Thermodynamic II Problems

Thermodynamic II Problems 1. Calculate the So at 25oC for the following reaction Al2O3 (s) + 3 H2 (g) → 2 Al(s) + 3 H2O (g) Answer: Using tables in text: So = Soproducts - Soreactants So = 179 J/K 2. Calculate the S surroundings for the following reactions, which were run at 1 atm and 25 oC. Sb2S3 (s) + 3 Fe → 2 Sb (s) + 3 FeS (s) H = -125 kJ Sb4O6 (s) + 6 C (s) → 4 Sb (s) + 6 CO (g) H = 778 kJ Ssurr = - ( Ho)/T T = 298K Sb2S3 (s) + 3 Fe →2 Sb (s) + 3 FeS (s) Answer: Sb4O6 (s) + 6 C (s) → 4 Sb (s) + 6 CO (g) 3.  Ssurr = - (778 kJ/298K) (1000J/1 kJ)  Ssurr = - 2610 J/K Calculate the Goreaction for graphite going to diamond Diamond(s) + O2 (g) →CO2 (g) Go = - 397 kJ Graphite (s) + O2 (g) → CO2 (g) Go = - 394 kJ Answer: Diamond(s) + O2 (g) → CO2 (g) CO2 (g) → Graphite (s) + O2 (g) Graphite → Diamond 4.  Ssurr = - (-125 kJ/298K) (1000J/1 kJ) Ssurr = 419 J/K Go = + 394 kJ Go = - 397 kJ Go = - 3 kJ At what temperature is the following process spontaneous at 1 atm? Br2 (l) → Br2 (g) Ho = 31.0 kJ and So = 93.0 J/K Answer: Vaporization is spontaneous at all temperatures when G is (-) & S is (+) which favors T; reason more disorder. At equilibrium Go =  Go = Ho - TSo Ho = 31.0 kJ or 3.10 x 104 J 4 0 = 3.10 x 10 J – T (93.0 J/K) T = 333K What does this mean? At 333K gas  → liquid (gas is in equilibrium with the liquid phase) At T > 333K the Br2 (g) dominant phase for So is the major contributor At T < 333K then the Br2 (l) is dominant phase for Ho is the major factor 5. Calculate the Ho, So and Go for the following reaction at 25.0 oC: H2 (g) + Cl2 (g) → 2 HCl (g) Answer: use tables to obtain values of Ho & So So & formula So = Soproducts - Soreactants Ho & formula Ho = Hoproducts - Horeactants So = + 20.0 J/K Ho = - 184.0 kJ Can calculate Go from tables using formula Go = Goproducts - Goreactants or can use Go = Ho - TSo Go & formula Go = Goproducts - Goreactants Go = Ho - TSo Go = - (-184 kJ) – 298K (0.200 kJ/K) 6. Go = - 190.7 kJ Go = - 191 kJ Calculate the Ho, So Ssurroundings, S universe and Go for the following reaction at 25.0 oC: 2 NiS (s) + 3 O2 (g) → 2 SO2 (g) + 2 NIO (s) Ho = - 890.0 kJ So SO2 (g) = 248 J/K mol So O2 (g) = 205 J/K mol So NiO (s = 38 J/K mol SoNiS (s) = 53 J/K mol Answer: So values from above & formula So = Soproducts - Soreactants So = - 149.0 J/K Ssurr = - (Ho)/T Ssurr = 2.99 kJ/K or 2990 J/K Ssurr = - (-890.0 kJ)/298K Go = Ho - TSo Go = -846 kJ Go = - (-890.0 kJ) kJ – 298K (2.99 kJ/K) Or can calculate Go from table values & formula Go = Goproducts - Goreactants Go = - 846 kJ S universe = So + Ssurroundings S universe = -149 J/K + 2990 J/K S universe = 2840 J/K 7. Calculate the S surroundings, S universe and Go for the following reaction at 25 oC: C (s) + O2 (g) (1 atm) → CO2 (g) (1 atm) Ho = - 393.5 kJ So = 3.05 J/K Answer: to calculate Go: Go = Ho - TSo Go = - 392.6 kJ Ssurr = - (Ho)/T Ssurr = 1.320 kJ/K or 1320 J/K Go = - (-393.5 kJ) – 298K (0.00305 kJ/K) Ssurr = - (-393.5 kJ)/ 298K S universe = So + Ssurroundings S universe = 3.05 J/K + 1320 J/K S universe = 1323.05 J/K 8. Given the following: N2 (g) + 2 H2 (g) ⇌ 2 NH3 (g) Where Go = -33.3 kJ per mole of N2 consumed at 25 .0oC Predict the direction in which the system will shift in order to reach equilibrium Set A: PNH3 = 1.00 atm, PN2 = 1.47 atm and PH2 = 0.01 atm Set B: PNH3 = 1.47 atm, PN2 = 0.01 atm and PH2 = 0.01 atm Answer: The K value was not given in this problem: can only determine if the reaction is at equilibrium or if the reaction is not spontaneous. Set A: G = Go + RTlnQ G = -33.3 kJ + {(8.314 x 10-3 kJ/K mol) (298 K) ln {(1.00)2/ (1.47) (0.01)2}} G = -11.1 kJ says spontaneous and not at equilibrium at these parameters. Since G is (-) can only speculate that the reaction needs to shift towards the reactant side to reach equilibrium (G = 0) Set B: G = Go + RTlnQ G = -33.3 kJ + {(8.314 x 10-3 kJ/K mol) (298 K) ln {(1.47)2/ (0.01) (0.01)2}} G = 3.14 kJ say not spontaneous and not at equilibrium at these parameters. Since G is (+) can only speculate that the reaction needs to shift towards the product side to reach equilibrium (G = 0) 9. Determine the G at 25.0 oC for the reaction given below where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol Reaction: CO (g) +2 H2 (g) → CH3OH (l) Answer: Calculate Go from table values & the formula Go = Goproducts - Goreactants Go = - 29.0 kJ G = Go + RTlnQ G = 38 kJ G = -29.0 kJ + (8.314 x 10-3 kJ/K mol) (298 K) ln {(1/ (5.00)(3.0)2} G > Go (-38 kJ > - 29 kJ) so the reaction is spontaneous