1 / 37

Chapter 17 Chemical Thermodynamics

Chapter 17 Chemical Thermodynamics. Chemical Thermodynamics. Chemical thermodynamics is the study of the energetics of chemical reactions. Chapter 5 Review. The system is the part of the universe under examination (for us, it would be a chemical reaction or other process).

talib
Download Presentation

Chapter 17 Chemical Thermodynamics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 17Chemical Thermodynamics

  2. Chemical Thermodynamics • Chemical thermodynamics is the study of the energetics of chemical reactions.

  3. Chapter 5 Review • The system is the part of the universe under examination (for us, it would be a chemical reaction or other process). • The surroundings are the rest of the universe. • The change in enthalpy (DH) is the heat absorbed or released by the system at constant temperature and pressure.

  4. Work and Heat • Under normal laboratory conditions, a reaction (the system) usually exchanges energy with the surroundings in two forms, as heat and as work. • Work is directed energy, and is associated with moving a part of the system. • Heat is random energy, and is associated with the temperature of the system.

  5. Work • Work is an application of a force through a distance: work = force × distance. • Work is also an application of a pressure causing a volume change: work = -PDV.

  6. Pressure-Volume Work • Express the work (in joules) when 20.0 L of an ideal gas at a pressure of 12.0 atm expands against a constant pressure of 1.50 atm. Assume constant temperature.

  7. First Law of Thermodynamics • First law of thermodynamics is the law of conservation of energy; energy can neither be created nor destroyed. • Internal energy, E, represents the total energy of the system and is a state function (something that depends only on the state of the system, not how the system got to that state).

  8. Internal Energy, E • When a change occurs in a closed system, the change in internal energy, DE, is given by DE = q + w • The sign convention for q and w are that they are positive if they transfer energy to the system from the surroundings and negative if they transfer energy from the system to the surroundings.

  9. Calculating ΔE, w, and q A 7.56g sample of gas in a balloon that has a volume of 10.5 L. Under an external pressure of 1.05 atm, the balloon expands to a volume of 15.00 L. Then the gas is heated from 0.0 °C to 25.0 °C. If the specific heat of the gas is 0.909 J/g*°C. Calculate the work, heat, and ΔE for the overall process.

  10. Energy and Enthalpy • Data about heats of reaction are tabulated as standard enthalpies of formation (DHf), as described in Chapter 5. • The equation that relates DH to DE is DH = DE + PDV • This equation is used to calculate the enthalpy of reaction from heats measured using constant-volume calorimetry.

  11. Energy, Enthalpy, and PV Work • The difference between changes in enthalpy and internal energy is PV work and is significant only for reactions that involve gases. • From the ideal gas law, PV = nRT, so DH = DE + DnRT

  12. Example: DH from DE • Calculate DH at 25C and 1.00 atm pressure for the following reaction. 2NO2(g) → N2O4(g) DE = -54.7 kJ

  13. Randomness • An increase in randomness (sometimes referred to as disorder) is an important driving force for many changes. • Two different gases, initially separated by a partition, will mix with the partition is removed, increasing the randomness of the system.

  14. Entropy and Spontaneity • Thermodynamics is able to relate spontaneity to a state function called entropy. • Entropy (S) is the thermodynamic state function that describes the amount of randomness. • A large value for entropy means a high degree of randomness.

  15. Entropy Change • An increase in randomness results in an increase in entropy. Some general guides are: • the entropy of a substance increase when solid becomes liquid, and when liquid becomes gas. • the entropy generally increases when a solute dissolves. • the entropy decreases when a gas dissolves in a solvent. • the entropy increases as temperature increases.

  16. Entropy • The entropy of a system generally increases when a molecular solid dissolves in a liquid.

  17. Second Law of Thermodynamics • The second law of thermodynamics states that in any spontaneous process, the entropy of the universe increases. • DSuniv = DSsys + DSsurr > 0 • If DSsys < 0 for a spontaneous process, then a larger positive change in DSsurr must occur.

  18. Spontaneity and DSuniv • When DSuniv > 0, the change occurs spontaneously. • When DSuniv < 0, the reverse change occurs spontaneously. • When DSuniv = 0, the change is not spontaneous in either direction (the process is at equilibrium).

