Transcript g - TeacherWeb

Chapter 6
Thermochemistry:
Energy Flow and Chemical Change
If you are doing this lecture “online” then print the lecture
notes available as a word document, go through this ppt
lecture, and do all the example and practice assignments for
discussion time.
Review of Physics
Velocity and acceleration: v = d/t, a = Dv/t
Force and work: force is what is required to
overcome inertia or to change the velocity of an
object; Newton = kg*m/s2
Work = force acting through distance,
w = f*d in Joules or kg*m2/s2
Energy is capacity to do work or produce heat,
unit is Joules
Potential (PE) and nonpotential energy (nPE)
Heat: energy transferred as result of temperature
difference; transfer always from warmer body
to colder body
LEARNING OBJECTIVES:
1. Distinguish between heat & work; temperature,
thermal energy & heat
2. Relate q, DT, cp, thru heat transfer equations
q = m * cp * DT
and q = DHphase * amount
3. Calculate DHrxn from calorimeter data
4. Understand how DHfo values are established for
compounds
5. Calculate DHrxn from either heats of formation
or Hess' law
6. Understand all the symbols:
DHrxno
DHfo DHvapo
DHfuso
DHcombo
Figure 6.1 from 4th ed, not in Prin
A chemical system and its surroundings.
the surroundings
the system
The system is the chemical reaction being studied.
The surroundings are everything else in the
universe!
Definitions
Thermodynamics is the study of heat and its transformations
Thermochemistry is a branch of thermodynamics that deals with the heat
involved with chemical and physical changes
Fundamental premise
When energy is transferred from one object to another, it appears as work
and/or as heat
For our work we must define a system to study; everything else then
becomes the surroundings
The system is composed of particles with their own internal energies (E or U).
Therefore the system has an internal energy. When a change occurs, the
internal energy changes
MORE ON ENERGY
First law of thermodynamics or the Law of
Conservation of Energy:
In an isolated system energy can't be created or
destroyed, so the total energy can't change
Energy is in two major forms - nonpotential (often
kinetic) and potential
Energy can be transferred between the two forms,
but the total, nPE + PE, has to remain the same
ENERGY CONTINUED:
1. POTENTIAL ENERGY - relative to position (stored)
2. TYPES OF NONPOTENTIAL ENERGY:
KINETIC ENERGY - MOVING OBJECT = ½mv2
HEAT - CHANGING TEMPERATURE – FROM ENERGY
TRANSER
Thermal energy from the motion of atoms or molecules
in s, l or g
Heat involves the transfer of energy between two
objects due to a temperature difference
OTHER FORMS OF nPE:
Radiant energy from the sun
Electrical energy generated by mechanical devices or by
redox reactions
ENERGY CONTINUED:
Chemical reactions either release or absorb
energy as:
heat - thermal energy**
light - radiant energy
electrical current - electrical energy
What does high thermal energy
mean?
One relative measure of thermal energy is
temperature, but temperature does not
equal thermal energy
Compare a cup of coffee at 102°F to a tub
of water at 102°F
Which contains more thermal energy?
WHY?
MORE THERMOCHEMISTRY:
SYSTEM = the substance being evaluated for
energy content in a thermodynamic process
SURROUNDINGS = everything outside the system
in the process
ENDOTHERMIC PROCESS = a thermodynamic
process in which energy transfers into a system
from its surroundings
EXOTHERMIC PROCESS = from system to
surroundings
ENTHALPY CHANGE = energy transferred at a
constant pressure, DH
"A bit of physics again"
There are 2 ways to transfer energy between
system and surroundings: work or heat
Work = action of a force that causes displacement
within the system in a measurable dimension as
“-DX” if linear or as “DV” if volume
Work = -F*Dh
Since Area*Dh = DV and Pressure = F/Area
Then w = - (F/Area) * Area*Dh = - PDV
for work done at constant pressure
Figure 6.5
Pressure-volume
work.
When the volume of a
system increases by
an amount DV against
an external pressure,
P, they system pushes
back and does PV
work on the surr (w=PDV).
