
1 M concentration for solutions
“f” stands for formation.
ΔH f° for an element in its most stable form is zero.
ΔH f° for many substances are tabulated in Appendix 2 of the
textbook.
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Standard Enthalpies of Formation
Worked Example 10.8
The standard enthalpy of reaction (ΔH °rxn) is defined as the
enthalpy of a reaction carried out under standard conditions.
aA + bB → cC + dD
ΔH °rxn = [cΔH f°(C) + dΔH f°(D) ] – [aΔH f°(A) + bΔH f°(B)]
Using data from Appendix 2, calculate ΔH °rxn for Ag+(aq) + Cl-(aq) → AgCl(s).
Strategy Use ΔH °rxn = ΣnΔH f°(products) – ΣmΔH f°(reactants) and ΔH f° values
from Appendix 2 to calculate ΔH °rxn. The ΔH °f values for Ag+(aq), Cl-(aq), and
AgCl(s) are +105.9, –167.2, and –127.0 kJ/mol, respectively.
Solution
ΔH °rxn = ΔH f°(AgCl) – [ΔH f°(Ag+) + ΔH f°(Cl-)]
= –127.0 kJ/mol – [(+105.9 kJ/mol) + (–167.2 kJ/mol)]
ΔH °rxn = ΣnΔH f°(products) – ΣmΔH f°(reactants)
n and m are the stoichiometric coefficients for the reactants and
products.
Worked Example 10.9
Think About It Watch out for misplaced or missing minus signs. This is an easy
place to lose track of them.
Worked Example 10.9 (cont.)
Given the following information, calculate the standard enthalpy of formation of
acetylene (C2H2) from its constituent elements:
C(graphite) + O2(g) → CO2(g)
1
H2(g) + O2(g) → H2O(l)
2
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)
= –127.0 kJ/mol – (–61.3 kJ/mol) = –65.7 kJ/mol
ΔH °rxn = –393.5 kJ/mol (1)
ΔH °rxn = –285.5 kJ/mol (2)
ΔH °rxn = –2598.8 kJ/mol (3)
Strategy Arrange the equations that are provided so that they will sum to the
desired equation. This may require reversing or multiplying one or more of the
equations. For any such change, the corresponding change must also be made to
the ΔH rxn
° value. The desired equation, corresponding to the standard enthalpy of
formation of acetylene, is
2C(graphite) + H2(g) → C2H2(g)
Solution We multiply Equation (1) and its ΔH °rxn value by 2:
2C(graphite) + 2O2(g) → 2CO2(g)
We reverse Equation (3) and divide it by 2 (i.e., multiply through by 1/2):
5
Think
About
It Remember
°rxn isΔH
only
°f whenkJ/mol
there is
°rxn a=ΔH
+1299.4
2CO2(g)
+ H2O(l)
→ C2H2(g)that
+ 2aOΔH
2(g)
just one
product,
just one
mole produced,
and all the reactants are
elements
in
their
standard
states.
Summing the resulting equations and the corresponding ΔH ° values:
rxn
2C(graphite) + 2O2(g) → 2CO2(g)
Bond Enthalpy and the Stability of Covalent
Molecules
The bond enthalpy is the enthalpy change associated with breaking
a bond in 1 mole of gaseous molecule.
ΔH °rxn = –787.0 kJ/mol
ΔH °rxn = –285.5 kJ/mol
1
H2(g) + 2 O2(g) → H2O(l)
5
2CO2(g) + H2O(l) → C2H2(g) + 2 O2(g)
2C(graphite) + H2(g) → C2H2(g)
10.7
ΔH °rxn = –787.0 kJ/mol
We include Equation (2) and its ΔH °rxn value as is:
1
ΔH °rxn = –285.5 kJ/mol
H2(g) + 2 O2(g) → H2O(l)
ΔH °rxn = +1299.4 kJ/mol
ΔH °f = +226.6 kJ/mol
Bond Enthalpy and the Stability of Covalent Molecules
Bond enthalpy change in an exothermic reaction.:
H2(g) → H(g) + H(g) ΔH° = 436.4 kJ/mol
The enthalpy for a gas phase reaction is given by:
ΔH° = ΣBE(reactants) – ΣBE(products)
ΔH° = total energy input – total energy released
bonds broken
bonds formed
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Bond Enthalpy and the Stability of Covalent Molecules
Bond enthalpy change in an endothermic reaction:
Worked Example 10.10
Use bond enthalpies from Table 10.4 to estimate the enthalpy of reaction for the
combustion of methane:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Think
About
It structures
Use equation
ΔH °rxn = ΣnΔH
Strategy
Draw
Lewis
to determine
what f°(products)
bonds are to–be broken and
ΣmΔHare
and data from Appendix 2 to calculate this
what bonds
to be formed.
f°(reactants)
enthalpy of reaction again; then compare your results using the two
approaches. The difference in this case is due to two things: most
tabulated bond enthalpies are averages and, by convention, we show
the product of combustion as liquid water–but average bond
Bonds enthalpies
to break: 4apply
C–H to
andspecies
2 O=Oin the gas phase, where there is little or
influence
exerted
neighboring molecules.
