Transcript Chapter 6- Thermochemistry

THERMOCHEMISTRY
CHAPTER 6
6.1- The Nature of Energy
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Ch. 6 Unit Essential Question:
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Lesson Essential Question:
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What does the thermodynamics branch of chemistry study and
how is it involved in chemical reactions?
What aspects must be considered when studying the area of
thermodynamics?
Temperature vs. heat

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Temperature: measure of average KE of particles.
Heat: E transfer between two objects at different
temperatures.
 Flows from higher energy (hotter object) to lower energy
(cooler object).
6.1- The Nature of Energy
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State function: depends on the current state of the
substance, not how the current state of the substance
was reached.
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Ex. 1: The net distance from A to B is shown, and does not
change (does not depend on the path taken)- state function.
But, the total distance travelled from A to B may change
depending upon the path taken- path function.
A
B
State Functions
According to this
diagram, does it
matter how sodium
chloride is formed with
respect to the net
amount of heat energy
involved (ΔH)?
Ex. 2
No matter the path
that’s taken to form a
compound, the net
amount of heat energy
involved will always be
the same for that
compound- state
function!
http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions
Thermodynamics
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Thermodynamics: study of heat E and its
transformations.
Examines internal energy changes of systems- sum of PE and
KE.
 Use simplest system to examine internal energy: ideal gas.
 No attraction between particles = no PE.
 Thus internal E is all KE, which we said in the last
chapter is directly related to T.
 We can say that the internal E of a gas increases with T.
 Recall: E = 3/2RT
Bonds in reactants and products have PE.
 Stronger bonds = more PE.
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Thermodynamics
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Recall system and surroundings.
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Heat involved is given in terms of the system (reaction).
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System = object(s) being studied- for us reactions
Surroundings = everything outside of the object(s) being
studied- for us reaction container
Signs always reflect what happens to the system!
Positive (+q) = system absorbs heat (takes in E).
Negative (-q) = system releases heat (lets E out).
Same notation applies for work.
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+w = surroundings do work on system (system ‘takes in’
work).
-w = system does work on surroundings (system ‘puts out’
work).
Thermodynamics
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First Law of Thermodynamics: The E of the universe is
constant (law of conservation of E).
 ΔEuniverse = ΔEsystem + ΔEsurroundings = 0
KE + PE of all particles in a system = internal E of a system.
 Internal E can be changed by doing work, heat flow, or
both: ΔE = q + w (take signs into account!)
Thermodynamics & Work
Work specification: V of system (gas!) can expand
during a reaction- the system is doing work on the
surroundings.
 w can be replaced with –PΔV
 Expansion: ΔV is positive, w is negative: gas
(system) does work.
 Compression: ΔV is negative, w is positive: gas
(system) has work done on it.
 ΔE = q – PΔV
To convert units: -PΔV will give you Latm, but q will be
in J so you need both units to be in J.
 101.3J = 1Latm  use as a conversion factor!
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AP Practice Question
The average ________ is the same for any ideal gas at a
given temperature.
a)
b)
c)
d)
Free energy
Lattice energy
Kinetic energy
Activation energy
AP Practice Question- Review
What is the energy released when the gaseous ions
combine to form an ionic solid?
a)
b)
c)
d)
Free energy
Lattice energy
Kinetic energy
Activation energy
AP Practice Question
When ammonium chloride dissolves in water, the
temperature drops. Which of the following conclusions may
be related to this?
a)Ammonium
chloride is more soluble in hot water.
b)Ammonium chloride produces an ideal solution.
c)The heat of solution for ammonium chloride is
exothermic.
d)Ammonium chloride has a low lattice energy.
*Ideal solution = follows Raoult’s Law (assumptions are
made for liquids like there were for gases)
6.1- The Nature of Energy
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HW: pg. 267 #6,10, 21, 24
Lesson Essential Questions
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What is enthalpy and how does it relate to chemical
reactions?
