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George Mason UniversityGeneral Chemistry 212
Chapter 20Thermodynamics
Acknowledgements
Course Text: Chemistry: the Molecular Nature of Matter and Course Text: Chemistry: the Molecular Nature of Matter and Change, 7 Change, 7thth edition, 2011, McGraw-Hill edition, 2011, McGraw-Hill Martin S. Silberberg & Patricia Amateis Martin S. Silberberg & Patricia Amateis
The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material.Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor.
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Thermodynamics Thermodynamics: Enthalpy, Entropy, Free Energy
The Direction of Chemical Reactions The First Law of Thermodynamics
Conservation of Energy Limitations of the First Law
The Sign of H and Spontaneous Change Freedom of Motion and Disposal of Energy
The Second Law of Thermodynamics Predicting Spontaneous Change Entropy and the Number of Microstates Entropy and the Second Law
The Third Law of thermodynamics Standard Molar Entropies
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Thermodynamics Calculating the Change in Entropy of a Reaction
The Standard Entropy of Reaction Entropy Changes in the Surroundings Entropy Change and the Equilibrium State Spontaneous Exothermic and Endothermic
Reactions Entropy, Free Energy, and Work
Free Energy Change (∆G) and Reaction Spontaneity Standard Free Energy Changes G and Work Temperature and Reaction Spontaneity Coupling of Reactions
Free Energy, Equilibrium, and Reaction Direction
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ThermodynamicsEnthalpy (∆H)
Sum of Internal Energy (E) plusProduct of Pressure & Volume(Endothermic vs. Exothermic)
( Hrxn = Hf prod - Hf react)
Entropy (S)Measure of system order/disorder
&the number of ways energy can
can be dispersed throughthe motion of its particles
All real processes occur spontaneously
in the direction that increases theEntropy of the universe
(universe = system + surroundings)Gibbs Free Energy (∆G)
Difference between Enthalpy andthe product of absolute temperature
and the Entropy
w P Δ V pΔE q w
pq ΔE P ΔV H E P V
pΔH q (Constant Pressure)
univ sys surrE = E + E
final initial finalsys
initial final initial
V P CΔS = R ln = R ln = R ln
V p C
rxn products reactantsΔS = m S - n So o o
univΔS > 0
univΔS = 0 sys surrΔS = - S(At Equilibrium)
ΔG ΔS - o o osys sys sysH T o o orxn f(products) f(reactants)ΔG = mΔG - nΔG
(Spontaneous)
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Thermodynamics Thermodynamics - study of relationships between
heat and other forms of energy in chemical reactions
The direction and extent of chemical reactions can be predicted through thermodynamics (i.e., feasibility)
In thermodynamics, a state variable is also called a state function
Examples include:
Temperature (T), Pressure (P), Volume (V),
Internal Energy (E), Enthalpy (H), and Entropy (S)
In contrast Heat (q) and Work (W) are not state functions, but process functions
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Thermodynamics Chemical reactions are driven by heat (Enthalpy)
and/or randomness (Entropy)
A measure of randomness (disorder) is Entropy (S)
An increase in disorder is spontaneous
Spontaneous reactions are moving toward equilibrium
Spontaneous reactions move in the direction where energy is lowered, and move to Q/K = 1 (equilibrium)
Thermodynamics is used to determine spontaneity(a process which occurs by itself) and the natural forces that determine the extent of a chemical reaction (i.e., Kc)
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Thermodynamics For a reaction to be useful it must be spontaneous
(i.e., goes to near completion, i.e., far to the right) Spontaneity of a reaction depends on:
Enthalpy - heat flow in chemical reactions Entropy - measure of the order or randomness of
a system (Entropy units - J/ oK) Entropy is a state function; S = Sfinal - Sinitial
Higher disorder equates to an increase in Entropy Entropy has positional and thermal disorder
There are three principal laws of thermodynamics, each of which leads to the definition of thermodynamic properties(state variables) which help us to understand and predict the operation of a physical system
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Laws of Thermodynamics The Laws of Thermodynamics define
fundamental physical quantities (temperature, energy, and entropy that characterize thermodynamic systems. The laws describe how these quantities behave under various circumstances, and forbid certain phenomena (such as perpetual motion)
The First Law of Thermodynamics is a statement of the conservation of energy
The Second Law is a statement about the direction of that conservation
The Third Law is a statement about reaching Absolute Zero (0° K)
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Laws of Thermodynamics First law of thermodynamics:
The first law, also known as the Law of Conservation of Energy, states that energy cannot be created or destroyed in a chemical reaction
Energy can only be transferred or changed from one form to another. For example, turning on a light would seem to produce energy; however, it is electrical energy taken from another source that is converted
It relates the various forms of kinetic and potential energy in a system to the work(W = -PΔV) which a system can perform and to the transfer of heat
It applies to the changes in internal energy (ΔE) when energy passes, as work (W), as heat (q), or with matter, into or out from a system
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Laws of Thermodynamics The first law is usually formulated by stating that
the change in the Internal Energy (E) of a closed system is equal to the amount of Heat (q) supplied to the system, minus the amount of Work (W = -PV) performed by the system on its surroundings
The law of conservation of energy can be stated
The Energy of an Isolated System is Constant
ΔE = q + w = q - P V
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Laws of Thermodynamics First Law of system Thermodynamics
Conservation of Energy, E (or U in some texts)
Any change in the energy of the systemcorresponds to the interchange of “heat” (work) with an “External” surrounding
Total Internal Energy (E) - The sum of the kinetic and potential energies of the particles making up a substance
Kinetic Energy (Ek) - The energy associated with an object by virtue of its motion,
Ek = ½mv2 (kgm2/s2) (joules) Potential Energy (Ep) - The energy an object
has by virtue of its position in a field of force,Ep = mgh (kg m/s2 m = kgm2/s2)
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Laws of Thermodynamics Work – The energy transferred is moved by a
force, such as the expansion of a gas in an open system under constant pressure
Pressure = kg/(ms2)
Volume = m3
Work (W) = kg/(ms2) m3 = kgm2/s2 = joules (J)
Internal Energy
●The Internal Energy of a system, E, is precisely defined as the heat at constant pressure (qp)plus the work (w) done by the system:
pΔE = q + w pΔE = q + (-P ΔV)
pq ΔE P ΔV
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Laws of Thermodynamics Enthalpy is defined as the internal energy plus
the product of the pressure and volume – work
The change in Enthalpy is the change in internal energy plus the product of constant pressure and the change in Volume
pΔH = q (At Constant Pressure) The change in Enthalpy equals the heat gained or
lost (heat of reaction) at constant pressure – the entire change in “internal energy” (E), minus any expansion “work” done by the system (PV) would have negative sign
Recall
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Laws of Thermodynamics E – total internal energy; the sum of kinetic and
potential energies in the system q – heat flow between system and surroundings
(-q indicates that heat is lost to surroundings) w – work (-w indicates work is lost to
surroundings) H – Enthalpy – extensive property dependent on
quantity of substance and represents the heat energy tied up in the chemical bonds (heat of reaction)
Useful Units in Energy expressions 1 J (joule) = 1 kgm2/s2
1 Pa (pascal) = 1 kg/ms2
1 atm = 1.01325 x 105 Pa 1 atm = 760 torr = 760 mm Hg
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Exchanges of Heat and Workwith the Surroundings
q<0q>0
w<0 by system
w>0on system
Pressure x VolumeWork = expansionof volume due toforming a gas
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Practice ProblemConsider the combustion of Methane (CH4) in Oxygen
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)
The heat of reaction (q) at 25 oC and 1.00 atm is -890.2 kJ. What is E for the change indicated by the chemical equation at 1 atm?
