Chapter 19 Free Energy and Thermodynamics - PowerPoint Presentation

Chapter 19 Free Energy and Thermodynamics

Spontaneous vs Nonspontaneous A goal of thermodynamics is to predict the spontaneity of a process Spontaneous process- occurs without ongoing outside intervention Mechanical systems– letting go of a ball, it will automatically drop to decrease the potential energy Chemical systems- Less intuitive, but if the overall chemical potential drops or the entropy (disorder) increases, the process is spontaneous Nonspontaneous process – requires ongoing outside intervention to occur

Spontaneous processes are automatic, but not necessarily instantaneous Speed is dependent on the rate of the reaction/process

Phase changes

Entropy (S)A thermodynamic function that increases with the number of energetically equivalent ways to arrange components of a system to achieve a statek = boltzmann constant = 1.38 x 10-23 J/KW = number of energetically equivalent ways to arrange components in a system

Macrostate vs microstate Macrostate A given set of conditions of a system that defines its energy, for example P, V, and T. If the conditions are constant the energy of the system remains the same. The sum of the energy of all the individual particles is the same Microstate An individual particle within that system can possibly have multiple different energies at different times The more possible energies = larger W value

Microstate E=2J E=3J E=1J System A System B Microstate 1 Microstate 1 Microstate 2 W = 1 W = 2 System B will have more entropy

Entropy and spontaneity Second law of thermodynamics – For any spontaneous process the entropy of the universe increases ΔS univ > 0, spontaneous process Like enthalpy , H, entropy, S, is a state function ΔS = S final - S initial A chemical system proceeds in a direction that will increase the entropy of the universe

Entropy and state changes Going from solid to liquid to gas increases entropy The number of equivalent energy states increases

Calculating Entropy In processes, like the evaporating of water, there is heat exchange that occurs from the system and the surroundings The heat exchange can be related to entropy at a constant temperature q rev = heat exchanged in a reversible process (phase change) Units are Joules T = some constant temperature Units are Kelvin

Calculating entropy for a phase changeWhat is the change in entropy for 1.00 mol of water at 273 K to melt from the solid to liquid phase?

Practice Calculate the change in entropy that occurs in a system when 25.0 g of water condenses from a gas to a liquid at 100.0 ° C. The enthalpy of condensation is -40.7 (kJ/mol).

Entropy and the Universe Δ S sys = entropy of system Δ S surr = entropy of surroundings Δ S sys can decrease and still be a spontaneous process At below 0 degrees Celsius water will spontaneously freeze To still have an increase in entropy of the universe the Δ S surr must be positive and larger than Δ S sys

Even though the ΔSsys decreases (the water becomes more ordered), the process is exothermic and gives off heat to the surrounding so the ΔSsurr increasesThe increase in ΔSsurr is larger than the decrease in ΔSsys , so the ΔSuniv is increased overallAn exothermic processes increases the ΔSsurr An endothermic processes decreases the ΔSsurr

Temperature dependence of ΔSsurr Although water freezing below 0 degrees Celsius is spontaneous and increases the entropy of the surroundings, above 0 degrees Celsius water freezing is non spontaneousΔSsurr is temperature dependentAt high enough temperatures, ΔSuniv is negative for the freezing of water, a non spontaneous process

Quantifying ΔSsurr The heat going out of the system goes into the surroundings and vice versaA process that emits heat into the surroundings (ΔH is negative) increases ΔSsurr A process that absorbs heat from the surroundings ( Δ H is positive) decreases Δ Ssurr

Practice1. Calculate the ΔSsurr when this reaction occurs at 25 °C.2. Is this process exothermic or endothermic3. Determine whether the ΔSsys is negative or positive4. Is the ΔSuniv negative or positive? 5. Is the process spontaneous or non spontaneous

Gibbs Free energy (G) Also called the chemical potential , similar to potential in mechanical systems Related to whether a process is spontaneous or nonspontaneous Δ G = Δ H - T Δ S If Δ G > 0 reaction is nonspontaneous If Δ G < 0 reaction is spontaneous

