Thermodynamics (Part 2) Chapter 19

Questions to be answered... Will a given reaction occur “by itself” at a particular temperature and pressure without the exertion of any outside force? Why does a particular process occur in a given direction?

Spontaneous Processes occurs without outside intervention -sometimes may require a stimulus to get started, but continues itself without further input of energy ex: combustion, steel rusts when exposed to moisture & air, water freezes below 0C, water boils above 100C, a gas fills a container uniformly -moves the system towards equilibrium -can be fast or slow -thermodynamics tells us if a reaction occurs; kinetics explains speed Nonspontaneous: do not occur at all, or require a constant supply of energy ex: electrolysis of water

Spontaneous Processes Predict whether each process is spontaneous as described, spontaneous in the reverse reaction or in equilibrium a.water at 40C gets hotter when a piece of metal heated to 150C is added b.water at room temperature decomposes into H 2 (g) and O 2 (g) c.benzene vapor, C 6 H 6 (g) at a pressure of 1atm condenses to liquid benzene at the normal boiling point of benzene (80.1C)

Why are some reactions spontaneous? At first, it was thought that reactions are spontaneous because they are exothermic -exothermic: energy is lost to surroundings and dH is negative -it is true that almost all exothermic reactions are spontaneous at 298K and 1atm What about reactions that are endothermic & nonspontaneous at one temp, but are spont. when temp is raised? ex: ice melting at temps above 0C -there must be another factor besides the loss of energy to predict the direction of spontaneity

Entropy - The Randomness Factor Nature tends to move from a more ordered to a more random state -tendency to move towards states that have higher probabilities of existing once the stopcock is opened, the bottom picture has a higher probability of occurring because there are more positions available

Entropy and Microstates Example: what are the possible outcomes of a toss of 4 coins? What are the possible combinations to achieve the macrostates? Which macrostate is most probable? Disorder is more probable than order because there are so many more ways of achieving it.

Entropy and Motion 3 types of motion 1.translation: movement from one location to another (all particles: monatomic and polyatomic) 2.rotation: polyatomic molecules only (although linear molecules can only have 2 types of rotation; others have 3) 3.vibrations: atoms periodically move toward and away from eachother (polyatomic: increases as molecules become more complicated/more atoms)

Putting it Together The number of possible microstates increases with: 1.an increase in volume 2.an increase in temperature 3.an increase in the number of molecules 4.an increase in the complexity of the molecule Increase in entropy is driven by the tendency of thermal energy to spread into as many microstates of the system and surroundings as possible

Molecular Motions In which phase are water molecules least able to have rotational motion?

Entropy Entropy is a state function -depends only by the final and initial states, not the pathway to those states (enthalpy is also a state function) ΔS = S final - S initial +ΔS = more entropy = more random -ΔS = less entropy = less random S solids <S liquids <S aqueous <S gases -generally, the higher the temperature, the greater the entropy -particles move around more freely -more gas molecules = more entropy

Predicting the Sign of ΔS Predict whether Δ S is positive or negative, assuming each occurs at constant temperature a.H 2 O(l) → H 2 O(g) b.Ag + (aq) + Cl - (aq) → AgCl(s) c.4Fe(s) + 3O 2 (g) → 2Fe 2 O 3 (s) d.N 2 (g) + O 2 (g) → 2NO(g)

Predicting Entropy Indicate whether each process produces an increase or decrease in the entropy of the system. a.CO 2 (s) → CO 2 (g) b.CaO(s) + CO 2 (g) → CaCO 3 (s) c.HCl(g) + NH 3 (g) → NH 4 Cl(s) d.2SO 2 (g) + O 2 (g) → 2SO 3 (g)

Predicting Entropy In each pair, choose the system that has greater entropy and explain your choice: a.1 mol of NaCl(s) or 1 mol of HCl(g) b.2 mol of HCl(g) or 1 mol of HCl(g) at 25C c.1 mol of HCl(g) or 1 mol of Ar(g) at 298K

Entropy & the 2nd Law of Thermodynamics 2nd Law of Thermodynamics: in any spontaneous process, there is always an increase in the entropy of the universe ΔS universe = (ΔS system + ΔS surroundings ) ΔS > 0 = spontaneous -entropy changes in surroundings are primarily determined by the direction of heat flow -exothermic: increase in the entropy of the surroundings -endothermic: decrease in entropy of surroundings, possible increase in entropy of the system

Calculating Δ S -can assign absolute entropy values 3rd Law of Thermodynamics: the entropy of a perfect crystal at 0K is zero (and has only one microstate) -entropy values are listed in Appendix C -elements have nonzero entropies -standard molar entropies of pure substances are always greater than zero