Heat Transfer and the Meaning of
Enthalpy Change, DH
As previously discussed, if no force is involved, then energy
flows spontaneously as heat from regions of higher T to
regions of lower T. Most rxns take place at constant
pressure. We also know DErxn = Ef - Ei = q + w, and
that w = -PDV.
At constant P,
DErxn = q + w
DErxn = qP - PDV
At constant P,
DHrxn = DErxn + PDV
substituting:
DHrxn = (qp - PDV) + PDV
DHrxn = qp
Therefore, Enthalpy of Rxn = Heat transfer DHrxn = qp
Heat Transfer and the Meaning
of Enthalpy Change, DH
If DH < 0 rxn is exothermic, if >0 rxn is
endothermic
For chemical reactions:
Hf = final enthalpy of products
Hi = initial enthalpy of reactants
DHrxn = Hproducts - Hreactants
Figure 6.1
Energy diagrams for the transfer of internal energy (E)
between a system and its surroundings.
When internal energy of a system
decreases, DE is lost to the
surroundings and is negative.
When internal energy of a system
increases, DE is gained by the
surroundings and is positive.
DE = Efinal - Einitial = Eproducts - Ereactants
Figure 6.2
A system transferring energy as heat only.
Ice water gains energy as heat, q,
from surr until Tsys = Tsurr; DE>0 so
q is pos.
Hot water transfers energy as
heat, q, to the surr until Tsys=Tsurr;
DE<0 so q is neg.
See notes in box below or in textbook.
Figure 6.3
A system losing energy as work only.
Zn(s) + 2H+(aq) + 2Cl-(aq)
Energy, E
DE<0
work done on
surroundings
H2(g) + Zn2+(aq) + 2Cl-(aq)
The internal energy of the system decreases as the reactants form products
because the H2(g) does work, w, on the surr by pushing back the piston. The
vessel is insulated, so q = 0. Here DE<0 and sign of w is positive.
Table 6.1 The Sign Conventions* for q, w and DE
(from perspective of the system)
q
+
w
=
DE
+
+
+
+
-
depends on sizes of q and w
-
+
depends on sizes of q and w
-
-
-
* For q: + means system gains heat; - means system loses heat.
* For w: + means work done on system; - means work done by system.
DEuniverse = DEsystem + DEsurroundings
Units of Energy
Joule (J)
1 J = 1 kg*m2/s2
Calorie (cal)
= energy to heat
exactly 1 g H2O
by exactly 1.00°C
at 14.5°C
1 cal = 4.184 J
1 kcal = 1000 cal = 1 Cal
(nutritional)
(British Thermal Unit 1 Btu = 1055 J)
Figure 6.5 in 4th ed.
only
Some interesting
quantities of energy.
Sample Problem 6.1
PROBLEM:
PLAN:
Determining the Change in Internal Energy of a
System
When gasoline burns in a car engine, the heat released causes
the products CO2 and H2O to expand, which pushes the pistons
outward. Excess heat is removed by the car’s cooling system.
If the expanding gases do 451 J of work on the pistons and the
system loses 325 J to the surroundings as heat, calculate the
change in energy (DE) in J, kJ, and kcal.
Define system and surroundings, assign signs to q and w and calculate
DE. The answer should be converted from J to kJ and then to kcal.
SOLUTION:
q = - 325 J
DE = q + w =
-776J
kJ
103J
w = - 451 J
-325 J + (-451 J) = -776 J
= -0.776kJ
-0.776kJ
kcal
4.184kJ
= -0.185 kcal
Figure 6.4
Two different paths for the energy change of a system.
The change in internal energy when a given amount of octane burns in air is
the same no matter how the energy is transferred. On left, fuel is burned in
open can and energy is lost almost entirely as heat. On right, fuel is burned
in car engine so a portion of energy is lost as work to move the car and less
energy is lost as heat.
Figure 6.5
Pressure-volume
work.