n = 3 mol converted to 1 mol = -2 mol
@ 25 oC and 1 atm, 1 mol of gas = 24.5 L, thus
V = -2(24.5) = -49 L (1m3/1000 L) = -0.049 m3
E = q - PV
E = -890.2 kJ – 1 atm x (-0.049 m3)
E = -890.2 kJ – (1.01 x 105 Pa)(-0.049 m3)
E = -890.2 kJ – (1.01 x 105 kg/ms2)(-0.049m3)
E = -890.2 kJ + (4949 J x 1 kJ/1000 J)
E = -890.2 kJ + 4.949 kJ = -885 kJ
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Laws of Thermodynamics Second Law of Thermodynamics:
The second law introduces a new state variable, Entropy (S)
Entropy is a measure of the number of specific ways in which the energy of a thermodynamic system can be dispersed through the motions of its particles
In a natural thermodynamic process, the sum of the entropies of the participating thermodynamic systems increases
The total entropy of a system plus its environment (surroundings) can not decrease; it can remain constant for a reversible process but must always increase for an irreversible process
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Laws of Thermodynamics Second Law of Thermodynamics:
According to the second law the entropy of an isolated system not in thermal equilibrium never decreases; such a system will spontaneously evolve toward thermodynamic equilibrium, the state of maximum entropy of the system
More simply put: the entropy of the world only increases and never decreases
A simple application of the second law of thermodynamics is that a room, if not cleaned and tidied, will invariably become more messy and disorderly with time - regardless of how careful one is to keep it clean. When the room is cleaned, its entropy decreases, but the effort to clean it has resulted in an increase in entropy outside the room that exceeds the entropy lost
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Laws of Thermodynamics The 2nd Law of Thermodynamics
Entropy is a state function; S = Sf - Si
Higher disorder equates to an increase in Entropy
Entropy has positional and thermal disorder
The Entropy, S, is conserved for a reversible process
The disorder of the system and thermal surroundings must increase for a spontaneous process
The total Entropy of a system and its surroundingsalways increases for a “Spontaneous” process
A process occurs spontaneously in the direction that increases the Entropy of the universe
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Laws of Thermodynamics A spontaneous change, whether a chemical or
physical change, or just a change in location is one that:
Occurs by itself under specified conditions
Occurs without a continuous input of energy from outside the system
In a non-spontaneous change, the surroundings must supply the system with a continuous input of energy
Under a given set of conditions, a spontaneous change in one direction is not spontaneous in the “other” direction
A limitation of the 1st Law of Thermodynamics
Spontaneous does not equate to “Instantaneous”
The first and second laws make it impossible to construct a perpetual motion machine.
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Laws of Thermodynamics Limitations of the 1st law of Thermodynamics
The 1st Law accounts for the energy involved in a chemical process (reaction)The internal energy (E) of a system, the sum of
the kinetic and potential energy of all its particles, changes when heat (q) and/or work (w= -PV) are gained or lost by the system
Energy not part of the system is part of the surroundings
ΔE = q + w = q - P V
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Laws of Thermodynamics The surroundings (sur) and the system
(sys) together constitute the “Universe (univ)”
Heat and/or work gained by system is lost by surroundings
The “total” energy of the Universe is constant
univ sys surE = E + E
sys surr(q + w) = - (q + w)
sys sur sys sur univΔE = - ΔE ΔE + ΔE = 0 = ΔE
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Laws of Thermodynamics The first Law, however, does not account for the
“direction” of the change in energy
Ex. The burning of gas in your car
Potential energy difference between chemical bonds in fuel mixture and those in exhaust is converted to kinetic energy to move the car
Some of the converted energy is released to the environmental surroundings as heat (q)
Energy (E) is converted from one form to another, but there is a “net” conservation of energy
1st law does not explain why the exhaust gas does not convert back into gasoline and oxygen
1st law does not account for the “direction” of a spontaneous change
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Laws of Thermodynamics Spontaneous Change and Change in Enthalpy (H)
It was originally proposed (19th Century) that the “sign” of the Enthalpy change (H) – the heat lost or gained at constant pressure (qp) – was the criterion of spontaneity
Exothermic processes (H < 0) were “spontaneous”
Endothermic processes (H > 0) were “nonspontaneous”
Ex. Combustion (burning) of Methane in Oxygen is
“Spontaneous” and “Exothermic” (H < 0)
When Methane burns in your furnace, heat is released
4 2 2 2CH (g) + 2O (g) CO (g) + 2H O(g) ΔH = - 802 kJ
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Laws of Thermodynamics The sign of the change in Enthalpy (H), however,
does not indicate spontaneity in all cases An Exothermic process can occur spontaneously
under certain conditions and the opposite Endothermic process can also occur spontaneously under other conditions
Ex. Water freezes below 0oC and melts above 0oC
Both changes are spontaneous
Freezing is Exothermic
Melting (& Evaporation) is Endothermic
Most Water-Soluble Salts have a positive Hsoln yet they dissolve spontaneously
The decomposition of N2O5 is Endothermic and spontaneous 2 5 2 2 rxnN O (s) 2NO (g) + 1 2O (g) ΔH = +109.5 kJ
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Entropy Freedom of motion & energy dispersion
Endothermic processes result in more particles (atoms, ions, molecules) with more freedom of motion – Entropy increases
During an Endothermic phase change, “fewer” moles of reactant produce “more” moles of product
The energy of the particles is dispersed over more quantized energy levels
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Entropy Endothermic Spontaneous Process
Less freedom of particle motion more freedom of motion
Localized energy of motion dispersed energy of motion
Phase Change: Solid Liquid GasDissolving of Salt: Crystalline Solid + Liquid Ions in SolutionChemical Change: Crystalline Solids Gases + Ions in Solution
In thermodynamic terms, a change in the freedom of motion of particles in a system, that is, in the dispersal of their energy of motion, is a key factor determining the direction of a spontaneous process
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Entropy Quantized Energy Levels
Electronic Kinetic - vibrational, rotational, translational
Microstate A single quantized state at any instant The total energy of the system is dispersed
throughout the microstate New microstates are created when system
conditions change At a given set of conditions, each microstate has
the same total energy as any other A given microstate is just as likely to occur as
any other microstate
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Entropy Microstates vs Entropy (Positional Disorder)
Boltzmann Equation
where k – Boltzmann Constant
where R = Universal Gas Constant
NA = Avogadro’s Number
where W = No. of Microstates
S = k lnW
-23
23
A
R 8.31447J / mol • Kk = = = 1.38 10 J / K
N 6.02214 10 / mole
-23
