Chapter 14: Thermodynamics

1. Chapter 14: Thermodynamics

2. Review of the Law • 1st law of Thermodynamics is a restatement of the law of conservation of energy • Energy can never be created nor destroyed, it may only change form • This is true only within a closed system • The universe is a closed system • So the amount of energy in the universe is constant • By default, the amount of heat energy is constant, but may exist in potential and kinetic heat energy.

3. Review of the Law • Consider the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + energy • The combustion of Methane produces a quantity of energy released as heat. • Potential energy stored in the bonds of CH4 andO2 is released as heat as they react to form water and CO2. • Potential energy has been converted to thermal energy, but the amount of energy in the system itself has not changed.

4. Review of the Law • The 1st law of Thermodynamics allows us to answer the following questions: • How much energy is involved in the change? • Does energy flow into or out of the system? • What form does the energy finally assume? • It DOES NOT, however, allow us to account for “why” a process occurs in a given direction. This is the main question to be addressed in this chapter.

5. Spontaneity • A process is considered spontaneous if it occurs without outside intervention • This process may be fast or slow • Thermodynamics allow us to determine the direction of the process (chemical change), but can tell us nothing about the speed of the process. • Speed of the process is dependent on rate of reaction. • Remember that rate of reaction =

6. Spontaneity • We can think of spontaneity in terms of a few simple scenarios: • A ball spontaneously rolls down a hill, but not back up. Why? • Heat flows from a hot object to a cooler one, never the opposite. Why? • Wood burns spontaneously to form CO2 and Water, but this is not reversible.

7. Spontaneity and Entropy • After many years of observation, one common characteristic is present in all spontaneous processes; Entropy. • Entropy (S) is the driving force for all spontaneous processes. • Most accurately defined as a measurement of randomness or disorder in a system. • Represents the multitude of possible arrangements available to a system that exists in a given state.

8. Spontaneity and Entropy • These two concepts combined together present a level of statistical probability that is beyond the scope of this course, however, we can simplify it greatly using a simple metaphor. • Think of your room, is it clean? Is it orderly? • Think of your clothes and personal items as gas molecules. If you expend no energy to confine these things to a certain area, what happens?

9. High Entropy, low energy.

10. Low entropy, high energy

11. In other words • Without work (energy), there is no order (entropy), it takes work to keep things in their place in a non-random state. • This is true of all things in nature. Think of the balance between different aspects of an ecosystem. Each part exerts work (energy) on another, thus keeping that other part in order. • Predator prey interactions, survival of the fittest, etc… • What happens when humans introduce outside forces into ecosystems? • Imbalance occurs, energy dynamics are un-equalized, entropy increases in certain areas.

12. Positional Probability of Gases • The most likely state of a gas in a container is the one that requires the less energy to obtain, as well as the one with the most microstates.

13. Positional Probability • Also explained in changes of state. • In a solid state, molecules of a substance are very close together, with very few positional microstates • In a liquid state, molecules of a substance are further apart, with more positional microstates • In a gas state, molecules of a substance are very far apart, with very many positional microstates.

14. Entropy and Positional Probability (entropy) Sgas > Sliquid > Ssolid • Gases posses the most Entropy (S) as a result of the most available microstates, it takes energy (from somewhere at some point in time) to produce a liquid or a solid; a gas is always the most likely state of matter when talking about entropy.

15. Positional Probability • With this all in mind, • Which has the higher positional entropy? • Solid Carbon Dioxide or Gaseous Carbon Dioxide? • Nitrogen Gas at 1atm or at 1.0 x 10-2atm? • Predict the sign (negative or positive) of the following processes (∆S=Sfinal– Sinitial) • Solid sugar dissolved in water? • Iodine Vapor condenses on a cold surface to form crystals.

16. 2nd Law of Thermodynamics • In any spontaneous process there is always an increase in the entropy of the universe ∆Suniverse= ∆Ssystem + ∆Ssurroundings • This would indicate that there is a decrease of energy in the universe, but remember that there is constant work being put into individual systems that keep everything constant.

17. 2nd Law of Thermodynamics • In any spontaneous process there is always an increase in the entropy of the universe ∆Suniverse= ∆Ssystem + ∆Ssurroundings • If ∆Suniverseis positive, the process is spontaneous in the direction it is written. • If∆Suniverseis negative, the process is spontaneous in the opposite direction. • If ∆Suniverseis zero, the process has no tendency to occur and is at equilibrium.