  19. Third Law of Thermodynamics • The third law of thermodynamics states that the entropy of a perfect crystal of a substance at absolute zero is equal to 0. • There is a minimum randomness in a perfect crystal at 0 K. • Unlike enthalpy and internal energy, absolute values of entropy can be determined.

  20. Units for Entropy • As heat is added to a perfect crystal at 0 K, the temperature rises and randomness begins to increase. • The change in entropy depends on the amount of heat and on the temperature, and is given by the equation DS = q/T • Therefore, entropy has units of J/K.

  21. Absolute Entropies • In Appendix G, absolute entropies are given for substances in their standard state at 298 K. • The entropy change for a reaction is DSrxn = SnS[prods] - SmS[reacts] where n and m are the coefficients of the products and reactants in the reaction. • S for the free elements in their standard states is not zero.

  22. Example: Calculate DSrxn • Calculate the standard entropy change for the reaction 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)

  23. Test Your Skill • The combustion of ethane is spontaneous, but DS is -620.21 J/K. Explain why this does not violate the second law of thermodynamics.

  24. Free Energy and DSuniv • J. W. Gibbs defined a state function called the Gibbs free energy, G: G = H – TS • At constant temperature and pressure, this becomes • DG = DH – TDS • DG is negative for a spontaneous change that occurs at constant temperature and pressure.

  25. Free Energy and Spontaneity • DG is a state function of the system. • For any spontaneous change, DG < 0.

  26. DG and Spontaneity • From the equation DG = DH – TDS, a negative DH and a positive DS favor spontaneity. • A minimum free energy occurs when the system is at equilibrium, and DG = 0.

  27. Standard Free Energy of Formation • The standard free energy of formation is the free energy change to form one mole of a compound from its elements in their standard states. • , where DSf must be calculated from absolute entropies. • DGrxn = SnGf[prods] - SmGf[reacts] where n and m are the coefficients of the products and reactants in the reaction.

  28. Temperature Dependence of DG • The dependence of DG on temperature arises mainly from the TDS term in the definition of DG. • The sign of DG is dominated by the sign of DH at low temperatures and by the sign of DS at high temperatures. • Negative values of DH and positive values of DS favor spontaneity.

  29. Direction of Spontaneous Reaction

  30. Example: Temperature Dependence • Assuming that DH and DS do not change with temperature, calculate the temperature for which DG is 0 for the reaction CS2(l) ⇌ CS2(g) At 298 K, DH = 27.66 kJ and DS = 86.39 J/K.

  31. Example: Calculate DG • Given the following chemical reaction and data at 298 K: N2(g) + 3H2(g)  2NH3(g)DHfo = 0 0 -46.1 kJ/molSo = 191.5 130.6 192.3 J/mol.K assuming that DHo andDSo do not change with temperature, calculate DGo at 1000 K.

  32. Concentration and Free Energy • Concentrations of reactants and products influence the free energy change of a reaction according to the equation DG = DG + RTlnQ where Q is the reaction quotient (see Chapter 14).

  33. Example: Concentration Dependence • Substance DGfo kJ/mol NO2(g) 51.29 N2O4(g) 97.82For the reaction 2NO2(g) ⇌ N2O4(g)(a) calculate DGo at 298 K.(b) calculate DG when PNO2 = 0.12 atm and PN2O4 = 0.98 atm.

  34. Free Energy and Keq • For a system at equilibrium, DG = 0, and Q = Keq, soDGo = -RTln Keq • DGo, calculated from the data in Appendix G, can be used to calculate the value of the equilibrium constant.

  35. Example: Keq Calculation • Evaluate the equilibrium constant at 298 K for 2NO(g) + Br2(g) ⇌2NOBr(g)using the standard free energies of formation at 298 K given below.

  36. Temperature and Keq • The temperature dependence of Keq is derived from two equations given earlier:DH - TDS = DG = - RTln KeqA graph of ln Keqvs. 1/T gives a line with a slope = -DH/R and an intercept of DS/R.

  37. Useful Work • The change in free energy is the maximum work that can be performed by a spontaneous chemical reaction at constant temperature and pressure: wmax = DG • This is a limit imposed by nature, as we strive to improve the efficiency of energy conversions • When DG > 0 (spontaneous in the reverse direction), it represents the minimum work that must be provided to cause the change.

More Related