A
RS = ln W = 1.38 x 10 lnW J / K
N
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Entropy The number of microstates (W) possible for a
given number of particles (n) as the volume changes is a function of the nth power of 2:
nfinal
initial
W = 2
W
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Entropy Compute Ssys
When n becomes NA , i.e. 1 mole
The Boltzman constant “k = R/NA” has become “R”
A system with fewer microstates (smaller Wfinal) has lower Entropy (Lower S)
A system with more microstates (larger Wfinal) has higher Entropy (higher S)
sys final initial final initialΔS = S - S = k ln W - k ln W
n nfinalsys
initial A
W RΔS = k ln = k × ln 2 = × ln 2
W N
AN
sys A
A A
R RΔS = ln 2 = N ln 2 = R ln 2
N N
sysΔS = R ln 2 8.314 J / mol • K 0.693 = 5.76 J / mol • K
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Entropy Entropy change – Volume, Pressure, Concentration
S = R ln(V, P, C)
sys final initial final initialΔS = S - S = R x ln (V, P, C) - R x ln (V, P, C)
Recall : V T or V = const × T
1 V or PV = const
P V [conc] or V n or V = const [conc] or V = const n
1( )
1 ( )
n
n
final final finalsys
initial initial
initial
final initial finalsys
initial final initial
V P CΔS = R ln = R ln = R ln
V Cp
V p CΔS = R ln = R ln = R ln
V P C
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Entropy Changes in Entropy
The change in Entropy of the system (Ssys) depends only on the difference between its final and initial values
(Ssys) > 0 when its value increases during a change
Ex. Sublimation of dry ice to gaseous CO2
(Ssys) < 0 when its value decreases during a change
Ex. Condensation of Water
sys final initialΔS = S - S
2 2CO (s) CO (g)
2 2H O(g) H O(l)
sys final initial gaseous 2 solid 2ΔS = S - S = S CO - S CO
sys liquid 2 gaseous 2ΔS = S H O - S H O
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Entropy Entropy Changes based on Heat Changes
The 2nd Law of Thermodynamics states that the change in Entropy for a gas expanding into a vacuum is related to the heat absorbed (qrev) and the temperature (T) at which the exchange occurs
Qrev refers to a “Reversible” process where the expansion of the gas can be reversed by the application of pressure (work, PV)
The heat absorbed by the expanding gas increases the dispersal of energy in the system, increasing the Entropy
If the change in Entropy, Ssys, is greater than the heat absorbed divided by the absolute temperature, the process occurs spontaneously
revsys
qΔS =
T
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Laws of Thermodynamics Determination of the Direction of a Spontaneous
Process
Second Law Restated
All real processes occur spontaneously in the direction that increases the Entropy of the universe (system + surroundings)
When changes in both the system and the surroundings occur, the universe must be considered
Some spontaneous processes end up with higher Entropy
Other spontaneous processes end up with lower Entropy
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Laws of Thermodynamics The Entropy change in the system or
surroundings can be positive or negative
For a spontaneous process, the “sum” of the Entropy changes must be positive
If the Entropy of the system decreases, the Entropy of the surroundings must increase, making the net increase to the universe positive
univ sys surrΔS = ΔS + ΔS > 0 ( )spontaneous process
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Laws of Thermodynamics The 3rd Law of Thermodynamics:
Entropy & Enthalpy are both “state” functions
Absolute Enthalpies cannot be determined, only changes i.e., No reference point
Absolute Entropy of a substance provides a reference point and can be determined
The Entropy of a system approaches a constant value as the temperature approaches zero
The entropy of a system at absolute zero is typically zero, and in all cases is determined only by the number of different ground states it has
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Laws of Thermodynamics Specifically, the entropy of a pure crystalline
substance (perfect order, where all particles are perfectly aligned with no defects of any kind) at absolute zero temperature is zero
This statement holds true if the perfect crystal has only one state with minimum energy
Ssys = 0 at 0oK
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Entropy Entropy values for substances are compared to
“standard” states
Standard States
Gases – 1 atmosphere (atm)
Concentrations – Molarity (M)
Solids – Pure Substance
Standard Molar Entropy
So (Units – J/molK @ 298oK)
Values available in Reference Tables (Appendix “B”)
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Entropy Predicting Relative So Values of a System
Temperature Changes
So increases as temperature increases
Temperature increases as “heat” is absorbed
(q > 0)
As temperature increases, the Kinetic Energies of gases, liquids, and solids increase and are dispersed over larger areas increasing the number of microstates available, which increases Entropy
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Entropy
At any T > 0o K, each particle moves about its lattice position
As temperature increases through the addition of “heat”, movement is greater
Total energy is increased giving particles greater freedom of movement
Energy is more dispersed
Entropy is increased
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Entropy Predicting Relative So Values of a System (Con’t)
Physical States and Phase Changes So increases for a substance as it changes
from a solid to a liquid to a gas Heat must be absorbed (q>0) for a change in
phase to occur Increase in Entropy from liquid to gas is much
larger than from solid to liquid Svapo >> Sfus
o
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Entropy Predicting Relative So Values of a System (Con’t)
Dissolving a solid or liquid
Entropy of a dissolved solid or liquid is greater than the Entropy of the “pure” solute
As the crystals breakdown, the ions have increased freedom of movement
Particle energy is more dispersed into more “microstates”
Entropy is increased
Entropy increase is “greater” for ionic solutes than “molecular” solutes – more particles are produced
The slight increase in Entropy for “molecular” solutes in solution arises from the separation of molecules from one another when mixed with the solvent
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Entropy Predicting Relative So Values of a System (Con’t)
Dissolving a Gas
Gases have considerable freedom of motion and highly dispersed energy in the gaseous state
Dissolving a gas in a solvent results in diminished freedom of motion
Entropy is “Decreased”
Mixing (dissolving) a gas in another gas
Molecules separate and mix increasing microstates and dispersion of energy
Entropy “Increases”
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Entropy Predicting Relative So Values of a System (Con’t)
Atomic Size
Multiple substances in a given phase will have different Entropies based on Atomic Size and Molecular Complexity
Down a “Periodic” group energy levels become “closer” together as the atoms get “Heavier”
No. of microstates and molar Entropy increase
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Entropy Predicting Relative So Values of a System (Con’t)
Molecular Complexity
Allotropes – Elements that occur in different forms have higher Entropy in the form that allows more freedom of motion
Ex. Diamond vs Graphite
Diamond bonds extend the 3 dimensions, allowing limited movement – lower Entropy
