27 Oct. 2010 - PHA Science

Report
1 November 2011
Objective: You will be able to:
 describe enthalpy and differentiate
between endothermic and exothermic
reactions and calculate the enthalpy of a
reaction.
 Homework: p. 263 #21, 22, 23, 24, 25:
tomorrow

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
Exothermic process 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 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)
Enthalpy of Chemical Reactions
Specifically, we’re interested in heat flow that
occurs under constant pressure
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
Enthalpy change (∆H) is the heat absorbed during a physical
or chemical process.
DH = H (products) – H (reactants)
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0
6.4
Enthalpies of Reaction




Expressed in kJ or kJ/mol
endothermic: ∆H is always positive
 endothermic changes absorb heat
exothermic: ∆H is always negative
 exothermic changes liberate heat
Calculating ∆H is the same for a wide variety
of processes (see table)
Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at
00C and 1 atm.
H2O (s)
H2O (l)
DHfus = 6.01 kJ
Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 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) DHcomb = -890.4 kJ
Thermochemical Equations
• The stoichiometric coefficients always refer to the
number of moles of a substance
H2O (s)
H2O (l) DH = 6.01 kJ/mol
ΔH = 6.01 kJ
• If you reverse a reaction, the sign of DH changes
H2O (l)
H2O (s) DH = -6.01 kJ
• If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 2 mol x 6.01 kJ/mol = 12.0 kJ
The physical states of all reactants and products must be specified in
thermochemical equations.
Note on “implied signs”
In a sentence, the sign of ∆H is often
implied.
 “6.9 kJ of heat was liberated from the
system” = −6.9 kJ
 “6.9 kJ of heat was absorbed by the
system” = +6.9 kJ

Phase transitions


solid to liquid; liquid to gas
 absorb heat
 endothermic
 positive ∆H
 H2O(l) → H2O(g) ∆Hvap= +44 kJ
gas to liquid; liquid to solid
 liberate heat
 exothermic
 negative ∆H
 H2O(g) → H2O(l) ∆Hvap= −44 kJ
Example 1
Given the thermochemical equation
2SO2(g) + O2(g)  2SO3(g) DH = -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) DH= -3013kJ/mol

Example 3

How many kJ of heat are absorbed by the
surroundings when 25.0 g of methane
(CH4) burns in air?
Problem
Determine the amount of heat (in kJ) given off
when 1.26x104 g of NO2 are produced
according to the equation
2NO(g) + O2(g) → 2NO2(g) ∆H = −114.6 kJ/mol

What scientific law requires that the
magnitude of the heat change for forward
and reverse processes be the same with
opposite signs? Explain.
Why are phase changes from solid to
liquid and from liquid to gas always
endothermic?
 Is the process of sublimation endothermic
or exothermic? Explain.

2 November 2011
Objective: You will be able to:
 describe calorimetry and calculate heat
change and specific heat.
 Homework Quiz: Week of Oct. 31
Determine the amount of heat (in kJ) given off
when 1.26x104 g of NO2 are produced
according to the equation
2NO(g) + O2(g) → 2NO2(g) ∆H = −114.6 kJ/mol

Agenda
Homework Quiz
II. Go over homework
III. Calorimetry calculations examples
IV. Practice Problems
V. Pre-lab questions and lab set up
Homework: Finish pre-lab questions, lab
notebook set up
p. #
I.
Calorimetry




Calorimetry: measurement of heat flow
Calorimeter: an apparatus that measures
heat flow
Heat capacity, C, heat required to raise the
temperature of an object by 1 K (units are
J/K)
Molar heat capacity, Cmolar: amount of heat
absorbed by one mole of a substance when it
experiences a one degree temperature change
(units are J/mol K)

Specific heat capacity: the heat capacity of
one gram of a substance (units: J/g K)
 water: 1 cal/g K = 4.184 J/ gK
Calorimetry
q=CΔt
 q = heat absorbed in Joules
 C = specific heat capacity of a substance: the
amount of heat required to raise the
temperature of a given quantity of a
substance by 1 degree. (=mass*specific heat)
 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?
Ex. 4
What is the molar heat of combustion of
liquid ethanol if the combustion of 9.03
grams of ethanol causes a calorimeter to
increase in temperature by 3.19 K?
 heat capacity of the calorimeter is 75.8
kJ/K

Pre-Lab
Molar heat of crystallization
 a.k.a. Latent heat of fusion
 Set up your lab notebook and answer the
pre-lab questions (refer to your textbook!)
 Make a procedure summary and set up a
data table.

