George Mason University

George Mason University
General Chemistry 212 Chapter 20 Thermodynamics Acknowledgements Course Text: Chemistry: the Molecular Nature of Matter and Change, 7th edition, 2011, McGraw-Hill 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.

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

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

Thermodynamics Enthalpy (∆H) Sum of Internal Energy (E) plus
Product of Pressure & Volume (Endothermic vs. Exothermic) ( Hrxn = Hf prod - Hf react) (Constant Pressure) Entropy (S) Measure of system order/disorder & the number of ways energy can can be dispersed through the motion of its particles All real processes occur spontaneously in the direction that increases the Entropy of the universe (universe = system + surroundings) (Spontaneous) (At Equilibrium) Gibbs Free Energy (∆G) Difference between Enthalpy and the product of absolute temperature and the Entropy

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

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)

(i.e., goes to near completion, i.e., far to the right)
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

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)

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

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 = -PV) performed by the system on its surroundings The law of conservation of energy can be stated The Energy of an Isolated System is Constant

First Law of system Thermodynamics Conservation of Energy, E (or U in some texts) Any change in the energy of the system corresponds 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 = ½mv (kgm2/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 = kgm2/s2)

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/(ms2) Volume = m3 Work (W) = kg/(ms2)  m3 = kgm2/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:

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 Recall  (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 (PV) would have negative sign

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 kgm2/s2 1 Pa (pascal) = 1 kg/ms2 1 atm = x 105 Pa 1 atm = 760 torr = 760 mm Hg

Exchanges of Heat and Work with the Surroundings
q<0 q>0 w<0 by system w>0 on system Pressure x Volume Work = expansion of volume due to forming a gas

Practice Problem Consider 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 kJ. What is E for the change indicated by the chemical equation at 1 atm? n = 3 mol converted to 1 mol = oC and 1 atm, 1 mol of gas = 24.5 L, thus V = -2(24.5) = -49 L  (1m3/1000 L) = m3 E = q - PV E = kJ – 1 atm x ( m3) E = kJ – (1.01 x 105 Pa)( m3) E = kJ – (1.01 x 105 kg/ms2)(-0.049m3) E = kJ + (4949 J x 1 kJ/1000 J) E = kJ kJ = -885 kJ

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

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

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 surroundings always increases for a “Spontaneous” process A process occurs spontaneously in the direction that increases the Entropy of the universe

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.

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

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

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

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

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

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

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  Gas Dissolving of Salt: Crystalline Solid + Liquid  Ions in Solution Chemical 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

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

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

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:

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)

Entropy Entropy change – Volume, Pressure, Concentration

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

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

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

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

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

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

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/molK @ 298oK) Values available in Reference Tables (Appendix “B”)

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

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

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 >> Sfuso

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

Entropy is “Decreased”
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”

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

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

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

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

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

Practice Problem s < l < g
Choose 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

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

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)

Practice Problem Predict the sign of S for each process:
Alcohol Evaporates ΔSsys positive, the process described is liquid alcohol becoming gaseous alcohol The gas molecules have greater Entropy than the liquid molecules A solid explosive converts to a gas ΔSsys 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 room ΔSsys positive, the perfume molecules have more possible locations in the larger volume of the room than inside the bottle A system that has more possible arrangements has greater Entropy

Practice Problem Without 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

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, Sorxn is usually negative

Practice Problem Calculate Sorxn 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

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

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

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

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

Practice Problem How 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)

Practice Problem What 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 ∆Sof Ans: Since Entropy values have a reference point (0 Entropy at 0 K), actual Entropy values can be determined, not just Entropy changes

Practice Problem Predict 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

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

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 K and 1 atm is not spontaneous

Practice Problem Calculate Suniv and state whether the reaction occurs spontaneously at 298oK for the following reaction

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

Practice Problem Calculate 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)

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

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

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

Practice Problem Acetone, 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 kJ/mol; the same quantity for the vapor is kJ/mol. What is the Entropy change when 1.00 mol of liquid acetone vaporizes at 25 oC? Endothermic reaction - Energy from surroundings is input to system to vaporize acetone (∆Hosys is positive) Energy (temperature) of surroundings is decreased, decreasing Entropy

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)

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

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

Practice Problem Calculate Gsyso (Grxno) at 25oC for the following reaction Calculate Hsyso from Hfo values from tables Con’t

Calculate Ssyso from So values from tables Calculate Gsyso at 298oK

Gibbs Free Energy Standard Free Energy of Formation (Gfo)
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 Hfo values Gfo of an element in its standard form is “zero” An equation coefficient (m or n) multiplies Gfo by that number Reversing a reaction changes the sign of Gfo Gfo values are obtained from tables

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”

“Positive” G switches to a “Negative” G
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:

Thermodynamics The Effect of Temperature on Reaction Spontaneity
Reactions Independent of Temperature Spontaneous Reaction at all Temperatures H < 0 (Exothermic) S > 0 (- TS) term is always negative G is always “negative” Nonspontaneous Reaction at all Temperatures H > 0 (Endothermic) S < 0 Both oppose spontaneity - TS is positive G is always positive

Practice Problem Predict spontaneity of the following reactions

Thermondynamics Temperature Dependent Reactions
When H & S have the same sign, the relative magnitudes of the –TS and H terms determine the sign of G Reaction is spontaneous at high Temperatures H > S > 0 S favors spontaneity (-TS) < 0) H does not favor spontaneity Spontaneity will occur only when -TS (generally high temperature) is large enough to make G negative

Practice Problem Predict spontaneity of the following reaction
2N2O(g) O2(g)  4NO(g) ∆H = kJ and ∆S = 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

Thermodynamics Temperature Dependent Reactions (Con’t)
When H & S have the same sign, the relative magnitudes of the (– TS) and H terms determine the sign of G Reaction is spontaneous at lower Temperatures H < S < 0 H favors spontaneity S does not favor spontaneity (- TS) > 0) G will only be negative when -TS is smaller the H term, usually at a lower temperature

Practice Problem Predict spontaneity of the following reaction
4Fe(s) O2(g)  2Fe2O3(s) ∆H = kJ and ∆S = 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

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 Temperature Nonspontaneous at Lower Temperatures Spontaneous at Lower Temperatures Nonspontaneous at Higher Temperatures

Practice Problem Predict the spontaneity of the following reaction

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

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”

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

Thermodynamics For expressing the free energy for nonstandard initial conditions Substitute Go equation into G equation

Practice Problem If 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

Practice Problem Calculate 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)

Practice Problem Obtain 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: N2O J/mol-K NO J/mol-K The standard molar entropies at 25 oC are: N2O J/mol-K NO J/mol-K Con’t

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

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

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

Summary Equations

Summary Equat Free 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

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