How is calorimetry performed and what does it study?
6.2- Enthalpy & Calorimetry
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Enthalpy is represented by H: heat involved in a reaction.
 ΔH = q at constant pressure; the change in enthalpy of a
system is equal to the E flow as heat.
For chemical reactions: ΔH = Hproducts – Hreactants
 Enthalpy of reaction shows the change in enthalpy from
reactants to products.
 This is the E flow as heat, what you feel as hot or cold!
Enthalpy and stoichiometry: ΔH for a reaction can be used
to develop conversion factors.
 The given ∆H is often the enthalpy of reaction for one
mole of a substance involved in the reaction. But it could
be for another amount (such as per gram)!
6.2- Enthalpy & Calorimetry
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Ex: When 1mol of CH4 burns at constant P, 890kJ of E is
released as heat. What is ΔH if 5.8g of CH4 is burned at
constant P?
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*If not told whether P is constant or not in a problem, assume P is
constant.*
Determine necessary conversion factor(s):
 890kJ/molCH4
Convert to moles and solve; remember signs!!:
 5.8gCH4 x (1molCH4/16.05gCH4) x (890kJ/molCH4) = -320kJ
 Negative because you’re told E is released.
Calorimetry
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Lab technique to measure heat involved in a chemical or
physical change.
Heat capacities:
Cp = q/ΔT (heat capacity)
 No specified amount.
c = q/mΔT (specific heat capacity)
 For 1 gram of a substance.
C = q/nΔT (molar heat capacity)
 For 1 mole of a substance.
Two types of calorimeters: coffee
cup and bomb.
Calorimetry at Constant Pressure
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Coffee cup calorimetry.
At constant P: ΔH = c x m x ΔT = q
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This is the formula for specific heat; now you’re solving for
q since ΔH = q at constant P.
May need concentration, D, etc. to find mass in grams.
Remember- AP questions like to combine various
topics/knowledge/calculations!
Calorimetry at Constant Pressure
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Example: Solid BaSO4 forms when 1.00L of 1.00M Ba(NO3)2 is
mixed with 1.00L of 1.00M Na2SO4, and the temperature
increases from 25.0°C to 28.1°C. If the amount of heat
absorbed by the calorimeter is negligible, the specific heat of the
mixed solutions is 4.18J/g°C, and the density of the mixed
solutions is 1.0g/mL, what is the enthalpy change per mole of
BaSO4 formed?
Answer: 26,000J/mol BaSO4
**Note stoichiometric relationships that can be made from reactions:
4Fe (s) + 3O2 (g)  2Fe2O3 (s)
ΔH = -1652kJ
(1) 4mol Fe/1652kJ
(3) 2molFe2O3/1652kJ
(2) 3mol O2/1652kJ
Calorimetry at Constant Volume
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Bomb calorimeter is used.
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Reactants placed in steel container (constant V) and
ignited.
Energy change determined by measuring T increase of water
and other calorimeter parts.
ΔH = ΔT  ccalorimeter + cwater  mwater  ΔT
Need to account for all heat gained- by the water and the
calorimeter!
AP Practice Question
A 1.5886g sample of glucose (C6H12O6) is ignited in a bomb
calorimeter. The temperature increased by 3.682°C. The
heat capacity of the calorimeter was 3.562kJ/°C and the
calorimeter had 1.000kg of water inside. Find the molar
heat of the reaction:
C6H12O6 (s) + 6O2 (g)  6CO2 (g) + 6H2O (l)
Answer: -3,234kJ/mol
6.2- Enthalpy & Calorimetry
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HW: pg. 267 #: 32, 34, 36, 42(a), 52, 54
Lesson Essential Question
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How is Hess’ Law used to determine the change in
enthalpy of a chemical reaction?
6.3- Hess’ Law
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Hess’ Law: for a chemical equation that can be written
as the sum of two or more steps, the enthalpy change for
the overall equation equals the sum of the enthalpy
changes for the individual steps.