Homework
3 November 2011
Objective: You will be able to:
 calculate the molar heat of crystallization
for a chemical handwarmer
 Homework Quiz: A 10.0 gram piece of
copper (specific heat = 0.385 J/g oC) has
been heated to 100oC. It is then added to a
sample of water at 22.0oC in a calorimeter. If
the final temperature of the water is 28.0oC,
calculate the mass of the water in the
calorimeter.
specific heat of water = 4.184 J/ g oC

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.
1.
7 November 2011
Objective: You will be able to
 calculate the standard enthalpy of
formation and reaction for compounds.
 Homework Quiz: (Week of Nov. 7)
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
Homework Quiz
II. Return Tests
III. Standard Enthalpy of Formation/Reaction
notes/problems
 Homework: p. 264 #46, 49, 51, 53, 56, 57,
59, 62, 64: Tuesday
 Test on Thermochem Thurs.
 Lab notebook: due tomorrow (checklist!)
 Test Corrections: Mon.
I.
8 November 2011
Objective: You will be able to
 review thermochemistry
 Homework Quiz: (Week of Nov. 7)
Calculate the heat of combustion for the
following reaction from the standard
enthalpies of formation in Appendix 3.

2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
Agenda
Homework quiz
II. Collect lab notebooks
III. Homework answers and more enthalpy
problems
IV. Problem set work time
 Homework: Problem Set: Thurs.
Test on Thermochem Thurs.
Test Corrections: Mon.
I.
9 November 2022
Objective: You will be able to:
 practice thermochemistry for a test.
 Homework Quiz:

Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0 = -393.5 kJ
S(rhombic) + O2 (g)
SO2 (g) DH0 = -296.1 kJ
CS2(l) + 3O2 (g)
CO2 (g) + 2SO2 (g)
DH0 = -1072 kJ
Agenda
Homework Quiz
 Problem Set work time
Homework: Problem set due tomorrow
Thermochemistry test tomorrow

Test Corrections
Required. 2nd quarter quiz grade
 Show work, including for multiple choice
questions.
 On a separate sheet of paper or using a
different color pen.
 Due Mon.

Standard Enthalpy of Formation
and Reaction
Formation Reaction
a reaction that produces one mole of a
substance from its constituent elements in
their most stable thermodynamic state
½H2(g) + ½I2(s) → HI(g) ∆H = +25.94 KJ

Standard heat of formation (DH0f ) is the heat
absorbed when one mole of a compound is
formed from its elements at a pressure of 1 atm at
25oC.
The standard heat of formation of any element
in its most stable form is zero.
DH0f (O2) = 0
DHf0 (O3) = 142 kJ/mol
DHf0 (C, graphite) = 0
DHf0 (C, diamond) = 1.90 kJ/mol
Example

Write the thermochemical equation
associated with the standard heat of
formation of AlCl3(s) using the values in
the table. (Hint: Be sure to make only one
mole of product!)

