27 Oct. 2010
Take out Homework
 Objective: SWBAT calculate work done on
a system and energy change for a heat
transfer.
 Do now: Describe how you would make
1.00 L of a 0.8 M LiCl solution from solid
LiCl.

Agenda
Do now
II. Homework answers
III. Thermochemistry notes and examples
Part 1
 Homework: p. 263 #13, 15, 17, 19 (ELS)
 Read p. 239-245
I.
Thermochemistry

Thermodynamics: study of
interconversion of heat and other kinds of
energy
 First Law of Thermodynamics: energy
can be converted from one form to another,
but can not be created or destroyed
 We can not accurately measure total energy
of a system
 Instead, we measure changes in energy ΔE



ΔEsys + ΔEsurr = 0
ΔEsys=- ΔEsurr
 energy gained in one place must be lost
somewhere else
In chem, we are usually interested in just the
ΔE to the system we are studying:
 ΔE=q+w
 q=heat exchange between systems
 w=work done on or by the system
Sign Conventions: heat and work
Process
Heat absorbed by the
system (endothermic)
Heat absorbed by the
surroundings (exothermic)
Work done by the system
on surroundings
Work done on the system
by the surroundings
Sign
+q
-q
-w
+w
Exothermic process (-q) is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process (+q) is any process in which heat has to
be supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
Work, Pressure, and Volume
w   PV
Expansion
+V (increase)
Compression
-V (decrease)
+w results
-w results
E
system
decreases
Work has been done by
the system on the
surroundings
E
system
increases
Work has been done on
the system by the
surroundings
Ex 1
A gas expands in volume from 2.0 L to 6.0 L
at a constant temperature. Calculate the
work done by the gas if it expands
a) against a vacuum
b) against a constant pressure of 1.2 atm
1 L·atm = 101.3 J
Joule = unit of heat
Ex 2
A gas expands from 264 mL to 971 mL at
constant temperature. Calculate the work
done (in J) by the gas if it expands
a) against a vacuum
b) against a constant pressure of 4.00 atm

Ex 3
The work done when gas is compressed in a
cylinder is 462 J. During this process,
there is a heat transfer of 128 J from the
gas to the surroundings. Determine the
signs for q and w. Calculate the energy
change for this process.
ΔE=q+w
Homework
p. 263 #13, 15, 17, 19 (ELS)
 Read p. 239-245

28 Oct. 2010
Objective: SWBAT calculate heat change to a
system given a thermochemical equation.
Do now: The work done when gas is
compressed in a cylinder is 462 J. During
this process, there is a heat transfer of 128 J
from the gas to the surroundings. Determine
the signs for q and w. Calculate the energy
change for this process.
ΔE=q+w
Agenda
Do now
II. Homework check
III. Thermochemistry part 2
Homework: p. 263 #24, 25, 27 (JMS)
Read p. 245-252, read lab handout and do
pre-lab questions
I.

A gas expands and does P-V work on the
surroundings equal to 279 J. At the same
time, it absorbs 216 J of heat from the
surroundings. What is the change in
energy of the system?
Enthalpy of Chemical Reactions
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
H = E + PV
H = H (products) – H (reactants)
H = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
H < 0
Hproducts > Hreactants
H > 0
6.4
Thermochemical Equations
Is H negative or positive?
System absorbs heat
Endothermic
H > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at
00C and 1 atm.
H2O (s)
H2O (l)
H = 6.01 kJ
Thermochemical Equations
Is H negative or positive?
System gives off heat
Exothermic
H < 0
890.4 kJ are released for every 1 mole of methane that is
combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) H = -890.4 kJ
Thermochemical Equations
• The stoichiometric coefficients always refer to the
number of moles of a substance
H2O (s)
H2O (l) H = 6.01 kJ/mol
ΔH = 6.01 kJ
• If you reverse a reaction, the sign of H changes
H2O (l)
H2O (s) H = -6.01 kJ
• If you multiply both sides of the equation by a factor n,
then H must change by the same factor n.
2H2O (s)
2H2O (l)
H = 2 mol x 6.01 kJ/mol = 12.0 kJ
The physical states of all reactants and products must be specified in
thermochemical equations.
Example 1
Given the thermochemical equation
2SO2(g) + O2(g)  2SO3(g) H  198.2kJ/mol
calculate the heat evolved when 87.9 g of SO2 is
converted to SO3.