Add together ΔH’s of each reaction involved.
 This works because H is a state function!
 Ex: making nitrogen dioxide.
 N2 + 2O2  2NO2
ΔH1 = 68kJ
VS.
Add reactions together
to obtain same reaction
 N2 + O2  2NO
ΔH2 = 180kJ
above; add ΔH together
+ 2NO + O2  2NO2 ΔH3 = -112kJ to get net ΔH.
N2 + 2O2  2NO2
ΔH = 180kJ – 112kJ = 68kJ
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Using Hess’ Law
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Two important characteristics about ΔH:
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1. If a reaction is reversed, ΔH’s sign is reversed.
 Ex: C(s) + O2(g)  CO2(g)
ΔH = -394kJ
CO2(g)  C(s) + O2(g) ΔH = 394kJ
2. Size of ΔH is directly proportional to amounts of
reactants and products. If a reaction is multiplied by some
integer, ΔH is also multiplied by the same integer.
 Ex: [C(s) + O2(g)  CO2(g)]x2
ΔH = (-394kJ)x2
2C(s) + 2O2(g)  2CO2(g) ΔH = -788kJ
 Watch for ∆H given as kJ/mol of a substance- if the
balanced reaction has more than 1mol of the substance,
multiply H by the appropriate coefficient!
 Ex: Combustion of C4H4 has ∆H = -585.2kJ/mol CO2.
C4H4(g) + 5O2(g)  4CO2(g) + 2H2O(g) ∆H = -2341kJ
AP Practice Question
Given the following information:
C (s) + O2 (g)  CO2 (g)
H2 (g) + 1/2O2 (g)  H2O
C2H2 (g) + 5/2 O2  2CO2 (g) + H2O (l)
ΔH = -393.5kJ
ΔH = -285.8kJ
ΔH = -1299.8kJ
Find the enthalpy change for: 2C (s) + H2 (g)  C2H2 (g)
a)454.0kJ
b)-227.0kJ
c)0.0kJ
d)227.0kJ
6.3- Hess’ Law
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HW: pg. 268 #: 57, 60, 62
AP Practice Question
Determine ΔH for the reaction if CH3OH (l) were formed
instead of CH3OH (g). The ΔH of vaporization for CH3OH
is 37 kJ/mol.
CO (g) + 2H2 (g)  CH3OH (g) ΔH = -91kJ
a)-128kJ
b)-54kJ
c)128kJ
d)54kJ
Lesson Essential Question
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How can standard enthalpies of formation be used to
calculate enthalpy changes in reactions?
6.4- Standard Enthalpies of Formation
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Standard enthalpy of formation: enthalpy change when
one mole of a substance is formed from its elements
when all substances are in their standard states.
 Represented by ∆H°f. Degree sign indicates standard
conditions, f indicates ‘formation’.
Standard states for compounds:
 Gases- at a pressure of 1atm.
 Solutions- concentration of 1M.
 Pure substances (elements & compounds)- most stable
form (how they’re typically found) at 1atm & 25°C.
 Ex: oxygen- O2 (g), mercury- Hg (l), sodium chlorideNaCl (s)
Standard Enthalpies of Formation
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Always given per mole of substance in its standard state;
formed from elements in standard states:
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Enthalpy of formation for an element in its standard state
is set to be zero.
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= -239kJ/mol
Ex: C(s) + 2H2(g) + ½O2(g)  CH3OH(l) ∆H°
f
The enthalpy of formation of methanol is -239kJ/mol.
The element is found this way; no energy is involved to put it
into this form!
Besides Hess’ Law, Enthalpy changes for reactions can also
be calculated by subtracting the enthalpies of formation
of reactants from the enthalpies of formation of products:
∆H°rxn = ∑n∆H°products
– ∑n∆H°reactants
f
f
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Note coefficients/moles (n) are taken into account!