When the volume of a
system increases by
an amount DV against
an external pressure,
P, they system pushes
back and does PV
work on the surr (w=PDV).
Figure 6.6
Enthalpy diagrams for exothermic and endothermic processes.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
H2O(l)
CH4 + 2O2
H2O(g)
DH < 0
heat out
Enthalpy, H
Enthalpy, H
Hinitial
DH > 0
CO2 + 2H2O
H2O(l)
Hfinal
A
Exothermic process
B
H2O(g)
Hfinal
heat in
Hinitial
Endothermic process
Sample Problem 6.2
PROBLEM:
Drawing Enthalpy Diagrams and Determining
the Sign of DH
In each of the following cases, determine the sign of DH, state
whether the reaction is exothermic or endothermic, and draw an
enthalpy diagram. These are thermochemical equations!
(a) H2(g) + 1/2O2(g)
(b) 40.7 kJ + H2O(l)
PLAN:
H2O(l) + 285.8 kJ (DH = -285.8 kJ)
H2O(g)
(or DH = +40.7 kJ
Determine whether heat is a reactant or a product. As a reactant,
the products are at a higher energy and the reaction is
endothermic. The opposite is true for an exothermic reaction
SOLUTION: (a) The reaction is exothermic.
H2(g) + 1/2O2(g) (reactants)
EXOTHERMIC
H2O(l)
DH = -285.8kJ
(products)
(b) The process is endothermic.
H2O(g)
ENDOTHERMIC
H2O(l)
(products)
DH = +40.7kJ
(reactants)
Phase Changes: Latent Heat
THREE MAJOR EXPRESSIONS FOR CHANGE IN
STATE:
DHfus for change from solid to liquid (= -DHfreeze
from liquid to solid)
DHvap for change from liquid to gas (= -DHcond from
gas to liquid)
DHsubl for change from solid to gas (= -DHdep from
gas to solid)
There is no temperature change, so we call it a
latent heat transfer:
q = DHphase * quantity
Enthalpy of Phase Change
Substance
Benzene, C6H6
Bromine, Br2
Ethanol, CH3CH2OH
Mercury, Hg
Sodium Chloride
Water
DHfus
126 J/g
67.8 J/g
104 J/g
11.6 J/g
517 J/g
333 J/g
DHvap
395 J/g
187 J/g
854 J/g
296 J/g
3100 J/g
2257 J/g
Practice heat transfer
1. For 237 mL of ice at 0.00oC find the heat
required to melt it and then the heat required to
warm it to 25.0oC. Finally find the total of the
two heat transfers. Dice is 0.917 g/mL.
qphase = DHfus * quantity
= 333 J/g * 237 mL * 0.917g/mL
= 72,371 J
qwarm = m*cp*DT = 237 mL * 0.917 g/mL
*4.184J/goC*(25.0-0.00)oC = 22,730 J
qtotal = 9.51x104 J
Practice heat transfer now you
do it!
2. Find the total heat transfer energy
required to heat 1.00 mole of ice from
below the freezing point at -10.0oC to
above the boiling point as steam at
110.0oC. (That includes two phase
changes!) Given this data: cp(ice) = 37.7
J/moloC; cp(water) = 75.3 J/moloC; cp(steam) =
36.4 J/moloC; DHfus = 333 J/g; DHvap =
2257 J/g. (Total q ~ 55000 J)
TO BEGIN TO UNDERSTAND
CALORIMETRY:
Heat capacity, cp, of a substance = quantity of heat that
can be transferred through the substance per
quantity/per change in temperature
Cp = quantity of heat transferred
DT * gram or mole
Specific heat capacity is the heat energy required to
warm up 1.00 g of a substance by 1.00°C
Ex: Water’s specific heat is 4.184 J/g-degree
Glass’ specific heat is 0.80 J/g-degree
TO BEGIN TO UNDERSTAND
CALORIMETRY:
Specific heat capacity Cp is a physical
property of a substance
Cp can also be expressed as molar heat
capacity by multiplying by molar mass
For example, for water:
4.184 J/g-oC * 18.015 g/mol =75.37 J/mol-oC
Thermal energy
Thermal energy transferred is called sensible heat
transfer, q, uses specific heat capacity in this
heat transfer equation:
q = m*cp*(Tf - Ti)
The direction of heat transfer is determined by
which is warmer body: heat flows from hot to
cold
Heat has to be transferred from something to
something. The warmer body loses heat; the
colder body gains heat.