Bonds no
to form:
2 C=O
and 4byH–O
Bond enthalpies from Table 10.4: 414 kJ/mol (C–H), 498.7 kJ/mol (O=O),
799 kJ/mol (C=O in CO2), and 460 kJ/mol (H–O).
Solution
[4(414 kJ/mol) + 2(498.7 kJ/mol)] – [2(799 kJ/mol) + 4(460 kJ/mol)] = –785 kJ/mol
10.8
Lattice Energy and the Stability of Ionic Compounds
A Born-Haber cycle is a cycle that relates the lattice energy of an
ionic compound to quantities that can be measured.
Na(s) +
Lattice Energy and the Stability of Ionic Compounds
A Born-Haber cycle is a cycle that relates the lattice energy of an
ionic compound to quantities that can be measured.
1
Cl (g) → Na+(g) + Cl-(g)
2 2
Lattice Energy and the Stability of Ionic Compounds
Na(s) +
1
Cl (g) → Na+(g) + Cl-(g)
2 2
Worked Example 10.11
Using data from Figure 4.8 and 4.10 and Appendix 2, calculate the lattice energy
of cesium chloride (CsCl).
Strategy Using the Born-Haber Cycle Figure as a guide, combine pertinent
thermodynamic data and Hess’s law to calculate the lattice energy.
Think4.8,
About
It Compare
this value
that for
NaCl
Figure
From Figure
IE1(Cs)
= 376 kJ/mol.
FromtoFigure
4.10,
EAin1(Cl)
= 349.0
-)
kJ/mol).
Both
compounds
contain
theΔH
same
anion =(Cl
kJ/mol.10.14
From(787
Appendix
2, ΔH
°f [Cs(g)]
= 76.50
kJ/mol,
°f [Cl(g)]
121.7
have cations
withkJ/mol.
the same
chargewe
(+1),
the relative
kJ/mol,and
ΔHboth
°f [CsCl(s)]
= – 422.8
Because
areso
interested
in magnitudes
sizes
thethe
cations
willvalues
determine
thethermodynamic
relative strengths
ofAnd,
their because
lattic
only, we
canofuse
absolute
of the
data.
+ is larger than Na+, the lattice energy of CsCl
energies.
only the
standardBecause
heat of Cs
formation
of CsCl(s) is a negative number, it is only one
is smaller
the lattice energy of NaCl.
for which
the signthan
changes.
Solution
{ΔH °f [Cs(g)] + ΔH °f [Cl(g)] + IE1(Cs) + ΔH °f [CsCl(s)]} – EA1(Cl) = lattice energy
= (76.50 kJ/mol + 121.7 kJ/mol + 376 kJ/mol + 422.8 kJ/mol) – 349.0 kJ/mol
= 648 kJ/mol
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Chapter Summary: Key Points
Group Quiz #21
Energy and Energy Changes
Forms of Energy
Energy Changes in Chemical
Reactions
Units of Energy
Introduction to Thermodynamics
States and State Functions
The First Law of Thermodynamics
Work and Heat
Enthalpy
Reactions Carried Out at Constant
Volume or at Constant Pressure
Enthalpy and Enthalpy Changes
Thermochemical Equations
Calorimetry
Specific Heat and Heat Capacity
Constant-Pressure Calorimetry
Constant-Volume Calorimetry
Hess’s Law
Standard Enthalpies of
Formation
• A hot piece of copper (at 98.7oC, specific
heat = 0.385 J/g•oC) weighs 34.6486 g.
When placed in room temperature water, it is
calculated that 915.1 J of heat are released
by the metal.
• What gains heat?
• What loses heat?
• What is the final temperature of the metal?
• Watch signs!!!!
System
Surroundings
Universe = System + Surroundings
Group Quiz #23
Group Quiz #22
• Given the following equations:
• 2CO2 (g)  O2 (g) + 2CO (g) DH = 566.0 kJ
• ½ N2 (g) + ½ O2 (g)  NO (g) DH = 90.3 kJ
• Use Standard Heat of Formation values
to calculate the enthalpy of reaction for:
• C6H12O6(s)  C2H5OH(l) + CO2(g)
• Hint: Is the equation balanced?
• Calculate the enthalpy change for:
• 2CO (g) + 2NO (g)  2CO2 (g) + N2 (g) DH = ?
• DHof (C6H12O6(s)) = -1260.0 kJ/mol
• DHof (C2H5OH(l)) = -277.7 kJ/mol
• DHof (CO2(g)) = -393.5 kJ/mol
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