* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download A(g) - wwphs
Survey
Document related concepts
Spinodal decomposition wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Stability constants of complexes wikipedia , lookup
Eigenstate thermalization hypothesis wikipedia , lookup
Heat transfer physics wikipedia , lookup
Marcus theory wikipedia , lookup
Determination of equilibrium constants wikipedia , lookup
George S. Hammond wikipedia , lookup
Gibbs paradox wikipedia , lookup
Maximum entropy thermodynamics wikipedia , lookup
Work (thermodynamics) wikipedia , lookup
Thermodynamics wikipedia , lookup
Equilibrium chemistry wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Transcript
Chapter 16 Spontaneity, entropy and free energy 1st Law of Thermodynamics When methane and oxygen react to form carbon dioxide and water, the products have lower potential energy than the reactants. This change in potential energy results in thermal energy flow (heat) to the surroundings. But the energy of the universe remains “constant” 2 1st Law of Thermodynamics Keeps track of the energy How much energy is involved in the change? What form is the energy in? Does the energy flow in or out of the system? ***But does not tell us why?** 3 Spontaneous A reaction that will occur without outside intervention. We can’t determine how fast. We need both thermodynamics and kinetics to describe a reaction completely. Thermodynamics compares initial and final states. Kinetics describes pathway between (activation energy, [concentration], T, catalysts. The rate of a reaction depends on the pathway from reactants to products; this is the domain of kinetics. Thermodynamics tells us whether a reaction is spontaneous based only on the properties of the reactants and products. The predictions of thermodynamics do not require knowledge of the pathway between reactants and products. Thermodynamics 1st Law- the energy of the universe is constant. Keeps track of thermodynamics doesn’t correctly predict spontaneity. What makes a process spontaneous? – A ball rolls down a hill, but never back up – Steel rusts – A gas fills a container uniformly – Why???? – One common characteristic in all… 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. 7 Defined Entropy in terms of probability. Substances take the arrangement that is most likely. The most likely is the most random. Calculate the number of arrangements for a system. Nature spontaneously proceeds toward states that have the highest probability of existing. The expansion of an ideal gas into an evacuated bulb. Why is this Spontaneous? Exist in more states, the more spread out the molecules. 9 Physical evidence of entropy 4 possible arrangements (microstates) 25% chance of finding the left empty 1/22=1/4 50 % chance of them being evenly dispersed 2 4 atoms ½ = 1/16 8% chance of finding the left empty 50 % chance of them being evenly dispersed 4 The greater the disorder the larger the entropy. 12 Entropy from physical evidence Ssolid Gases <Sliquid <<Sgas have a huge number of positions possible. (Positional probability) Pure solid dissolves in a solvent, the entropy of the substance increases – Except:Carbonates Entropy increases with increasing molecular complexity (more e- moving) Increase # moles, increase entropy Concept Check Predict the sign of S for each of the following, and explain: + a) The evaporation of alcohol – b) The freezing of water – c) Compressing an ideal gas at constant temperature + d) Heating an ideal gas at constant pressure + e) Dissolving NaCl in 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. Suniv= Ssys + Ssurr 15 Entropy Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate. ∆Suniv = ∆ Ssys + ∆ Ssurr If ∆ Suniv is positive the process is spontaneous. If ∆ Suniv is negative the process is spontaneous in the opposite direction. Entropy Changes in Chemical Reactions N2(g) +3H2(g) 2NH3(g) We go from 4 to 2 gas molecules. What happens to the ∆S, (+,-)? ∆S – The change in positional entropy is dominated by the relative number of molecules of gaseous reactants and products. (Although change is important we can assign absolute entropy values.) Predict the sign of ΔS° a. CaCo3(s) →CaO(s) + CO2(g) + b. 2SO2(g) + O2(g) → 2SO3(g) - What if the # of molecules is the same? Generally, the more complex the molecule, the higher the standard entropy value. 18 Third Law of Thermo The entropy of a pure crystal at 0 K is 0. Gives us a starting point for finding values of S at other temperatures which must be > 0. Remember Entropy is a state function. Figure 16.5: (a) A perfect crystal of hydrogen chloride at 0 K. (b) As the temperature rises above 0 K, lattice vibrations allow some dipoles to change their orientations, producing some disorder and an increase in entropy. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 20 S reaction n p S products nr S reac tan ts Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. ∆S is also an extensive property, so it is dependent on the amount. 21 Calculate ΔS ° Al2O3(s) + 3H2(g) → 2Al(s) + 3H20(g) S reaction Substance Al2O3 H2 Al H20 n p S products nr S reac tan ts S°(J/K·mol) 51 131 28 189 (56J/K + 567J/K) – (393J/K + 51J/K) = 179J/K The more complex the molecule, the higher Is the standard entropy value. 22 Figure 16.6: The H2O molecule can vibrate and rotate in several ways, some of which are shown here. Effect of Temperature ∆Suniv = ∆ Ssys + ∆ Ssurr exothermic processes ∆H∆Ssurr is (+) endothermic processes ∆H+ ∆ Ssurr is (-) Consider this process: H2O(l) H2O(g) These are in ∆ Ssys is positive Opposition Which controls ∆ Ssurr is negative The situation? ∆ Suniv depends on temperature. At 1atm water changes from liquid to gas: Above 100°C it is spontaneous. Below 100°C the opposite process is spontaneous The lower the T, the greater the impact. Δssys = heat transferred T at which the transfer occured ΔSuni= ΔSsys+ ΔSsurr ΔSsurr= + when E is added (exo) = - when E is used (endo) Express ΔSsurr in terms of enthalpy ΔH Heat flow(q)(constant P) = change in enthalpy= ΔH Into and out of the system + endothermic/- exothermic At constant temperature and pressure: S surr H T (-)Sign changes the point of View from the sys. to the surr. 26 2H2 (g) + 2O2(g) H2O(g) Ignite & very fast! (spontaneous) Ssys = -88.9 J/K why??? Hsys = -483.6 kJ H S surr Ssurr = ? 1620 J/K T Suniv = ? 1530J/K Even though the entropy of the system declines, the entropy of the surroundings is so large that the overall change is positive. Bottom line: A process is spontaneous in spite of a negative entropy change as long as it is extremely exothermic. Sufficient exothermicity offsets system ordering. Ssys -H/T Ssurr Suniv + + + No, Reverse + - ? At High temp. - + ? Spontaneous? Yes At Low temp. Free Energy G = H TS (from the standpoint of the system) A process (at constant T, P) is spontaneous in the direction in which free energy decreases: **S (refers to the system)** G means +Suniv 30 G = H TS To see how this equation relates to spontaneity, Divide by –T G H H S remember S surr T T T At constant T and P G H S S surr S S uni T T S uni G T 31 This means; A process at constant T and P is only spontaneous if ΔG is negative. (decreasing) ΔG > 0 entropy of universe decreases ΔG < 0 entropy of universe increases Lets use the free energy equation to predict the spontaneity of melting of ice: H2O(s)→H2O(l) 32 H2O(s)→H2O(l) Based on your daily experiences you would say melting ice is spontaneous at T> 0,and not spontaneous at T< 0. G = H TS ENTROPY DECREASING HEAT BEING RELEASED ENTROPY INCREASING 33 Example At what temperatures is the following process spontaneous at 1 atm? Br2(l)→Br2(g) ΔH°=31.0 kJ/mol and ΔS°=93.0 J/K ·mol What is the normal boiling point of liquid Br2? 