2 Chemical Reactions: Why ? Why are some reactions reversible and others are not?HC2H3O2 < > C2H3O2 + HWhat does it mean that the Ka = 1.8 x 105 ?Not all reactions that you can write on paper will occur. CO2 (g) ---> C (s) + O2 (g) does not, but the reverse does.What determines if a reaction will occur?Most chemical reactions are exothermic. Why?Thermodynamics answers these questions.
3 Spontaneous Processes Certain processes will always occur.ice melts on a table at 20 Can iron nail left outside will rustSpontaneous Processes occur without outside interventionMay be fast or slow (combustion vs. conversion of coal to a diamond)
4 A spontaneous process has a definite direction, while energy is conserved in both forward and reverse directions.Temperature plays a roleExothermic processes tend to occur spontaneously.Spontaneous processes produce “Free Energy”.
5 Reversible vs. Irreversible Processes A reversible process:the system can be restored to it’s exact state by simply reversing the process which led to the change+QExample: ice ----> liquid waterQice < liquid water
6 Reversible vs. Irreversible Processes An irreversible process:the system cannot be restored by simplyreversing the procedure.It must be restored to its original state by a new pathway which changes the surroundings.Example: Freezing water at 30C,it takes work to keep it from freezingbetween 0 and 30 C.
7 Reversible vs. Irreversible Processes When a chemical system is in equilibrium, reactants and products can move back and forth reversibly.For any spontaneous process, the path between reactants and products is irreversible.Keep in Mind . . .a spontaneous reaction always has a certain direction, while energy is conserved in both forward and reverse directions.Some spontaneous reactions are VERY slow.
8 First Law of Thermodynamics Energy is neither created or destroyedThe energy of the universe is constantEnergy can be transferred between the system and the surroundingsDE = q + wDE = energy absorbed by the systemq = heat absorbed by systemw = work don on the system by the surroundings
9 Second Law of Thermodynamics The Entropy of the universe is increasing.Entropy, S, is disorder or randomness.The more disorder, the more randomnessUnits: J/K∆S = Sf Si+S = final state more disorderedS = initial state more disordered
10 Entropy (S)What would drive spontaneous forces? An increase or decrease in disorder?An increase in spontaneous forcesUnlike energy, entropy is NOT conserved.The Suniv is continually increasing.No process that produces order within the system can proceed without producing an equal or greater disorder in its surroundings.Entropy is defined in terms of probability.Substances take the arrangement that is most likely.The most likely is the most random
11 2 possible arrangements 50 % chance of finding the left empty Calculate the number of arrangements for a system.1 molecule2 possible arrangements50 % chance of finding the left empty
12 2 molecules4 possible arrangements25% chance of finding the left empty50 % chance of them being evenly dispersed
13 12 possible arrangements 8% chance of finding the left empty 4 molecules12 possible arrangements8% chance of finding the left empty50 % chance of them being evenly dispersedSubstances takethe most randomarrangement
14 Ssolid < Sliquid <<< Sgas EntropyWhich State of matter will have the most entropy? The least? Explain.Ssolid < Sliquid <<< Sgasthere are many more ways for gas molecules to be arranged than as a liquid than a solid.Gases have a huge number of positions possible.Solids are very orderly.
15 Entropy increases when: Temperature increases.Gases are formed from solids or liquids.The number of molecules of gas increases during a reaction.Solutions are formed from solids or liquids.Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate.
17 Entropy DSuniv = DSsys + DSsurr If DSuniv is positive the process is spontaneous.If DSuniv is negative the process is spontaneous in the opposite direction.For exothermic processes:DSsurr is +.For endothermic processes:DSsurr is .Consider this process: H2O(l)® H2O(g)DSsys is positiveDSsurr is negativeDSuniv depends on temperature.
18 Temperature and Spontaneity Entropy changes in the surroundings are determined by the heat flow.An exothermic process is favored because by giving up heat the entropy of the surroundings increases.The size of DSsurr depends on temperatureDSsurr = -DH/T
19 DSuniv = DSsys + DSsurr Yes No, Reverse ? At High Temp At Low Temp ? + Spontaneous?+++Yes--No, Reverse-+-?At High TempAt Low Temp-+?DSuniv = DSsys + DSsurr
20 Entropy Changes in Chemical Reactions Standard Molar Entropies, So: the molar entropy values of substances in their standard states.Where standard enthalpies of elements are 0, standard entropies are not 0.What state would have the highest standard molar entropies? The lowest?Gases = Highest, Solids = LowestSo increase or decrease with molar mass?IncreasesSo increase or decrease with increasing number of molecules involved in the formula?
21 Third Law of Thermodynamics 3rd Law- The entropy of a pure crystal at absolute zero ( 0 K) is 0.Gives us a starting point. All others must be > 0.
