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Understanding Spontaneous Processes in Thermodynamics

Explore the concepts of spontaneous processes, entropy, and the Second Law of Thermodynamics. Learn about reversible and irreversible processes and the relationship between entropy and system disorder. Dive into the molecular interpretation of entropy.

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Understanding Spontaneous Processes in Thermodynamics

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  1. Chemical Thermodynamics Review First Law

  2. Spontaneous Processes • Thermodynamics is concerned with the question: can a reaction occur? • First Law of Thermodynamics: energy is conserved. • Any process that occurs without outside intervention is spontaneous. • When two eggs are dropped they spontaneously break. • The reverse reaction is not spontaneous. • We can conclude that a spontaneous process has a direction.

  3. Spontaneous Processes • A process that is spontaneous in one direction is not spontaneous in the opposite direction. • The direction of a spontaneous process can depend on temperature: Ice turning to water is spontaneous at T > 0C, Water turning to ice is spontaneous at T < 0C. Reversible and Irreversible Processes • A reversible process is one that can go back and forth between states along the same path.

  4. Spontaneous Processes

  5. Spontaneous Processes Reversible and Irreversible Processes • Chemical systems in equilibrium are reversible. • In any spontaneous process, the path between reactants and products is irreversible. • Thermodynamics gives us the direction of a process. It cannot predict the speed at which the process will occur.

  6. Entropy and the Second Law of Thermodynamics The Spontaneous Expansion of a Gas • Why do spontaneous processes occur? • Consider an initial state: two flasks connected by a closed stopcock. One flask is evacuated and the other contains 1 atm of gas. • The final state: two flasks connected by an open stopcock. Each flask contains gas at 0.5 atm. • The expansion of the gas is isothermal (i.e. constant temperature). Therefore the gas does no work and heat is not transferred.

  7. Entropy and the Second Law of Thermodynamics The Spontaneous Expansion of a Gas • Why does the gas expand?

  8. Probability of having one player (out of four) getting 13 cards of same suit is: Entropy and the Second Law of Thermodynamics The Spontaneous Expansion of a Gas

  9. Entropy and the Second Law of Thermodynamics Entropy • Entropy, S, is a measure of the disorder of a system. • Spontaneous reactions proceed to lower energy or higher entropy. • In ice, the molecules are very well ordered because of the H-bonds. • Therefore, ice has a low entropy.

  10. Entropy and the Second Law of Thermodynamics Entropy • Generally, when an increase in entropy in one process is associated with a decrease in entropy in another, the increase in entropy dominates. • Entropy is a state function. • For a system, S = Sfinal - Sinitial. • If S > 0 the randomness increases, if S < 0 the order increases.

  11. Entropy and the Second Law of Thermodynamics Entropy • Suppose a system changes reversibly between state 1 and state 2. Then, the change in entropy is given by • at constant T where qrev is the amount of heat added reversibly to the system. (Example: a phase change occurs at constant T with the reversible addition of heat.) The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9oC, and its molar enthalpy of fusion is 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? MW(Hg) = 200.59 g/mol

  12. The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9oC, and its molar enthalpy of fusion is 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? MW(Hg) = 200.59 g/mol -2.44 J K-1

  13. Entropy and the Second Law of Thermodynamics The Second Law of Thermodynamics • The second law of thermodynamics explains why spontaneous processes have a direction. • In any spontaneous process, the entropy of the universe increases. • Suniv = Ssys + Ssurr • Entropy is not conserved: Suniv is increasing.

  14. Entropy and the Second Law of Thermodynamics • The Second Law of Thermodynamics • For a reversible process: Suniv = 0. • For a spontaneous process (i.e. irreversible): Suniv > 0. • Note: the second law states that the entropy of the universe must increase in a spontaneous process. It is possible for the entropy of a system to decrease as long as the entropy of the surroundings increases. • For an isolated system, Ssys = 0 for a reversible process and Ssys > 0 for a spontaneous process. Calculate ΔSsys , ΔSsurr , ΔSuniv for the reversible melting of 1 mole of Ice in a large, isothermal water bath at 0oC and 1 atm. Heat of Fusion for water is 6.01 kJ/mol.

