1. Definitions
   1. Entropy (S)
      1. measure of molecular randomness or disorder
   2. First Law of Thermodynamics
      1. energy can neither be created or destroyed (∆U=Q-W)
   3. Second Law of Thermodynamics
      1. for any spontaneous reaction the entropy of the universe must increase
   4. Third Law of Thermodynamics
      1. the entropy of a perfect crystal is zero at 0K
   5. Zeroth Law of Thermodynamics
      1. if two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other
   6. Spontaneous Reactions
      1. occurs without being driven by some outside force
   7. Non-spontaneous Reactions
      1. needs energy input to proceed, spontaneous in the opposite direction
   8. Gibbs Free Energy
      1. amount of work a thermodynamic system can perform
   9. Standard Free Energy Change
      1. change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states
   10. Standard Free Energy of Formation
       1. change in free energy that accomopanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states
   11. Video
2. Learning Objectives
   1. 2.15
      1. The student is able to describe the relationships qualitatively (including using representations) to describe the interactions of ions and the attractions between ions and solvents to explain the factors that contribute to the solubility of ionic compounds.
   2. 5.12
      1. The student is able to use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes.
   3. 5.13
      1. The student is able to predict whether or not a physical or chemical process is thermodynamically favored by determination of (either quantitatively or qualitatively) the signs of both ∆H° and ∆S°, can calculation or estimation of ∆G° when needed.
   4. 5.14
      1. The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy.
   5. 5.15
      1. The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable.
   6. 5.16
      1. The student can use Le Chatelier's principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product.
   7. 5.17
      1. The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction.
   8. 5.18
      1. The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.
   9. 6.25
      1. The student is able to express the equilibrium constant in terms of ∆G° and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process.
3. 17.1 Spontaneous Processes and Entropy
   1. Spontaneous Reaction
      1. occurs without outside intervention
      2. may be fast or slow
      3. shows direction of reaction, but now speed to reaction
      4. Video
   2. Entropy (S)
      1. measure of molecular randomness or disorder
      2. describes the positional probability of a substance
         1. positional probability/entropy
            1. the number of possible arrangements of states of a substance
         2. as positional probability increases, entropy increases
         3. Arrangement vs. Microstates
         4. Arrangement III has the highest positional probability, thus nature spontaneously moves toward the state with the most evenly spread out molecules
      3. Video
   3. All spontaneous reactions increase the entropy of the universe
   4. Nature spontaneously proceeds towards the states that have the highest probabilities of existing
   5. Entropy of States From Lowest to Highest
      1. Solid
      2. Liquid
      3. Gas
   6. Example 1: Which has higher entropy (per mole)?
      1. a) solid CO2 and gaseous CO2
         1. Answer: gaseous CO2
         2. Explanation: gaseous state has a higher entropy than solid state
      2. b) N2 gas at 1 atm and N2 gas at 1.0e-2 atm?
         1. Answer: N2 gas at 1.0e-2 atm
         2. Explanation: volume of container at 1.0e-2 atm is 100x the volume of the container at 1 atm
   7. Example 2: Predict the sign of entropy change for the following processes
      1. a) solid sugar is added to water to form a solution
         1. Answer: ∆S = "+"
         2. Explanation: liquid state has a higher entropy than solid state; sugar particles in a solution have a larger positional probability than in solid
      2. b) iodine vapor condenses on a cold surface to form crystals
         1. Answer: ∆S = "-"
         2. Explanation: gaseous state has a higher entropy than solid state; iodine particles in vapor state have a larger positional probability than iodine particles in solid state
   8. difference in entropy between gas and liquid is much greater than the difference in entropy between liquid and solid.
4. 17.2 Entropy and Second Law of Thermodynamics
   1. 2nd Law of Thermodynamics
      1. for any spontaneous reaction, the entropy of the universe must increase
   2. Signs of Entropy of Universe
      1. ∆S universe = "+"
         1. entropy increases, process is spontaneous
      2. ∆S universe = "-"
         1. entropy decreases, process is spontaneous in opposite direction
      3. ∆S universe = 0
         1. no tendency to occur, system is at equilibrium
   3. ∆S universe = ∆S system + ∆S surroundings
5. 17.3 The Effect of Temperature on Spontaneity
   1. Sign of ∆S system
      1. "+"
         1. if there is more internal disorder in the products
            1. ex: solid to liquid
      2. "-"
         1. if there is less internal disorder in products
            1. ex: liquid to solid
   2. Sign of ∆S surroundings
      1. "+"
         1. exothermic; adding energy to surroundings produces more disorder
            1. = - "-∆H"/T
      2. "-"
         1. endothermic; taking energy from surroundings creates less disorder
            1. = - "+∆H"/T
   3. For the reaction: H2O (l) ⟶ H2O (gas)
      1. ∆ S system = "+"
         1. because of moving from liquid state to gaseous state
      2. ∆S surroundings = "-"
         1. is negative because the reaction is endothermic
   4. Relative magnitude of ∆S system vs. ∆S surroundings is based on temperature
      1. 50J of energy transferred to high temperature surroundings has little effect
      2. compared to...
