Chemistry lecture 28

Thermodynamics:
Enthalpy, Entropy
& Gibbs Free
Energy
Thermo 2
Thermodynamics: thermo = heat (energy)
dynamics = movement, motion
Some thermodynamic terms chemists use:
System: the portion of the universe that we are
considering
open system: energy & matter can transfer
closed system: energy transfers only
isolated system: no transfers
Surroundings: everything else besides the system
Isothermal: a system that is kept at a constant
temperature by adding or subtracting heat from the
surroundings.
Heat Capacity: the amount of heat energy required to
raise the temperature of a certain amount of material by
1°C (or 1 K).
Specific Heat Capacity: 1 g by 1°C
Molar Heat Capacity: 1 mole by 1°C
Thermo 3
Calorie: the amount of heat required to raise the
temperature of 1g of water by 1°C.
specific heat of water = 1 cal/g °C
1 calorie = 4.18 joules
Specific Heats and Molar Heat Capacities
Substance Specific Heat (J/°C•g) Molar Heat (J/°C•mol)
Al 0.90 24.3
Cu 0.38 24.4
Fe 0.45 25.1
CaCO3 0.84 83.8
Ethanol 2.43 112.0
Water 4.18 75.3
Air 1.00 ~ 29
important to: engineers chemists
EXAMPLE: How many joules of energy are needed to raise
the temperature of an iron nail (7.0 g) from 25°C to 125°C?
The specific heat of iron is 0.45 J/°C•g.
Heat energy = (specific heat)(mass)(ΔT)
Heat energy = (0.45 J/°C•g)(7.0 g)(100°C) = 315 J
Note that ΔT can be ºC or K, but NOT ºF. When just T is being used
in a scientific formula it will usually be kelvin (K).
Thermo 4
Problem: How much energy does it take to raise
the body temperature 2.5ºC (a fever of just over
103ºF) for someone who weighs 110 pounds (50 kg).
Assume an average body specific heat capacity of
3 J/ºC.g.
Problem: What would be more effective at melting
a frozen pipe – hot water or a hair dryer (hot air
gun). Why?
Thermo 5
Energy: "The capacity to do work
and/or transfer heat"
Forms of Energy:
Kinetic (Ekinetic = ½mv2)
Potential
Heat
Light (Electromagnetic)
Electricity
Chemical
Nuclear
Matter (E = mc2)
􀃗􀃘
WORK
Thermo 6
First Law of Thermodynamics:
The total amount of energy (and
mass) in the universe is constant.
In any process energy can be
changed from one form to
another; but it can never be
created nor destroyed.
"You can't get something for
nothing"
Thermo 7
Enthalpy (Heats) of Reaction
The amount of heat released or absorbed by a
chemical reaction at constant pressure (as one would
do in a laboratory) is called the enthalpy or heat or
reaction. We use the symbol ΔH to indicate
enthalpy.
Sign notation (EXTREMELY IMPORTANT!!):
+ΔH indicates that heat is being absorbed in the
reaction (it gets cold) endothermic
−ΔH indicates that heat is being given off in the
reaction (it gets hot) exothermic
Standard Enthalpy = ΔH° (° is called a “not”)
􀃐
Occurring under Standard Conditions:
Pressure 1 atm (760 torr)
Concentration 1.0 M
Temperature is not defined or part of Standard
Conditions, but is often measured at 298 K (25°C).
Thermo 8
Standard Enthalpy of Formation -- ΔH f
°
The amount of heat absorbed (endothermic) or
released (exothermic) in a reaction in which one
mole of a substance is formed from its elements in
their standard states, usually at 298 K (25°C).
Also called heat of formation.
ΔH f
° = 0 for any element in its standard state (the
natural elemental form at 1 atm or 1 M) at 298 K.
EXAMPLES:
C(graphite, s) + O 2 (g) CO2 (g)
ΔHr x n = 393.5 kJ/mol
° 0 kJ/mol 0 kJ/mol
elements in their
standard states
product
(one mole)
negative sign
heat released -- exothermic rxn ΔHf (CO ) =
° 2 393.5 kJ/mol
Thermo 9
2H (g) + O 2 (g) 2H 2 O (g)
ΔHr x n = 483.6 kJ/ 2 mol
° 0 kJ/mol 0 kJ/mol
elements in their
standard states
product
(two moles)
negative sign
heat released -- exothermic rxn
2
ΔHf (H O) =
° 2 241.8 kJ/mol
divide by 2 to put
on per mole basis!!