Determining G and spontaneityG determines spontaneity, to determine the sign of G we must know H and S, sometimes T is also relevantΔG = ΔH - TΔS Δ HΔSΔ G Spontaneous? Case 1 positive negative positive No Case 2 negative positive negative Yes Case 3 positive positive T dependant T dependant Case 4 negative negative T dependant T dependant

Determining Enthalpy and Entropy sign Exothermic reaction = negative enthalpy Endothermic reaction = positive enthalpy Process that increases in entropy, for example going from 2 moles of gas to 4 moles of gas, will = positive entropy Process that decreases in entropy, for example going from gas phase to liquid phase = negative entropy

Temperature dependence ΔG = ΔH - TΔSCase 3 – positive H and positive SThe magnitude of T determines if the G value is negative or positive.Higher Temperatures, G is negativeLower temperatures, G is positiveCase 4 – negative H and negative S The magnitude of T determines if the G value is negative or positive.Higher Temperatures, G is positiveLower temperatures, G is negative

Practice CCl4(g)  C(s) + 2Cl 2 (g) ΔH = +95.7kJ, ΔS = +142.2 J/K Calculate ΔG at 25 ° C and determine whether the reaction is spontaneous. Determine the temperature at which the reaction turns from spontaneous to nonspontaneous

Part 1 Δ G = Δ H - T Δ S Δ H = +95.7 kJ or 95700 J Δ S = +142.2 J/K T = 25 ° C or 298 K Δ G = (95700 J) – (298 K)(142.2 J/K) Δ G = +53300 J Δ G > 0, non spontaneous reaction

Part 2 To find the temperature where it changes from spontaneous to nonspontaneous, set G = 0 and then solve for T This is the point at which G switches between negative and positive Δ G = Δ H – T Δ S 0 = Δ H – T Δ S T Δ S = Δ H T = Δ H/ Δ S T = (95700 J)/ (142.2 J/K) T = 673 K

Summary

Problems C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g) Δ H = -137.5kJ, Δ S = -120.5 J/K Calculate Δ G at 25 ° C and determine whether the reaction is spontaneous. If you increase the temperature, will Δ G increase or decrease?

Calculating standard entropy for reactions Δ S° rxn Standard conditions Gas –pure gas at 1 atm of pressure Liquid or solid- pure substance in stable form, 1atm pressure and usually 25 ° C Substance in solution- 1M solution Δ S ° rxn = S ° products - S ° reactants

Standard molar entropies (S°) Third law of thermodynamics – The entropy of a perfect crystal at absolute zero (0 K) is zero Serves as a reference point to calculate all other

Trends in relative standard entropies Phase Molar Mass AllotropesMolecular complexity Dissolution

Phase More dispersed or disordered state has more entropySolid < liquid < gas

Molar Mass Larger molar mass corresponds to a bigger atom/molecule, which typically leads to a larger entropy

Allotropes More ordered allotrope has less entropyLess ordered allotrope has more entropy

Molecular Complexity More atoms in a molecule increases complexity of molecule, increasing the entropy, more significant than molar mass

Dissolution When solids are dissolved, there is generally an increase in entropy

Additional Note Whether reactant or product, the value of the molar entropy will be the same

Calculating ΔS°rxn ΔS°rxn = S°products - S°reactantsΔS°rxn = ΣnpS°products - Σnr S°reactantsnp = stoichiometric coefficient of productnr = stoichiometric coefficient of reactantS° of products and reactants determined from table of values

Practice4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)Calculate the ΔS°rxn for the balanced chemical equation. Reactant or product S° (J/(mol*K))NH3(g)192.8 O 2 (g) 205.2 NO(g) 210.8 H 2 O(g) 188.8 ΔS° rxn = Σ n p S° products - Σ n r S° reactants

Calculating ΔG°rxn Using known values of S ° of reactants and products we can calculate Δ S ° rxn Using known values of H ° of reactants and products we can calculate Δ H ° rxn