      3. 50J of energy transferred to low temperature surroundings, which has huge effect
   5. ∆S surroundings = -∆H (J) / T (K)
   6. Example 1: Calculate ∆S surroundings for this reaction at 25°C and 1 atm: Sb2S3(s) + 3Fe(s) ⟶ 2Sb(s) + 3FeS(s) ∆H= -125kJ
      1. Answer: 419 J/K
      2. Explanation: ∆S surroundings = -∆H/T = - (-125kJ)/(25+273)K = 0.419 kJ/K = 419 J/K
6. 17.4 Free Energy
   1. Free Energy (G)
      1. amount of work a thermodynamic system can perform
      2. Video
   2. G = H - TS
      1. ∆G = ∆H - T∆S
         1. H = enthalpy
         2. T = temperature
         3. S = entropy
         4. Sign of ∆G
            1. "+"
            2. non-spontaneous
            3. "-"
            4. spontaneous
   3. ∆S = -∆G/T
      1. *at constant temperature and pressure
   4. A process (constant temperature and pressure) is spontaneous when free energy decreases
   5. Example 1: At what temperature is the following spontaneous at 1 atm? Br2(l) ⟶ Br2(g) ∆H = 31.0 kJ/mol ∆S = 93.0 J/K*mol
      1. Answer: Temperature is...
         1. > 333K spontaneous
         2. < 333K spontaneous in opposite direction
         3. = 333K normal boiling point
      2. Explanation:
         1. 0 > ∆H - T∆S ⟶ 0 > 3.10e4 J/mol - T*93.0 J/K*mol ⟶ T > 333K
         2. 0 < ∆H - T∆S ⟶ 0 < 3.10e4 J/mol - T*93.0 J/K*mol ⟶ T< 333K
         3. 0 = ∆H - T∆S ⟶ 0 = 3.10e4 J/mol - T*93.0 J/K*mol ⟶ T = 333K
7. 17.5 Entropy Changes in Chemical Reactions
   1. If there are more gaseous molecules on the products side than on the reactants side, the positional probability increases, and thus, ∆S will be positive for the reaction
   2. Example 1: Predict the sign of ∆S° for the following reaction: CaCo3 (s) ⟶ CaO (s) + CO2
      1. Answer: ∆S° "+"
      2. Explanation: Since in this reaction a gas is produced form a solid reactant, the positional entropy increases, and ∆S° is positive
   3. Third Law of Thermodynamics
      1. The entropy of a perfect crystal at 0 K is zero
   4. Entropy is a state function, and therefore, not pathway dependent
   5. ∆S reaction = sumof ∆S products - sumof ∆S reactants
   6. Example 2: Calculate ∆S at 25°C for the reaction: 2NiS (s) + 3O2 (g) ⟶ 2SO2 (g) + 2NiO (s)
      1. Answer: -149 J/K
      2. Explanation: ∆S reaction = sumof ∆S products - sumof ∆S reactants ⟶ ∆S° = 2S° SO2 + 2S° NiO - 2S°NiS - 2S°O2 ⟶ ∆S° = 2 mol (249 J/K*mol) + 2 mol (38 J/K*mol) - 2 mol (53 J/K*mol) - 3 mol (205 J/K*mol) ⟶ ∆S° 496 J/K + 76 J/K - 106 J/K - 615 J/K ⟶ ∆S° = -149 J/K We would expect ∆S* to be negative because the number of gaseous molecules decreases in this reaction.