Note that we usually will not have you calculate ΔHfº on homeworks or
tests – so you generally don’t have to worry about normalizing your
answer to a per mole basis.
Hess's Law -- Adding Reactions
The overall heat of reaction (ΔHrxn) is equal to the
sum of the ΔHf (products) minus the sum of the ΔHf
(reactants):
ΔH r x n = ° Σ ΔH f (products) ° Σ ΔH f (reactants) (# eqiv) (# eqiv) °
Therefore, by knowing ΔHf of the reactants and
products, we can determine the ΔHrxn for any
reaction that involves these reactants and products.
Thermo 10
EXAMPLE: CO2 is used in certain kinds of fire extinguishers
to put out simple fires. It works by smothering the fire with
"heavier" CO2 that replaces oxygen needed to maintain a fire.
CO2 is not good, however, for more exotic electrical and
chemical fires.
2Mg(s) + CO (g) 2MgO(s) + C(s)
ΔH f = ° 0 kJ/mol
2
- 393 kJ/mol - 602 kJ/mol 0 kJ/mol
ΔH r x n = ° Σ ΔH f (products) ° Σ ΔH f (reactants) °
ΔH r x n = ° Σ(2 eqiv)(-602 kJ/mol) + (1 eqiv)(0 kJ/mol)
Σ(2 eqiv)(0 kJ/mol) + (1 eqiv)(-393 kJ/mol)
REACTANTS PRODUCTS
ΔH r x n = ° (-1204 kJ/mol) (-393 kJ/mol)
ΔH r x n = ° (-1204 kJ/mol) + 393 kJ/mol
ΔH r x n = ° 811 kJ/mol } highly exothermic rxn !!
(# eqiv) (# eqiv)
Therefore, Mg will "burn" CO2 !
Thermo 11
You can also add two reactions together to get the ΔHrxn for
another new reaction:
C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l)
2 5
C 2 H 4 (g) + 3O 2 (g)
Given these two reactions and thermodynamic data:
Calculate Δ H r x n for the following reaction: °
ΔH r x n = -1367 kJ/mol °
ΔH r x n = -1411 kJ/mol °
C H OH(l) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(l)
2CO 2 (g) + 2H 2 O(l)
ΔH r x n = ?? °
1) C 2 H 5 OH is on the product side of the first reaction -- so we want to
a)
b)
switch equation a) around to get C 2 H 5 OH also on the product side:
2 5 ΔH r x n = +1367 kJ/mol 2CO2 (g) + 3H 2 O(l) C H OH(l) + 3O 2 (g) °
How to solve:
note that when we
reverse the reaction, ΔHrxn °
*changes sign!!!
2) Now we can add the two reactions together to give us the desired
net reaction:
2 5 ΔH r x n = +1367 kJ/mol 2CO2 (g) + 3H 2 O(l) C H OH(l) + 3O 2 (g) °
C 2 H 4 (g) + 3O 2 (g) ΔH r x n = -1411 kJ/mol 2CO 2 (g) + 2H 2 O(l) °
C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) ΔH r x n = -44 kJ/mol °
1
+
If we have to multiply one (or more) of the reactions by some
constant to get them to add correctly, then we also would have
to multiply ΔHrxn for that reaction by the same amount.
Thermo 12
Chemists use bomb calorimeters to measure
Enthalpies of formation or reaction.
Thermocouple
to measure
temperature Stirrer
Electrical
contacts
to initate
sample
combustion
Water
Highly insulated outside container
Thick-walled inner container
(bomb) to contain combustion
of sample (pressurized with O2)
Bomb Calorimeter
Sample
placed
inside
inner
container
In order to use this effectively one must know the
heat capacity of the bomb (inner part) and water
bath. By measuring the temperature increase of the
water one can calculate the amount of heat given off
during the combustion process.
Thermo 13
Problem: Calculate ΔHrxn for the following
reactions given the following ΔHf ° values:
ΔHf ° (SO2, g) = −297 kJ/mol ΔHf ° (SO3, g) = −396 kJ/mol
ΔHf ° (H2SO4, l) = −814 kJ/mol ΔHf ° (H2SO4, aq) = −908 kJ/mol
ΔHf ° (H2O, l) = −286 kJ/mol ΔHf ° (H2S, g) = −20 kJ/mol
a) S(s) + O2(g) SO2(g)
b) 2SO2(g) + O2(g) 2SO3(g)
c) SO3(g) + H2O(l) H2SO4(l)
d) 2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(l)
Thermo 14
Entropy
The final state of a system is more energetically
favorable if:
1. Energy can be dispersed over a greater number
and variety of molecules.
