# Entropy and Free Energy

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Entropy and Free Energy

Driving Forces of Reactions
So far we have seen that reactions are spontaneous if they give off heat – exothermic There is a natural tendency in the universe for systems to get to lowest energy state Why do endothermic reactions happen? Demo with barium Ammonium chloride dissolving in H2O - endothermic

Natural direction Scientists notice that there is a natural direction for processes Balls roll down hill, not up hill Ice melts above 0º C, never refreezes above 0 C A gas fills its container uniformly, never collects in one area Heat flows from hot to cold; hotter object doesn’t get hotter when exposed to a colder object Wood burns spontaneously, but CO2 and H2O don’t form wood when heated

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 before Analogy: Why don’t we have fire drills during lunch? Less “information” about where students are during lunch compared to when they are in class Measure of “information” about a system = ENTROPY

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 system Symbol: S Units: J/K No matter what the process, entropy (of universe) is always increasing…

Entropy Analogies Throw 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 entropy MORE MOLES = MORE ENTROPY

Given the following reaction, how is entropy changing: N2 (g) + 3 H2(g)  2NH3(g)
Increasing Decreasing Stays the same Need more information 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

What causes entropy to increase?
Statistics… Boltzmann Bucks Demo At first everyone has \$1, we play rock-paper-scissors for awhile. Some people have more money than others, but no one has all the money and Not everyone has exactly \$1 Why not? There is only one way for the money to be arranged so that everyone has \$1 There 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 C Nature spontaneously proceeds to the state that has the highest probability of existing. Highest probability = most disordered

Entropy Changes Increase moles Dissolving and mixing
Increasing temperature Increase volume Solid to liquid or liquid to gas (or S to G) More complicated molecules have higher S than simpler molecles

Which one of the following does not generally lead to an increase in entropy of a system?
Increase in total number of moles or particles Formation of a solution Formation of a gas Formation of a solid 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Given the reaction below, how is entropy changing? Br2(l)  Br2(g)
Increasing Decreasing Stays the same Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Given the following reaction, how is entropy changing: Ag+1(aq) + Cl-1(aq)  AgCl(s)
Increasing Decreasing Stays the same Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Given the following reaction, how is entropy changing: 2NO2(g)  N2O4(g)
Increasing Decreasing Stays the same Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Given the following reaction, how is entropy changing: CO(g) + H2O(g)  CO2(g) + H2(g)
Increasing Decreasing Stays the same Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Given the following reaction, how is entropy changing: H2(g) + F2(g)  2HF(g)
Increasing Decreasing Stays the same Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Given the following reaction, what is the sign for ΔS: NaCl(s)  NaCl(aq)
positive negative Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Given the following reaction, how is entropy changing: 2OH-(aq) + CO2(g)  H2O(l) + CO32- (aq)
Increasing Decreasing Stays the same Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

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) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Spontaneity Spontaneous change: Spontaneous reactions occur when
Occurs w/o continuous input of energy Spontaneous reactions occur when Reaction is exothermic ( ΔH < 0) Increase in entropy for the system (ΔS >0) But which is more important? ΔH or ΔS ?

Total Entropy System vs. Surroundings vs. Universe
Suniverse = Ssystem + Ssurroundings Suniv is always increasing. Two spontaneous processes: CaCl2(s)  Ca2+(aq) + Cl-1(aq) ΔH = -66kJ Ssys is increasing Ssurr is increasing because heat is released to surroundings NH4Cl(s)  NH4+ + Cl-(aq) ΔH = 15 kJ Ssurr is decreasing b/c surroundings are losing heat Non-spotaneous Na(s)  Na (l) ΔH =2.59 kJ Ssys = increasing Ssurr = decreasing

NH4Cl(s)  NH4+ + Cl-(aq) ΔH = 15 kJ Na(s)  Na (l) ΔH =2.59 kJ
Na+(g) + Cl-(g)  NaCl(s) ΔH = -771 kJ Why is A spontaneous, but not B? Entropy is much greater for A Why is C spontaneous? Enthalpy is large

How do you know if enthalpy or entropy will make the reaction more spontaneous?
Remember: Suniverse = Ssystem + Ssurroundings Ssurr depends on temperature The lower the surrounding temperature, the more significant adding heat is The higher the surrounding temp, the less significant adding heat is. Analogy Imagine you give \$1 to someone w/ only \$10 to their name? Imagine the effect of giving \$1 to a millionaire. Who is affected more?

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 + Ssurroundings Entropy of system Determined by ΔH of system

Substitute: - ΔHsys/T for Ssurr Multiply by (-T) Define new quantity
Suniv = Ssys + Ssurr Substitute: - ΔHsys/T for Ssurr Multiply by (-T) Define new quantity Free energy: ΔG = -T ΔSuniv Free energy tells whether a rxn will be spontaneous at a given temperature. Measures the maximum energy available to do useful work Reactions at equilibrium have ΔG = 0 -T ΔSuniv = ΔHsys - T ΔSsys ΔG = ΔHsys - T ΔSsys

Based on the previous slides and derivation of ΔG, rxns will be spontaneous if ΔG is
Less than 0 Greater than 0 Equal to 0 Spontaneity has nothing to do with ΔG. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Under which conditions will reactions ALWAYS be spontaneous?
ΔH > 0, ΔS >0 ΔH > 0, ΔS < 0 ΔH < 0, ΔS >0 ΔH < 0, ΔS <0 ΔG = ΔHsys - T ΔSsys 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Under which conditions will reactions NEVER be spontaneous?
ΔH > 0, ΔS >0 ΔH > 0, ΔS < 0 ΔH < 0, ΔS >0 ΔH < 0, ΔS <0 ΔG = ΔHsys - T ΔSsys 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

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 >0 ΔH > 0, ΔS < 0 ΔH < 0, ΔS >0 ΔH < 0, ΔS <0 ΔG = ΔHsys - T ΔSsys 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

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 ΔH Reaction 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

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 >0 ΔH > 0, ΔS < 0 ΔH < 0, ΔS >0 ΔH < 0, ΔS <0 ΔG = ΔHsys - T ΔSsys 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

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 ΔH Reaction 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

Calculating ΔG For 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!

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

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 )reactants Standard Gibbs Free Energy of Formation Tables (ΔG°f ) Σ n(ΔG°f )products - Σ n(ΔG°f )reactants Only good for standard conditions! For ΔG at non-standard conditions, use ΔG = ΔH - TΔS

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)

CH3CH2OH(l) + O2(g)  CH3COOH(l) + H2O(l)
DGf°, kJ/mol – – –237.2 n, mol nDGf°, kJ – –392.5 –237.2 –174.8 kJ –629.7 kJ DG° = –629.7 – (–174.8) DG° = –454.9 kJ

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.

2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO(g)
DHf°, kJ/mol –947.7 – –241.8 –393.5 n, mol nDHf°, kJ – – –241.8 –393.5 – kJ – kJ DH° = kJ Sf°, J/mol  K nSf°, J/K 204 J/K J/K DS° = J/K

Use ΔG to get K Equilibrium position represents the lowest free energy value available to a particular reaction system

ΔG and K Standard 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 Q At equilibrium: Δ G = 0, because there is no ability to do any more useful work and Q = K So we get: Δ G° = –RT ln K

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).

DG° = 2 mol(51.30 kJ/mol) – 1 mol(97.82 kJ/mol)
DG° = kJ – kJ DG° = 4.78 kJ