18. Another Example from Biology • In living cells, large molecules are made from simpler ones, is this process consistent with the 2nd law? • Making large molecules from simpler ones gives a negative value for ∆Ssystem • Remember that ∆Suniverse= ∆Ssystem + ∆Ssurroundings • Only if the value for ∆Ssurroundings is positive and larger can this be spontaneous process, and in cells it is.

19. Effect of Temperature on Spontaneity • To illustrate this, lets consider the following process: H2O(l) →H2O(g) • Consider the water to be the system and everything else as the surroundings. • Since we are moving from a liquid to a gas, entropy (Ssystem) increases and is positive.

20. Effect of Temperature on Spontaneity H2O(l) →H2O(g) • Lets assume this process absorbs 50 joules (J) of heat energy to the surroundings; it takes heat to evaporate water. • ∆Ssurroundingsis negative(endothermic). • Thus, ∆ Ssystemand ∆Ssurroundings are in opposition. • Since we don’t have actual values, how will be determine whether the process is spontaneous?

21. Effect of Temperature on Spontaneity • ∆Ssurroundingsdepends on the temperature at which the heat is transferred. • The sign of ∆Ssurroundings (-/+) depends on the direction of the heat flow. Therefore: heat=magnitude of entropy change=

22. ∆S and Enthalpy • Exothermic process: ∆Ssurroundings=+ • Endothermic process: ∆Ssurroundings= • In terms of Enthalpy (H) ∆Ssurr= - • The negative sign is necessary because ∆H is determined in terms of the reaction system, not the surroundings. So a negative ∆H indicates a exothermic reaction because the reaction system is losing heat, but then ∆Ssurr would be positive …and vice versa.

23. Sample Problem • Calculate the Entropy change in the reaction system of: • Sb2S3(s) + 3Fe(s) → 2Sb(s) + 3FeS(s) ∆H = -125 kJ • Sb4O6(s) + 6C(s) → 4Sb(s) + 6CO(g) ∆H = 778 kJ both at 25°C and 1atm • use the equation: ∆Ssurr= -

24. Sample Problem • Sb2S3(s) + 3Fe(s) → 2Sb(s) + 3FeS(s) ∆H = -125 kJ • ∆Ssurr= - ∆Ssurr= 419 J/K - exothermic, surroundings absorb heat • Sb4O6(s) + 6C(s) → 4Sb(s) + 6CO(g) ∆H = 778 kJ • ∆Ssurr= - ∆Ssurr= - - endothermic, surroundings lose heat

25. Effect of ∆H, ∆S and T on Spontaneity

26. Section II:Entropy Changes in Chemical Reactions • Key definitions: • ∆Hrxn - standard change in enthalpy for a reaction. “standard heat of reaction” • ∆Hf-Standard Enthalpy of formation “heat of formation” • Compounds to elements: negative ∆Hf • Elements to compounds: positive ∆Hf • ∆Srxn = Standard entropy change for a reaction • ∆Srxn= Sproducts + Sreactants

27. Section II:Entropy Changes in Chemical Reactions • Key definitions: • ∆Hrxn = ∆H1 + ∆H2

28. Section II:Entropy Changes in Chemical Reactions • Key definitions: • Third Law of thermodynamics • The entropy of a perfect crystal at absolute zero is “zero”. • A crystal at absolute zero has only one possible way to arrange its molecules or atoms. The positional probability is 1, and therefore there is no level of disorder.

29. Section II:Entropy Changes in Chemical Reactions • Key definitions: • Third Law of thermodynamics • The entropy of a perfect crystal at absolute zero is “zero”. • A crystal at absolute zero has only one possible way to arrange its molecules or atoms. The positional probability is 1, and therefore there is no level of disorder.

30. Entropy is related to complexity of structure • Many substances can exist in different forms in the same state. These are called allotropes. • Example is carbon. • Graphite is carbon structure as sheets • Diamonds are carbon structred in three-dimensional crystals. • Which has greater entropy?

