Spontaneity, Entropy, and Free Energy Chapter 16

1st Law of Thermodynamics The first law of thermodynamics is a statement of the law of conservation of energy: energy can neither be created nor destroyed. The energy of the universe is constant, but the various forms of energy can be interchanged in physical and chemical processes. 1st law does not give a hint as to WHY a process occurs in a given direction.

Spontaneous Processes and Entropy Thermodynamics lets us predict whether a process will occur but gives no information about the amount of time required for the process. A spontaneous process is one that occurs without outside intervention. SPONTANEOUS DOES NOT MEAN FAST! Tells us the information about the direction of the reaction AP exam wants you to use the phrase “thermodynamically favorable”

Kinetics & Thermodynamics Chemical kinetics focuses on the pathway between reactants and products--the kinetics of a reaction depends upon activation energy, temperature, concentration, and catalysts. Thermodynamics only considers the initial and final states. To describe a reaction fully, both kinetics and thermodynamics are necessary.

The rate of a reaction depends on the pathway from reactants to products. Thermodynamics tells whether the reaction is thermodynamically favorable and depends upon initial & final states only.

Entropy The driving force for a spontaneous process is an increase in the entropy of the universe. Entropy, S, can be viewed as a measure of randomness, or disorder. Nature spontaneously proceeds toward the states that have the highest probabilities of existing. Think of your bedroom…it naturally tends to get messy. An ordered room requires everything to be in its place…There are many more ways for things to be out of place than in place

Entropy Entropy is a thermodynamic function that describes the number of arrangements (positions and/or energy levels) that are available to a system existing in a system Is closely linked to probability Think of the expansion of an ideal gas, more possibility of it being spread out than concentrated in one area

The expansion of an ideal gas into an evacuated bulb.

Ssolid < Sliquid << Sgas Positional Entropy A gas expands into a vacuum because the expanded state has the highest positional probability of states available to the system. Therefore, Ssolid < Sliquid << Sgas

Positional Entropy Which of the following has higher positional entropy? a) Solid CO2 or gaseous CO2? b) N2 gas at 1 atm or N2 gas at 1.0 x 10-2 atm?

Entropy What is the sign of the entropy change for the following? a) Solid sugar is added to water to form a solution? S is positive b) Iodine vapor condenses on a cold surface to form crystals? S is negative

SUniverse Suniverse is positive -- reaction is spontaneous. Suniverse is negative -- reaction is spontaneous in the reverse direction. Suniverse = 0 -- reaction is at equilibrium.

Concept Check The evaporation of alcohol The freezing of water Predict the sign of S for each of the following, and explain: The evaporation of alcohol The freezing of water Compressing an ideal gas at constant temperature Heating an ideal gas at constant pressure Dissolving NaCl in water + – a) + (a liquid is turning into a gas) b) - (more order in a solid than a liquid) c) - (the volume of the container is decreasing) d) + (the volume of the container is increasing) e) + (there is less order as the salt dissociates and spreads throughout the water)

The Second Law of Thermodynamics . . . in any spontaneous process there is always an increase in the entropy of the universe. Suniv > 0 for a spontaneous process.

Second Law of Thermodynamics In any spontaneous process there is always an increase in the entropy of the universe. The entropy of the universe is increasing. The total energy of the universe is constant, but the entropy is increasing. Suniverse = ΔSsystem + ΔSsurroundings Copyright © Cengage Learning. All rights reserved

ΔSsurr

∆Ssys depends on what we have already talked about Increases when Going from solid to liquid to gas Solution formation occurs Pressure descreases (increases in volume) # of molecules going from reactants to products increases ∆Ssurr sign depends on direction of heat flow & temperature Higher temp-lower ∆S surr; Lower temp- higher ∆S Surr Consider: H2O(l)  H2O (g) ∆S sys: + ∆S surr: - (endothermic process)

Ssurroundings Ssurr = - Hsystem T Ssurr is positive -- heat flows into the surroundings out of the system. (EXOTHERMIC) Ssurr is negative -- heat flows out of the surroundings and into the system. (ENDOTHERMIC) Ssurr = - Hsystem T

Ex. Is this rxn thermodynamically favored at 25˚C? N2(g) + 3 H2(g)  2NH3 (g) ∆H= -92.6 kJ, ∆Ssys= -199 J/K Suniverse = ΔSsystem + Δssurroundings = -199 + 310.74 = 111.74 J/K YES is THERMODYNAMICALLY FAVORED because + Ssurr = - Hsystem T Ssurr = - (-92.6) (298) Ssurr = .31074 kJ/K = 310.74 J/K

Ssurroundings Calculations Sb2S3(s) + 3Fe(s) ---> 2Sb(s) + 3FeS(s) H = -125 kJ Sb4O6(s) + 6C(s) ---> 4Sb(s) + 6CO(g) H = 778 kJ What is Ssurr for these reactions at 250C & 1 atm. Ssurr = - Hsystem T Ssurr = -(-125kJ/298K) Ssurr = 419 J/K Ssurr = - Hsystem T Ssurr = -(778kJ/298K) Ssurr = -2.61 x 103 J/K

G -- Free Energy Two tendencies exist in nature: tendency toward higher entropy -- S tendency toward lower energy -- H If the two processes oppose each other (e.g. melting ice cube), then the direction is decided by the Free Energy, G, and depends upon the temperature.