2. The particles of the system can be more
dispersed (more disordered).
The dispersal of energy and matter is described by
the thermodynamic state function entropy, S.
The greater the dispersal of energy or matter in a
system, the higher is its entropy. The greater the
disorder (dispersal of energy and matter, both in
space and in variety) the higher the entropy.
Adding heat to a material increases the disorder.
Ice - well ordered structure
water - more disordered
water vapor - most disordered
Thermo 15
Unlike ΔH, entropy can be defined exactly because
of the Third Law of Thermodynamics:
Third Law of Thermodynamics: Any
pure crystalline substance at a
temperature of absolute zero (0.0 K) has
an entropy of zero (S = 0.0 J/K•mol).
Sign notation (EXTREMELY IMPORTANT!!):
+ΔS indicates that entropy is increasing in the
reaction or transformation (it's getting more
disordered -- mother nature likes)
−ΔS indicates that entropy is decreasing in the
reaction or transformation (it's getting less
disordered {more ordered} -- mother nature doesn't
like, but it does happen)
Thermo 16
Qualitative "Rules" About Entropy:
1) Entropy increases as one goes from a solid to a
liquid, or more dramatically, a liquid to a gas.
Solid
Liquid
Gas
phase transitions
0
50
100
150
200
250
Temperature (K)
Entropy (J/mol)
2) Entropy increases if a solid or liquid is dissolved
in a solvent.
3) Entropy increases as the number of particles
(molecules) in a system increases:
N2O4(g) 2NO2(g)
S° = 304 J/K (1 mole) S° = 480 J/K (2 moles)
The first 3 above are most important for evaluating
ΔSrxn. The rules below are for comparing the
entropy of individual molecules or materials.
Thermo 17
4) The Entropy of any material increases with
increasing temperature
5) Entropy increases as the mass of a molecule
increases
S°(Cl2(g)) > S°(F2(g))
S° = 165 J/K•mol S° = 158 J/K•mol
6) Entropy is higher for weakly bonded compounds
than for compounds with very strong covalent
bonds
S°(graphite) > S°(diamond)
S° = 5.7 J/K•mol S° = 2.4 J/K•mol
Thermo 18
7) Entropy increases as the complexity (# of atoms,
# of heavier atoms, etc.) of a molecule increases
Entropy of a Series of Gaseous Hydrocarbons
C H
H
H
H
Methane
S° = 186 J/K•mol
H C
Acetylene
C H S° = 201 J/K•mol
C
H
C Ethylene
H
H
H
S° = 220 J/K•mol
C
H
H
H
Ethane
C
H
H
H S° = 230 J/K•mol
C
H
H
H
Propane
C
H
H
C H
H
H
S° = 270 J/K•mol
What are the biggest factors for evaluating ΔSrxn for
a chemical rxn?
1) phase change 2) change in # of molecules
Thermo 19
Problem: For the following reactions, is the entropy
of the reaction increasing or decreasing?
a) Ag+(aq) + Cl-(aq) AgCl(s)
b) H2CO3(aq) H2O + CO2(g)
c) Ni(s) + 4CO(g) Ni(CO)4(l)
d) H2O(s) H2O(l)
e) graphite diamond
f) 2Na(s) + 2H2O 2Na+(aq) + 2OH-(aq) + H2(g)
g) H2S(g) + O2(g) H2O(l) + SO(g)
h) 2H2O(l) 2H2(g) + O2(g)
i) CO2(g) + CaO(s) CaCO3(s)
j) CaCl2(s) + 6H2O(l) CaCl2•6H2O(s)
k) 2NO2(g) N2O4(g)
Thermo 20
Just as with enthalpies, one can calculate entropies
of reaction.
