Enthalpy (H)
• The heat transferred sys ↔ surr during a
chemical rxn @ constant P
• Can’t measure H, only ΔH
• At constant P, ΔH = q = mCΔT, etc.
• Literally, ΔH = Hproducts - Hreactants
• ΔH = + (endothermic)
• Heat goes from surr into sys
• ΔH = - (exothermic)
• Heat leaves sys and goes into surr
In this example, the energy of the system (reactants and products) ↑, while
the energy of the surroundings ↓
Notice that the total energy does not change
Reactant + Energy
Product
Endothermic Reaction
Surroundings
Energy
Surroundings
System
System
Myers, Oldham, Tocci, Chemistry, 2004, page 41
Before
reaction
After
reaction
Notice
that E
must be
added,
and thus
is like a
reactant
In this example, the energy of the system (reactants and products) ↓, while
the energy of the surroundings ↑
Notice again that the total energy does not change
Reactant
Product + Energy
Energy
Surroundings
Myers, Oldham, Tocci, Chemistry, 2004, page 41
System
Before
reaction
Exothermic Reaction
Surroundings
System
After
reaction
Notice
that E is
released
and thus
is like a
product
Burning of a Match
Potential energy
System
Surroundings
(Reactants)
D(PE)
Energy released to the surrounding as heat
(Products)
Exothermic Reaction
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293
Reaction Coordinate Diagrams:
Endothermic Reaction
Activation Energy
Ea
D(PE)
PE
Products
Reactants
Progress of the Reaction
ΔHrxn= +
Reaction Coordinate Diagrams:
Exothermic Reaction
Activation Energy
Ea
ΔHrxn= -
PE
Reactants
D(PE)
Products
Progress of the Reaction
Reaction Coordinate Diagrams
Draw the reaction coordinate diagram for the following rxn:
C(s) + O2(g)  CO2 + 458.1kJ
Activation Energy
EXOTHERMIC
Ea
ΔHrxn= -458.1 kJ
PE
C + O2
D(PE)
CO2
Progress of the Reaction
Enthalpies of Reaction
• All reactions have some ΔH associated with it
H2(g) + ½ O2(g) → H2O(l)
ΔH = - 483.6 kJ
• How can we interpret this ΔH?
• Amount of energy released or absorbed per specific
reaction species
• Use balanced equation to find several definitions
- 483.6 kJ
1 mol H2
or
- 483.6 kJ
½ mol O2
or
- 483.6 kJ
1 mol H2O
Able to use like conversion factors in stoichiometry
Enthalpies of Reaction
• Formation of water
H2(g) + ½ O2(g) → H2O(l)
ΔH = - 483.6 kJ
• ΔH is proportional to amount used and will
change as amount changes
2H2(g) + O2(g) → 2H2O(l)
ΔH = - 967.2 kJ
• For reverse reactions, sign of ΔH changes
2H2O(l) → 2H2(g) + O2(g)
ΔH = + 967.2 kJ
• Treat ΔH like reactant or product
H2(g) + ½ O2(g) → H2O(l)
ΔH = - 483.6 kJ
H2(g) + ½ O2(g) → H2O(l) + 483.6 kJ (exo)
Enthalpies of Reaction Practice
Consider the following rxn:
C(s) + 1/2O2(g)  CO + 458.1kJ
Is the ΔH for this reaction positive or negative?
NEGATIVE (E released as a product)
What is the ΔH for 2.00 moles of carbon, if all the carbon
is used? 2.00 mol C - 458.1 kJ
1 mol C
= - 916 kJ
What is the ΔH if 50.0g of oxygen is used?
50.0 g O2
1 mol O2
32.0 g O2
- 458.1 kJ
= -1430 kJ
0.5 mol O2
1 mol CO
28.0 g CO
458.1 kJ
= 818 kJ
1 mol CO
What is the ΔH if 50.0 g of carbon monoxide decompose,
in the reverse reaction?
50.0 g CO
Hess’s law
• Hess’s Law states that the enthalpy of a whole
reaction is equivalent to the sum of it’s steps.
• For example: C + O2  CO2
This can occur as 2 steps
C + ½O2  CO
DH = – 110.5 kJ
CO + ½O2  CO2
DH = – 283.0 kJ
C + CO + O2
 CO + CO2
DH = – 393.5 kJ
I.e. C + O2  CO2
DH = – 393.5 kJ
• Hess’s law allows us to add equations.
• We add all reactants, products, & DH values.