22 Entropy Changes in Chemical Reactions DSo = SnSo (products) – SmSo (reactants)S = “sum of”n and m are the coefficients in the chemical equation.Calculate the DSo for the synthesis of ammonia from N2 (g) and H2 (g) at 298 KN2 (g) + 3H2 (g) 2NH3 (g)DSo for N2 = J/mol KDSo for H2 = J/mol KDSo for NH3 = J/mol KDSo = 2 (192.5) – [(191.6) + 3(130.7)DSo = J/K … spontaneous or not?
23 Gibb's Free EnergyFree Energy= energy available to do work, (however, much is in the form of heat)All spontaneous processes produce free energy.Whether a process is spontaneous or not depends on 2 thermodynamic concepts:1. Enthalpy, ∆H and 2. Entropy, ∆S
24 Gibbs Free Energy DG = DH -TDS at constant temperature (use Kelvin!) Formula: G = H-TSNever used this way. But rather as:DG = DH -TDS at constant temperature (use Kelvin!)Divide by -T-DG/T = -DH/T-DS-DG/T = DSsurr + DS-DG/T = DSunivIf DG is negative at constant T and P, the process is spontaneous.If ∆G is positive, the process is nonspontaneous.If ∆G is 0, the process is at equilibrium.
25 - + - ? + + - - ? - + + DG = DH -TDS DH DS Spontaneous? ∆G At all TemperaturesAt high temperatures,“entropy driven”?++At low temperatures,“enthalpy driven”--?Not at any temperature,Reverse is spontaneous-++
26 Is this process spontaneous? For the reaction H2O(s) ® H2O(l)DSº = 22.1 J/K mol DHº =6030 J/molCalculate DG at 10ºC and -10ºCDG=DH-TDS∆G = 6030 J/mol - (283K)(22.1J/K mol)∆G =∆G = yes, at 10CDG= 6030 J/mol - (263K)(22.1J/K mol)∆G =∆G = no, at -10C
27 Free Energy and Work Free energy is that energy free to do work. The maximum amount of work possible at a given temperature and pressure.Never really achieved because some of the free energy is changed to heat during a change, so it can’t all be used to do work.
28 If ∆G = , reaction is spontaneous In SummaryIf ∆G = , reaction is spontaneousIf ∆G = +, reaction is Non-spontaneous;work MUST be done to reversethe reactionIf ∆G = 0, reaction is at equilibriumIf ∆G = very large (-) #, then a lot of energy is released to surr. and is capable of doing much work. (burning of gasoline)
29 Entropy in ReactionsEnthalpy, ∆H, is measured experimentallyNo easy way to measure Entropy, ∆SUse absolute Entropy, S, values from tables in back of textbookThese values are based on difficult calculations using absolute zero data.
30 Free Energy in Reactions DGº = standard free energy change.DGº = DHº - TDSºCan’t be measured directly, must be calculated from other measurements.Free energy change that will occur if reactants in their standard state turn to products in their standard state.∆G = ∑ n Gf products ∑ m Gf reactants
31 Free energy in Reactions DGo = DHo –TDSoExample:The Haber process for the production of ammonia involves the following equillibrium: N2 + 3H2 2NH3Assume DHo and DSo don’t change with temperature.Predict the direction in which DGo changes for this reaction with increasing temperature.Expect DSo to be (–) because number of moles of gas is smaller on product side, thus –TDS term will be positive and get bigger as temperature increase.
32 Calculate the values of DGo for the reaction at 25oC and 500oC. First we need to Calculate the DHo and DSo.DHo values = N2 (0); H2 (0); NH3 ( kJ/mol)DSo values = N2 (191.6 J/mol K); H2 (130.7 J/mol K); NH3 (192.5 J/mol K)At 298K, DGo = kJAt 773K, DGo = 61 kJ
33 Free Energy and Equilibrium For a system at equilibrium, ∆G = 0.we can find the ∆G for any conditions, not only ∆G, with this formula:DG = DGº +RT ln (Q)where Q is the reaction quotients (pressure of the products / pressure of the reactants), and R is the ideal gas law constant, 8.31 J/mol KAt equilibrium, Q = KeqDG = DGº +RT ln (Q)0 = DGº + RT ln (Keq)DGº = RT ln (Keq)
35 How far will an equilibrium reaction continue to go? During the course of a reaction, the free energy always decreases.DG tells us spontaneity at current conditions. When will it stop?It will go to the lowest possible free energy which may be an equilibrium.Analogy of boulder rolling down a hill
36 Equilibrium/Free Energy analogy ReactantsFree EnergyProductsPEEquilibriumposition invalleyMixture atEquilibriumCourse of ReactionPositionEquilibrium Position = Minimum free energy available to the system
37 Free Energy and Work “Free Energy” is the energy “free” to do “work” The amount of work obtained is always less than the max.Henry Bent’s First two Laws of Thermodynamics:“You can’t win, you can only break even”“ You can’t even break even”The amount of free energy of our system (Earth) is decreasing