qlost = -qgained
Thermal energy example
You drink 1 cup of coffee, 250. mL, at 60.0°C & your body
is at 37.0°C. Assume 0.997 g/mL density, and water’s cp
can be used for both coffee and your body with 60.0 kg
mass. Heat is transferred from the coffee to your body.
What is your body’s final temperature?
q(coffee)= 249.25 g * 4.184 J/g-degree (Tf-60.0)
q(body)= 60.0 kg (103 g/kg) 4.184 J/g-degree (Tf – 37.0)
Set q(coffee) = -q(body)
249.25 g * 4.184 J/g-degree (Tf-60.0) =
-[60.0 kg (103 g/kg) 4.184 J/g-degree (Tf – 37.0)
Solve for Tf
Tf = 37.1°C in sig figs
Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and
Materials
Substance
Specific Heat
Capacity (J/g*K)
Elements
Substance
Specific Heat
Capacity (J/g*K)
Materials
aluminum, Al
0.900
wood
1.76
graphite,C
0.711
cement
0.88
iron, Fe
0.450
glass
0.84
copper, Cu
0.387
granite
0.79
gold, Au
0.129
steel
0.45
Compounds
water, H2O(l)
4.184
ethanol, C2H5OH(l)
2.46
ethylene glycol, (CH2OH)2(l)
2.42
carbon tetrachloride, CCl4(l)
0.864
Figure 6.7
Coffee-cup calorimeter.
Still used in school labs
to measure heat at
constant pressure, qp.
Surroundings actually
include water, cups,
thermometer, and
stirrer.
Sample Problem 6.4
Determining the Heat of a Reaction
PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at
25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C.
After stirring, the final temperature is 27.210C. Calculate qsoln (in J) and
DHrxn (in kJ/mol). (Assume the total volume is the sum of the individual
volumes and that the final solution has the same density and specfic
heat capacity as water: d = 1.00 g/mL and c = 4.184 J/g*K)
PLAN:
Find mass of solution: m = 75.0 mL * 1.00 g/mL = 75.0 g
Find DT = Tf – Ti = 27.21oC – 25.00oC = 2.21oC
qsol’n = m*cp*DT = 75.0 g * 4.184 J/g-deg * 2.21oC = 693 J
Sample Problem 6.4
Determining the Heat of a Reaction
continued
SOLUTION:
HCl(aq) + NaOH(aq)
H+(aq) + OH-(aq)
NaCl(aq) + H2O(l)
H2O(l)
For NaOH
0.500 M x 0.0500 L = 0.0250 mol OH-
For HCl
0.500 M x 0.0250 L = 0.0125 mol H+
HCl is the limiting reactant.
0.0125 mol of H2O will form during the rxn.
Find DHrxn = - qsol’n = -693 J (opposite of q from previous slide)
(-693 J/0.0125 mol H2O)(kJ/103 J) = -55.4 kJ/ mol H2O formed
CALORIMETRY:
The bomb calorimeter is used for combustible substances that are solid
or liquid
SYSTEM = SUBSTANCE + OXYGEN
DHcalor = qbomb + qwater
SURROUNDINGS = BOMB + WATER
DHrxn = - DHcalor
There is no DV, therefore qrxn = -(qbomb + qwater)
Note: bomb and water have different heat capacities even if change in
temperature is the same. Must know or determine heat capacity of
calorimeter
Figure 6.8
A bomb calorimeter
Combustion of octane in bomb
calorimeter:
1. Balanced eq'n*: C8H18(l) + 25/2 O2(g)  8 CO2(g) + 9 H2O(g)
Given this data: 1.00 g of octane, 1.20 kg water,
initial T of bomb + water is 25.00°C
final T is 33.20°C
Heat capacity of water is 4.184 J/g-deg
Heat capacity of calorimeter is determined to be 837 J/deg
2. Calculations
qbomb = Cbomb * DT = 837 J/deg * (33.20 - 25.00) = +6.86 x 103 J
qwater = Cw * m * DT = 4.184 J/g-deg * 1200 g * (33.20 - 25.00)
= +41.2 x 103 J
total q = -(qbomb + qw) = -(6.86 + 41.2) kJ
= -48.1 kJ/gram octane
3. Find molar DH: -48.1 kJ/gram * 114.2 g/mol = -5490 kJ/mol
Note: in thermochemistry the substance of interest has a coefficient of
one, and other coefficients can be fractions!