AP Practice Question
The decomposition of CaCO3 is shown in the equation
below. Using the data below, which of the following values is
closest to the ∆Hrxn of the decomposition of CaCO3?
CaCO3 (s)  CaO (s) + CO2 (g)
∆H°f (kJ/mol): -1,207.1
-635.5
-393.5
a)-2,240
kJ/mol
b)-180 kJ/mol
c)180 kJ/mol
d)1,207 kJ/mol
e)2,240 kJ/mol
Bond Energies & Enthalpy
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Heats of reaction can also be calculated using bond
energies.
Similar formula is used as seen with standard enthalpy of
formations:
∆H°rxn = ∑nbonds broken– ∑nbonds formed
Bonds broken are the reactants, and bonds formed are
the products.
Take into account ALL bonds broken and formed- in
other words multiply each bond type by how many there
are. Drawing out molecules can help! Watch out for
single, double, and triple bonds!
Ex: CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
∆H°rxn = [4(C-H) + 2(O-O)] – [2(C=O) + 4(H-O)]
AP Practice Question
Use the values of average bond energies to find the
enthalpy change for the following reaction:
2H2 (g) + O2 (g)  2H2O (g)
Average bond energies (in kJ/mol): H-H: 436, O=O: 499, and
H-O: 464.
a)
b)
c)
d)
0kJ
485kJ
-485kJ
464kJ
Note: the actual enthalpy of reaction for
this reaction is -572kJ. Why such a big
difference?
Bond Energies & Enthalpy Cont.
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Note that the same bond in different molecules impacts
bond energy too. Thus ∆H°rxn is approximate here.
Ex: O-H bond in H2O does not have the exact same E as
the O-H bond in CH3OH.
Note also that using bond energies to calculate enthalpies
of reaction are only sufficient if all species are in the
gaseous state!
Why might this be?
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Only in the gaseous state can we assume interactions between
particles are negligible (ideal gases), and so only bond energies
need to be considered for energy changes.
In other states the interactions between particles are not
negligible and cannot be ignored since these too contribute to
energy changes.
6.4- Standard Enthalpies of Formation
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HW: pg. 269 #: 65, 67
Three Laws of Thermodynamics
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Three laws of thermodynamics:
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1. Total E of the universe is constant (law of conservation!):
ΔEuniverise = ΔEsystem + ΔEsurroundings = 0
2. All processes that occur spontaneously move in the direction
of an increase in entropy of the universe (system +
surroundings): ΔSuniverse = ΔSsystem + ΔSsurroundings > 0
3. The entropy of a pure, perfect crystal at 0K is zero.
Note: the third law has more to it, but this is all you need
to take away. There is also a zeroth law that we will not
cover.
We’ve already investigated #1, let’s look at #2!
Entropy Changes
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List all of the examples you can think of that would result
in an increase in entropy.
1)
Entropy increases when the number of molecules
increases from reactants to products.
Entropy increases when T increases.
Entropy increases when a gas is formed from a liquid or
a solid.
Entropy increases when a liquid is formed from a solid.
2)
3)
4)
AP Practice Question
Choose the reaction expected to have the greatest increase
in entropy.
a)
b)
c)
d)
H2O (g)  H2O (l)
2KClO3 (s)  2KCl (s) + 3O2 (g)
Ca (s) + H2 (g)  CaH2 (s)
N2 (g) + 3H2 (g)  2NH3 (g)
AP Practice Question
Which of the following combinations is true when sodium
chloride melts?
> 0 and ΔS > 0
b)ΔH = 0 and ΔS > 0
c)ΔH > 0 and ΔS < 0
d)ΔH < 0 and ΔS < 0
a)ΔH
AP Practice Question
Which of the following reactions would be accompanied by
the greatest decrease in entropy?
a)
b)
c)
d)
N2 (g) + 3H2 (g)  2NH3 (g)
C (s) + O2 (g)  CO2 (g)
2H2 (g) + O2 (g)  2H2O (g)
2Na (s) + Cl2 (g)  2NaCl (s)
Note: according to the practice book the reaction that
produces the most gas will have the greatest increase in S
and the one losing the most gas will have the greatest
decrease in S.
