Transcript Thermodynamics

Thermodynamics
The universe is in a state of constant change, the only invariant is Energy
Consider ….
Gravity causes molecules of water move
turbine blades
turbines move coils of
wire in magnetic fields
moving magnetic fields move
electrons
moving electrons drive
chemical reactions in a
battery
chemical reactions power your phone
creating light and sound
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Consider ….
 In this example gravity is responsible for a working
smartphone




Gravity does work on the blades
The turning blades do work on electrons
Electrons do work in chemical reactions
Chemical reactions do work on a speaker and power LEDs
in the screen, power radios etc
 The ability of something to do work on something
else is transferred from gravity to water to electrons
to chemical reactions to moving magnetics and
moving air and light from the screen
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Consider ….
Gravity causes hydrogen atoms fuse to make
Helium and a little bit of mass is converted into
light and heat etc
Photons are absorbed
by chlorophyll and
used to power
photosynthesis
We extract the oil
make biofuel
ignite the biofuel and excess energy
released gives the molecules to power to
move pistons
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Consider ….
 In this example Gravity causes fusion reactions in the
sun are responsible for a working car
 Mass is converted to light
 light moves electrons
 Electrons do work in chemical reactions to create sugars oils
etc
 Oils react with oxygen to create fast moving CO2 and H2O
 Molecules push pistons and drive the car
 The ability of something to do work on something
else is transferred from the sun to chlorophyll to
electrons to chemical reactions to moving molecules
and moving pistons and moving wheels
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Energy: the capacity to do work
 In each step in the previous examples a capacity to do work is
transferred from one thing to another, this is called energy
 Gravitational energy is transferred into kinetic energy, into
electrical energy, chemical energy, sound and light energy etc
 All dynamic processes in the universe are due to the flow of
energy
 Thermodynamics is the study of heat flow and the laws that
govern it
 Since we want to understand chemical transformation we need
to understand energy transformation
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 Energy is a universal invariant
 It can change from one form to another but cannot be created or
destroyed
 It is measured in Joules (J)
Energy
 There is potential energy (energy that something has
because of where and what it is) and kinetic energy (the
energy is has because of how fast it is moving)
 The lower the potential energy the more stable something
is. Potential energy can be negative
 When some process happens, generally it is to lower the
potential energy
 The study of energy helps us to predict whether a process
is spontaneous or not
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What is Thermodynamics?
 Thermodynamics is a branch of physics concerned with energy flow.
Historically it had an emphasis on heat, temperature and their
relation to energy and work.
 Study of energy changes accompanying chemical and physical
changes to a system
 Defines systems using a few macroscopic (measurable) variables,
such as internal energy, entropy, temperature and pressure
 Statistical treatment of microstates (atom positions and velocities) to
obtain macrostates
 In chemistry, thermodynamics predicts if reactions occur, how the
equilibrium constant changes with temperature
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First Law of Thermodynamics
 you can’t get something for nothing
 First Law of Thermodynamics: Energy cannot be Created
or Destroyed
 the total energy of the universe cannot change
 though you can transfer it from one place to another

Euniv = 0 = Esys + Esurr
(1)
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Conservation of Energy
First Law of Thermodynamics
For an exothermic reaction, “lost” heat from the system goes into the
surroundings
two ways energy “lost” from a system,
 converted to heat, q
 used to do work, w
Energy conservation requires that the internal energy E change in the
system equal the heat released (q) + work done (w)
E = q + w
(2)
E = H + PV
(3)
E is the total energy of everything in the system (the kinetic and
potential energy of the atoms)
E (U)is a state function
 internal energy change independent of how this change occurs
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The first law and time reversal
 The first law tells us that only processes where there is no net
change in the total energy are allowed (energy is conserved)
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The first law and spontaneity
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
In all observed phenomena the total energy is always the same
The energy at t, E(t) is equal to the energy at time time t+dt,
E(t) = E(t+dt)
So if that is the case why do we always see some processes
only going one way?
✓
✗
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The first law and spontaneity
 Clearly the first law isn’t the end of the story
regarding energy and what happens in processes
✓
✗
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Factors Affecting Whether a
Reaction Is Spontaneous
 It turns out that there are two factors that determine the thermodynamic
favorability are the enthalpy H and the entropy S.
 The enthalpy is a comparison of the bond energy of the reactants to the
products.
 bond energy = amount needed to break a bond.
 statistical model of collective behavior
 ΔH
 The entropy factors relates to the randomness/orderliness of a system
 ΔS
 The enthalpy factor is generally more important than the entropy factor
 Let’s look at these
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Enthalpy

related to the internal energy E, the energy change measured at constant P is
ΔH = generally kJ/mol)