Example 2
Calculate the heat evolved with 266 g of white
phosphorus (P4) burns in air:
P4(s) + 5O2(g)  P4O10(s) H 3013kJ/mol

Homework
p. 263 #24, 25, 26
 Read LAB, DO prelab, read in book….

1 Nov. 2010
Take out pre-lab questions
 Objective: SWBAT calculate the amount
of heat supplied to the surroundings (q)
by a reusable chemical hand warmer.
 Do now: For the chemical hand warmer,
predict the sign convention for heat (q)
and work (w).

Lab
With your partner, carefully follow the
directions.
Hand warmer pouch and metal disk: 3.42
grams
2. Collect all data, results and answers to
questions in your lab notebook.
3. Lab notebooks collected tomorrow: 2 lab
grades.
1.
2 Nov. 2010
Objective: SWBAT relate ΔE to ΔH and
complete calorimetric calculations.
 Do now: Type your second trial results
into the spreadsheet.
Homework: Lab notebooks due tomorrow
– 2 lab grades
p. 263 #27, 28, 31, 32, 34, 36 (JR)

Comparing ΔH and ΔE
ΔE = ΔH – RTΔn
 R = 8.314J/K·mol
 T = temperature in Kelvin
 n = number of moles
 For gas reactions, the change in volume
can be very large because hot gases
expand.
 For reactions not involving gas, ΔH and
ΔE are often very similar.

Ex 1
Calculate the change in internal energy when 2
moles of CO are converted to 2 moles of CO2 at
1 atm and 25oC
2CO(g) + O2(g)  2CO2(g) ΔH = -566.0 kJ/mol

Ex 2
What is ΔE for the formation of 1 mole of CO
at 1 atm and 25oC?
C(graphite) + ½O2(g)  CO(g) ΔH = -110.5 kJ/mol

Calorimetry
the measurement of heat changes
 q=CΔt
 C = heat capacity of a substance: the
amount of heat required to raise the
temperature of a given quantity of a
substance by 1 degree Celsius. (=m*s)
 q=msΔt
Ex. 1: A 466 g sample of water is heated from
8.50oC to 74.60oC. Calculate the amount of
heat absorbed (in kJ) by the water.

Ex. 2

1.435 grams of naphthalene (C10H8) was
burned in a constant-volume bomb
calorimeter. The temperature of the water
rose from 20.28oC to 25.95oC. If the heat
capacity of the bomb plus water was 10.17
kJ/oC, calculate the heat of combustion of
naphthalene on a molar basis (find the
molar heat of combustion).
Ex. 3

A lead pellet having a mass of 26.47 g at
89.98oC was placed in a constant pressure
calorimeter of negligible heat capacity
containing 100.0 mL of water. The water
temperature rose from 22.50oC to 23.17oC.
What is the specific heat of the lead pellet?
Homework
Lab notebooks due tomorrow – 2 lab grades
p. 263 #27, 28, 31, 32, 34, 36 (JR)
3 Nov. 2010
Take out Homework and Lab notebook
 Objective: SWBAT calculate the standard
enthalpy of formation and reaction for
compounds.
 Do now: An iron bar of mass 869 g cools
from 94oC to 5oC. Calculate the heat
released (in kJ) by the metal.
specific heat of iron=0.444 J/g·oC

Agenda
Do now
II. Homework check
III. Standard Enthalpy of Formation/Reaction
notes/problems
Homework: p. 264 #42, 43, 46, 51, 53, 58, 63
(SR)
Read p. 258-261: Heats of solution/dilution
Tomorrow: Practice problems/review
Bring your book to class!
Test Monday (ch. 6)
I.
Announcement
Tutor needed for 10th grade chem student.
 You will earn community service hours.
 Interested? See Ms. Kern.