Hess’s Law
Reactants  Products
The change in enthalpy is the same whether the
reaction takes place in one step or a series of steps
Victor Hess
Why? Because enthalpy is a state function
To review:
1. If a reaction is reversed, ΔH is also reversed
2 CH4 + O2  2 CH3OH
ΔHrxn = -328 kJ
2 CH3OH  2 CH4 + O2
ΔHrxn = +328 kJ
2. If the coefficients of a reaction are multiplied by an integer,
ΔH is multiplied by that same integer
CH4 + 2 O2  CO2 + 2 H2O
2(CH4 + 2 O2  CO2 + 2 H2O)
ΔHrxn = -802.5 kJ
ΔHrxn = -1605 kJ
Example: Methanol-Powered Cars
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g)
2 CH4(g) + O2(g)  2 CH3OH(l)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
ΔHrxn = ?
ΔHrxn = -328 kJ
ΔHrxn = -802.5 kJ
2 CH3OH(l)  2 CH4(g) + O2(g)
ΔHrxn = +328 kJ
2(CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)) ΔHrxn = -1605 kJ
2 CH3OH(l) + 2 CH4(g) + 4 O2(g)  2 CH4(g) + O2(g) + 2 CO2(g) + 4 H2O(g)
3
2 CH3OH + 3 O2  2 CO2 + 4 H2O
ΔHrxn = -1277 kJ
Tips for applying Hess’s Law…
Look at the final equation that you are
trying to create first…
• Find a molecule from that eq. that is only in
one of the given equations
• Make whatever alterations are necessary to those
• Once you alter a given equation, you will not alter it again
• Continue to do this until there are no other
options
• Next, alter remaining equations to get things to
cancel that do not appear in the final equation
1. Given the following data:
S(s) + 3/2O2(g) → SO3(g)
ΔH = -395.2 kJ
2SO2(g) + O2(g) → 2SO3(g)
ΔH = -198.2 kJ
.
Calculate ΔH for the following reaction:
S(s) + O2(g) → SO2(g)
S(s) + 3/2O2(g) → SO3(g)
2SO2(g) + O2(g) → 2SO3(g)
2SO3(g) → O2(g) + 2SO2(g)
SO3(g) → ½ O2(g) + SO2(g)
ΔH = -395.2 kJ
ΔH = -198.2 kJ
ΔH = +198.2 kJ
ΔH = +99.1 kJ
S(s) + O2(g) → SO2(g)
ΔH = -296.1kJ
2. Given the following data:
C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1300 kJ
C(s) + O2(g) → CO2(g)
ΔH = -394 kJ
H2(g) + 1/2O2(g) → H2O(l)
ΔH = -286 kJ
Calculate ΔH for the following reaction:
2C(s) + H2(g) → C2H2(g)
C(s) + O2(g) → CO2(g)
2C(s) + 2O2(g) → 2CO2(g)
C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l)
2CO2(g) + H2O(l) → C2H2(g) + 5/2O2(g)
H2(g) + 1/2O2(g) → H2O(l)
2C(s) + H2(g) → C2H2(g)
ΔH = -394 kJ
ΔH = -788 kJ
ΔH = -1300 kJ
ΔH = +1300 kJ
ΔH = -286 kJ
ΔH = +226kJ
3. Given the following data:
2O3(g) → 3O2(g)
O2(g) → 2O(g)
NO(g) + O3(g) → NO2(g) + O2(g)
ΔH = - 427 kJ
ΔH = + 495 kJ
ΔH = - 199 kJ
Calculate ΔH for the following reaction:
NO(g) + O(g) → NO2(g)
NO(g) + O3(g) → NO2(g) + O2(g)
O2(g) → 2O(g)
O(g) → ½ O2(g)
2O3(g) → 3O2(g)
3/2 O2(g) → O3(g)
NO(g) + O(g) → NO2(g)
ΔH = - 199 kJ
ΔH = + 495 kJ
ΔH = - 247.5 kJ
ΔH = - 427 kJ
ΔH = + 213.5 kJ
ΔH = +233kJ
Heats of Formation, ΔH°f
The enthalpy change when one mole of a compound is
formed from the elements in their standard states
° = standard conditions
• Gases at 1 atm pressure
• All solutes at 1 M concentration
• Pure solids and pure liquids
f
= a formation reaction
• 1 mole of product formed
• From the elements in their standard states (1 atm, 25°C)
For all elements in their standard states, ΔH°f = 0
What’s the formation reaction for adrenaline, C9H12NO3(s)?
