2 Driving Forces of Reactions So far we have seen that reactions are spontaneous if they give off heat – exothermicThere is a natural tendency in the universe for systems to get to lowest energy stateWhy do endothermic reactions happen?Demo with bariumAmmonium chloride dissolving in H2O - endothermic
3 Natural directionScientists notice that there is a natural direction for processesBalls roll down hill, not up hillIce melts above 0º C, never refreezes above 0 CA gas fills its container uniformly, never collects in one areaHeat flows from hot to cold; hotter object doesn’t get hotter when exposed to a colder objectWood burns spontaneously, but CO2 and H2O don’t form wood when heated
4 What makes these processes irreversible? All of the processes have the following in common:in all cases you have less information about how the particles are organized than you did beforeAnalogy: Why don’t we have fire drills during lunch?Less “information” about where students are during lunch compared to when they are in classMeasure of “information” about a system = ENTROPY
5 Entropy Measure of the amount of randomness or disorder in a system Entropy was introduced in 1865 by Rudolf J. E. Clausius, a German physicist. Clausius said he derived the term from the Greek words en trope, which means “in the transformation” He used it to describe the dissipation or apparent loss of energy available to do work as energy is transformed in a system.Measure of the amount of randomness or disorder in a systemSymbol: SUnits: J/KNo matter what the process, entropy (of universe) is always increasing…
6 Entropy AnalogiesThrow a card into the air – 2 possible positions (up or down)Throw a deck of cards into the air – how many possible positions?One possibility is that they land organized in a stack. How probable is that?The more cards = the more entropyMORE MOLES = MORE ENTROPY
7 Given the following reaction, how is entropy changing: N2 (g) + 3 H2(g) 2NH3(g) IncreasingDecreasingStays the sameNeed more information123456789101112131415161718192021222324
8 What causes entropy to increase? Statistics…Boltzmann Bucks DemoAt first everyone has $1, we play rock-paper-scissors for awhile. Some people have more money than others, butno one has all the money andNot everyone has exactly $1Why not?There is only one way for the money to be arranged so that everyone has $1There are only 9 ways for the money to be arranged so that one person has $9 (assuming class size of 9)There are many ways for some people to have no money, some to have $1 and some to have $2 or $3.So there is a higher probability that the $ will be arranged as described in CNature spontaneously proceeds to the state that has the highest probability of existing.Highest probability = most disordered
9 Entropy Changes Increase moles Dissolving and mixing Increasing temperatureIncrease volumeSolid to liquid or liquid to gas (or S to G)More complicated molecules have higher S than simpler molecles
10 Which one of the following does not generally lead to an increase in entropy of a system? Increase in total number of moles or particlesFormation of a solutionFormation of a gasFormation of a solid123456789101112131415161718192021222324
11 Given the reaction below, how is entropy changing? Br2(l) Br2(g) IncreasingDecreasingStays the sameNeed more info123456789101112131415161718192021222324
12 Given the following reaction, how is entropy changing: Ag+1(aq) + Cl-1(aq) AgCl(s) IncreasingDecreasingStays the sameNeed more info123456789101112131415161718192021222324
13 Given the following reaction, how is entropy changing: 2NO2(g) N2O4(g) IncreasingDecreasingStays the sameNeed more info123456789101112131415161718192021222324
14 Given the following reaction, how is entropy changing: CO(g) + H2O(g) CO2(g) + H2(g) IncreasingDecreasingStays the sameNeed more info123456789101112131415161718192021222324
15 Given the following reaction, how is entropy changing: H2(g) + F2(g) 2HF(g) IncreasingDecreasingStays the sameNeed more info123456789101112131415161718192021222324252627282930
16 Given the following reaction, what is the sign for ΔS: NaCl(s) NaCl(aq) positivenegativeNeed more info123456789101112131415161718192021222324252627282930
17 Given the following reaction, how is entropy changing: 2OH-(aq) + CO2(g) H2O(l) + CO32- (aq) IncreasingDecreasingStays the sameNeed more info123456789101112131415161718192021222324252627282930
18 Which of the following has the largest increase in entropy? Pb(NO3)2(s) Pb(NO3)2(aq)CaCO3(s) CaO(s) + CO2(g)2NH3 (g) 2H2(g) + N2(g)H2(g) + Br2(g) 2HBr(g)123456789101112131415161718192021222324252627282930
19 Spontaneity Spontaneous change: Spontaneous reactions occur when Occurs w/o continuous input of energySpontaneous reactions occur whenReaction is exothermic ( ΔH < 0)Increase in entropy for the system (ΔS >0)But which is more important?ΔH or ΔS ?
20 Total Entropy System vs. Surroundings vs. Universe Suniverse = Ssystem + SsurroundingsSuniv is always increasing.Two spontaneous processes:CaCl2(s) Ca2+(aq) + Cl-1(aq) ΔH = -66kJSsys is increasingSsurr is increasing because heat is released to surroundingsNH4Cl(s) NH4+ + Cl-(aq) ΔH = 15 kJSsurr is decreasing b/c surroundings are losing heatNon-spotaneousNa(s) Na (l) ΔH =2.59 kJSsys = increasingSsurr = decreasing
21 NH4Cl(s) NH4+ + Cl-(aq) ΔH = 15 kJ Na(s) Na (l) ΔH =2.59 kJ Na+(g) + Cl-(g) NaCl(s) ΔH = -771 kJWhy is A spontaneous, but not B?Entropy is much greater for AWhy is C spontaneous?Enthalpy is large
22 How do you know if enthalpy or entropy will make the reaction more spontaneous? Remember: Suniverse = Ssystem + SsurroundingsSsurr depends on temperatureThe lower the surrounding temperature, the more significant adding heat isThe higher the surrounding temp, the less significant adding heat is.AnalogyImagine you give $1 to someone w/ only $10 to their name?Imagine the effect of giving $1 to a millionaire.Who is affected more?