ΔS r x n = ° Σ(# eqiv)S ° (products) Σ(# eqiv)S ° (reactants)
EXAMPLE:
2Mg(s) + CO (g) 2MgO(s) + C(s)
ΔSf = ° 32 J/K·mol
2
215 J/K·mol 27 J/K·mol 6 J/K·mol
ΔS r x n = ° Σ S ° (products) Σ S ° (reactants)
ΔS r x n = ° Σ(2 eqiv)(27 J/K·mol) + (1 eqiv)(6 J/K·mol)
Σ(2 eqiv)(32 J/K·mol) + (1 eqiv)( 214 J/K·mol)
REACTANTS PRODUCTS
ΔS r x n = ° (60 J/K.mol) (278 J/K.mol)
ΔS r x n = ° 218 J/K.mol } entropy is decreasing
(# eqiv) (# eqiv)
(reaction is becoming more ordered)
Thermo 21
Spontaneous Processes
A process that takes place without the net input of
energy from an external source is said to be
spontaneous (not instantaneous).
1) Rxn of sodium metal with water:
2Na(s) + 2H2O 2Na+(aq) + 2OH-(aq) + H2(g)
2) Combustion rxns:
2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)
2H2(g) + O2(g) 2H2O(l)
3) Expansion of a gas into a vacuum
xCO2(g) yCO2(s) + zCO2(g) (x = y + z)
4) A salt dissolving into solution:
NH4NO3(s) + H2O(l) NH4+(aq) + NO3-(aq)
Thermo 22
Second Law of Thermodynamics: In
any spontaneous process the entropy
of the universe increases
ΔSuniverse = ΔSsystem + ΔSsurroundings
Second Law (variant): in trying to do
work, you always lose energy to the
surroundings.
You can't even break even!
Neither entropy (ΔS) or enthalpy (ΔH)
alone can tell us whether a chemical
reaction will be spontaneous or not.
An obvious (?) conclusion is that one
needs to use some combination of the two.
Thermo 23
Gibbs Free Energy
The combination of entropy, temperature
and enthalpy explains whether a reaction is
going to be spontaneous or not. The symbol
ΔG is used to define the Free Energy of a
system. Since this was discovered by J.
Willard Gibbs it is also called the Gibbs
Free Energy. "Free" energy refers to the
amount of energy available to do work once
you have paid your price to entropy. Note
that this is not given simply by ΔH, the heat
energy released in a reaction.
ΔGº = ΔHº − TΔSº
When ΔG is negative, it indicates that a
reaction or process is spontaneous. A
positive ΔG indicates a non-spontaneous
reaction.
Thermo 24
ΔG = ΔH − TΔS
ΔH
ΔS
Δ G = negative
spontaneous
at all temperatures
Δ G = ??
Δ G = ?? Δ G = positive
non-spontaneous
at all temperatures
spontaneous
at low temperatures
spontaneous
at high temperatures
0
+
- +
-
Spontaneous = exoergic (energy releasing)
Non-spontaneous = endoergic (energy releasing)
Thermo 25
Remember that entropies are
given in units of J/K•mol while
enthalpies and free energies are in
kJ/mol.
DON'T forget to convert all units to
kJ or J when using both ΔS and ΔH
in the same equation!!
DANGER!!
Common
mistake!!
Thermo 26
ΔGº vs. ΔG: Standard vs. Non-Standard Conditions
Remember that the º (“not”) on ΔGº indicates that the
numerical value of ΔGº is based on the reaction at
standard conditions (1 M solution concentration, 1 atm
gas pressure). Temperature is NOT part of standard
conditions!
As soon as one has a concentration different than 1 M or
1 atm pressure, the º “not” goes away and one has ΔG.
Consider the reaction:
Initial: 1 atm 1 atm 1 atm
2SO2(g) + O2(g) 2SO3(g)
ΔGºrxn= −142 kJ/mol
The ΔGºrxn of −142 kJ/mol is for when each gas is
present with a concentration of 1 atm. This indicates
that the reaction under these conditions will proceed to
make products (spontaneous).
As the reactants start reacting, however, their
concentrations decrease (SO2 twice as fast as O2) and
ΔGº turns into ΔG and becomes less negative.
When ΔG = 0 the reaction has reached equilibrium.
Although for this rxn, SO2 is probably the limiting
reagent (not enough present to complete the rxn).
Thermo 27
Example: Calculate ΔGºf for CO2 at 298 K. ΔHf °
(CO2) = −393 KJ/mol, S° (O2) = 205 J/mol•K, S°
(C) = 6 J/mol•K, S° (CO2) = 213 J/mol•K
C(graphite) + O2(g) CO2(g)
ΔGºf = ΔHºf − TΔSºf
ΔSºf = Σ Sºprod − Σ Sºreact
ΔSºf = (213 J/mol•K) − (205 + 6 J/mol•K)
ΔSºf = 2 J/mol•K)
ΔGºf = (−393 KJ/mol) − (298 K)(0.002 KJ/mol•K)
ΔGºf = (−393 KJ/mol) − (1 KJ/mol)
ΔGºf = −394 KJ/mol
Problem: Calculate ΔGºf for CO at 298 K. ΔHf °
(CO) = −110 KJ/mol, S°(O2) = 205 J/mol•K,
S°(C) = 6 J/mol•K, S°(CO) = 198 J/mol•K
2C(graphite) + O2(g) 2CO(g)
Note change in
units – J to KJ
DANGER!!