Based on the definition of a formation
reaction, explain why the standard heat of
formation of an element in its most stable
thermodynamic state is zero. Write a
chemical equation to illustrate your
answer.
Hess’s Law
Hess’s Law: if a reaction is carried out in a series of
steps, DH of the overall reaction is equal to the sum of the
DH’s for each individual step.
(Enthalpy is a state function. It doesn’t matter how you
get there, only where you start and end.)
Example
Calculate DH for the following reaction
2S(s) + 3O2(g) → 2SO3(g)
from the enthalpies of these related
reactions
S(s) + O2(g) → SO2(g)
DH = −296.9 kJ
2SO2(g) + O2(g) → 2SO3(g) DH = −196.6 kJ

Standard Heat of Reaction
0 ) is the heat change
The standard heat of reaction (DHrxn
of a reaction carried out at 1 atm.
aA + bB
cC + dD
0
DHrxn
= [cDH0f (C) + dDH0f (D) ] - [aDH0f (A) + bDH0f (B) ]
0
DHrxn
= SDH0f (products) - SDH0f (reactants)
Example 1
Calculate the standard enthalpy change for
the combustion of one mole of liquid
ethanol.
(Hint: Be really careful with your math, and
the + and − signs!)
Example 2
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)
DH0 = S DH0 (products)
rxn
f
- S DH0 (reactants)
f
DH0 = [ 12DH0 (CO2) + 6DH0 (H2O)] - [ 2DH0 (C6H6)]
rxn
f
f
f
DH0 = [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ
rxn
-6535 kJ
= - 3267 kJ/mol C6H6
2 mol
Practice Problem 1
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.

Problem 2
The thermite reaction involves solid
aluminum and solid iron (III) oxide
reacting in a single displacement reaction
to produce solid aluminum oxide and
liquid iron.
 This reaction is highly exothermic and the
liquid iron formed is used to weld metals.
Calculate the heat released in kJ/g of Al.
 ∆Hof for Fe(l) is 12.40 kJ/mol.

What if the reaction is not a single
step? Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l)
given that:
0 = -393.5 kJ
C(graphite) + O2 (g)
CO2 (g) DH
rxn
S(rhombic) + O2 (g)
SO2 (g) DH
rxn0 = -296.1 kJ
CS2(l) + 3O2 (g)
CO2 (g) + 2SO2 (g)
0 = -1072 k
DH
rxn
Calculate the standard enthalpy of formation of CS2 (l) given
that:
C(graphite) + O2 (g)
CO2 (g) rxnDH0 = -393.5 kJ
0
DH
rxn = -296.1 kJ
0 = -1072 kJ
CS2(l) + 3O2 (g)
CO2 (g) + 2SO2 (g) DHrxn
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)
DH0 = -393.5rxn
kJ
CO2 (g)
DH0 = -296.1rxn
kJ
SO2 (g)
CO2 (g) + 2SO2 (g)
DH0 = -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)
DH0 = -393.5rxn
kJ
2SO2 (g)
DH0 = -296.1x2rxn
kJ
CS2 (l) + 3O2 (g)
C(graphite) + 2S(rhombic)
DH0 = +1072 kJ rxn
CS2 (l)
DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
rxn
Practice Problem 4
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) DH0=-393.5 kJ/mol
H2(g) + 1/2O2(g)  H2O(l) DH0=-285.8 kJ/mol
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l)
DH0= -2598.8 kJ/mol
Homework

p. 264 #46, 49, 51, 53, 56, 57, 59, 62, 64:
Tuesday
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) DH0=-393.5 kJ/mol
H2(g) + 1/2O2(g)  H2O(l) DH0=-285.8 kJ/mol
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l)
DH0= -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.
9 Nov. 2010
Take out homework: Bonus problem
 Objective: SWBAT predict and calculate
entropy changes to systems.
 Do now: Give one example of a
spontaneous process. Is it spontaneous in
the opposite direction?

Agenda
Do now
II. Notes on spontaneous processes and
entropy
III. Practice Problems
Homework: p. 829 # 10-14 (SR)
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)
74
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.
77
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)
Homework

p. 829 # 10-14 (SR)
10 Nov. 2010
Objective: SWBAT calculate Gibbs free
energy and predict the spontaneity of a
reaction.
 Do now: Give one example of a change
that increases entropy.