   7. The more complex the molecule, the higher the standard entropy value
8. 17.6 Free Energy and Chemical Reactions
   1. Standard Free Energy Change (∆G)
      1. change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states
   2. ∆G = ∆H - T∆S
   3. ∆G° = ∆H° - T∆S°
   4. Example 1: Calculate ∆H, ∆S, and ∆G for the reaction at 25°C and 1 atm: 2SO2 (g) + O2 (g) ⟶ 2SO3 (g)
      1. Answer: -142 kJ
      2. Explanation:
      3. ∆H reaction = sumof ∆H products - sumof ∆H reactants ⟶ ∆H° = 2∆H°f(so2) - 2∆H°f(o2) ⟶ ∆H° = 2 mol (-396 kJ/mol) - 2 mol (-297 kJ/mol) - 0 ⟶ ∆H° = -792 kJ + 594 kJ ⟶ ∆H° = -198 kJ
      4. ∆S° = sumof S* products - sumof S° reactants ⟶ ∆S° = 2S° so3 - 2S° so2 - S° o2 ⟶ ∆S° = 514 J/K - 497 J/K - 205 J/K ⟶ ∆S° = -187 J/K
   5. Free Energy is also a state function
   6. ∆G reaction = sumof ∆G products - sumof ∆G reactants
   7. Standard Free Energy of Formation (∆Gf°)
      1. change in free energy that accomopanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states
         1. ∆G° = sumof ∆G°f products - sumof ∆G°f reactants
         2. the standard free energy of formation of an element in its standard state is zero
9. 17.7 The Dependence of Free Energy
   1. Equilibrium position represesnts the lowest free energy value available to a particular reaction system because spontaneous reactions always proceed in the direction that lowers its free energy
   2. The reaction system achieves lowest possible free energy by going towards equilibrium, but not necessarily, completion
   3. S low pressure > S high pressure
   4. G = G° + RT*ln(P)
      1. G is the free energy of the gas
      2. P is the pressure of the gas in atm
      3. R is the universal gas constant
      4. T is the temperature in Kelvin
      5. Variation: ∆G = ∆G° + RT*ln (Q)
         1. Q is the reaction quotient
         2. R is the gas law constant
         3. ∆G° is the free energy change for the reaction at a pressure of 1 atm
   5. Example 1: Calculate ∆G at 25°C for the reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol CO (g) + 2H2 (g) ⟶ CH3OH (l)
      1. Answer: -38 kJ/mol rxn
      2. Explanation: ∆G = ∆G° + RT*ln (Q)
      3. ∆G° = -166kJ - (-137 kJ) - 0 = -29kJ
      4. ∆G° = -2.9e4 J/mol rxn , R = 8.3145 J/K*mol , T = 273 + 25 = 298K , Q = 1/(P co)(P H2)^2 = 1/(5.0)(3.0)^2 = 2.2e-2
      5. ∆G = -38 kJ/mol rxn
10. 17.8 Free Energy and Equilibrium
    1. When a reaction proceeds towards equilibrium: G products = G reactants or ∆G = G products - G reactants = 0
    2. Sign of ∆G°
       1. ∆G° = 0 reaction is at equilibrium
       2. ∆G° < 0 reaction will shift to the right
       3. ∆G° > 0 reaction will shift to the left
    3. Example 1: Find the equilibrium constant for the reaction at 25°C: 4Fe (s) + 3O2 (g) ⇆ 2FeO3 (s)
       1. Answer: e^601
       2. Explanation:
       3. ∆G° = -RTln(K)
       4. ∆G° = ∆H° - T∆S°
       5. ∆H° = 2 mol (-826 kJ/mol) - 0 - 0 = -1.652e6 J , ∆H° = 2 mol (-826 kJ/mol) - 0 - 0 = -1.652e6 JS° = 2 mol (90 J/K*mol) -3 mol (205 J/K*mol) - 4 mol (27 J/K*mol) = -543 J/K
       6. -1.490e6 = -(8.3145 J/K*mol)(298 K)ln(K)
       7. K = e^601
11. 17.9 Free Energy and Work
    1. The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy
    2. w max = ∆G
    3. The amount of work we actually obtain from a spontaneous process is always less than the maximum possible amount; maximum work available can only be achieved via a hypothetical pathway, any real pathway wastes energy.
    4. All real processes are irreversible, meaning the universe is different after a cyclic process (charging and discharging) has occurred.
    5. As we use energy, we degrade its usefulness; concentrated energy becomes more spread out over surroundings, more disordered and less useful.
12. MC Examples
    1. Answer Key
13. FR Examples
    1. Answer Key