23 Suniverse = Ssystem + Ssurroundings Exothermic reactions that release heat to the surroundings are a a stronger driver of reactions when the surrounding temperature is low.At high temperatures, an exothermic reaction isn’t such a strong driving force, entropy is more important.Negative sign b/c ΔH defined in terms of system: exothermic reaction from system’s perspective causes increase in entropy of surroundings.Suniverse = Ssystem + SsurroundingsEntropy of systemDetermined by ΔH of system
24 Substitute: - ΔHsys/T for Ssurr Multiply by (-T) Define new quantity Suniv = Ssys + SsurrSubstitute: - ΔHsys/T for SsurrMultiply by (-T)Define new quantityFree energy: ΔG = -T ΔSunivFree energy tells whether a rxn will be spontaneous at a given temperature.Measures the maximum energy available to do useful workReactions at equilibrium have ΔG = 0-T ΔSuniv = ΔHsys - T ΔSsysΔG = ΔHsys - T ΔSsys
25 Based on the previous slides and derivation of ΔG, rxns will be spontaneous if ΔG is Less than 0Greater than 0Equal to 0Spontaneity has nothing to do with ΔG.123456789101112131415161718192021222324
26 Under which conditions will reactions ALWAYS be spontaneous? ΔH > 0, ΔS >0ΔH > 0, ΔS < 0ΔH < 0, ΔS >0ΔH < 0, ΔS <0ΔG = ΔHsys - T ΔSsys123456789101112131415161718192021222324
27 Under which conditions will reactions NEVER be spontaneous? ΔH > 0, ΔS >0ΔH > 0, ΔS < 0ΔH < 0, ΔS >0ΔH < 0, ΔS <0ΔG = ΔHsys - T ΔSsys123456789101112131415161718192021222324
28 Under which conditions will reactions be spontaneous at high temps Under which conditions will reactions be spontaneous at high temps? (other than when they are always spontaneous)ΔH > 0, ΔS >0ΔH > 0, ΔS < 0ΔH < 0, ΔS >0ΔH < 0, ΔS <0ΔG = ΔHsys - T ΔSsys123456789101112131415161718192021222324
29 ΔH > 0, ΔS >0 ΔG = ΔHsys - T ΔSsys A large positive value for the term (TΔS) can make ΔG negative if it is bigger than ΔHReaction is endothermic so the entropy of surroundings is decreasing. At high temps, this won’t make as big of a difference as it would at lower temps.ΔG = ΔHsys - T ΔSsys
30 Under which conditions will reactions be spontaneous at low temps Under which conditions will reactions be spontaneous at low temps? (other than when they are always spontaneous)ΔH > 0, ΔS >0ΔH > 0, ΔS < 0ΔH < 0, ΔS >0ΔH < 0, ΔS <0ΔG = ΔHsys - T ΔSsys123456789101112131415161718192021222324
31 ΔH < 0, ΔS < 0 ΔG = ΔHsys - T ΔSsys A small negative value for the term (TΔS) can still make ΔG negative if it is smaller than the absolute value of ΔHReaction is exothermic so the entropy of surroundings is increasing. At low temps, this will make a bigger difference than it would at higher temps.ΔG = ΔHsys - T ΔSsys
32 Calculating ΔGFor a reaction at 25 C, ΔH = 100 kJ and ΔS = 80 J/K, determine if the reaction is spontaneous.For a reaction with a ΔH = 100 kJ and a ΔS of 80 J/K, at what temperature will the reaction become spontaneous?Watch your units!! Put temps in Kelvin and make sure you aren’t trying to add Joules to Kilojoules!
33 100 KJ – (298*80/1000) = 76.2 kJ = not spontaneous 0 = (x*80/1000) 100 = x(. 08)x = 1250 K, reaction becomes spontaneous at temperatures above 1250 K
34 Calculations of ΔS°rxn, ΔH°rxn and ΔG°rxn Standard Entropy of Formation Tables (ΔS°f )Σ n(ΔS°f )products - Σ n(ΔS°f )reactantsStandard Gibbs Free Energy of Formation Tables (ΔG°f )Σ n(ΔG°f )products - Σ n(ΔG°f )reactantsOnly good for standard conditions!For ΔG at non-standard conditions, use ΔG = ΔH - TΔS
35 Calculate the free-energy change, DG°, for the oxidation of ethyl alcohol to acetic acid using standard free energies of formation.CH3CH2OH(l) + O2(g) CH3COOH(l) + H2O(l)
37 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Sodium carbonate, Na2CO3, can be prepared by heating sodium hydrogen carbonate, NaHCO3:2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)Estimate the temperature at which the reaction proceeds spontaneously at 1 atm. See Appendix C for data.
40 Use ΔG to get KEquilibrium position represents the lowest free energy value available to a particular reaction system
41 ΔG and KStandard free energy change is related to the thermodynamic equilibrium constant, K, at equilibrium.IF a reaction is NOT at equilibrium, it is proceeding in some direction (forward or reverse) depending on Q, reaction quotient.That means there exists energy to do work (make reaction proceed)ΔG = Δ G° + RT ln QAt equilibrium:Δ G = 0, because there is no ability to do any more useful workand Q = KSo we get:Δ G° = –RT ln K
42 Calculate the value of the thermodynamic equilibrium constant at 25°C for the reaction N2O4(g) NO2(g)The standard free energy of formation at 25°C is kJ/mol for NO2 and kJ/mol for N2O4(g).