Agenda
Do now
II. Homework: SR
III. Gibbs Free Energy and Predicting
Spontaneity
HW: p. 829 #15, 16, 17, 18, 19, 20 (TTL)
(Tues.)
Read lab and answer pre-lab questions in
lab notebook (Mon.)
AP Test Thermo questions (Tues.)
I.
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
91
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
Homework
p. 829 #15, 16, 17, 18, 19, 20 (TTL) (Tues.)
Read lab and answer pre-lab questions in
lab notebook (Mon.)
AP Test Thermo questions (Tues.)
16 Nov. 2010
Objective: SWBAT calculate the
temperature at which a reaction is
spontaneous and review entropy and free
energy problems.
 Do now: If enthalpy is positive and
entropy is positive, at what temperatures
will a reaction proceed spontaneously?

Agenda
Do now
II. Homework solutions (TTL)
III. AP Test question solutions - collected
IV. Thermochemistry review problems
Homework: Lab notebook: Thurs.
Come after school to mass your filter paper
+ M2CO3!
I.
CaCO3(s)  CaO(s) + CO2(g)



CaO (quicklime) is used in steelmaking, producing
calcium metal, paper making, water treatment
and pollution control.
It is made by decomposing limestone (CaCO3) in a
kiln at high temperature.
But, the reaction is reversible! CaO(s) readily
combines with CO2 to form CaCO3.
 So, in the kiln, CO2 is constantly removed to
shift equilibrium to favor the formation of CaO.







At what temperature does this reaction favor
the formation of CaO?
Calculate ∆Ho and ∆So for the reaction at 25oC.
Then plug in to ∆Go=∆Ho – T∆So to find ∆Go
What sign is ∆Go? What magnitude?
Set ∆Go=0 to find the temperature at which
equilibrium occurs.
At what temperature can we expect this
reaction to proceed spontaneously?
Choose a temperature, plug in and check.
Keep in mind that this does not mean that
CaO is only formed above 835oC.
 Some CaO and CO2 will be formed, but the
pressure of CO2 will be less than 1 atm (its
standard state).

Practice Problems
1.
Find the temperatures at which reactions
will the following ∆H and ∆S will be
spontaneous:
a. ∆H = -126 kJ/mol
∆S = 84 J/K·mol
b. ∆H = -11.7 kJ/mol
∆S = -105 J/K·mol

Solutions to AP test problems
17 Nov. 2010
Objective: SWBAT review enthalpy and
entropy for a quiz.
 Do now:
The reaction between nitrogen and hydrogen to
form ammonia is represented below.
N2(g) + 3 H2(g) → 2 NH3(g) ∆H˚ = –92.2 kJ
a. Predict the sign of the standard entropy change,
∆S˚, for the reaction. Justify your answer.
b. The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but
positive at high temperatures. Explain.

Do now solutions


(a) (–); the mixture of gases (high entropy) is
converted into a pure gas (low entropy) and
the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ – T∆S˚; enthalpy favors
spontaneity (∆H < 0), negative entropy
change does not favor spontaneity. Entropy
factor becomes more significant as
temperature increases. At high temperatures
the T∆S factor becomes larger in magnitude
than ∆H and the reaction is no longer
spontaneous (∆G > 0).
Thermochemistry review
p. 831 #42, 52, 54, 56, 57, 60, 78
 Quiz Thursday
 Entropy (∆S)
 Free energy (∆G)
 Study homework assignments from
chapter 18, AP test questions, these review
problems.


stop
During the course of a chemical reaction,
not all the reactants and products will be
at their standard states (1.0 atm and
25oC).
 How do you calculate free energy (and
thus, the spontaneity of the reaction)?

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
What is Q?

Reaction quotient
What is K?

equilibrium constant
∆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 = msDt
q = CDt
Dt = tfinal - tinitial
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 (DHsub) is the
energy required to sublime 1 mole of a
solid.
Deposition
H2O (g)
Sublimation
H2O (s)
DHsub = DHfus + DHvap
( Hess’s Law)
11.8
Molar heat of fusion (DHfus) 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

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