Standard Molar Entropies
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Standard molar entropies: entropy associated with 1mol
of a substance in its standard state.
 Represented by S°. The change in entropy of a
reaction can be calculated like ΔH° was:
 ∆S° = ∑nS° products – ∑nS° reactants
Standard states are the same as previously discussed.
BUT entropies of elements in standard states are not
zero!
 Everything will have entropy values!
AP Practice Question
Calculate ΔS° for the following given the following S°
values.
a. H2 (g) + ½ O2 (g)  H2O (g) *Do you expect a positive
or negative ΔS° value? Why?
S° (J/molK): H2: 131.0 O2: 205.0
H2O: 188.7
ΔS° = -44.8J/molK
b. CaCO3 (s) + H2SO4 (l)  CaSO4 (s) + H2O (g) + CO2 (g)
*Do you expect a positive or negative ΔS° value? Why?
S° (J/molK): CaCO3: 92.9 H2SO4: 157
CaSO4: 107
H2O: 188.7
CO2 213.6
ΔS° = 259J/molK
Gibbs Free Energy
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Are ΔH and ΔS the only thermodynamic quantities that
determine if a reaction will occur? What if one is
favorable and the other is not?
 Recall favorable signs for each: -ΔH and +ΔS.
What do you recall about ΔG?
It combines ΔH and ΔS, and determines whether or not a
reaction will occur or not.
Favorable sign: -ΔG.
In terms of spontaneity:
-ΔG = spontaneous; +ΔG = nonspontaneous (energy
needs to be added for the reaction to occur); ΔG = 0 the
reaction is at equilibrium.
AP Practice Question
A sample of Ga metal is sealed inside a well-insulated, rigid
container. The T inside the container is at the melting point
of the Ga metal. What can be said about the E and the S of
the system after equilibrium has been established? Assume
the insulation prevents any E change with the surroundings.
> 0 and ΔS > 0
b)ΔH = 0 and ΔS > 0
c)ΔH > 0 and ΔS < 0
d)ΔH < 0 and ΔS < 0
a)ΔH
Recall ΔG Formula
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ΔG = ΔH -TΔS
ΔG°rxn = ΔH°rxn -TΔS°rxn
Remember- you can use these formulas to calculate
values of ΔG, but you can also use it with signs for ΔH
and ΔS to see if a reaction will occur and the effect T has
on the reaction.
Ex: A salt is dissolved in a beaker of water and the
temperature drops by 4.3°C. What are the signs of ΔH,
ΔS, and ΔG for this reaction?
+ΔH, +ΔS, -ΔG
AP Practice Question
A certain reaction is nonspontaneous under standard
conditions, but becomes spontaneous at higher
temperatures. What conclusions may be drawn under
standard conditions?
< 0, ΔS > 0, ΔG > 0
b)ΔH > 0, ΔS < 0, ΔG > 0
c)ΔH > 0, ΔS > 0, ΔG > 0
d)ΔH < 0, ΔS < 0, ΔG > 0
a)ΔH
Gibbs Free Energies of Formation
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Just like with ΔH and S, ΔG also has energies of formation:
ΔG°f
ΔG°f for elements in their standard states are zero, just
like they were for ΔH°f.
Same equation is used for ΔG°rxn as was used for ΔH°rxn:
∆G°rxn = ∑n∆G°f products – ∑n∆G°f reactants
This is another way to calculate ΔG, in addition to using
the formula previously discussed.
AP Practice Question
Calculate ΔG° for the following reaction.
2NH4Cl (s) + CaO (s)  CaCl2 (s) + H2O (l) + 2NH3 (g)
ΔG°f (kJ/mol) : -203.9
Answer: -8.6kJ/mol
-604.2
-750.2
-237.2
-16.6
Ch. 17 Homework
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Pg. 783 # 8, 20, 24, 25, 27(a&b), 28(a), 33, 36, 37(a), 43,
45(a), 52
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Note: problems in red are technically work for this
information, but due to the large quantity these will be
your class work for Friday when I’m out.