Standard Enthalpy of Formation and
Reaction
0) is the
Standard enthalpy of formation (H
f
heat change that results when one mole of a
compound is formed from its elements at a
pressure of 1 atm.
The standard enthalpy of formation of any
element in its most stable form is zero.
Hf0 (O2) = 0
Hf0 (O3) = 142 kJ/mol
Hf0 (C, graphite) = 0
Hf0 (C, diamond) = 1.90 kJ/mol
0 ) is the enthalpy
The standard enthalpy of reaction (Hrxn
of a reaction carried out at 1 atm.
aA + bB
cC + dD
0
Hrxn
= [cH0f (C) + dH0f (D) ] - [aH0f (A) + bH0f (B) ]
0
Hrxn
= SH0 f(products) - SH0f (reactants)
Hess’s Law: When reactants are converted to products,
the change in enthalpy is the same whether the reaction
takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you
get there, only where you start and end.)
Benzene (C6H6) burns in air to produce carbon
dioxide and liquid water. How much heat is
released per mole of benzene combusted? The
standard enthalpy of formation of benzene is
49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid
water. How much heat is released per mole of benzene
combusted? The standard enthalpy of formation of benzene is
49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
H0 = S H0 (products)
rxn
f
- S H0 (reactants)
f
H0 = [ 12H0 (CO2) + 6H0 (H2O)] - [ 2H0 (C6H6)]
rxn
f
f
f
H0 = [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ
rxn
-6535 kJ
= - 3267 kJ/mol C6H6
2 mol
Practice Problem
Calculate the heat of combustion for the
following reaction:
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
The standard enthalpy of formation of
ethylene (C2H4) is 52.3 kJ/mol.
Then, calculate per mole of ethylene
combusted.

What if the reaction is not a single
step?
Calculate the standard enthalpy of formation of CS2 (l)
given that:
0 = -393.5 kJ
C(graphite) + O2 (g)
CO2 (g) H
rxn
S(rhombic) + O2 (g)
SO2 (g) H
rxn0 = -296.1 kJ
CS2(l) + 3O2 (g)
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
H
rxn
Calculate the standard enthalpy of formation of CS2 (l) given
that:
C(graphite) + O2 (g)
CO2 (g) rxnH0 = -393.5 kJ
0
H
rxn = -296.1 kJ
0 = -1072 kJ
CS2(l) + 3O2 (g)
CO2 (g) + 2SO2 (g) Hrxn
1. Write the enthalpy of formation reaction for CS2
S(rhombic) + O2 (g)
SO2 (g)
2. Add the given rxns so that the result is the desired rxn.
Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g)
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
H0 = -393.5rxn
kJ
CO2 (g)
H0 = -296.1rxn
kJ
SO2 (g)
CO2 (g) + 2SO2 (g)
H0 = -1072 kJ rxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
CO2 (g)
2S(rhombic) + 2O2 (g)
+
CO2(g) + 2SO2 (g)
H0 = -393.5rxn
kJ
2SO2 (g)
H0 = -296.1x2rxn
kJ
CS2 (l) + 3O2 (g)
C(graphite) + 2S(rhombic)
H0 = +1072 kJ rxn
CS2 (l)
H0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
rxn
Calculate the standard enthalpy of formation of
acetylene (C2H2) from its elements:
2C(graphite) + H2(g)  C2H2(g)

Here are the equations for each step:
C(graphite) + O2(g)  CO2(g) H0=-393.5 kJ/mol
H2(g) + 1/2O2(g)  H2O(l) H0=-285.8 kJ/mol
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l)
H0= -2598.8 kJ/mol
Homework
p. 264 #42, 43, 46, 51, 53, 58, 63
 Bring your book to class!