34 Br2(l)→Br2(g) ΔH°=31.0 kJ/mol and ΔS°=93.0 J/K ·mol 1. Vaporization will occur at all T where ΔG is negative 2. ΔS° favors vaporization due to increase in positional entropy. 3. ΔH° favors the opposite process (exothermic) 4. The opposite tendencies will balance at the boiling pt. At this BP temp. liquid and gaseous Br2 are in equilibrium ΔG °= 0. ΔG °= ΔH °- TΔS ° 0 = ΔH °- T ΔS ° ΔH °= T ΔS ° T= ΔH / ΔS ° 3.1x104J/mol/93.0J/K·mol = 333K 35 T> 333K, TΔS° has larger magnitiude than ΔH °; ΔG ° is negative. Above 333K , vaporization is spontaneous; opposite occurs spontaneously below this T. At 333K, liquid and gaseous Br2 coexist in equilibrium. 36 Summary 1. 2. 3. T > 333K. ΔS° controls. Increase in entropy when Br2(l) is vaporized. T < 333K. The process is spontaneous in the direction in which it is exothermic. ΔH° controls. T = 333K. ΔS° and ΔH° are balanced (ΔG° = 0), liquid and gas coexist. Normal BP. 37 G=H-TS S H + - At all Temperatures + At high temperatures, “entropy driven” - At low temperatures, “enthalpy driven” + Not at any temperature, Reverse is spontaneous + - Spontaneous? Free Energy in Reactions Gº = standard free energy change. Free energy change that will occur if reactants in their standard state turn to products in their standard state. Why Free energy changes? Tells us relative tendancy of a reaction to occur. The more - ΔG° , the more to the right the reaction will proceed. Use standard state because energy varies w/ pressure or concentration We must compare all reactions under the same pressure or concentration conditions. 40 ΔG° not measured directly (calculated 3 ways) 1. 2. 3. G ° = H ° TS ° (constant T) Free energy is also a state function, therefore, we can use a process similar to finding H using Hess’s Law. Standard free energy of formation (Gf) SG = SnpGf(products) SnrGf(reactants) 41 G ° = H ° TS ° (constant T) Consider the reaction 2SO2(g) + O2(g) → 2SO3(g) Carried out at 25C and 1 atm. Calculate ΔH°, ΔS°& ΔG° using the following data: Substance SO2(g) O2(g) SO3(g) ΔH°f (kJ/mol) -297 -396 0 S°(J/K.mol) 248 257 205 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 42 ΔH°? From enthalpies of formation. ΔH°= 2mol(-396kJ/mol) – 2mol(-297kJ/mol) – 0 = -198 kJ ΔS°? Standard entropy values ΔS°= 2mol(257) – 2mol(248) – 1mol(205) = -187J/K & ΔG° using the following data: 43 ΔG°? G ° = H ° TS ° = -198kJ – (298K)(-7J/K)(1kJ/1000J) = -142kJ 44 Free energy is also a state function, therefore, we can use a process similar to finding ΔH using Hess’s Law. Use the following data (at 25C) Cdiamond(s) + O2(g) → CO2(g) Cgraphite(s) + O2(g) → CO2(g) ΔG°= -397kJ ΔG°= -394kJ Calculate ΔG° for the reaction Cdiamomd (s)→ Cgraphite(s) 45 Cdiamond(s) + O2(g) → CO2(g) ΔG°=-397kJ CO2(g) → Cgraphite(s) + O2(g) ΔG°=394kJ =-3kJ The reaction is spontaneous but very slow at 25C and 1atm. Reverse occurs at high T and P 46 SG = SnpGf(products) nrGf(reactants) Change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products. Standard free energy of formation of an element in its standard state is zero. 47 Methanol is a high-octane fuel used in high-performance racing engines. Calculate ΔG° for the reaction, 2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g) Free energies of formation are: Substance ΔG°(kJ/mol) CH3OH -163 O2 0 CO2 -394 H 2O -229 = 2mol(-394kJ/mol) + 4mol(-229kJ/mol) – 3(0)+ 2mol(-163kJ/mol) = -1378 kJ 48 Free Energy in Reactions are tables of Gºf . Products-reactants because it is a state function. The standard free energy of formation for any element in its standard state is 0. Remember- Spontaneity tells us nothing about rate. There Dependence of Free Energy on Pressure • • • A system at constant T and P will proceed spontaneously in the direction to lower G. The equilibrium position represents the lowest G value available to a reaction system. G changes as a reaction proceeds because G is dependent on P and concentration. 