9 Cgr + 6 H2(g) + 1/2 N2(g) + 3/2 O2(g)  C9H12NO3(s)
Thermite Reaction
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l)
ΔHrxn = ?
Welding railroad tracks
Thermite Reaction
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l)
Reactants
Elements
Products
(standard states)
Fe2O3(s)
2 Al(s)
2 Fe(s)
3/
2
O2(g)
2 Fe(l)
Al2O3(s)
2 Al(s)
ΔHrxn = 2ΔH°f(Fe(l)) + ΔH°f(Al2O3(s)) - ΔH°f(Fe2O3(s)) - 2ΔH°f(Al(s))
ΔHrxn = [2(15 kJ) + (-1676 kJ)] – [(-822 kJ) – 2(0)]
ΔHrxn = -824 kJ
ΔHrxn = nΔH°f(products) - nΔH°f(reactants)
ΔH°f Example Problems
∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants
1. CH4(g) + 2 Cl2(g)  CCl4(g) + 2 H2(g)
ΔHrxn = ?
2 (0)
(-106.7)
(-74.8) 2 (0)
∆H = [(-106.7) + 0] – [(-74.8)+0]
= -106.7 + 74.8
= -31.9 kJ/mol
2. 2 KCl(s) + 3 O2(g)  2KClO3(s)
2 (-435.9)
3 (0)
2 (-391.2)
∆H = [(2)(-391.2)] – [(2)(-435.9) + (3)(0)]
= -782.4 + 871.8
= 89.4 kJ/mol
ΔHrxn = ?
ΔH°f Example Problems
∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants
3. AgNO3(s) + NaCl (aq)  AgCl(s) + NaNO3(aq) ΔHrxn = ?
(-124.4)
(-407.1)
(-127.0)
(-446.2)
∆H = [(-127.0) + (-446.2)] – [(-124.4) + (-407.1)]
= -573.2 + 531.5
= - 41.7 kJ/mol
4. C2H5OH(l) + 7/2 O2(g)  2CO2(g) + 3H2O(g) ΔHrxn = ?
(-277.7)
(7/2)(0)
(2)(-393.5) (3)(-241.8)
∆H = [(2)(-393.5) + (3)(-241.8)] – [(-277.7) + (7/2)(0)]
= -1512.4 + 277.7
= -1234.7 kJ/mol
THIS MEANS THAT
THINGS FALL.
THEY FALL FROM HEIGHTS
OF ENERGY AND STRUCTURED
INFORMATION INTO
MATTER IS ENERGY.
ENERGY IS INFORMATION.
MEANINGLESS, POWERLESS
DISORDER.
EVERYTHING IS INFORMATION.
PHYSICS SAYS THAT
STRUCTURES... BUILDINGS,
SOCIETIES, IDEOLOGIES...
THIS IS CALLED
ENTROPY.
WILL SEEK THEIR POINT OF
LEAST ENERGY.
Entropy (S) = a measure of randomness or disorder
Entropy: Tendency toward disorder
Entropy: Tendency toward disorder
Second Law of Thermodynamics
occurs without outside intervention
• In any spontaneous process, the entropy of
the universe increases or is +
ΔSuniverse > 0
ΔSuniverse = ΔSsystem + ΔSsurroundings
Entropy of the Universe
ΔSuniverse = ΔSsystem + ΔSsurroundings
Positional disorder
ΔSuniverse > 0
Energetic disorder
spontaneous process
Both ΔSsys and ΔSsurr positive
spontaneous process
Both ΔSsys and ΔSsurr negative
nonspontaneous process
ΔSsys negative, ΔSsurr positive
depends
ΔSsys positive, ΔSsurr negative
depends
ΔSsys: Positional Disorder and Probability
Probability of
1
particle in left bulb
"
2
particles both in left bulb = (½)(½) = ¼
"
3
particles all in left bulb
"
4
"
all
"
= (½) 4 = 1/16
"
10
"
all
"
= (½)10 = 1/1024
"
20
"
all
"
= (½)20 = 1/1048576
"
a mole of
"
all
"
=½
= (½)(½)(½) = 1/8
=
23
6.02
x10
(½)
The arrangement with the greatest entropy is the one
with the highest probability (most “spread out”).