Graphite bonds extend only within two-dimensional sheets, which move relatively easy to each other – higher Entropy
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Entropy Predicting Relative So Values of a System (Con’t)
Molecular Complexity (Con’t)
Compounds
Entropy increases as the number of atoms (or ions) in a formula unit of a molecule increases
The trend is based on types of movement and the number of microstates possible
NO (Nitrous Oxide) in the chart below can vibrate only toward and away from each other
The 3 atoms of the NO2 molecule have more virbrational motions
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Entropy Predicting Relative So Values of a System (Con’t)
Molecular Complexity (Con’t) Compounds of large molecules
A long organic hydrocarbon chain can rotate and vibrate in more ways than a short chain
Entropy increase with “Chain Length”
A ring compound with the same molecular formula as a corresponding chain compound has lower Entropy because a ring structure inhibits freedom of motion
cyclopentane (C5H10) vs pentene (C5H10)
Scyclopentane < Spentene
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Entropy Predicting Relative So Values of a System
(Con’t)
Physical State vs Molecular Complexity
When gases are compared to liquids:
The effect of physical state (g, l, s) usually dominates that of molecular complexity, i.e., the No. atoms in a formula unit or chain length
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Practice ProblemChoose the member with the higher Entropy in each of the following pairs, and justify the choice
1 mol of SO2(g) or 1 mol SO3(g)
SO3 has more types of atoms in the same state, i.e., more types of motion available
More Entropy
1 mol CO2(s) or 1 mol CO2(g)
Entropy increases in the sequence:s < l < g
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Practice Problem 3 mol of O2(g) or 2 mol of O3(g)
The two samples contain the same number of oxygen atoms (6), but different numbers of molecules
O3 is more complex, but the greater number of molecules of O2 dominates – more moles of particles produces more microstates
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Practice Problem (Con’t) 1 mol of KBr(s) or 1 mol KBr(aq)
Both molecules have the same number of ions (2)
Motion in a crystal is more restricted and energy is less dispersed
KBr(aq) has higher Entropy
Sea Water at 2oC or at 23oC
Entropy increases with increasing temperature
Seawater at 23oC has higher Entropy
1 mol CF4(g) or 1 mol CCl4(g)
For similar compounds Entropy increases with increasing molar moss
S(CF4)(g) < S(CCl4)(g)
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Practice Problem Predict the sign of S for each process:
Alcohol EvaporatesΔSsys positive, the process described is liquid alcohol becoming gaseous alcoholThe gas molecules have greater Entropy than the liquid molecules
A solid explosive converts to a gasΔSsys positive, the process described is a change from solid to gas, an increase in possible energy states for the system
Perfume vapors diffuse through a roomΔSsys positive, the perfume molecules have more possible locations in the larger volume of the room than inside the bottleA system that has more possible arrangements has greater Entropy
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Practice ProblemWithout using Appendix B predict the sign of S for:
2K(s) + F2(g) → 2KF(s)
ΔSsys negative – reaction involves a gaseous reactant and no gaseous products, so Entropy decreases
The number of particles also decreases, indicating a decrease in Entropy
NH3(g) + HBr(g) → NH4Br(s)
ΔSsys negative – gaseous reactants form solid product and number of particles decreases, so Entropy decreases
NaClO3(s) → Na+(aq) + ClO3-
ΔSsys positive – when a solid salt dissolves in water, Entropy generally increases
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Entropy Calculating Change in Entropy
Gases The sign of the Standard Entropy of Reaction
(Sorxn) of a reaction involving gases can often
be predicted when the reaction involves a change in the number of moles that occurs and all the reactants and products are in their “standard” states
Gases have great freedom of motion and high molar Entropies If the number of moles of gas increases,
Sorxn is usually positive
If the number of moles of gas decreases, So
rxn is usually negative
rxn products reactantsΔS = m S - n So o o
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Practice ProblemCalculate So
rxn for the combustion of 1 mol of Propane
at 25oC
Calculate Δn to determine if the change in moles from reactant to product indicates increased or decreased Entropy
Calculate Sorxn , using values from Appendix B
3 8 2 2 2C H (g) + 5O (g) 3CO (g) + 4H O(l)o o o
productsrxn reactantsΔS = mS - nS
o o orxn 2 2 2 2
o o3 8 3 8 2 2
ΔS = [(3 mol CO )(S of CO ) + (4 mol H O)(S of H O)]
- [(1 mol C H )(S of C H ) + (5 mol O )(S of O )]
orxnΔS = [(3 mol )(213.7 J / mol K) + (4 mol)(69.9 J / mol K)]
- [(1 mol)(269.9 J / mol K) + (5 mol)(205.0 J / mol K)]
orxnΔS = - 374 J / K < 0
Δn = 3 mol (product) - 6 mol (reactant) = - 3o
rxnEntropy should decrease (ΔS < 0)
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Entropy Entropy Changes in the Surroundings
2nd Law – For a spontaneous process, a decrease in Entropy in the system, Ssys, can only occur if there is an increase in Entropy in the surroundings, Ssys
Essential role of the surroundings is to either add heat to the system or remove heat from the system – surroundings act as a “Heat Sink”
Surroundings are generally considered so large that its temperature essentially remains constant even though its Entropy will change through the loss or gain of heat
univ sys surΔS = ΔS + ΔS > 0 (spontaneous process)
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Thermodynamics Surroundings participate in two (2) types of
Enthalpy changes
Exothermic Change
Heat lost by system is gained by surroundings
Increased freedom of motion from temperature increase in surroundings leads to Entropy increase
Endothermic Change
Heat gained by system is lost by surroundings
Heat lost reduces freedom of motion in surroundings, energy dispersal is less, and Entropy decreases
sys sur surExothermic Change : q < 0 q > 0 ΔS > 0
sys sur surEndothermic Change : q > 0 q < 0 ΔS < 0
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Thermodynamics Temperature of the Surroundings
At Lower Temperatures Little random motion Little energy Fewer energy levels Fewer microstates Transfer of heat from system has larger effect
on how much energy is dispersed At Higher Temperatures
Surroundings already have relatively large quantity of energy dispersal
More energy levels More available microstates Transfer of heat from system has much smaller
effect on the total dispersion of energy
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Thermodynamics Temperature of the Surroundings
The change in Entropy of the surroundings is “greater” when heat is added at lower temperatures
Recall 2nd Law – The change in Entropy of the surroundings is directly related to an “opposite” change in the heat (q) of the system and “inversely” related to the temperature at which the heat is transferred
Recall that for a process at “Constant Pressure”, the heat (qp) = H
syssur
qΔS = -
T
syssurΔS = -
T
H
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Practice ProblemHow does the Entropy of the surroundings change during an Exothermic reaction?