50 G = H - TS • How does pressure affect the thermodynamic of free energy? • For an ideal gas H is not P dependent • S depends on P, because of the dependence on V. • Therefore G depends on P. 51 Consider 1 mol of ideal gas at a given Temp • At V=10.0L there are more positions available than at V=1.0L Slarger V > Ssmall V Slow P > Shigh P (P and V are inversely proportional.) 52 G = G° + RT ln(P) G°= P at 1atm R= gas constant G = P at P atm T=K To demonstrate how the G is pressure dependent: N2(g) + 3H2(g) → 2NH3(g) G = SnpGf(products) SnrGf(reactants) 53 N2(g) + 3H2(g) → 2NH3(g) G = SnpG(products) SnrG(reactants) G = S2GNH3 S1GN2 + 3GH2 GNH3 = G°NH3 + RT ln(PNH3) GN2 = G°N2 + RT ln(PN2) GH2 = G°H2 + RT ln(PH2) 54 G = 2[G°NH3 + RT ln(PNH3)] [G°N2 + RT ln(PN2)] -3[G°H2 + RT ln(PH2)] G =(2G°NH3- G°N2 -3G°H2) + RT[2ln(PNH3)ln(PN2)- 3ln(PH2)] G= G° + RT[2ln(PNH3)- ln(PN2)- 3ln(PH2)] 55 N2(g) + 3H2(g) → 2NH3(g) G= G° + RT[2ln(PNH3)- ln(PN2)- 3ln(PH2)] PNH3 2 G G RT ln ( PN 2 )( PH2 ) 2lnPNH3= lnPNH32 3 -3ln(PH2)= ln(1/PH2 ) PNH3 2 ( PN 2 )( PH2 ) -ln(PN2)= ln(1/PN2 ) Q Reaction quotient Q=K no shift Q>K left Q<K right 56 Free Energy and Pressure G = G + RT ln(Q) 57 One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases: CO(g) +2H2(g) → CH3OH(l) Calculate ΔG at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol. 58 CO(g) +2H2(g) → CH3OH(l) G = G + RT ln(Q) Gf(CH3OH)= -166kJ Gf(H2)= 0 Gf(CO)= -137kJ = -166 – (-137) – 0= -29kJ = -2.9x104J ΔG°= -2.9x104J/mol rxn. R= 8.3145J/K mol T=298K Q=1/(P(co))(PH22)= 1/(5.0)(3.02)= 2.2x10-2 ΔG= -2.9x104J/mol rxn + (8.3145J/K mol)(298K)ln(2.2x10-2) = -38kJ/mol rxn **G is more negative than G° 59 •We have seen that the formation of CH3OH is spontaneous. •But will it really form in a flask under these conditions? -no! -it is true that 1mol of CH3OH has lower G than CO or H2. However, when they are mixed under these conditions there is an even lower free energy available to this system than 1.0 mol of pure CH3OH. **The system can achieve lowest possible free energy by going to equilibrium, not going to completion. 60 Figure 16.7: Schematic representations of balls rolling down two types of hills. Analogous to Phase change Chemical Change Lowest possible Free energy The system can achieve lowest possible free energy by going to equilibrium, not going to completion 61 Free Energy and Equilibrium Equilibrium- when the forward and reverse reaction rates are equal. Thermodynamics view (equilibrium point) occurs at the lowest value of free energy available to the reaction system. 62 A(g)↔ B(g) 1.0 mol of A placed in a reaction vessel at a pressure of 2.0 atm. Free energy of A = GA= G°A + RTln(PA) Free energy of B = GB= G°B + RTln(PB) Total free energy of system =G =GA + GB 63 Figure 16.8: (a) The initial free energies of A and B. (b) As A(g) changes to B(g), the free energy of A decreases and that of B increases. (c) Eventually, pressures of A and B are achieved such that GA = GB, the equilibrium position. Suppose, minimum free energy is reached when 75% of A has been changed to B. At this point, PA=.25 its original P. (.25)(2.0atm)=0.50atm PB= (0.75)(2.0atm)= 1.5atm Since this is the equilibrium position, we use P to find K. PBe 15 . atm K e 3.0 0.50atm PA 65 Figure 16.9: (a) The change in free energy to reach equilibrium, beginning with 1.0 mol A(g) at PA = 2.0 atm. (b) The change in free energy to reach equilibrium, beginning with 1.0 mol B(g) at PB = 2.0 atm. (c) The free energy profile for A(g) → B(g) in a system containing 1.0 mol (A plus B) at PTOTAL = 2.0 atm. Each point on the curve corresponds to the total free energy of the system for a given combination of A and B. 66 Summary The reaction proceeds to the minimum G (equilibrium), which corresponds to the point where: Gprod.