Gsys means +Suniv Free Energy G = H  TS (from the standpoint of the system) A process (at constant T, P) is spontaneous in the direction in which free energy decreases: Gsys means +Suniv Entropy changes in the surroundings are primarily determined by the heat flow. An exothermic process in the system increases the entropy of the surroundings.

G, H, & S Thermodynamically favored, Spontaneous reactions are indicated by the following signs: G = negative H = negative S = positive

Temperature Dependence Ho & So are not temperature dependent. Go is temperature dependent. G = H  TS

G = negative -- spontaneous Free Energy G G = H  TS G = negative -- spontaneous G = positive -- spontaneous in opposite direction G = 0 -- at equilibrium

Effect of H and S on Spontaneity

Calculations showing that the melting of ice is temperature dependent. The process is spontaneous above 0oC.

Concept Check A liquid is vaporized at its boiling point. Predict the signs of: w q H S Ssurr G Explain your answers. – (gas expanding) + (endothermic process) + (endothermic process) + (going from liquid to gas) – ( -∆H/T, ∆H is positive) 0 (∆H favors reverse process, ∆S favors forward process; these tendencies will exactly balance out at boiling point, and system will be at equilbrium) As a liquid goes to vapor, it does work on the surroundings (expansion occurs). Heat is required for this process. Thus, w = negative; q = H = positive. S = positive (a gas is more disordered than a liquid), and Ssurr = negative (heat comes from the surroundings to the system); G = 0 because the system is at its boiling point and therefore at equilibrium.

Concept Check Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The bond energy of AB is much greater than that of either reactant. Predict the signs of: H Ssurr S Suniv Explain. – + 0 + Since the average bond energy of the products is greater than the average bond energies of the reactants, the reaction is exothermic as written. Thus, the sign of H is negative; Ssurr is positive; S is close to zero (cannot tell for sure); and Suniv is positive.

Since the average bond energy of the products is greater than the average bond energies of the reactants, the reaction is exothermic as written. Thus, the sign of H is negative; Ssurr is positive; S is close to zero (cannot tell for sure); and Suniv is positive.

Boiling Point Calculations What is the normal boiling point for liquid Br2? Br2(l) ---> Br2(g) Ho = 31.0 kJ/mol & So = 93.0 J/Kmol At equilibrium, Go = 0 Go = Ho  TS0 = 0 Ho = TS0 T = Ho/S0 T = 3.10 x 104 J/mol/(93.0J/Kmol) T = 333K

Go = Ho  TS0 At temperatures above 333 K, TS has a larger magnitude than H, and G would be negative (thermodynamically favored) At temperatures below 333 K, TS is smaller than H, and G would be positive (reverse reaction favored)

Thermodynamically favored? S G Thermodynamically favored? - + YES - high temps +low temps YES @ high temps -low temps +high temp YES @ low temps

The Third Law of Thermodynamics . . . the entropy of a perfect crystal at 0 K is zero. Because S is explicitly known (= 0) at 0 K, S˚ values at other temps can be calculated. See Appendix 4 for values of S0.

Standard Entropy Values (S°) Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure. ΔS°reaction = ΣnpS°products – ΣnrS°reactants

Soreaction Calculate S at 25 oC for the reaction 2NiS(s) + 3O2(g) ---> 2SO2(g) +2NiO(s) S = npS(products)  nrS(reactants) S = [(2 mol SO2)(248 J/Kmol) + (2 mol NiO)(38 J/Kmol)] - [(2 mol NiS)(53 J/Kmol) + (3 mol O2)(205 J/Kmol)] S = 496 J/K + 76 J/K - 106 J/K - 615 J/K S = -149 J/K # gaseous molecules decreases!

So So increases with: solid ---> liquid ---> gas greater complexity of molecules (have a greater number of rotations and vibrations) greater temperature (if volume increases) lower pressure (if volume increases)

2 SO2(g) + O2(g) ----> 2 SO3(g) G Calculations Calculate H, S, & G for the reaction 2 SO2(g) + O2(g) ----> 2 SO3(g) H = npHf(products)  nrHf(reactants) H = [(2 mol SO3)(-396 kJ/mol)]-[(2 mol SO2)(-297 kJ/mol) + (0 kJ/mol)] H = - 792 kJ + 594 kJ H = -198 kJ

G Calculations Continued S = npS(products)  nrS(reactants) S = [(2 mol SO3)(257 J/Kmol)]-[(2 mol SO2)(248 J/Kmol) + (1 mol O2)(205 J/Kmol)] S = 514 J/K - 496 J/K - 205 J/K S = -187 J/K

G Calculations Continued Go = Ho  TSo Go = - 198 kJ - (298 K)(-187 J/K)(1kJ/1000J) Go = - 198 kJ + 55.7 kJ Go = - 142 kJ The reaction is spontaneous at 25 oC and 1 atm.