ΔHrxn is related to the breaking and forming of chemical bonds. Stronger
bonds = more stable molecules
DH °rxn = å DH °(bonds broken) –
i
å DH °(bonds formed)
j

if products more stable than reactants, energy released
 exothermic
 H = negative

if reactants more stable than products, energy absorbed
 endothermic
 H = positive

The enthalpy is favorable for exothermic reactions and unfavorable for
endothermic reactions.
 Hess’ Law
DH °rxn = å (ni × DH ° prod,i ) –
i
å (n × DH °
j
j
(5')
react, j
)
(5)
Spontaneity: Enthalpy Driven Processes
• All transformations have accompanying energy changes.
• Can we tell which transformations will occur spontaneously by studying the
energy change?
• In many cases, the direction of spontaneity can be determined by comparing the
potential energy of the system at the start and the end
• Cellulose and O2 have a bigger
potential energy than the equivalent
amount of carbon dioxide and water
• The transformation lowers the overall
potential energy, C-O and H-O bonds
are more stable than C-C and C-H
bonds
• exothermic reactions are
spontaneous
• The extra energy leaves as heat
Spontaneity: Entropy Driven Processes
• But some processes are spontaneous but not exothermic!
• These are entropy driven processes
Entropy S
 Entropy, S, is a thermodynamic function that increases as the number
of equivalent ways of arranging the atoms/molecules (positions and
velocities) in a system to give the appropriate V, U and T increases
 S generally J/(K.mol)
S = k ln W = Q/T
(6)
 k = Boltzmann Constant = 1.38 x 10-23 J/K
 W is the number of energetically equivalent ways accessible, unitless
(measure of our lack of knowledge about the system)
 Entropy is the energy dispersal per unit temperature
 Random systems require less energy than ordered systems
 Measure of the unavailability of a system to do work
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W
Energetically Equivalent
States for the Expansion of a
Gas
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Macrostates → Microstates
These microstates all
have the same
macrostate
So there are 6 different
particle arrangements
that result in the same
macrostate This macrostate can be achieved through
several different arrangements of the particles
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Macrostates and Probability
There is only one possible arrangement
that gives State A and one that gives
State B
There are 6 possible arrangements that
give State C
Therefore State C has higher entropy
than either State A or State B
The macrostate with the highest entropy
also has the greatest dispersal of energy
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Changes in Entropy, S
 entropy change is favorable when the result is a more random system.
 S is positive
 Some changes that increase the entropy are:
 reactions whose products are in a more disordered state.
 (solid > liquid > gas)
 reactions which have larger numbers of product molecules than
reactant molecules.
 increase in temperature
 solids dissociating into ions upon dissolving
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Increases in Entropy
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The 2nd Law of Thermodynamics: Spontaneity
 "Energy spontaneously disperses from being localized to becoming
spread out if it is not hindered from doing so.”
 The total entropy change of the universe must be positive for a
process to be spontaneous
 for reversible process Suniv = 0,
 for irreversible (spontaneous) process Suniv > 0
Suniv = Ssys + Ssurr
(7)
 if the entropy of the system decreases, then the entropy of the
surroundings must increase by a larger amount
 when Ssys is negative, Ssurr is positive
 the increase in Ssurr often comes from the heat released in an
exothermic reaction
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Temperature Dependence of Ssurr
 when a system process is exothermic, it adds heat to the
surroundings, increasing the entropy of the surroundings
 when a system process is endothermic, it takes heat from the
surroundings, decreasing the entropy of the surroundings
 the amount the entropy of the surroundings changes depends on the
temperature it is at originally
 the higher the original temperature, the less effect addition or
removal of heat has
DSsurr =
-DHsys
T
or DHsys = -T DSsurr
(8)
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Gibbs Free Energy, G

For a spontaneous process ΔSuniv > 0

maximum amount of energy from the system available to do work on the
surroundings at constant temperature T
DSuniv = DSsys + DSsurr
DH sys
= DH sys >0
T
or
T DSuniv = -DH sys + T DSsys > 0
or
DGsys = DH sys - T DSsys < 0

(9)
when G < 0, there is a decrease in free energy of the system that is released
into the surroundings; therefore a process will be spontaneous when G is
negative
N
p
i=1
j=1
DGrxn = å ni × DG prod,i – å n j × DGreact , j
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Thermodynamics and Spontaneity
Free Energy
 spontaneity is determined by comparing the free energy G of the
system before the reaction with the free energy of the system after
reaction, it includes both the enthalpy and entropy change of a
process
ΔG = ΔH – T∙ΔS
(9)
 if the system after reaction has less free energy than before the
reaction, the reaction is thermodynamically favorable
 spontaneity ≠ fast or slow
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Gibbs Free Energy, G
 process will be spontaneous when G is negative