= Greact. ΔG = G prod – Greact. = 0 67 Free Energy and Equilibrium G = G RT ln(Q) at equilibrium G =0 and Q = K So G =0 =G +RT ln(K) G = RT ln(K) K = equilibrium constant 68 G = RT ln(K) 1. G = 0. The free energies of the reactants and products are equal when all components are in the standard states (1 atm for gases). The system is at equilibrium when the Preact and prod.=1atm. K=1 69 Case 2 G< 0 G = (Gprod- Greact) is negative Gprod < Greact the system will adjust to the right to reach equilibrium. K>1 since the Pprod. at equilibrium is >1atm and Preact.<1atm 70 Case 3 G>0 G = (Gprod- Greact) is positive Greact < Gprod the system will adjust to the left to reach equilibrium. K<1 since the Preact.>1atm and Pprod <1atm at equilibrium 71 72 Consider the ammonia synthesis reaction N2(g) + 3H2(g) ↔ 2NH3(g) Where ΔG°= -33.3kJ/mol of N2 consumed at 25C. For each of the following mixtures of reactants and products at 25C, predict the direction in which the system will shift to reach equilibrium. a. b. PNH3= 1.00atm, PN2=1.47atm, PH2=1.00x102atm PNH3= 1.00atm, PN2=1.00atm, PH2=1.00atm 73 N2(g) + 3H2(g) ↔ 2NH3(g) G = G RT ln(Q) PNH3= 1.00atm, PN2=1.47atm, PH2=1.00x102atm Q=(1.00)2/(1.47)(1.00x10-2)3= 6.80x105 T=298K R = 8.3145 J/K mol ΔG° = -33.3kJ/mol ΔG= -3.33x104J/mol + (8.3145 J/K mol ) (298K)(ln 6.80x105) = 0 At equilibrium no shift 74 N2(g) + 3H2(g) ↔ 2NH3(g) G = G RT ln(Q) PNH3= 1.00atm, PN2=1.00atm, PH2=1.00atm Q=(1.00)2/(1.00)(1.00)3= 1.00atm T=298K R = 8.3145 J/K mol ΔG° = -33.3kJ/mol ΔG= G RT ln(1) = G 0 ΔG= G 33.3kJ/mol Since it’s (-), shift right, K>1 75 The overall reaction for the corrosion (rusting) of iron by oxygen is + 3O2(g)↔ 2Fe2O3(s) Using the following data, calculate the equilibrium constant for this reaction at 25C. 4Fe(s) Substance Fe2O3 ΔHf° (kJ/mol) -826 S° (J/K . Mol) 90 Fe 0 27 O2 0 205 76 Temperature Dependence of K G RT ln( K ) H T S H S - H S ln( K ) (1 / T ) RT R R R y = mx + b y=ln(K), m=-ΔH/R=slope, xx=1/T, and b= ΔS/R=intercept ; will be linear if plotted (H and S independent of temperature over a small temperature range) Copyright©2000 by Houghton Mifflin Company. All rights reserved. 77 Free Energy and Work using a chemical process to do work Maximum possible useful work obtainable from a process at constant T and P is equal to the change in free energy. wmax = ΔG ΔG for a spontaneous process: energy that is free to do useful work ΔG for a non-spontaneous process: minimum amount of work that must be expended to make the process occur. 78 Knowing ∆G tells us about efficiency. Amount of work actually obtained from a spontaneous process is always less than the max possible amount. Energy is always wasted. 79 Chemical change in a battery can do work by sending current to a starter motor. ( Hypothetical Reversible process) Force current in opposite direction To recharge battery Current flowing through a wire: Energy is wasted as heat 80 Reversible v. 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). Surroundings have less of an ability to do the work. Entropy is increasing!!!! 81 At 1500°C for the reaction CO(g) + 2H2(g) → CH3OH(g) the equilibrium constant is Kp = 1.4 x 10-7. Is H° at this temperature: A. positive B. negative C. zero D. can not be determined The standard free energy (Grxn0 for the reaction N2(g) + 3H2(g) → 2NH3(g) is -32.9 kJ. Calculate the equilibrium constant for this reaction at 25oC. A. 13.3 B. 5.8 x 105 C. 2.5 D. 4.0 x 10-6 E. 9.1 x 108