Practice
Work on problems 23, 25 and 32.
23: Calc q when 0.10 g ice is cooled from
10.0oC to -75.0oC, given cp of ice is 2.087
J/g-degree.
25: A 27.7 g sample of ethylene glycol loses
688 J of heat. Given Tf is 32.5oC and cp of
2.42 J/g-degree, determine Ti.
32: When 25.0 mL of 0.500 M H2SO4 is added to
25.0 mL of 1.00M KOH in a calorimeter at
23.50°C, the temperature rises to 30.17°C.
Figure 6.9
AMOUNT (mol)
of compound A
Summary of the relationship between
amount (mol) of substance and the heat
(kJ) transferred during a reaction.
AMOUNT (mol)
of compound B
molar ratio from
balanced equation
HEAT (kJ)
DHrxn (kJ/mol)
gained or lost
DH is function of quantity of reactants
based on mass or moles
Burning 441 g propane in backyard grill, how
much heat is transferred to surroundings,
including meat, air, grill, etc.?
1. Always have a balanced chemical equation:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
2. Look up DHcomb : -2220 kJ/mol propane
3. How many moles consumed?
441 g/44.09 g/mol = 10.0 mol
4. How much heat?
-2220 kJ/mol * 10.0 mol = -22,200 kJ
Writing a Thermochemical
Equation and an example
A thermochemical equation is a balanced
chemical equation that includes the
enthalpy of reaction.
Intro to Hess’ Law
Given the combustion reaction:
2 H2(g) + O2(g)  2 H2O(g) DHrxn = -484 kJ
If we cut in half the reaction in half to make the
formation reaction of one mol of H2O:
H2(g) + 1/2 O2(g)  H2O(g)
Then we have to cut the value of the first reaction
in half: DHrxn = - 242 kJ/mol H2O = DHf
If we do some multiple of a known reaction we
have to apply the same multiplier to the DHrxn
Intro to Hess’ Law
Given the vaporization of liquid water to steam:
H2O(l)  H2O(g) DHvap = 44 kJ/mol
Then what is DHcond ?
H2O(g)  H2O(l)
Reverse of vaporization is condensation:
DHcond = -44 kJ/mol = - DHvap
If a reaction is reversed the sign of DH is changed:
DHrev rxn = - DHfwd rxn
Intro to Hess’ Law
We can use the formation of water and the
condensation of steam data to determine for the
combustion reaction if the product is liquid
water.
Rxn 1: H2(g) + 1/2 O2(g)  H2O(g) DHrxn = - 242 kJ/mol
Rxn 2: H2O(g)  H2O(l) DHcond = -44 kJ/mol
Add the two reactions to obtain this reaction:
H2(g) + ½ O2(g)  H2O(l)
DHrxn = -242kJ + -44 kJ = -286 kJ/mol H2O(l)
Intro to Hess’ Law
HESS' LAW OF CONSTANT HEAT
SUMMATION:
If a reaction is the sum of 2 or more other
reactions, then DH for overall process
must be the sum of DH of constituent
reactions
Hess’ Law Example
Find DHrxn for 2 SO2(g) + O2(g) 2 SO3(g),
given the info for the following two rxns:
SO2(g)  S(s) + O2(g)
DHrxn = +297 kJ
2 S(s) + 3 O2(g)  2 SO3(g) DHrxn = -791 kJ
Double rxn 1 and add to rxn 2, cancel
common species. Sum of DH's is -197kJ
For practice see problems 45-48!