Additional noteThrough this method we can estimate what the ΔG°rxn will be at other temperatures besides 25°CThis is due to the change in ΔS°rxn and ΔH°rxn at different temperatures is small in comparison to the value of the temperature itself

Calculating ΔG ° rxn again

Calculating Δ G ° rxn again and again If multiple reactions with known Δ G° rxn values can sum up to an overall reaction with an unknown Δ G° rxn , then by summing up the known Δ G° rxn ’s the unknown Δ G° rxn for the overall net reaction can be determined Depending on how individual reactions are used to build the overall reaction, Δ G° rxn values are summed up with certain rules Same method to solve Δ H° rxn from individual steps (previously learned)

Rules for summing up equations and ΔG°rxn If a chemical equation is multiplied by some factor, then ΔG°rxn is also multiplied by the same factorA + B  C Δ G ° rxn = 5kJ 2A + 2B  2C Δ G ° rxn =10kJ If a chemical equation is reversed, the sign of Δ G ° rxn is also reversed A + B  C Δ G ° rxn = 5kJ C  A + B Δ G ° rxn = -5kJ If a chemical equation is expressed as a sum of a series of steps, then Δ G ° rxn for the overall equation is the sum of the free energies of reactions for each step A + B  C Δ G ° rxn = 5kJ C  D ΔG°rxn = 7kJA + B  D ΔG°rxn = 12kJ

Example Overall rxn: F  2G Reaction 1: F + E  X + Y ΔG°rxn = RReaction 2: X  G ΔG°rxn = S Reaction 3: E + X  Y ΔG° rxn = T Use reactions to build overall reaction ……. Reaction 1: F + E  X+ Y Δ G ° rxn = R 2(Reaction 2): 2X  2G 2( Δ G ° rxn )= 2S -(Reaction 3): Y  E + X -( Δ G ° rxn )=- T Overall rxn 1: F  2G Δ G° rxn = R + 2S - T

Summary Use known reactions to build overall reaction.Multiply or reverse reactions if necessaryKeep track of which reaction you multiplied or reversedSum the ΔG °rxn values of all the reactions used to make the overall reactionIf you multiply a chemical reaction, multiply the ΔG°rxn for that step before adding to the totalIf you reversed a reaction, reverse the sign of the ΔG°rxn for that step before adding to the total

Problem 3C(s) + 4H2(g)  C3 H8(g) ΔG°rxn = ????Use known ΔG° rxn values to calculate the Δ G °rxn of the above reaction and determine whether the reaction is spontaneous. C 3 H 8 (g) +5O 2 (g)  3CO 2 (g) + 4H 2 O(g) Δ G ° rxn = -2074 kJ C(s) + O 2 (g)  CO 2 (g) Δ G ° rxn = -394.4 kJ 2H 2 (g) +O 2 (g)  2H 2 O(g) Δ G ° rxn = - 457.1 kJ

Calculating Δ G rxn : Non standard conditions Δ G rxn = Δ G ° rxn + (RT) ln Q Q = reaction quotient Q c = concentrations at any point in the reaction Q p = partial pressure at any point in reaction P = partial pressure of a gas

Example2NO(g) + O2(g)  2NO2(g)Determine ΔGrxn with the following partial pressures and ΔG°rxn = -71.2kJ NO = 0.100 atm, O2 = 0.100 atm, NO2 = 2.00 atm

Relationship between Δ G ° rxn and K Δ G rxn = Δ G ° rxn + (RT) ln Q At equilibrium Q = K, Δ G rxn = 0 0 = Δ G ° rxn + (RT) ln K - (RT) ln K = Δ G ° rxn

Relationship between Δ G ° rxn and K Δ G ° rxn = - (RT) ln K K<1, Δ G ° rxn is positive, Under standard conditions(when Q=1) the reaction is spontaneous in the reverse direction K>1, Δ G ° rxn is negative, Under standard conditions(when Q=1) the reaction is spontaneous in the forward direction K=1, Δ G ° rxn is 0, Under standard conditions(when Q=1) the reaction is at equilibrium

Why Free Energy is Free

Chapter 18 Summary