Common
mistake!!
Thermo 28
Just as with enthalpies and entropies, one can
calculate free energies of reaction.
ΔGr x n = ° Σ ΔGf (products) ° Σ ΔGf (reactants) (# eqiv) (# eqiv) °
EXAMPLE:
ΔGr x n = ° 746 kJ } highly exothermic rxn !!
Compare to Δ rxn H° which was -811 kJ for the same rxn.
The "missing" 65 kJ of energy went to ENTROPY!
2Mg(s) + CO (g) 2MgO(s) + C(s)
ΔGf = ° 0 kJ/mol
2
- 394 kJ/mol - 570 kJ/mol 0 kJ/mol
ΔGr x n = ° Σ ΔGf (products) ° Σ ΔGf (reactants) °
ΔGr x n = ° Σ(2 mol)(-570 kJ/mol) + (1 mol)(0 kJ/mol)
Σ (2 mol)(0 kJ/mol) + (1 mol)(-394 kJ/mol)
REACTANTS PRODUCTS
ΔGr x n = ° (-1140 kJ) (-394 kJ)
ΔGr x n = ° (-1140 kJ) + 394 kJ
(# mol) (# mol)
SPONTANEOUS rxn!
Thermo 29
Example: To make iron, a steel mill takes Fe2O3 (rust
or iron ore) and reacts it with coke (a complex, impure
form of carbon) to make iron and CO2. Based on the
data below, this is a non-spontaneous reaction at room
temperature, but it becomes spontaneous at higher
temperatures. Assuming that ΔH° and ΔS° do not
change much with temperature, calculate the temperature
above which the reaction becomes spontaneous
(i.e., ΔG°rxn = 0).
ΔH°rxn = +465 kJ/mol
ΔS°rxn = +552 J/molK (or 0.552 kJ/molK)
ΔG°rxn = +301 kJ/mol (at 298 K)
ΔG°rxn = ΔH°rxn − TΔS°rxn
as we raise the temperature, ΔG° will eventually reach 0 and
then go negative & spontaneous, so let ΔG° = 0 and solve for
T, the temperature at which this will happen:
0 = ΔH°rxn − TΔS°rxn
rearranging to solve for T gives:
T = (ΔH°rxn) / (ΔS°rxn)
T = (465 kJ/mol) / (0.552 kJ/molK)
T = 842 K
(above this temperature ΔG°rxn will be negative – we will have
a spontaneous reaction)
Thermo 30
Problem: Calculate ΔGºrxn for the following.
ΔGºf (SO2, g) = −300 kJ/mol ΔGºf (SO3, g) = −371 kJ/mol
ΔGºf (H2SO4, l) = −690 kJ/mol ΔGºf (H2SO4, aq) = −742 kJ/mol
ΔGºf (H2O, l) = −237 kJ/mol ΔGºf (H2S, g) = −34 kJ/mol
a) S(s) + O2(g) SO2(g)
b) 2SO2(g) + O2(g) 2SO3(g)
c) SO3(g) + H2O(l) H2SO4(l)
d) 2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(l)
Thermo 31
Comparisons of ΔHºrxn and ΔGºrxn
S(s) + O2(g) SO2(g)
ΔHºrxn = −297 kJ/mol
ΔGºrxn = −300 kJ/mol
2SO2(g) + O2(g) 2SO3(g)
ΔHºrxn = −198 kJ/mol
ΔGºrxn = −142 kJ/mol
SO3(g) + H2O(l) H2SO4(l)
ΔHºrxn = −132 kJ/mol
ΔGºrxn = −82 kJ/mol
2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(l)
ΔHºrxn = −1126 kJ/mol
ΔGºrxn = −1006 kJ/mol
ΔSºrxn = +11 J/mol K
Thermo 32
Thermo 33
From “General Chemistry”, 7th Ed, by Whitten, Davis, Peck & Stanley. Thomson Brooks/Cole Publisher.