Enthalpy of Formation:
DHo is standard quantity: 1 atm pressure,
1.000 M, 298.15 K
DHfo is standard enthalpy of formation of
compound, ion, etc., at standard
conditions from the elements in their
standard states (recall diatomics, etc.)
Elements by definition have DHfo = 0.0
Enthalpy of Formation:
Another way to use Hess’ Law:
DHorxn = DHfo(products) - DHfo(reactants)
Example: Find DHorxn for combustion:
H2(g) + ½ O2(g)  H2O(l)
Use Reference to find DHfo for the reactants and
products.
DHorxn = [DHfo(H2O(l))] - [DHfo(H2(g)) + ½ DHfo(O2(g))]
= [-285.8 kJ/mol * 1mol] - [0.0 kJ/mol * 1 mol
+ 0.0 kJ/mol * ½ mol]
= -285.8 kJ
Table 6.3 Selected Standard Heats of Formation at 250C(298K)
Formula
calcium
Ca(s)
CaO(s)
CaCO3(s)
DH0f(kJ/mol)
0
-635.1
-1206.9
carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(l)
HCN(g)
CSs(l)
chlorine
Cl(g)
0
1.9
-110.5
-393.5
-74.9
-238.6
135
87.9
121.0
Formula DH0f(kJ/mol)
Formula DH0f(kJ/mol)
0
-92.3
silver
Ag(s)
AgCl(s)
hydrogen
H(g)
H2(g)
218
0
sodium
nitrogen
N2(g)
NH3(g)
NO(g)
0
-45.9
90.3
oxygen
O2(g)
O3(g)
H2O(g)
0
143
-241.8
H2O(l)
-285.8
Cl2(g)
HCl(g)
Na(s)
Na(g)
NaCl(s)
0
-127.0
0
107.8
-411.1
sulfur
S8(rhombic) 0
S8(monoclinic) 2
SO2(g)
-296.8
SO3(g)
-396.0
Example
Find DHorxn for the decomposition of calcium carbonate to
calcium oxide and carbon dioxide gas.
1. Write balanced chemical equation of formation and look
up heats of formation: CaCO3(s)  CaO(s) + CO2(g)
CaCO3(s) -1206.9 kJmol CaO(s) -635.1 kJ/mol CO2(g) -393.5 kJ/mol
2. Set up DHf in equation
DHorxn = DHfo(products) - DHfo(reactants)
[-635.1 kJ/mol*1mol + -393.5 kJ/mol*1mol] – (-1206.9
kJ/mol*1mol) = +178.8 kJ/mol
3. Write the finished thermochemical equation:
CaCO3(s)  CaO(s) + CO2(g)
DHrxn = + 178.8kJ/mol
For more practice, see problems 53-56.
Sample Problem 6.8
PROBLEM:
Writing Formation Equations
Write balanced equations for the formation of 1 mol of the following
compounds from their elements in their standard states and include
DH0f.
(a) Silver chloride, AgCl, a solid at standard conditions.
(b) Calcium carbonate, CaCO3, a solid at standard conditions.
(c) Hydrogen cyanide, HCN, a gas at standard conditions.
PLAN:
Use the table of heats of formation for values.
SOLUTION:
(a) Ag(s) + 1/2Cl2(g)
DH0f = -127.0 kJ
AgCl(s)
(b) Ca(s) + C(graphite) + 3/2O2(g)
(c) 1/2H2(g) + C(graphite) + 1/2N2(g)
CaCO3(s)
HCN(g)
DH0f = -1206.9 kJ
DH0f = 135 kJ
Figure 6.10
Elements
-DH0f
formation
Reactants
decomposition
Enthalpy, H
The general process for determining DH0rxn from DH0f values.
