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Chapter 19 Chemical Thermodynamics

Chapter 19 Chemical Thermodynamics. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Troy Wood University of Buffalo Buffalo, NY  2006, Prentice Hall.

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Chapter 19 Chemical Thermodynamics

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  1. Chapter 19Chemical Thermodynamics Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Troy Wood University of Buffalo Buffalo, NY  2006, Prentice Hall

  2. Melting a cube of water ice at 10°C and atmospheric pressure is a ________ process. • Spontaneous • Nonspontaneous

  3. Correct Answer: Melting of an ice cube at 1 atm pressure will only be spontaneous above the freezing point (0°C). • Spontaneous • Nonspontaneous

  4. The sign of entropy change, DS, associated with the boiling of water is _______. • Positive • Negative • Zero

  5. Correct Answer: • Positive • Negative • Zero Vaporization of a liquid to a gas leads to a large increase in volume and hence entropy; DS must be positive.

  6. Decomposition of CaCO3 has a H° = 178.3 kJ/mol and S° = 159 J/mol K. At what temperature does this become spontaneous? 121°C 395°C 848°C 1121°C

  7. Correct Answer: 121°C 395°C 848°C 1121°C T = H°/S° T = 178.3 kJ/mol/0.159 kJ/mol K T = 1121 K T (°C) = 1121 – 273 = 848

  8. Which of the following would be expected to have the highest entropy? 1 mole N2(l) 1 mole N2(g) 1 mole Ar(g) 1 mole Ar(s)

  9. Correct Answer: 1 mole N2(l) 1 mole N2(g) 1 mole Ar(g) 1 mole Ar(s) Given the same amounts, gases have higher entropy than liquids, which have higher entropy than solids. Of the gases, N2 has greater entropy than Ar because N2 is a molecule that can store energy in ways not possible for Ar.

  10. Calculate DS° for the equation below using the standard entropy data given: 2 NO(g) + O2(g) 2 NO2(g) DS° values (J/mol-K): NO2(g) = 240, NO(g) = 211, O2(g) = 205. 47 J/mol 76 J/mol +176 J/mol +147 J/mol

  11. Correct Answer: 147 J/mol 76 J/mol +176 J/mol +147 J/mol ( ) ( )     D  = D  D S S products S reactants DS° = 2(240)  [2(211) + 205] DS° = 480  [627] DS° =  147

  12. Predict whether entropy increases, decreases, or stays constant for: 2 NO2(g) 2 NO(g) + O2(g) Increases Decreases Stays constant

  13. Correct Answer: Increases Decreases Stays constant The entropy increases because two molecules of gas react to form three molecules of gas, leading to more disorder.

  14. Calculate DG° for the equation below using the Gibbs free energy data given: 2 SO2(g) + O2(g) 2 SO3(g) DG° values (kJ): SO2(g) = 300.4, SO3(g) = 370.4 +70 kJ +140 kJ 140 kJ 70 kJ

  15. Correct Answer: +70 kJ +140 kJ 140 kJ 70 kJ ( ) ( )   ° °  ° =   G G products G reactants  f f DG° = (2  370.4)  [(2  300.4) + 0] DG° = 740.8  [600.8] = 140

  16. Calculate DG° for the equation below using the thermodynamic data given: N2(g) + 3 H2(g) 2 NH3(g) DH° (NH3) = 46 kJ; DS° values (J/mol-K) are NH3 = 192.5, N2 = 191.5, H2 = 130.6. 66 kJ 33 kJ +33 kJ +66 kJ

  17. Correct Answer: +33 kJ +66 kJ 66 kJ 33 kJ DH° = 92 kJ DS° = 198.7J/mol  K DG° = DH° TDS° DG° = (92)  (298)(0.1987) DG° = (92) + (59.2) = 33

  18. Both the entropy,DS° and the enthalpy DH° have positive values. Near absolute zero (0 K), will the process be spontaneous or nonspontaneous? • Spontaneous • Nonspontaneous

  19. Correct Answer: • Spontaneous • Nonspontaneous At low temperatures, the enthalpy term dominates the entropy term, leading to a positive DG and a nonspontaneous process.

  20. Suppose G° is a large, positive value. What then will be the value of the equilibrium constant, K? K = 0 K = 1 0 < K < 1 K > 1

  21. Correct Answer: G° = RTlnK Thus, large positive values of DG° lead to large negative values of lnK. The value of K itself, then, is very small, approaching zero. K = 0 K = 1 0 < K < 1 K > 1

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