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25-4-2011

25-4-2011. At what temperature will the reaction below starts to be non-spontaneous: N 2 (g) + 3H 2 (g) D 2NH 3 (g) D H° - T D S° = D G° - 92 kJ -198.5 J/K ( D S is -ve) @ T Low D H° dominates D G° = -92000 –(-198.5) = -91802J a Spontaneous

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25-4-2011

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  1. 25-4-2011

  2. At what temperature will the reaction below starts to be non-spontaneous: N2 (g) + 3H2(g) D 2NH3 (g) DH° - T DS° = DG° - 92 kJ -198.5 J/K (DS is -ve) @ T Low DH° dominates DG° = -92000 –(-198.5) = -91802J a Spontaneous @ T High -TDS° dominates DG° (+) a Non-spontaneous To go from spontaneous to non-spontaneous, DG° = 0 T = - 92kJ = 463.5 K -0.198 kJ / K

  3. CaCO3(s) CaO (s) + CO2(g) At what temperature will the reaction below starts to be spontaneous: DH0 = 177.8 kJ/mol Equilibrium Pressure of CO2 DS0 = 160.5 J/K·mol DG0 = DH0 – TDS0 At 25 oC, DG0 = 177.8–(298*160.5*10-3) = 130.0 kJ/mol, Suggesting a non-spontaneous process at this temperature. DG0 = 0 at 1108 K or 835 oC, where the reaction becomes possible

  4. The Significance of the Sign and Magnitude of DGo DGprod << DGreac DGo is a large, negative quantity and equilibrium is very far to the right (towards products)

  5. The Significance of the Sign and Magnitude of DGo DGprod >> DGreac DGo is a large, positive quantity and equilibrium is very far to the left (towards reactants)

  6. The Significance of the Sign and Magnitude of DGo DGprod DGreac The equilibrium lies more toward the center of the reaction profile When Q is smaller than K (Q<Keq ) this means that reactants have a large value. This shifts the equilibrium towards product formation, and vice versa.

  7. [NH3]2 Q = [H2]3[N2] When Q is smaller than K this means that reactants have a large value. This shifts the equilibrium towards product formation, and vice versa.

  8. Phase Transition Phase transitions can be looked at as an equilibrium process, in which both phases will be present. Therefore, DG = 0 and: 0 = DH – TDS DS = DH/T We can then write: DSfusion = DHfusion/Tmp and DSvap = DHvap/Tbp

  9. The heat of fusion and vaporization of benzene are 10.9kJ/mol and 31.0kJ/mol, respectively. Calculate the entropy changes from solid to liquid and liquid to vapor. At 1 atm, benzene melts at 5.5 oC and boils at 80.1 oC. DSfusion = DHfusion/Tmp (solid g liquid) DSfusion = (10.9*103J/mol)/(273 + 5.5)K DSfusion = 39.1 J/K mol DSvap = DHvap/Tbp (liquid g gas) DSvap = (31.0*103J/mol)/(273 + 80.1)K DSvap = 87.8 J/K mol

  10. Gibbs Free Energy and the Equilibrium Constant • Recall that G and K (equilibrium constant) apply to standard conditions • Recall that G and Q (equilibrium quotient) apply to any conditions. • It is useful to determine whether substances under any conditions (non-standard conditions) will react:

  11. Calculate DG at 298K for the reaction below, where 1.0 atm N2(g), 3.0 atm H2(g) and 1.0 atm NH3(g) are present in the mixture. DGo at 298K = -33.3 kJ N2(g) + 3H2(g) D 2NH3(g) Solution Q = P2NH3/(PN2*P3H2) Q = 1.02 / (1.0 * 3.03) Q = 3.7 x 10-2 DG = DGo + RT ln Q DG = -33.3 kJ + 8.314 J/K * 298K * ln (3.7 x 10-2) DG = -33.3 kJ - 8.17 kJ = -41.5 kJ

  12. Is the reaction 2 SO2 (g) + O2 (g) g 2 SO3 (g) expected to be spontaneous at 25 oC and 1 atm. DGfo (SO2 (g)) = -300 kJ/mol and DGfo (SO3 (g)) = -370 kJ/mol. First, you should remember that DGfo (O2 (g)) = 0 since the element in its standard state has both DGfo and DHfo = 0 DGo = 2 DGfo (SO3 (g)) – { 2 DGfo (SO2 (g)) + DGfo (O2 (g))} DGo = 2*(-370 kJ/mol) – { 2*(-300 kJ/mol) + 0 }= -140 kJ/mol The value of DGo is very negative suggesting that the reaction is far to the right. Thus the reaction is spontaneous

  13. Free Energy and Equilibrium Under non-standard conditions, we need to use DG instead of DG°. Q is the reaction quotiant. Note: at equilibrium: DG = 0. away from equil, sign of DGtells which way rxn goes spontaneously.

  14. Gibbs Free Energy and the Equilibrium Constant • At equilibrium, Q = K and G = 0, so • We can conclude: • If G < 0, then K > 1 (products are favored). • If G = 0, then K = 1 (equilibriunm state). • If G > 0, then K < 1 (reactants are favored). This equation is one of the most important relationships in thermodynamics because it allows us to calculate K from standard free energy changes.

  15. Calculate the equilibrium constant, Kp, at 25 oC, for the reaction: 2H2O (l) D 2H2 (g) + O2 (g) Where DGof (H2O(l)) = 237.2 k/mol, DGof (H2(g)) = 0.0 k/mol DGof (O2(g)) = 0.0 k/mol DGorxn = {(2*0.0 + 1*0.0)} –(- 2*237.2) =474 kJ DGorxn = -RTlnKp 474.4*103 J/mol = - 8.314 J/K mol*298K lnKp Kp = 7*10-84asince Kp is very small it is not wise to use partial pressures to calculate it.

  16. DG Equations DG° DH° - TDS° - RT ln Keq S n DG°prod - S n DG° reaction DG - RT ln Q

  17. Coupled Reactions A thermodynamically unfavored reaction can be made to proceed in the forward direction by coupling to a thermodynamically favored one: Example: Zn(s) g Zn(s) + S(s) has a DGorxn = 198 kJ/mol However, the reaction: S(s) + O2(g) g SO2(g) has a DGorxn = -300 kJ/mol Coupling the two reactions gives: ZnS(s) + O2(g) g Zn(s) + SO2(g) where DGorxn = (- 300 –(+198)} = -112 kJ/mol, which is favorable

  18. Selected Problems 1, 4, 5, 7, 8, 10-14, 17, 19-25, 27, 31

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