DH0f
Hinitial
DH0rxn
Products
DH0rxn = S mDH0f(products) - S nDH0f(reactants)
See notes below or in textbook.
Hfinal
Sample Problem 6.9
Calculating the Heat of Reaction from Heats of
Formation
Nitric acid, whose worldwide annual production is about 8 billion
kilograms, is used to make many products, including fertilizer, dyes,
and explosives. The first step in the industrial production process is
the oxidation of ammonia:
PROBLEM:
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
Calculate DH0rxn from DH0f values.
PLAN:
Look up the DH0f values and use Hess’s Law to find DHrxn.
SOLUTION:
DHrxn = S mDH0f (products) - S nDH0f (reactants)
DHrxn = [4(DH0f NO(g) + 6(DH0f H2O(g)] - [4(DH0f NH3(g) + 5(DH0f O2(g)]
= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]
DHrxn = -906 kJ
Practice
Work on problems 36, 45, 48, 52, and 54.
36. Given MgCO3(s)  MgO(s) + CO2(g) with
DHrxn = +117.3kJ.
a. Is heat absorbed or released?
b. *What is DHrxn when 5.50mol of the reactant
decompose?
You look up the rest!
Figure 9.16
Using bond energies to calculate DH0rxn.
Enthalpy, H
DH0rxn = DH0reactant bonds broken + DH0product bonds formed
BOND BREAKING
DH01 = + sum of BE
DH02 = - sum of BE
BOND FORMATION
DH0rxn
See notes below or in textbook.
Figure 9.17
Using bond energies to calculate DH0rxn of
methane.
BOND BREAKING
4BE(C-H)= +1652kJ
2BE(O2)= + 996kJ
DH0(bond breaking) = +2648kJ
BOND FORMATION
Enthalpy,H
2[-BE(C O)]= -1598kJ
4[-BE(O-H)]= -1868kJ
DH0(bond forming) = -3466kJ
DH0rxn= -818kJ
Treating the combustion of methane as a hypothetical two-step process (see Figure
9.16) means breaking all the bonds in the reactants and forming all the bonds in the
products.
SAMPLE PROBLEM 9.3
PROBLEM:
Calculating Enthalpy Changes from Bond
Energies
Use Table 9.2 (button at right) to calculate DH0rxn for the
following reaction:
CH4(g) + 3Cl2(g)
PLAN:
CHCl3(g) + 3HCl(g)
Write the Lewis structures for all reactants and products and
calculate the number of bonds broken and formed.
SOLUTION:
H
Cl
H C H
+
3
Cl
H
Cl
H C Cl
+
3 H
Cl
bonds broken
bonds formed
Cl
SAMPLE PROBLEM 9.3
Calculating Enthalpy Changes from Bond
Energies
continued
bonds broken
4 C-H
bonds formed
= 4 mol(413 kJ/mol) = 1652 kJ
3 C-Cl = 3 mol(-339 kJ/mol) = -1017 kJ
3 Cl-Cl = 3 mol(243 kJ/mol) = 729 kJ
1 C-H = 1 mol(-413 kJ/mol) = -413 kJ
DH0bonds broken = 2381 kJ
3 H-Cl = 3 mol(-427 kJ/mol) = -1281 kJ
DH0bonds formed = -2711 kJ
DH0reaction = DH0bonds broken + DH0bonds formed = 2381 kJ + (-2711 kJ) = - 330
kJ
This is just so you can see how it’s done – won’t be quiz or
test.
Practice
Do problem 37 in chp 9: Use the table of
bond energies to calculate DHrxn for:
CO2(g) + 2 NH3(g)  NH2CONH2(l) + H2O(l)
YOU MUST draw their Lewis structures and
count the number of bonds of each type!
Summary of symbols used in
Chp 6:
DHorxn - any reaction at standard
conditions
DHof - formation of compound from
elements, standard conditions
DHovap - vaporization of 1 mol at constant
P&T
DHofus - melting of 1 mol at constant P & T
DHocomb - combustion reaction