Entropy of the System: Positional Disorder
Ludwig Boltzmann
Ludwig Boltzmann
Ordered
state
Low probability
(few ways)
Low S
Disordered
state
High probability
(many ways)
High S
• Ssystem is proportional to positional disorder
• S increases with increasing # of possible positions
Ssolid
< Sliquid <<<<
Sgas
Entropy of the Surroundings
(Energetic Disorder)
System
Heat
Entropy
ΔSsurr > 0
ΔHsys < 0 Surroundings
Surroundings
System
ΔS surr  
Heat
Entropy
ΔSsurr < 0
ΔHsys > 0
ΔHsys
Low T → large entropy change (surroundings)
T
High T → small entropy change (surroundings)
The Third Law of Thermodynamics
The Third Law:
The entropy of a perfect
crystal at 0 K is zero
• Everything locked into
place
• No molecular motion
whatsoever
Crystallization of Water into Ice
Entropy Curve
Solid
Liquid
Gas
ΔHvap (l ↔ g)
S
(J/K)
Δfus (s ↔ l)
0
0
Temperature (K)
S° (absolute entropy) can be calculated for any substance
Entropy Increases with...
• Melting (fusion)
Sliquid > Ssolid
ΔHfus/Tfus = ΔSfus
• Vaporization
Sgas > Sliquid
ΔHvap/Tvap = ΔSvap
• Increasing ngas in a reaction
• Heating
ST2 > ST1 if T2 > T1
• Dissolving (usually)
Ssolution > (Ssolvent + Ssolute)
• Molecular complexity
more bonds, more entropy
• Atomic complexity
more e-, protons, neutrons
Entropy Practice
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)
For the above reaction, predict the sign of ΔSsys
We can predict S values based on phases…
Remember that Ssolid< Sliquid <<<<<< Sgas and that
ΔSsys = Sproducts – Sreactants
ΔSsys. = S (12 mol gas) – S (6 mol gas + 1 mol solid)
Solid is negligible, more gas products, ↑ disorder
Therefore…ΔSsys = +
Entropy Practice
4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
For the above reaction, predict the sign of ΔSsys
We can predict S values based on phases…
Remember that Ssolid< Sliquid <<<<<<< Sgas
ΔSsys = Sproducts – Sreactants
ΔSsys = S (2 mol solid) – S (4 mol solid + 3 mol gas)
Solid is negligible, gas R → solid P , ↓ disorder
Therefore…ΔSsys = -
Predicting ΔSsys sign summary
• Use relative S values for phases
Ssolid< Saqueous< Sliquid < Sgas
• Gases always have greater entropy
• Consider number of moles, especially with
gases
• You cannot predict for some reactions
H2(g) + Cl2(g) → 2HCl(g)
2 mol gas → 2 mol gas
Unable to predict sign ΔSsys for this reaction
Calculating Entropy Quantitatively
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
Calculate the value of ΔS°sys:
Compound
C6H12O6(s)
O2(g)
CO2(g)
H2O(g)
S° (J/mol K)
212
205
214
189
ΔS°sys = ΣnS°(products) - ΣnS°(reactants)
= [6 S°(CO2(g)) + 6 S°(H2O(g))] – [S°(C6H12O6(s)) + 6 S°(O2(g))]
= [6(214) + 6(189)] – [(212) + 6(205)] J/K
ΔS°sys = 976 J/K
Is this reaction spontaneous at 298K?
YES!
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
ΔSuniverse = ΔSsys + ΔSsurr
Compound
C6H12O6(s)
O2(g)
CO2(g)
H2O(g)
and
ΔSsurr = - ΔH/T
ΔH°f (kJ/mol)
-1275
0
-393.5
-242
S° (J/mol K)
212
205
214
189
ΔSsys = 976 J/K from last problem
ΔH°rxn = ΣnΔH°f (products) - ΣnΔH°f(reactants)
= [6 ΔH°f(CO2(g)) + 6 ΔH°f(H2O(g))] – [ΔH°f(C6H12O6(s)) + 6 ΔH°f(O2(g))]
= [6(-393.5) + 6(-242)] – [(-1275) + 6(0)] kJ
ΔH°rxn = -2538 kJ
ΔSuniverse = 0.976 kJ/K + -(-2538 kJ)/298K = 9.49 kJ/K
Entropy (S) Review
• ΔSuniverse > 0 for spontaneous processes
• ΔSuniverse = ΔSsystem + ΔSsurroundings
positional
•
energetic
We can find the absolute entropy value for a substance
• S° values for elements & compounds in their standard
states are tabulated (Thermodynamic Appendix)
•
For any chemical reaction, we can calculate ΔS°rxn:
ΔS°rxn = ΣnS°(products) - ΣnS°(reactants)
Recap: Characteristics
of Entropy
• S is a state function (we can use final – initial)
• S is extensive (more stuff, more entropy)
• At 0 K, S = 0 (we can determine absolute entropy)
• S > 0 for elements and compounds in their standard states
• Raise T  increase S
• Increase ngas  increase S
• More complex systems  larger S
Gibbs Free Energy (G)
G = H – TS
ΔG = ΔH – TΔS
At constant temperature,
Divide both sides by –T
ΔG = ΔH – TΔS
-ΔG/T = -ΔH/T + ΔS
(system’s point of view)
ΔSuniverse = ΔS – ΔH/T
–ΔG means +ΔSuniv
A process (at constant T, P) is
spontaneous if free energy decreases
Josiah Gibbs
ΔG and Chemical Reactions
ΔG = ΔH – TΔS
• If ΔG < 0, the reaction is spontaneous
• If ΔG > 0, the reaction is not spontaneous
(The reverse reaction is spontaneous)
• If ΔG = 0, the reaction is at equilibrium
• Neither the forward nor the reverse reaction is favored
• Both reactions are occurring simultaneously and at
equal rates
• ΔG is an extensive, state function
Depends on how much stuff
Depends on final and initial states only
Is the following reaction spontaneous at 298 K?