4 Nov. 2010

Take out homework: p. 264 #42, 43, 46, 51, 53, 58, 63
Objective: SWBAT review thermochemistry
for a test on Monday!
 Do now: Calculate the standard enthalpy of
formation of acetylene (C2H2) from its
elements:
2C(graphite) + H2(g)  C2H2(g)
Here are the equations for each step:

C(graphite) + O2(g)  CO2(g) H0=-393.5 kJ/mol
H2(g) + 1/2O2(g)  H2O(l) H0=-285.8 kJ/mol
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l)
H0= -2598.8 kJ/mol
Agenda
Do now
II. Homework
III. Chapter 6 Problem Set Work Time
Homework: Finish problem set: Mon.
Test: Mon.
Bonus: Tues.
I.
Chapter 6 Problem Set
Due Mon.
 p. 263 #16, 20, 33, 37, 38, 52, 54b, 57, 64,
70, 73, 80
 Please show all your work!
 BONUS: p. 269 #119 Tues. (Up to +5% on
your quarter grade!)


Read p. 258-261: Heats of solution and
dilution, on your own
8 Nov. 2010
Hand in Ch. 6 Problem Set
 Objective: SWBAT show what you know
about thermochemistry on a test!
 Do Now: Review the sign conventions for q
and w.
 A reaction absorbs heat from its
surroundings and has work done on it by
the surroundings. Is it endothermic or
exothermic? What are the signs for q and
w?

Agenda
Collect ch. 6 problem set
II. Questions?
III. Thermodynamics Test
Homework: Bonus problem due tomorrow!
p. 269 #119
Read p. 801-814
p. 829#1-5
I.
Thermodynamics Part 2

1st Law of Thermodynamics: energy can be
converted from one form to another, but
can not be created or destroyed.
 How to measure these changes?
 ∆H: Change in enthalpy: amount of heat
given off or absorbed by a system

2nd Law of Thermodynamics: explains why
chemical processes tend to favor one
direction
Spontaneous Processes

Chemists need to be able to predict
whether or not a reaction will occur when
reactants are brought together under a
specific set of conditions (temperature,
pressure, concentration, etc.)
Spontaneous Physical and Chemical Processes
• A waterfall runs downhill
• A lump of sugar dissolves in a cup of coffee
• At 1 atm, water freezes below 0 0C and ice melts
above 0 0C
• Heat flows from a hotter object to a colder object
• A gas expands in an evacuated bulb
spontaneous
• Iron exposed to oxygen and water forms rust
nonspontaneous
•Processes that occur spontaneously in one direction
can not, under the same conditions, also take place
spontaneously in the opposite direction!
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a
reaction proceeds spontaneously?
Spontaneous reactions
CH4 (g) + 2O2 (g)
H+ (aq) + OH- (aq)
H2O (s)
CO2 (g) + 2H2O (l) ∆H0 = -890.4 kJ/mol
H2O (l) ∆H0 = -56.2 kJ/mol
H2O (l) ∆H0 = 6.01 kJ/mol
NH4NO3 (s) H2O NH4+(aq) + NO3- (aq) ∆H0 = 25 kJ/mol




Does a decrease in enthalpy mean a reaction
proceeds spontaneously?
No!
Being exothermic (“exothermicity”) favors
spontaneity, but does not guarantee it!
 An endothermic reaction can be
spontaneous.
 Not all exothermic reactions are
spontaneous.
So how do we predict if a reaction is
spontaneous under given conditions if ∆H
doesn't help us?!
Entropy (S) is a measure of the randomness or
disorder of a system.
order
S
disorder
S
If the change from initial to final results in an
increase in randomness, ∆S > 0
For any substance, the solid state is more ordered
than the liquid state and the liquid state is more
ordered than gas state Ssolid < Sliquid << Sgas
H2O (s)
H2O (l) ∆S > 0
Processes that lead to an increase in entropy (∆S > 0)
63
Example: Br2(l)
∆S > 0
Br2(g)
Example: I2(s)
∆S > 0
I2(g)
Entropy
State functions are properties that are determined by the
state of the system, regardless of how that condition was
achieved.
Examples:
energy, enthalpy, pressure, volume, temperature, entropy
Review
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
66
How does the entropy of a system change for each of the
following processes?
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the
following processes?
(a) Condensing water vapor
Randomness decreases
Entropy decreases (∆S < 0)
(b) Forming sucrose crystals from a supersaturated solution
Randomness decreases
Entropy decreases (∆S < 0)
(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S > 0)
(d) Subliming dry ice
Randomness increases
Entropy increases (∆S > 0)
Practice