Cdiamond(s) ---> Cgraphite(s) Hess’s Law & Go Cdiamond(s) + O2(g) ---> CO2(g) Go = -397 kJ Cgraphite(s) + O2(g) ---> CO2(g) Go = -394 kJ Calculate Go for the reaction Cdiamond(s) ---> Cgraphite(s) CO2(g) ---> Cgraphite(s) + O2(g) Go = +394 kJ Cdiamond(s) ---> Cgraphite(s) Go = -3 kJ Diamond is kinetically stable, but thermodynamically unstable.

Free Energy Change and Chemical Reactions G = standard free energy change that occurs if reactants in their standard state are converted to products in their standard state. G = npGf(products)  nrGf(reactants) The more negative the value of G, the further a reaction will go to the right to reach equilibrium.

Go & Temperature Go depends upon temperature. If a reaction must be carried out at temperatures higher than 25 oC, then Go must be recalculated from the Ho & So values for the reaction.

Thermodynamics & Kinetics A reaction may be favorable thermodynamically, but scientists need to study the kinetics of the reaction to see if it will proceed at a fast enough rate to be useful

G is pressure dependent S is pressure dependent Free Energy & Pressure The equilibrium position represents the lowest free energy value available to a particular system (reaction). G is pressure dependent S is pressure dependent H is not pressure dependent

Free Energy and Pressure G = G + RT ln(P) or G = G + RT ln(Q) Q = reaction quotient from the law of mass action.

Free Energy Calculations CO(g) + 2H2(g) ---> CH3OH(l) Calculate G for this reaction where CO(g) is 5.0 atm and H2(g) is 3.0 atm are converted to liquid methanol. G = G + RT ln(Q) G = npGf(products)  nrGf(reactants) G = [(1 mol CH3OH)(- 166 kJ/mol)]-[(1 mol CO)(-137 kJ/mol) + (0 kJ)] G = - 166 kJ + 137 kJ G = - 2.9 x 104 J

Free Energy Calculations Continued = 2.2 x 10-2

Free Energy Calculations Continued G = G + RT ln(Q) G = (-2.9 x 104 J/mol rxn) + (8.3145 J/Kmol)(298 K) ln(2.2 x 10-2) G = (- 2.9 x 104 J/mol rxn) - (9.4 x 103 J/mol rxn) G = - 38 kJ/ mol rxn Note: G is significantly more negative than G, implying that the reaction is more spontaneous at reactant pressures greater than 1 atm. Why? (Le Chatelier’s Princ)

A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.

As A is changed into B, the pressure and free energy of A decreases, while the pressure and free energy of B increases until they become equal at equilibrium.

Graph a) represents equilibrium starting from only reactants, while Graph b) starts from products only. Graph c) represents the graph for the total system.

Free Energy and Equilibrium G = RT ln(K) K = equilibrium constant This is so because G = 0 and Q = K at equilibrium.

G < 0 K > 1 (favored) Go & K Go = 0 K = 1 G < 0 K > 1 (favored) G > 0 K < 1 (not favored)

Equilibrium Calculations 4Fe(s) + 3O2(g) <---> 2Fe2O3(s) Calculate K for this reaction at 25 oC. Ho = - 1.652 x 106 J (from values in table) So = -543 J/K (from values in table) Go = - 1.490 x 106 J (from ∆ G= ∆ H-T∆S) G = RT ln(K) K = e - G/RT K = e 601 or 10261 K is very large because G is very negative.

Temperature Dependence of K y = mx + b (H and S  independent of temperature over a small temperature range) If the temperature increases, K decreases for exothermic reactions, but increases for endothermic reactions.

Free Energy & Work The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy: wmax = G

Reversible vs. Irreversible Processes Reversible: The universe is exactly the same as it was before the cyclic process. Irreversible: The universe is different after the cyclic process. All real processes are irreversible -- (some work is changed to heat).  w < G Work is changed to heat in the surroundings and the entropy of the universe increases.

Laws of Thermodynamics First Law: You can’t win, you can only break even. Second Law: You can’t break even.