Ba(OH)2(s) + 2NH4Cl(s)  BaCl2(s) + 2NH3(g) + 2 H2O(l)
ΔH°rxn = 50.0 kJ (per mole Ba(OH)2)
ΔS°rxn = 328 J/K (per mole Ba(OH)2)
ΔG = ΔH - TΔS
ΔG° = 50.0 kJ – 298 K(0.328 kJ/K)
ΔG° = – 47.7 kJ
Spontaneous
At what T does the reaction stop being spontaneous?
The T where ΔG = 0
ΔG = 0 = 50.0 kJ – T(0.328 kJ/K)
50.0 kJ = T(0.328 kJ/K)
T = 152 K
not spontaneous below 152 K
Effect of ΔH and ΔS on Spontaneity
ΔG = ΔH – TΔS
ΔG (-) → spontaneous reaction
ΔH
ΔS
Spontaneous?
–
+
Spontaneous at all temps
+
+
Spontaneous at high temps
–
+
–
–
• Reverse reaction spontaneous at low temps
Spontaneous at low temps
• Reverse reaction spontaneous at high temps
Not spontaneous at any temp
Ways to Calculate ΔG°rxn
1. ΔG° = nΔG°f(products) - nΔG°f(reactants)
• ΔG°f = free energy change when forming 1 mole of
compound from elements in their standard states
(see Thermodynamics Appendix for values)
2. ΔG° = ΔH° - TΔS°
3. ΔG° can be calculated by combining ΔG°
values for several reactions
Using Hess’s Law! Your favorite!
Calculate ΔG° for the following reaction:
2 H2(g) + O2(g)  2 H2O(g)
1. ΔG° = ΔG°f(products) - ΔG°f(reactants)
ΔG°f(O2(g)) = 0
ΔG°f(H2(g)) = 0
ΔG°f(H2O(g)) = -229 kJ/mol
ΔG° = (2(-229 kJ) – 2(0) – 0) kJ = -458 kJ
2. ΔG° = ΔH° - TΔS°
ΔH° = -484 kJ
ΔS° = -89 J/K
ΔG° = -484 kJ – 298 K(-0.089 kJ/K) = -457 kJ
2H2(g) + O2(g)  2 H2O(g)
3. ΔG° = combination of ΔG° from other reactions
(using Hess’s Law)
2H2O(l)  2H2(g) + O2(g)
ΔG°1 = 475 kJ
H2O(l)  H2O(g)
ΔG°2 = 8 kJ
ΔG° = - ΔG°1 + 2(ΔG°2)
ΔG° = -475 kJ + 16 kJ = -459 kJ
Method 1:
Method 2:
Method 1: -458 kJ
Method 3:
Method 2: -457 kJ
Method 3: -459 kJ
What is Free Energy, Really?
• NOT just “another form of energy”
• Free Energy is the energy available to do useful work
• If ΔG is negative, the system can do work (wmax =
ΔG)
• If ΔG is positive, then ΔG is the work required to
make the process happen
– Example: Photosynthesis
6 CO2 + 6 H2O  C6H12O6 + 6 O2
ΔG = 2870 kJ/mol of glucose at 25°C
Thus, 2870 kJ of work is required to photosynthesize
1 mole of glucose