Ans: In an Exothermic process, the system releases heat to its surroundings. The Entropy of the surroundings increases because the temperature of the surroundings increases (ΔSsur > 0)
How does the Entropy of the surroundings change during an Endothermic reaction?
Ans: In an Endothermic process, the system absorbs heat from the surroundings and the surroundings become cooler. Thus, the Entropy of the surroundings decreases (ΔSsur < 0)
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Practice ProblemWhat is the Entropy of a perfect crystal at 0 oK
Ans: According to the Third Law, the Entropy is zero
Does the Entropy increase or decrease as the temperature rises?
Ans: Entropy will increase with temperature
Why is ∆Hof = 0 but So > 0 for an element?
Ans: The third law states that the Entropy of a pure, perfectly crystalline element or compound may be taken as zero at zero Kelvin. Since the standard state temperature is 25°C and Entropy increases with temperature, S° must be greater than zero for an element in its standard state
Why does Appendix B list ∆Hof values but not ∆So
f
Ans: Since Entropy values have a reference point (0 Entropy at 0 K), actual Entropy values can be determined, not just Entropy changes
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Practice ProblemPredict the spontaneity of the following:
Water evaporates from a puddle
Spontaneous, evaporation occurs because a few of the liquid molecules have enough energy to break away from the intermolecular forces of the other liquid molecules and move spontaneously into the gas phase
A lion chases an antelope
Spontaneous, a lion spontaneously chases an antelope without added force
An isotope undergoes radioactive decay
Spontaneous, an unstable substance decays spontaneously to a more stable substance
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Practice Problem Earth moves around the sun
Spontaneous
A boulder rolls up a hill
The movement of a boulder against gravity is nonspontaneous
Sodium metal and Chlorine gas form solid Sodium Chloride
The reaction of an active metal (Sodium) with an active nonmetal (Chlorine) is spontaneous
Methane burns in air
Spontaneous, with a small amount of energy input, Methane will continue to burn without additional energy (the reaction itself provides the necessary energy) until it is used up
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Practice Problem A teaspoon of sugar dissolve in hot coffee
Spontaneous, the dissolved sugar molecules have more states they can occupy than the crystalline sugar, so the reaction proceeds in the direction of dissolution
A soft-boiled egg become raw
Not spontaneous, a cooked egg will not become raw again, no matter how long it sits or how many times it is mixed
Water decomposes to H2 & O2 at 298 oK
Water is a very stable compound; its decomposition at 298 K and 1 atm is not spontaneous
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Practice ProblemCalculate Suniv and state whether the reaction occurs spontaneously at 298oK for the following reaction
o2 2 3 sysN (g) + 3H (g) 2NH (g) ΔS = - 197 J / K
For the reaction to react spontaneously :
osurr sysTo find ΔS , determine ΔH
o osys rxnΔH = ΔH
osys 3 2 2ΔH = [(2 mol NH )(-45.9 kJ / mol)]-[(3 mol H )(0 kJ / mol) + (1 mol N (0 kJ / mol)]
rxn prod reactH = ΔH - ΔHo o o
osysΔH = - 91.8 kJ
syssur o
1000 J-91.8 kJ x ΔH kJΔS = - = - = 308 J / K
T 298 K
univ sys surΔS = ΔS + ΔS = -197 J / K + 308 J / K = 111 J / K
ounivΔS > 0 Reaction proceeds spontaneously at 298 K
univ surrΔS > 0 and S > +197 J / Krxn(From ΔH values in Appendix B)
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Thermodynamics Entropy Change and the Equilibrium State
For a process “spontaneously” approaching equilibrium, the change in Entropy is positive
At equilibrium, there is no net change in the flow or energy to either the system or the surroundings
Any change in Entropy in the system is exactly balanced by an opposite Entropy change in the surroundings
univΔS > 0
univΔS = 0
univ sys surAt Equilibrium : ΔS = ΔS + ΔS = 0
sys surΔS = - S
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Practice ProblemCalculate Suniv for the vaporization of 1 mol water at 100oC (373oK)
Entropy of System is increasing as Heat is absorbed from surroundings changing the liquid to a gas
Compute Ssys from Standard Molar Entropies (from Appendix B)
Compute Ssurr from Hosys and Temperature (T = 373oK)
o2 2H O(l) H O(g) @ 373 K)
o o o o o oprod reac 2 2ΔS = mS - nS = S of H 0(g, 373 K) - S of H O(l, 373 K)o
sys
ΔS = 195.9 J / K - 86.8 J / K = 109.1 J / Kosys
sur
ΔHΔS = -
T
osys
o 3sys
sur
ΔH 40.7 x 10 JΔS = - = - = -109 J / K
T 373 K
univ sys surrΔS = ΔS + ΔS = 109.1 - 109 0
o o o 3sys vapwhere ΔH = ΔH at 373 K = 40.7 kJ / mol = 40.7 x 10 J / mol
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Thermodynamics Summary – Spontaneous Exothermic &
Endothermic Reactions Exothermic Reaction (Hsys < 0)
Heat, released from system, is absorbed by surroundings
Increased freedom of motion and energy dispersal in surroundings (Ssurr > 0)
Ex. Exothermic where Entropy change:(Ssys) > 0
orxn products reactantsΔS = m S - n S > 0o o
6 12 6 2 2 2C H O (s) + 6O (g) 6CO (g) + 6H O(g) + Heat
6 moles gas yields 12 moles gas and heat
sys sur univ sys surΔS > 0 ΔS > 0 then ΔS = ΔS + ΔS > 0
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Thermodynamics Summary – Spontaneous Exothermic &
Endothermic Reactions (Con’t)
Exothermic Reaction (Hsys < 0)
Ex. Exothermic where Entropy change (Ssys) < 0
Entropy in surroundings must increase even more (Ssurr > > 0) to make the total S positive 2 3CaO(s) + CO (g) CaCO + Heat
Entropy of system decreases because :
The amount (mol) of gas decreases
Heat released increases Entropy of surroundings even more