Thermodynamically Unfavorable Processes that readily occur Light or electricity may be used to cause a process to occur that is thermodynamically unfavorable Examples: Charging a battery, process of electrolysis (next chapter) Photosynthesis Rxn has a ∆G0 of +2880 kJ/mol but light helps the multistep process occur through the absorption of several photons in the Visible Spectrum range (400-700 nm) Coupling Reactions may also lead to reactions that are not thermodynamically favorable occur

Thermodynamically unfavorable reaction can be driven by a favorable reaction Review Free Energy Change The free energy change (ΔG) of a chemical process is a measure of the energy available to do work. Standard free energy change ΔGo is when reaction conditions are standard: T is 250 C, P = 1 atm, and conc of all reactants is 1M. For biochemical reactions, the pH has to be 7. The standard free energy change at pH 7 is designated as ΔGo’ ΔG = Gproducts – Greactants If ΔG <0 , the forward reaction is favored relative to the reverse reaction. The products are more stable than the reactants. As reaction proceeds, energy is released, it can be used to do work If ΔG = 0, reactants and products are at equilibrium If ΔG >0, reactants are at lower energy than products; energy needs to be supplied for the reaction to proceed.

Thermodynamically unfavorable reaction can be driven by a favorable reaction ΔGo’ is directly dependent on the equilibrium constant K’eq. For a reaction: A + B C + D ΔG = ΔGo’ + RT. ln [C] [D]/[A] [B] ΔG depends on the nature of the reactants (expressed in the ΔGo’ term) as well as the concentration of the reactant and products At equilibrium, ΔG = 0, and K’eq = [C] [D]/[A] [B] ΔGo’ = - RT.ln K’eq or ΔGo’ = - 2.303 RT log K’eq since R= 1.987 x 10-3 kcal.mol-1.deg-1 and T = 298K, ΔGo’ = - 1.36 log K’eq or K’eq = 10 –ΔGo’ /1.36

******************************* Thermodynamically unfavorable reaction can be driven by a favorable reaction Two or more reactions can be coupled when one or more products of one reaction are reactants of the next reaction. When multiple reactions are coupled, it becomes a pathway. A pathway must satisfy minimally two criteria: The individual reactions must be specific, yielding only one particular product or set of products. The entire set of reactions in a pathway must be thermodynamically favored An important thermodynamic fact: the overall free energy change for a chemically coupled series of reactions is equal to the sum of the free-energy changes of the individual steps Thus, a thermodynamically unfavorable reaction can be driven by a thermodynamically favorable reaction to which it is coupled. A  B + C G0’ = + 5 kcal mol-1 B  D G0’ = - 8 kcal mol-1 ******************************* A  C + D G0’ = - 3 kcal mol-1

ATP is the universal currency of free energy Metabolism is facilitated by the use of a common energy currency, ATP. Part of the free energy derived from the oxidation of foodstuffs and from light is transformed into ATP - the energy currency. ATP: The role of ATP in energy metabolism is paramount! Nucleotide consisting of an adenine, a ribose, and a triphosphate unit Is an energy rich molecule triphosphate unit contains 2 phosphoanhydride bonds. A large amount of free energy is liberated when ATP is hydrolyzed to ADP & Pi, or ATP to AMP & PPi Under typical cellular conditions, the actual G for these hydrolysis is approximately -12 kcal mol-1 ATP + H2O  ADP + Pi G0’ = -7.3 kcal mol-1 ATP + H2O  AMP + PPi G0’ = -10.9 kcal mol-1

Adenine Ribose Phosphate units

ATP hydrolysis drives metabolism by shifting the equilibrium of coupled reactions Thermodynamically unfavorable reactions can proceed when coupled with ATP hydrolysis. For example: A B ΔGo’ = 4 kcal.mol-1, K’eq = 10 –ΔGo’ /1.36 then K’eq = 1.15 x 10-3 and equilibrium is reached at very low [B]. If this reaction is coupled to another reaction that has a very low (negative) ΔGo’, then overall ΔGo’ of coupled reaction is lowered such that Keq is increased significantly. Biochemical reactions are usually coupled to the hydrolysis of ATP: ATP ADP + Pi ΔGo’= -7.3 kcal.mol-1 Overall ΔGo’ = 4 + (-7.3) = -3.3 kcal.mol-1 Overall Keq’ = [B] /[A] x [ADP][Pi]/[ATP] = 267 In the cell, [ATP]/[ADP][Pi] is maintained at 500 M-1, therefore Keq’ = [B] /[A] x (1/500) =267 or [B]/[A] = 267 x 500 = 1.34 x 105 Coupling with 1 ATP changed equilibrium ratio by a factor of about 108. For nATPs it changes by 108n

Relaxed muscle + energy  contracted muscle Example Glycolysis (breaking down glucose) Protein Synthesis Contraction / Relaxation of Muscles Relaxed muscle + energy  contracted muscle ATP + H2O  ADP + P + energy

YOUTUBE video on ATP Coupling http://www.youtube.com/watch?v=4O6DFv2p5js&feature=player_detailpage