Calculating Δ S for Reactions Standard entropy change for a reaction: ΔS° = ∑S°products - ∑S°reactants -entropy is an extensive property and depends on the number of moles so coefficients must be used Generally… -a reaction that results in an increase in the number of moles of a gas is accompanied by an increase in entropy -the more complex a molecule, the higher the entropy

Calculating ΔS Practice 1 Calculate the change in the standard entropy of the system, ΔS°, for the synthesis of ammonia from N 2 (g) and H 2 (g) at 298K

Calculating ΔS Practice 2 Calculate the standard entropy change, ΔS° for the following reaction at 298K: Al 2 O 3 (s) + 3H 2 (g) → 2Al(s) + 3H 2 O(g)

Gibbs Free Energy relates the enthalpy (H) and entropy (S) of a system -for a reaction taking place at constant pressure and temperature, ΔG represents the portion of the total energy change that is free to do useful work ΔG° = ΔH° - TΔS° ΔG° < 0 reaction is spontaneous ΔG° > 0 reaction is not spontaneous (reverse reaction is) ΔG° = 0 reaction is at equilibrium (no tendency for the reaction to occur in either direction)

Concept Comparison ΔH, change in enthalpy -a negative value tends to make a reaction spontaneous (maybe) -exothermic ΔS, change in entropy -a positive value tends to make a reaction spontaneous (maybe) -more disorder or randomness ΔG, change in free energy -a reaction at constant temperature and pressure will be spontaneous if ΔG is negative (ALWAYS) -decrease the free energy of a system

More on Free Energy ΔG can be used to determine: -the effect of temperature on reaction spontaneity -the effect of pressure on reaction spontaneity -the effect of concentration on reaction spontaneity -ΔG can be used to calculate the equilibrium constant for a reaction -when comparing reactions, the more negative ΔG, the further the reaction will go to the right to reach equil. -ΔG can be used to determine whether coupled reactions are spontaneous (Hess’s Law)

Calculating ΔG° - Method 1 Determine ΔH and ΔS using reference tables, then use the Gibbs equation to find ΔG Example: Calculate ΔG° for a reaction for which ΔH° = 24.6kJ and ΔS° = 132J/K at 298K. Is the reaction spontaneous under these conditions?

Calculating ΔG° - Method 1 Calculate the standard free-energy change for the formation of NO(g) from N 2 (g) and O 2 (g) at 298K -given: S° O2 = 205J/molK, S° N2 = 191.5J/molK, S° NO = J/molK; ΔH° f for O 2 and N 2 = 0, ΔH° f for NO = 90.37kJ/mol

Calculating ΔG° - Method 2 Use standard free energies of formation from tables in appendix C ΔG° rxn = ∑ mΔG° products - ∑ nΔG° reactants Example: Use ΔGf values to calculate ΔG° rxn for this reaction: 4KClO 3 (s) → 3KClO 4 (s) + KCl(s)

Calculating ΔG° - Method 2 Use Δ G ° f values to calculate Δ G ° rxn at 298K: a.2NO(g) + Cl 2 (g) → 2NOCl(g) b.3H 2 (g) + Fe 2 O 3 (s) → 2Fe(s) + 3H 2 O(g)

Calculating ΔG° - Method 3 Use procedures to find Δ G like those used to find Δ H using Hess’s Law. -this is possible because free energy is a state function remember: reverse equation → reverse sign; equation multiplied → multiply Δ G

Free Energy and Pressure How does pressure affect the thermodynamic functions that comprise free energy? -for an ideal gas: -enthalpy (H) is not pressure dependent -entropy is because it depends on volume -S large volume > S small volume SO: S low pressure > S high pressure -more room to move in large volumes

Free Energy and Pressure To calculate free-energy at pressures other than 1atm: Δ G ° rxn = Δ G ° + RTlnQ R = gas constant: J/Kmol Q = reaction quotient

Free Energy and Equilibrium Even though Δ G is negative, the reaction does not go to completion -the lowest possible free energy is at the equilibrium point At equilibrium Δ G = 0 and Q = K SO… Δ G ° = -RTlnK -we will use this equation for reactions at equilibrium -it’s in your reference tables

Free Energy and Equilibrium Qualitative relationship between Δ G ° f and the K for a given reaction Δ G ° f = 0andK = 0then: Δ G ° fp = Δ G ° fr Δ G ° f 1then: Δ G ° fp < Δ G ° fr system will shift right Δ G ° f > 0andK Δ G ° fr system will shift left