31. Entropy is related to molecular complexity • In general, with increasing molecular complexity, entropy increases. • At the same time, dissolving a solid into an aqueous solution increases entropy

32. Entropy Changes in Chemical Reactions • The major premise of this section is calculating the standard entropy change for a reaction (∆Srxn) • Simply stated, we can calculate this by subtracting the entropy of products from reactants: • ∆Srxn= ∑npS(products) - ∑nrS(reactants) • Where n is the mole coefficients, p and r represent “products” and “reactants” respectively. • S is the entropy of each portion of the reaction.

33. Entropy Changes in Chemical Reactions • ∆Srxn= ∑npS(products) - ∑nrS(reactants) • YOU WILL HAVE TO FIND THE VALUES FOR “S” IN THE APPENDIX IN THE BACK OF YOUR BOOKS. ON THE EXAM, THEY SHOULD PROVIDE A TABLE OR GIVE YOU THE VALUES IN THE QUESTIONS. • Make sure you pay close attention to the state of matter when extracting the values for S.

34. Example • Calculate the ∆Srxn for the equation: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆Srxn= ∑npS(products) - ∑nrS(reactants)

35. Example • Calculate the ∆Srxn for the equation: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) • ∆Srxn= [4mol(210.8 J/K*mol) + 6mol (188.8 J/K*mol)]- [4mol (192.8 J/K*mol) + 5mol (205.2 J/K*mol)] = 1976 J/K – 1797.2 J/K = 178.8 J/K

36. Example #2 • Calculate ∆Srxn for the equation: H2S(g) + O2(g) → H2O(g) + SO2(g) Step 1: Balance the equation Step 2: extract values of “S” for each reactant and product Step 3: insert into equation: ∆Srxn= ∑npS(products) - ∑nrS(reactants)

37. Gibbs Free Energy (G) • Free energy (G)-energy available to do work • Josiah Gibbs-a guy they named free energy after. G= H – TS Change in free energy for a constant temperature process: ∆G = ∆H - T∆S

38. Gibbs Free Energy (G) • ∆G<0-a reaction is accompanied by a release of unusable energy. Reaction is spontaneous • ∆G>0-a reaction is not spontaneous as written, it is spontaneous in the opposite direction • ∆G=0-equilibrium • Don’t forget…

39. Effect of ∆H, ∆S and T on Spontaneity

40. Standard free energy of reaction • ∆G°rxn-free energy change for a reaction when it occurs under standard state conditions, when reactants in their standard states are converted to products in their standard states.

41. Standard free energy of reaction • ∆Grxn= ∑np∆G°f(products) - ∑nr∆G°f(reactants) • “Same” equation as standard enthalpy change • Where p and r are mole coefficients, ∆Gf is the standard free energy of formation • ∆G°f--The free energy change that occurs when 1 mole of the compound is synthesized from its elements in their standard states. • ∆Gf – found in table in the appendices of your book

42. Difference between ∆G and ∆G° • Under non-standard state conditions we must use ∆G to predict direction of reaction. • Remember the table of effects of H, S, and T on ∆G. • Under equilibrium, ∆G° tells us whether products or reactants are favored. • Negative ∆G° favors product formation • Positive ∆G° favors reactant formation.

43. Examples • Calculate standard free energy changes for the following reactions at 25°C • CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) • 2MgO(s) → 2Mg(s) + O2(g)

44. Examples • CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) • -818.0 kJ/mol • 2MgO(s) → 2Mg(s) + O2(g) • 1139 kJ/mol

45. Phase Transitions • At the temperature at which a phase transition occurs, the system is at equilibrium, so ∆G=0. Therefore; ∆G = ∆H - T∆S 0 = ∆H - T∆S ∆S

46. Phase Transition Example • When 1 mole of ice melts at 0°C, Calculate the change in entropy (S). • ∆H is represented by the molar heat of fusion, T is the dual melting and freezing point of 273°K So: ∆Sice →water= -22.0 J/K * mol And ∆Swater→ice= 22.0 J/K * mol

47. Free energy and chemical Equilibrium • During the course of a chemical reaction, not all the reactants and products will be at their standard states. Under this condition, the relationship between ∆G and ∆G° can be described as: ∆G = ∆G° + RTlnQ Where Q is the equilibrium (reaction) quotient.

48. Free energy and chemical Equilibrium ∆G = ∆G° + RTlnQ • At equilibrium, ∆G = 0, and Q = K, so under equilibrium conditions: ∆G = ∆G° + RTlnK 0= ∆G° + RTlnK ∆G°= - RTlnK

49. Key equations for this Chapter