G will be negative when
 H is negative and S is positive
 exothermic and more random
 H is negative and large and S is negative but small
 H is positive but small and S is positive and large
 or high temperature

G will be positive when H is + and S is −
 never spontaneous at any temperature

when G = 0 the reaction is at equilibrium
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G, H, and S
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Chemical Potential Energy
The chemical potential – is a form of free energy used for
chemical reactions, in spontaneous reactions the chemical
potential decreases
Thermodynamics vs. Kinetics
• Kinetics describes how fast things change
• Thermodynamics is concerned if they will change and if so what
changes we will see in internal energy, temperature, pressure etc
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Example: Diamond → Graphite
Graphite is more stable than diamond, so the conversion of diamond into graphite
is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into
pencil lead in your lifetime (or through many of your generations).
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Reversibility of Process
 any spontaneous process is irreversible
 it will proceed in only one direction
 a reversible process will proceed back and forth between the two end
conditions
 equilibrium
 results in no change in free energy
 if a process is spontaneous in one direction, it must be
nonspontaneous in the opposite direction
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Entropy Change and State Change
Phase changes, melting boiling etc these are endothermic changes
driven by entropy concerns not enthalpy concerns
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Entropy Change in State Change
 when materials change state, the number of macrostates it
can have changes as well
 for entropy: solid < liquid < gas
 because the degrees of freedom of motion increases
solid → liquid → gas
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Heat Flow, Entropy, and the 2nd Law
Heat must flow from water to ice in order
for the entropy of the universe to increase
But why that way round? The 1st law is not
violated if more ice was formed?
Flowing hot to cold we increase energy
randomization.
Heat flowing into the hot concentrated
energy so S decreases
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The reaction C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) has Hrxn = -2044 kJ at
25°C.
Calculate the entropy change of the surroundings.
Given:
Hsystem = -2044 kJ, T = 298 K
Find:
Ssurroundings, J/K
Concept Plan:
T, H
S
DS surr =
Relationships:
Solution:
DS surr =
DS surr
Check:
-DH sys
T
= 6.86
kJ
K
=
-DH sys
T
-(-2044 kJ )
298 K
= 6.86 ´10 3
J
K
combustion is largely exothermic, so the entropy of the
surrounding should increase significantly
Free Energy Change and Spontaneity
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The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has
H = +95.7 kJ and S = +142.2 J/K at 25°C.
Calculate G and determine if it is spontaneous.
Given:
H = +95.7 kJ, S = 142.2 J/K, T = 298 K
Find:
G, kJ
Concept Plan:
T, H, S
G
Relationships:
Solution:
DG = DH - TDS
DG = DH - TDS
(
)
(
= +95.7 ´10 3 J - ( 298 K) +142.2 KJ
)
= +5.33 ´10 4 J
Answer:
Since G is +, the reaction is not spontaneous at this
temperature. To make it spontaneous, we need to
increase the temperature.
The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has
H = +95.7 kJ and S = +142.2 J/K.
Calculate the minimum temperature it will be spontaneous.
Given:
H = +95.7 kJ, S = 142.2 J/K, G < 0
Find:
T,K
Concept Plan:
G, H, S
Relationships:
Solution:
DG = DH - TDS
DG = ( DH - TDS) < 0
(+95.7 ´ 10 J) <
(+142.2 )
3
(
) (
)
(+95.7 ´10 J) < ( T)(+142.2 )
+95.7 ´10 3 J - ( T) +142.2 KJ < 0
3
Answer:
T
J
K
J
K
T
673 K < T
The temperature must be higher than 673K for the
reaction to be spontaneous
The 3rd Law of Thermodynamics
Absolute Entropy
 the absolute entropy of a substance is the
amount of energy it has due to dispersion of
energy through its particles
 the 3rd Law states that for a perfect crystal at
absolute zero, the absolute entropy = 0 J/mol∙K
 therefore, every substance that is not a
perfect crystal at absolute zero has some
energy from entropy
 therefore, the absolute entropy of
substances is always +
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Standard Entropies
 S°
 Extensive (depends on the system size)
 entropies for 1 mole at 298 K for a particular state, a particular
allotrope, particular molecular complexity, a particular molar
mass, and a particular degree of dissolution
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Relative Standard Entropies
States
 the gas state has a larger entropy than the liquid state at a particular
temperature
 the liquid state has a larger entropy than the solid state at a particular
temperature