Thermodynamics & Equilibrium
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What if substances in a reaction are not under standard
conditions?
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The new conditions must be taken into account:
ΔG = ΔG° + RTlnQ
Q is the reaction quotient; used to indicate where a
reaction is in terms of reaching equilibrium (we will do
more with this later):
Q = [products]x/[reactants]y
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Pressures and concentrations are no longer equal to one.
Note: x and y superscripts come from coefficients!
Ex: 2NO (g) + O2 (g)  2NO2 (g)
Q = [NO2]2/([NO]2[O2])
Example
Calculate ΔG for the following reaction at 500. K.
2NO (g) + O2 (g)  2NO2 (g)
The concentrations of the species are as follows:
NO: 2.00M, O2: 0.500M, and NO2: 1.00M.
 First find ΔG° using ΔGf° values and the equation used
before: ∆G°= ∑n∆G°f products – ∑n∆G°f reactants
 ΔGf° values (kJ/mol): NO: 86.71, O2: 0.000, NO2: 51.84
 ∆G° = -69.74 kJ/mol
 Now use the previous formula to find ΔG.
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
Note: be sure all units are kJ or J!
ΔG = -7.262 x 104 J/mol
Example #2- Gases & Pressures
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Calculate ΔG at 25°C for the reaction where CO gas at
5.0atm and H2 gas at 3.0atm form liquid methanol
(CH3OH).
CO (g) + 2H2 (g)  CH3OH (l)
ΔGf° values (kJ/mol): CH3OH: -166, CO: -137, H2: 0
ΔG = ΔG° + RTlnQ
Notes: (1) Reaction quotients are written the same way for
pressures as for concentrations.
(2) Pure solids and liquids have a value of 1.
Q = 1/(5.0 x 3.02) = 0.022
ΔG = -38kJ/mol, or -38,000 J/mol
Thermodynamics & Equilibrium
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Only use the previous equation if the reaction is not
under standard conditions and if you have not been told
the reaction is at equilibrium.
If the reaction is not under standard conditions but IS at
equilibrium:
ΔG° = - RTlnK
K is the equilibrium constant for the reaction and can
only be used if the reaction is at equilibrium (again, we’ll
do more with this later).
Recall that at equilibrium ΔG = 0, so it no longer appears
in the equation used above.
Ex: Find ΔG° for 2O3 (g)
3O2 (g) Kp = 4.17 x 1014
ΔG° = -8.34 x 104 J/mol
Thermodynamics & Equilibrium
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A few more things about K & Q:
Q is used when you are not told the reaction is at
equilibrium; if it is, use K.
Recall that Q = [products]x/[reactants]y
Also true for K: Keq = [products]x/[reactants]y
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K >1 means the reaction favors the products; spontaneous in
the direction of the products, mostly all products present.
K < 1 means the reaction favors the reactants; spontaneous in
the direction of the reactants (essentially nonspontaneous),
mostly all reactants present.
K = 1 means the reaction is at equilibrium; both reactants and
products are present.
AP Practice Question
Under standard conditions, calcium metal reacts readily
with chlorine gas. What conclusions may be drawn from the
fact?
a) Keq <1 and ΔG° > 0
b) Keq >1 and ΔG° = 0
c) Keq <1 and ΔG° < 0
d) Keq >1 and ΔG° < 0
AP Practice Question
What is the minimum energy required to force a
nonspontaneous reaction to occur?
a) free energy
b) lattice energy
c) kinetic energy
d) activation energy
AP Practice Question
Which of the following is the minimum energy required to
initiate a reaction?
a) free energy
b) lattice energy
c) kinetic energy
d) activation energy
Homework
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Pg. 786 #65