Predict whether the entropy change is
greater or less than zero for each of the
following processes:
a. freezing ethanol
b. evaporating a beaker of liquid bromine
at room temperature
c. dissolving glucose in water
d. cooling nitrogen gas from 80oC to 20oC
First Law of Thermodynamics
Energy can be converted from one form to another
but energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a
spontaneous process and remains unchanged in an
equilibrium process.
Spontaneous process:
∆Suniv = ∆Ssys + ∆Ssurr > 0
Equilibrium process:
∆Suniv = ∆Ssys + ∆Ssurr = 0
Entropy Changes in the System (∆Ssys)
The standard entropy of reaction (∆S0rxn) is the entropy
change for a reaction carried out at 1 atm and 250C.
aA + bB
cC + dD
∆S0rxn= [ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ]
∆S0 = Σ nS0(products) - Σ mS0(reactants)
rxn
What is the standard entropy change for the following
reaction at 250C? 2CO (g) + O2 (g)
2CO2 (g)
S0(CO) = 197.9 J/K•mol
S0(O2) = 205.0 J/K•mol
S0(CO2) = 213.6 J/K•mol
Entropy Changes in the System (∆Ssys)
The standard entropy of reaction (∆S0rxn) is the entropy
change for a reaction carried out at 1 atm and 250C.
aA + bB
∆S0rxn=
cC + dD
[ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ]
∆S0 = Σ nS0(products) - Σ mS0(reactants)
rxn
What is the standard entropy change for the following
reaction at 250C? 2CO (g) + O2 (g)
2CO2 (g)
S0(CO) = 197.9 J/K•mol
S0(O2) = 205.0 J/K•mol
S0(CO2) = 213.6 J/K•mol
DS0 = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
rxn
DS0 = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol
rxn
Practice

From the standard entropy values in
Appendix 3, calculate the standard
entropy changes for the following
reactions at 25oC:
a. CaCO3(s)
CaO(s) + CO2(g)
b. N2(g) + 3H2(g)
2NH3(g)
c. H2(g) + Cl2(g)
2HCl(g)
Entropy Changes in the System (∆Ssys)
When gases are produced (or consumed)
•
If a reaction produces more gas molecules than it
consumes, ∆S0 > 0.
•
If the total number of gas molecules diminishes,
∆S0 < 0.
•
If there is no net change in the total number of gas
molecules, then ∆S0 may be positive or negative
BUT ∆S0 will be a small number.
What is the sign of the entropy change for the following
reaction? 2Zn (s) + O2 (g)
2ZnO (s)
The total number of gas molecules goes down, ∆S is negative.
Practice

Predict (not calculate) whether the entropy
change of the system in each of the
following reactions is positive or negative:
a. 2H2(g) + O2(g)
2H2O(l)
b. NH4Cl(s)
NH3(g) + HCl(g)
c. H2(g) + Br2(g)
2HBr(g)
d. I2(s)
2I(g)
e. 2Zn(s) + O2(g)
2ZnO(s)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process
∆Ssurr > 0
Endothermic Process
∆Ssurr < 0
Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the absolute zero
of temperature.
S = k ln W
W=1
S=0
77
Gibbs Free Energy
Spontaneous process:
∆Suniv = ∆Ssys + ∆Ssurr > 0
Equilibrium process:
∆Suniv = ∆Ssys + ∆Ssurr = 0
For a constant-temperature process:
Gibbs free
energy (G)
∆G = ∆Hsys -T ∆Ssys
∆G < 0
The reaction is spontaneous in the forward direction.
∆G > 0
The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
∆G = 0
The reaction is at equilibrium.
The standard free-energy of reaction (∆G0rxn) is the
free-energy change for a reaction when it occurs under
standard-state conditions.
aA + bB
cC + dD
∆G0 (A) + b ∆G0 (B) ]
∆G0 = [c ∆G0 (C) +d ∆G0 (D) ] - a
[
rxn
f
f
f
f
∆G0 = S n ∆G0 (products)- S m ∆G0 (reactants)
rxn
f
f
Standard free energy of formation (∆G0f ) is the freeenergy change that occurs when 1 mole of the compound is
formed from its elements in their standard states.
∆G0 of any element in its stable form is zero.
f
What is the standard free-energy change for the following
reaction at 25 0C?
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following
reaction at 25 0C?
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DG0 = S nDG0 (products)- S mDG0 (reactants)
rxn
f
f
DG0 = [12DG0 (CO2) +6DG0 (H2O)] - [2DG0 (C6H6)]
rxn
f
f
f
DG0 = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ/mol
rxn
Is the reaction spontaneous at 25 0C?
DG0 = -6405 kJ/mol< 0
spontaneous
Practice