Substance
S°,
(J/mol∙K)
H2O (l)
70.0
H2O (g)
188.8
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Relative Standard Entropies
Molar Mass
 the larger the molar mass, the larger
the entropy
 available energy states more closely
spaced, allowing more dispersal of
energy through the states
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Relative Standard Entropies
Allotropes
 the less constrained the
structure of an allotrope
is, the larger its entropy
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Relative Standard Entropies
Molecular Complexity
 larger, more complex molecules
generally have larger entropy
 more available energy states,
allowing more dispersal of energy
through the states
S°,
Substance
Molar
Mass
Ar (g)
39.948
154.8
NO (g)
30.006
210.8
(J/mol∙K)
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Relative Standard Entropies
Dissolution
 dissolved solids generally
have larger entropy
 distributing particles
throughout the mixture
Substance
S°,
(J/mol∙K)
KClO3(s)
143.1
KClO3(aq)
265.7
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Calculate S for the reaction
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l)
Given:
Find:
Concept Plan:
Substance
S, J/mol/K
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
188.8
standard entropies look up in appendix to textbook or google
S, J/K
SoNH3, SoO2, SoNO, SoH2O,
S
Relationships:
Solution:
Check:
S is +, as you would expect for a reaction with more gas
product molecules than reactant molecules
Calculating Go
 at 25oC:
Goreaction = ΣnGof(products) - ΣnGof(reactants)
 at temperatures other than 25oC:
 assuming the change in Horeaction and Soreaction is negligible
Goreaction = Horeaction – TSoreaction
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Go
25oC
Calculate
at
for the reaction
CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4 O3(g)
Given:
Find:
Concept Plan:
Relationships:
Solution:
Substance
Gof, kJ/mol
CH4(g)
-50.5
O2(g)
0.0
CO2(g)
-394.4
H2O(g)
-228.6
O3(g)
163.2
standard free energies of formation from Appendix of
textbook or google Go, kJ
Gof of prod & react
Go
The reaction SO2(g) + ½ O2(g)  SO3(g) has
Ho = -98.9 kJ and So = -94.0 J/K at 25°C.
Calculate Go at 125oC and determine if it is spontaneous.
Given:
Ho = -98.9 kJ, So = -94.0 J/K, T = 398 K
Find:
Go, kJ
Concept Plan:
T, Ho, So
Go
Relationships:
Solution:
Answer:
Since G is -, the reaction is spontaneous at this temperature,
though less so than at 25oC
G Relationships
 if a reaction can be expressed as a series of reactions, the sum of the
G values of the individual reaction is the G of the total reaction
 G is a state function
 if a reaction is reversed, the sign of its G value reverses
 if the amounts of materials is multiplied by a factor, the value of the G
is multiplied by the same factor
 the value of G of a reaction is extensive
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Free Energy and Reversible Reactions
 the change in free energy is a theoretical limit as to the amount of work
that can be done
 if the reaction achieves its theoretical limit, it is a reversible reaction
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Real Reactions
 in a real reaction, some of the free energy is “lost” as heat
 if not most
 therefore, real reactions are irreversible
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G under Nonstandard Conditions
G = Go only when the reactants and products are in their standard
states
there normal state at that temperature
partial pressure of gas = 1 atm
concentration = 1 M
•
•
•
under nonstandard conditions, G = Go + RTlnQ
Q is the reaction quotient
•
at equilibrium G = 0
Go = ─RTlnK
•
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Example - G
 Calculate G at 427°C for the reaction below if the PN2 = 33.0 atm,
PH2= 99.0 atm, and PNH3= 2.0 atm
N2(g) + 3 H2(g)
2 NH3(g)
PNH32
Q=
PN21 x PH23
(2.0 atm)2
=
(33.0 atm)1 (99.0)3
= 1.2 x 10-7
H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
G° = -92380 J - (700 K)(-198.2 J/K)
G° = +46400 J
G = G° + RTlnQ
G = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
G = -46300 J = -46 kJ
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Example - K
 Estimate the equilibrium constant and position of equilibrium for the
following reaction at 427°C
N2(g) + 3 H2(g)
2 NH3(g)
H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
G° = -92380 J - (700 K)(-198.2 J/K)
G° = +46400 J
G° = -RT lnK
+46400 J = -(8.314 J/K)(700 K) lnK
lnK = -7.97
K = e-7.97 = 3.45 x 10-4
since K is << 1, the position of equilibrium favors reactants
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Temperature Dependence of K
 for an exothermic reaction, increasing the temperature decreases the
value of the equilibrium constant
 for an endothermic reaction, increasing the temperature increases the
value of the equilibrium constant
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