sys sur univ sys surΔS < 0 ΔS > > 0 then ΔS = ΔS + ΔS > 0
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Thermodynamics Summary – Spontaneous Exothermic &
Endothermic Reactions (Con’t) Endothermic Reaction (Hsys > 0)
Heat lost by surroundings decreases the molecular freedom of motion and dispersal of energy
Entropy of surroundings decreases (Ssurr) < 0 Only way an Endothermic reaction can occur
spontaneously is if (Ssys) > 0 and large enough to outweigh the negative Ssurr
Ex. Solution Process for many ionic compounds Heat is absorbed to form solution Entropy of surroundings decreases However, when crystalline solids become free-
moving ions, the Entropy increase in the system is quite large (Ssys) > > 0
Ssys increase far outweighs negative Ssurr
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Practice ProblemAcetone, CH3COCH3, is a volatile liquid solvent (it is used in nail polish, for example). The standard Enthalpy of formation of the liquid at 25 oC is -247.6 kJ/mol; the same quantity for the vapor is -216.6 kJ/mol. What is the Entropy change when 1.00 mol of liquid acetone vaporizes at 25 oC?
o3 3 3 3CH COCH (l) CH COCH (g) @ 298 K)
o o o o o oprod reacΔH = mH - nH = H of Acetone(g, 298 K) - H of Acetone(l, 298 K)o
sys
ΔH = - 216.6 J - (-247.6 J) = 31.0 Josys
osys
sur o
ΔH 31.0 JΔS = - = - = - 0.104 J / K
T 298 K
Endothermic reaction - Energy from surroundings is input to system to vaporize acetone (∆Ho
sys is positive)Energy (temperature) of surroundings is decreased, decreasing Entropy
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Gibbs Free Energy Entropy, Free Energy and Work
Gibbs Free Energy (G) Using Hsys & Ssurr , it can be predicted
whether a reaction will be “Spontaneous” at a particular temperature
J. Willard Gibbs developed a single criterion for spontaneity The Gibbs “Free Energy” (G) is a function
that combines the system’s Enthalpy (H) and Entropy (S)
G = H - TS
- sys sysΔG ΔSsysH T
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Gibbs Free Energy Gibbs Free Energy Change and Reaction
Spontaneity
The Free Energy Change (G) is a measure of the spontaneity of a process and of the useful energy available from it univ sys surAt Equilibrium : ΔS = ΔS + ΔS = 0
syssurΔS = -
T
H
sysuniv sys sur sys
ΔHΔS = ΔS + ΔS = ΔS -
T
sys sysTΔS ΔS - univ T H
univ sys sys-TΔS = ΔH - TΔS
univ sys sys sys-TΔS = ΔG = ΔH - TΔS
o o osys sys sysΔG = ΔH - TΔS "Standard Free Energy Change"
Recall :
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Gibbs Free Energy Gibbs Free Energy Change and Reaction Spontaneity
The sign of G tells if a reaction is spontaneous From the 2nd Law of Thermodynamics
Suniv > 0 for spontaneous reaction Suniv < 0 for nonspontaneous reaction Suniv = 0 for process in “Equilibrium”
Absolute Temperature (K) is “always positive”
G < 0 for a spontaneous process G > 0 for a nonspontaneous process G = 0 for a process in equilibrium
sys sys sys-TΔS ΔS = G - univ H T
univ univTΔS > 0 or - TΔS < 0 for spontaneous process
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Practice ProblemCalculate Gsys
o (Grxno) at 25oC for the following
reaction
Calculate Hsyso from Hf
o values from tables
Δ3 44KClO (s) 3KClO (s) + KCL(s)
o o o osys rxn f(products) f(reactants)ΔH = ΔH = m ΔH - ΔHn
o o osys 4 f 4 f
o3 f 3
ΔH = [(3 mol KClO )(ΔH of KClO ) + (1 mol KCl)(ΔH of KCl)]
-[(4 mol KClO )(ΔH of KClO )]
osysΔH = [(3 mol)(- 432.8 kJ / mol) + (1 mol)( - 436.7 kJ / mol)]
- [(4 mol)( - 397.7 kJ / mol)]
osysΔH = -144 kJ Con’t
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Practice Problem (Con’t)Calculate Ssys
o from So values from tables
Calculate Gsyso at 298oK
o o o osys rxn products reactantsΔS = ΔS = mΔS - ΔSn
o o osys rxn 4 4
o3 3
ΔS = ΔS = [(3 mol KClO )(S of KClO ) + (1 mol KCl)(S of KCl)]
- [(4 mol KClO )(S of KClO )]
o osys rxnΔS = ΔS = [(3 mol)(151.0 J / mol K) + (1 mol)(82.6 J / mol )]
- [(4 mol)(143.1 J / mol K)]
K
- ΔG ΔSo o osys sys sysH T
osys
J 1 kJΔG = - 144 kJ - (298 K)(-36.8 ) = - 133 kJ
K 1000 J
o osys rxnΔS = ΔS = - 36.8 J / K
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Gibbs Free Energy Standard Free Energy of Formation (Gf
o)
Gfo is the free energy change that occurs when
1 mole of compound is made from its “elements” and all of the components are in their “standard” states
Gfo values have properties similar to Hf
o values
Gfo of an element in its standard form is
“zero” An equation coefficient (m or n) multiplies Gf
o by that number
Reversing a reaction changes the sign of Gfo
Gfo values are obtained from tables
o o orxn f(products) f(reactants)ΔG = mΔG - ΔGn
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Thermodynamics G and the Work (w) a System Can Do
For a Spontaneous process (G < 0) at constant Temperature (T) and Pressure (P), G is the “Maximum” of useful work obtainable from the system as the process takes place
For a Nonspontaneous process (G > 0) at constant T & P, G is the “Minimum” work that must be done to the system to make the process take place
In any process, neither the “maximum” or the “minimum” work is achieved because some “Heat” is lost
A reaction at equilibrium, which includes phase changes (G = 0), can no longer do “any work”
maxΔG = W
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Thermodynamics The Effect of Temperature on Reaction Spontaneity
When the signs of H & S are the same, some reactions that are non-spontaneous at one temperature become spontaneous at another, and vice versa
The temperature at which a reaction becomes spontaneous is the temperature at which a
“Positive” G switches to a “Negative” G This occurs because of the changing magnitude
of the
-T S term This cross-over temperature (reaction at
equilibrium) occurs when G = 0 Thus:
ΔG = 0 = ΔH - TΔSΔH = TΔS
ΔHT =