  15. Calculate ΔSsys , ΔSsurr , ΔSuniv for the reversible melting of 1 mole of Ice in a large, isothermal water bath at 0oC and 1 atm. Heat of Fusion for water is 6.01 kJ/mol. Ssys = +22.0 J mol-1 K-1

  16. The Molecular Interpretation of Entropy • A gas is less ordered than a liquid that is less ordered than a solid. • Any process that increases the number of gas molecules leads to an increase in entropy. • When NO(g) reacts with O2(g) to form NO2(g), the total number of gas molecules decreases, and the entropy decreases. HyperChem

  17. The Molecular Interpretation of Entropy

  18. The Molecular Interpretation of Entropy • Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero. • Entropy changes dramatically at a phase change. • As we heat a substance from absolute zero, the entropy must increase. • If there are two different solid state forms of a substance, then the entropy increases at the solid state phase change.

  19. The Molecular Interpretation of Entropy • Boiling corresponds to a much greater change in entropy than melting. • Entropy will increase when • liquids or solutions are formed from solids, • gases are formed from solids or liquids, • the number of gas molecules increase, • the temperature is increased.

  20. ENTROPY OF THE UNIVERSE Die Enegie der Welt ist Konstant; die entropie der Welt einen Maximum Zu Entropy is the Arrow of Time A Brief History of Time S. Hawking’s Grand Design: The Meaning of Life (19 min)

  21. Entropy Changes in Chemical Reactions • Absolute entropy can be determined from complicated measurements. • Standard molar entropy, S: entropy of a substance in its standard state. Similar in concept to H. • Units: J/mol-K. Note units of H: kJ/mol. • Standard molar entropies of elements are not zero. • For a chemical reaction which produces n moles of products from m moles of reactants:

  22. Calculate ΔS for the system, the surroundings, and the universe for the synthesis of ammonia at 298 K. [ΔHorxn = -92.38 kJ/mol]

  23. Calculate ΔS for the system, the surroundings, and the universe for the synthesis of ammonia at 298 K. [ΔHorxn = -92.38 kJ/mol] Answer

  24. Gibbs Free Energy • For a spontaneous reaction the entropy of the universe must increase. • Reactions with large negative H values are spontaneous. • How do we balance S and H to predict whether a reaction is spontaneous? • Gibbs free energy, G, of a state is • For a process occurring at constant temperature

  25. Gibbs Free Energy • There are three important conditions: • If G < 0 , forward reaction is spontaneous. • If G = 0 , reaction is at equilibrium. • If G > 0 , then forward reaction is not spontaneous. If G > 0, work must be supplied from the surroundings to drive the reaction. • For a reaction the free energy of the reactants decreases to a minimum (equilibrium) and then increases to the free energy of the products.

  26. Gibbs Free Energy

  27. Calculate ΔS for the system, the surroundings, and the universe for the synthesis of ammonia at 298 K. [ΔHorxn = -92.38 kJ/mol] Srxn = -198.7 J mol-1 K-1 . Calculate the Gibbs Free Energy change for the reaction. Answer for Suniv -33.14 kJ/mol

  28. Gibbs Free Energy Standard Free-Energy Changes • We can tabulate standard free-energies of formation, Gf (c.f. standard enthalpies of formation). • Standard states are: pure solid, pure liquid, 1 atm (gas), 1 M concentration (solution), and G = 0 for elements. • G for a process is given by Calculate ΔGorxn for the combustion of methane. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) ΔGof (kJ/mol): -50.8 0 -394.4 -228.6

  29. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) ΔGof (kJ/mol): -50.8 0 -394.4 -228.6 -800.8 kJ/mol

  30. Free Energy and Temperature

  31. Free Energy and The Equilibrium Constant • At equilibrium, G = 0 : • From the above we can conclude: • If G < 0, then K > 1. • If G = 0, then K = 1. • If G > 0, then K < 1. Given ΔGo = -33.3 kJ/mol for the ammonia formation from nitrogen and hydrogen, calculate Keq at 25.0oC.

  32. Given ΔGo = -33.3 kJ/mol for the ammonia formation from nitrogen and hydrogen, calculate Keq at 25.0oC. Aurora + Entropy 6.82x105

  33. Chemical Thermodynamics S. Hawking’s Grand Design: The Meaning of Life

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