Calculate the standard free-energy
changes for the following reactions at
25oC
a. CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
b. 2MgO(s)
2Mg(s) + O2(g)
c. H2(g) + Br2(l)
2HBr(g)
∆G = ∆H - T ∆S
Recap: Signs of Thermodynamic Values
Negative
Enthalpy (ΔH) Exothermic
Positive
Endothermic
Entropy (ΔS)
Less disorder More disorder
Gibbs Free
Energy (ΔG)
Spontaneous Not
spontaneous
Gibbs Free Energy and Chemical Equilibrium
G = G0 + RT lnQ
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Q is the reaction quotient
At Equilibrium
G = 0
Q=K
0 = G0 + RT lnK
G0 =  RT lnK
18.6
G0 =  RT lnK
18.6
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q)
required to raise the temperature of one gram of the substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat (q) required to raise the
temperature of a given quantity (m) of the substance by one degree Celsius.
C = ms
Heat (q) absorbed or released:
q = mst
q = Ct
t = tfinal - tinitial
6.5
How much heat is given off when an 869 g iron bar cools from 940C to 50C?
s of Fe = 0.444 J/g • 0C
t = tfinal – tinitial = 50C – 940C = -890C
q = mst
= 869 g x 0.444 J/g • 0C x –890C
= -34,000 J
6.5
Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = mst
qcal = Ccalt
Reaction at Constant P
H = qrxn
No heat enters or leaves!
6.5
6.5
Phase Changes
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is
equal to the external pressure.
The normal boiling point is the temperature at which a liquid boils when the external
pressure is 1 atm.
11.8
The critical temperature (Tc) is the temperature above which the gas cannot be made to
liquefy, no matter how great the applied pressure.
The critical pressure (Pc) is the
minimum pressure that must be
applied to bring about
liquefaction at the critical
temperature.
11.8
Can you find…
Where’s Waldo?
The Triple Point?
Critical pressure?
Critical temperature?
Where fusion occurs?
Where vaporization occurs?
Melting point
(at 1 atm)?
Boiling point
(at 6 atm)?
Carbon Dioxide
H2O (s)
H2O (l)
Freezing
Melting
The melting point of a solid or the freezing
point of a liquid is the temperature at
which the solid and liquid phases coexist in
equilibrium
11.8
Molar heat of sublimation (Hsub) is the
energy required to sublime 1 mole of a
solid.
Deposition
H2O (g)
Sublimation
H2O (s)
Hsub = Hfus + Hvap
( Hess’s Law)
11.8
Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance.
11.8
11.8
Sample Problem

How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C?
Step 1: Heat the ice
Q=mcΔT
Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ
Step 2: Convert the solid to liquid
ΔH fusion
Q = 2.0 mol x 6.01 kJ/mol = 12 kJ
Step 3: Heat the liquid
Q=mcΔT
Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem

How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C?
Step 4: Convert the liquid to gas
Q = 2.0 mol x 44.01 kJ/mol =
ΔH vaporization
88 kJ
Step 5: Heat the gas
Q=mcΔT
Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ
Now, add all the steps together
0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ
= 118 kJ