ΔS
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Thermodynamics The Effect of Temperature on Reaction
Spontaneity
Reactions Independent of Temperature Spontaneous Reaction at all Temperatures
H < 0 (Exothermic) S > 0 (- TS) term is always negative G is always “negative”
Nonspontaneous Reaction at all Temperatures H > 0 (Endothermic) S < 0 Both oppose spontaneity - TS is positive G is always positive
- sys sysΔG ΔS sys H T
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Practice ProblemPredict spontaneity of the following reactions
2 2 2 22H O (l) 2H O(l) + O (g)
ΔH = -196 kJ ΔS = 125 J / K
ΔH = < 0 ΔS > 0 - TΔS < 0
ΔG = ΔH - TΔS < 0
Reaction is spontaneous at all temperatures
2 33O (g) 2O (g)
ΔH = 286 kJ ΔS = -137 J / K
ΔH = > 0 ΔS < 0 - TΔS > 0
ΔG = ΔH - TΔS > 0
Reaction is not spontaneous at any temperature
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Thermondynamics Temperature Dependent Reactions
When H & S have the same sign, the relative magnitudes of the –TS and Hterms determine the sign of G
Reaction is spontaneous at high Temperatures
H > 0 S > 0
S favors spontaneity (-TS) < 0)
H does not favor spontaneity
Spontaneity will occur only when -TS (generally high temperature) is large enough to make G negative
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Practice Problem Predict spontaneity of the following reaction
2N2O(g) + O2(g) 4NO(g)
∆H = 197.1 kJ and ∆S = 198.2 J/K
With a Positive ∆H, the reaction will be spontaneous only when - T∆S is large enough to make ∆G negative
This would occur at “Higher” temperatures
The oxidation of N2O occurs spontaneously at
T > 994 K
ΔG = ΔH - TΔS < 0
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Thermodynamics Temperature Dependent Reactions (Con’t)
When H & S have the same sign, the relative magnitudes of the (– TS) and Hterms determine the sign of G
Reaction is spontaneous at lower Temperatures
H < 0 S < 0
H favors spontaneity
S does not favor spontaneity (- TS) > 0)
G will only be negative when -TS is smaller the H term, usually at a lower temperature
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Practice Problem Predict spontaneity of the following reaction
4Fe(s) + 3O2(g) 2Fe2O3(s)
∆H = -1651 kJ and ∆S = -549.4 J/K
∆H favors spontaneity, but ∆S does not
With a negative ∆H, the reaction will occur spontaneously only if the -T ∆S term is smaller than the ∆H term.
This happens only at lower temperatures
The production of Iron(III) oxide occurs spontaneously at any T < 3005 K
ΔG = ΔH - TΔS > 0
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Thermodynamics Summary – Reaction Spontaneity and the Sign of
∆H ∆S -T∆S ∆G Description
— + — — Spontaneous at all Temperatures
+ — + + Nonspontaneous at all Temperatures
+ + ——+
Spontaneous at Higher TemperatureNonspontaneous at Lower
Temperatures
— — + —+
Spontaneous at Lower TemperaturesNonspontaneous at Higher
Temperatures
- ΔG ΔSo o osys sys sysH T
sys sysΔG ΔS - sysH T
ΔHRecall : At G = 0 T =
ΔS
ΔH, ΔS, and ΔG
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Practice ProblemPredict the spontaneity of the following reaction
2 22N O(g) + O (g) 4NO(g)
ΔH = 197.1 kJ ΔS = 198.2 J / K
ΔH = > 0 ΔS > 0 - TΔS = ?ΔG = ΔH - TΔS = ?
Reaction will be spontaneous when Temperature
is high enough to make ΔG negative
2 2 34Fe(s) + 3O (g) 2Fe O (s)
ΔH = -1651 kJ ΔS = - 549.4 J / K
ΔH = < 0 ΔS < 0 - TΔS = ?
ΔG = ΔH - TΔS = ?
With the negative H, the reaction will be spontaneous
only if - T S is smaller than the H to make G negative
This would have to occur at lower temperatures
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Thermodynamics Free Energy, Equilibrium, and Reaction Direction
From Chapter 17
Q < K (Q/K < 1) – reaction proceeds “Right”
Q > K (Q/K > 1) – reaction proceeds “Left”
Q = K (Q/K = 1) – Reaction has reached “Equilibrium”
Energy & Spontaneity
Exothermic (H < 0) – reaction proceeds “Right”
Endothermic (H > 0) – reaction proceeds “Left”
Free Energy & Spontaneity
G < 0 for a spontaneous process
G > 0 for a nonspontaneous process
G = 0 for a process in equilibrium
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Thermodynamics Relationship between Q/K and G
If Q/K < 1, then ln(Q/K) < 0 and if G < 0
Then: Reaction is Exothermic and spontaneous If Q/K > 1, then ln(Q/K) > 0 and if G > 0
Then: Reaction is Endothermic and nonspontaneous If Q/K = 1, then ln(Q/K) = 0 and if G = 0
Then: Reaction has reached equilibrium In each case the signs of ln(Q/K) and G are the
same for a given reaction direction Gibbs noted that ln(Q/K) and G are proportional
to each other and are related (made equal) by the proportionality constant “RT”
QΔG = RT ln = RT ln Q - RT ln K
K
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Thermodynamics Recall: Q represents the concentrations (or pressures)
of systems components at any time during the reaction, whereas, K represents the concentrations when the reaction has reached “equilibrium”
G depends on how the Q ratio of the concentrations differs from the equilibrium ratio, K
Expressing G when “Q” is at standard state conditions All concentrations are = 1 M (pressures = 1 atm) Q = 1
Standard Free Energy (Go) can be computed from the Equilibrium constant (K)
Logarithmic relationship means a “small” change in Go has a large effect on the value of K
oΔG = RT ln 1 - RT ln KoΔG = RT* 0 - RT ln K = - RT ln K
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Thermodynamics For expressing the free energy for nonstandard
initial conditions
Substitute Go equation into G equationoΔG = RT ln Q - RT ln K = RT ln Q - (-ΔG )
oΔG = ΔG RT ln Q
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Practice ProblemIf the partial pressures of all species in the reaction below are 0.50 atm, what is G (kJ) for the reaction at 25oC?
