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Chemistry 100 Chapter 19

Chemistry 100 Chapter 19 . Spontaneity of Chemical and Physical Processes: Thermodynamics. What Is Thermodynamics?. Study of the energy changes that accompany chemical and physical processes. Based on a set of laws.

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Chemistry 100 Chapter 19

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  1. Chemistry 100 Chapter 19 Spontaneity of Chemical and Physical Processes: Thermodynamics

  2. What Is Thermodynamics? Study of the energy changes that accompany chemical and physical processes. Based on a set of laws. In chemistry, a primary application of thermodynamics is as a tool to predict the spontaneous directions of a chemical reaction.

  3. What Is Spontaneity? Spontaneity refers to the ability of a process to occur on its own! Can the Niagara Falls suddenly reverse? “Ice will melt, water will boil,” Neil Finn, Tim Finn of Crowded House/Plant ‘It’s Only Natural’. Water spontaneously freezes on a cold winter day!

  4. The First Law of Thermodynamics The First Law deals with the conservation of energy changes. E = q + w The First Law tells us nothing about the spontaneous direction of a process.

  5. Entropy and Spontaneity • Need to examine • the entropy change of the process as well as its enthalpy change (heat flow). • Entropy – the degree of randomness of a system. • Solids – highly ordered  low entropy. • Gases – very disordered  high entropy. • Liquids – entropy is variable between that of a solid and a gas.

  6. Entropy Is a State Variable Changes in entropy are state functions S = Sf – Si Sf = the entropy of the final state Si = the entropy of the initial state

  7. Entropy Changes for Different Processes S > 0 entropy increases (melting ice or making steam) S < 0 entropy decreases (examples freezing water or condensing steam)

  8. The Solution Process Highly ordered – low entropy Disordered or random state – high entropy The formation of a solution is always accompanied by an increase in the entropy of the system! For the dissolution of NaCl (s) in water NaCl (s)  Na+(aq) + Cl-(aq)

  9. The Entropy Change in a Chemical Reaction • Burning ethane! C2H6 (g) + 7/2O2 (g)  2CO2 (g) + 3H2O (l) • The entropy change rS   np S (products) -  nr S (reactants) • np and nr represent the number of moles of products and reactants, respectively.

  10. Finding S Values Appendix C in your textbook has entropy values for a wide variety of species. Units for entropy values  J / (K mole) Temperature and pressure for the tabulated values are 298.2 K and 1.00 atm.

  11. Finding S Values Note – entropy values are absolute! Note – the elements have NON-ZERO entropy values! e.g., for H2 (g) fH = 0 kJ/mole (by def’n) S = 130.58 J/(K mole)

  12. Some Generalizations For any gaseous reaction (or a reaction involving gases). ng > 0, rS > 0 J/(K mole). ng < 0, rS < 0 J/(K mole). ng = 0, rS  0 J/(K mole). For reactions involving only solids and liquids – depends on the entropy values of the substances.

  13. The Second Law of Thermodynamics • The entropy of the universe (univS) increases in a spontaneous process. • univS unchanged in an equilibrium process

  14. What is univS? univS = sysS + surrS sysS = the entropy change of the system. surrS = the entropy change of the surroundings.

  15. How Do We Obtain univS? We need to obtain estimates for both the sysS and the surrS. Look at the following chemical reaction. C(s) + 2H2 (g)  CH4(g) The entropy change for the systems is the reaction entropy change, rS. How do we calculate surrS?

  16. Calculating surrS Note that for an exothermic process, an amount of thermal energy is released to the surroundings!

  17. Calculating surrS Note that for an endothermic process, thermal energy is absorbed from the surroundings!

  18. Connecting surrS to sysH For a constant pressure process qp = H surrS surrH = -sysH surrS = -sysH / T For a chemical reaction sysH = rH surrS = -rH/ T

  19. The Use of univS to Determine Spontaneity • Calculation of TunivS  two system parameters • rS • rH • Define a systemparameter that determines if a given process will be spontaneous?

  20. The Definition of the Gibbs Energy The Gibbs energy of the system G = H – TS For a spontaneous process sysG = Gf – G i Gf = the Gibbs energy of the final state Gi = the Gibbs energy of the initial state

  21. Gibbs Energy and Spontaneity Note that these are the Gibbs energies of the system under non-standard conditions sysG < 0 - spontaneous process sysG > 0 - non-spontaneous process (note that this process would be spontaneous in the reverse direction) sysG = 0 - system is in equilibrium

  22. Standard Gibbs Energy Changes • The Gibbs energy change for a chemical reaction? • Combustion of methane. CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l) • Define • rG =  np fG (products) -  nr fG (reactants) • fG = the formation Gibbs energy of the substance

  23. Gibbs Energy Changes fG (elements) = 0 kJ / mole. Use tabulated values of the Gibbs formation energies to calculate the Gibbs energy changes for chemical reactions.

  24. The Third Law of Thermodynamics Entropy is related to the degree of randomness of a substance. Entropy is directly proportional to the absolute temperature. Cooling the system decreases the disorder.

  25. The Third Law of Thermodynamics The Third Law - the entropy of any perfect crystal is 0 J /(K mole) at 0 K (absolute 0!) Due to the Third Law, we are able to calculate absolute entropy values.

  26. At a very low temperature, the disorder decreases to 0 (i.e., 0 J/(K mole) value for S). The most ordered arrangement of any substance is a perfect crystal!

  27. Applications of the Gibbs Energy • The Gibbs energy is used to determine the spontaneous direction of a process. • Two contributions to the Gibbs energy change (G) • Entropy (S) • Enthalpy (H) G = H - TS

  28. Spontaneity and Temperature

  29. Gibbs Energies and Equilibrium Constants rG < 0 - spontaneous under standard conditions rG > 0 - non-spontaneous under standard conditions

  30. The Reaction Quotient Relationship between QJ and Keq Q < Keq - reaction moves in the forward direction Q > Keq - reaction moves in the reverse direction Q = Keq - reaction is at equilibrium

  31. rG° refers to standard conditions only! For non-standard conditions - rG rG < 0 - reaction moves in the forward direction rG > 0 - reaction moves in the reverse direction rG = 0 - reaction is at equilibrium

  32. Relating Keq to rG rG = rG +RT ln Q rG = 0  system is at equilibrium rG = -RT ln Qeq rG = -RT ln Keq

  33. Phase Equilibria At the transition (phase-change) temperature only - trG = 0 kJ tr = transition type (melting, vapourization, etc.) trS = trH / Ttr

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