Kp = 0.16 3 2 5PCl (g) + Cl (g) PCl (g)
5
3 2
PClp
Cl Cl
P 0.50 0.50Q = = = = 2.0
P P (0.50)(0.50) 0.25
Q 2.0ΔG = RT ln = (8.314 J / mol K)(298 K) ln
K 0.16
3
1 kJJ
1000 JΔG = 6.3 10 = 6.3 kJ / molmol
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Practice ProblemCalculate the value of the thermodynamic equilibrium constant (K) at 25 oC for the reaction given below:
The values of standard free energy of formation of the substances in kJ/mol at 25 oC are NO2, 51.30; N2O4, 97.82)
2 4 2N O (g) 2NO (g) o o orxn f(products) f(reactants)ΔG = mΔG - nΔG
42 2
o o orxn f(NO ) f(N O )ΔG = mΔG - nΔG = (2 51.30) - (97.82)
orxnΔG = 4.78 kJ
oΔG = - RT ln K
o1000
4.78 kJ / molΔGln K = - = - = -1.93
JRT 8.314 298Kmol • K
JkJ
-1.93K = e = 0.15
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Practice ProblemObtain the Kp at 35oC for the reaction in the previous problem
The standard enthalpies of formation of the substances in kJ/mol at 25oC are:
N2O4 9.16 J/mol-K NO2 33.2 J/mol-K
The standard molar entropies at 25 oC are:
N2O4 304.3 J/mol-K NO2 239.9 J/mol-K 2 4 2N O (g) 2NO (g)
2 42
o o o osys rxn NO N OΔS = ΔS = 2ΔS - ΔS
o o o oprod reac 2 2 4ΔH = mH - nH = 2H of NO ) - H of N O )o
sys
ΔH = (2 33.2) - (9.16) = 57.24 kJ / molosys
o osys rxnΔS = ΔS = (2 239.9) - (304.3) = 175.5 J / mol K
o o orxn products reactantsΔS = m S - n S
Con’t
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Practice Problem (Con’t)
- ΔG ΔSo o osys sys sysH T
oΔG = - RT ln K
osysΔG = 57.24 kJ / mol - 308 K 175.5 J / mol • K
o 3
oo
ΔG 3.19 x 10 J / molln K = - = - = - 1.25
JRT 8.314 x 308 Kmol • K
o 3sys
1000 JkJ
kJΔG = 57.24 - 308 K x 175.5 J / mol • K = 3.19 x 10 J / molmol
-1.25K = e = 0.29
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Summary Laws of Thermodynamics
1st Law – The change in the Internal Energy of a closed system (E) is equal to the amount of Heat (q) supplied to the system, minus the amount of Work (w) performed by the system on its surroundings
Limitations of the 1st Law of Thermodynamics
The 1st Law accounts for the energy involved in a chemical process (reaction)
The first Law, however, does not account for the “direction” of the change in energy
ΔE = q + w = q - P V
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Summary 2nd Law of Thermodynamics
The total Entropy of a system and its surroundings always increases for a “Spontaneous” process
All real processes occur spontaneously in the direction that increases the Entropy of the universe (system + surroundings)
The 2nd Law of Thermodynamics states that the change in Entropy for a gas expanding into a vacuum is related to the heat absorbed (qrev) and the temperature (T) at which the exchange occurs rev
sys
qΔS =
T
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Summary 3rd Law of Thermodynamics
The Entropy of a system approaches a constant value as the temperature approaches zero
The Entropy of a perfect crystal at “absolute zero” is zero
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Summary Equations
w P Δ V
pΔE q w pΔE q (-P ΔV)
pq ΔE P ΔV
H E P V
pΔH = q (at constant Pressure)
univ sys surrE = E + E
sys surr(q + w) = - (q + w)
sys surr sys surr univΔE = - ΔE ΔE + ΔE = 0 = ΔE
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Summary EquationsS = k lnW
-23
23
A
R 8.31447J / mol • Kk = = = 1.38 x 10 J / K
N 6.02214 x 10
-23
A
RS = ln W = 1.38 x 10 x lnW
N
final final finalsys
initial initial initial
V P CΔS = R x ln = R x ln = R x ln
V p C
S = R ln(V, P, C)
sys final initial final initialΔS = S - S = R x ln (V, P, C) - R x ln (V, P, C)
revsys
qΔS =
T
syssurr
qΔS = -
T
syssurrΔS = -
T
H
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Summary Equationsrxn products reactantsΔS = m S - n So o o
sys surr surrExothermic Change : q < 0 q > 0 ΔS > 0
sys surr surrEndothermic Change : q > 0 q < 0 ΔS < 0
univ sys surrAt Equilibrium : ΔS = ΔS + ΔS = 0
sys surrΔS = - S
- sys sysΔG ΔS sys H T
QΔG = RT ln = RT ln Q - RT log K
K
o oproducts reactantsΔH = mH - nHo
sys
o o orxn f(products) f(reactants)ΔG = mΔG - nΔG
univ sys surrΔS = ΔS + ΔS > 0 (spontaneous process)
oΔG = - RT ln K (Q = 1 at standard state)
oΔG = ΔG + RT ln Q (at any non - standard state)
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Summary EquatFree Energy, Equilibrium, and Reaction Direction
Q < K (Q/K < 1) – reaction proceeds “Right”
Q > K (Q/K > 1) – reaction proceeds “Left”
Q = K (Q/K = 1) – reaction has reached “Equilibrium”
Energy & Spontaneity
Exothermic (H < 0) – reaction proceeds “Right”
Endothermic (H > 0) – reaction proceeds “Left”
Free Energy & Spontaneity
G < 0 for a spontaneous process
G > 0 for a nonspontaneous process
G = 0 for a process in equilibrium
