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23-4-2011

23-4-2011. Gibb’s Free Energy. The second law of thermodynamics suggests that there will always be an increase in entropy of the universe for a spontaneous process to occur. This is related to two terms, enthalpy and entropy changes in the system, as can be seen in the manipulation below.

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23-4-2011

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  1. 23-4-2011

  2. Gibb’s Free Energy The second law of thermodynamics suggests that there will always be an increase in entropy of the universe for a spontaneous process to occur. This is related to two terms, enthalpy and entropy changes in the system, as can be seen in the manipulation below. Suniv = Ssys + Ssurr > 0

  3. At Constant T and P DG = - TDSUniverse (DSuniv must be +ve for a spontaneous process, as suggested by the second law of thermodynamics) • Therefore, DG must be –ve for the reaction to be spontaneous

  4. Free Energy Change If DG < 0 (negative), a process is spontaneous (products are favored). If DG > 0 (positive), a process is non-spontaneous (reactants are favored). If DG = 0, neither the forward nor the reverse process is favored; there is no net change, and the process is at equilibrium.

  5. Gibbs’ Free Energy can be used to determine the Standard free energy (°) of formation DG = DH - TDS DG°f = DH°f - TDS°f °Standard State f -formation from elements If data is not for formation process, then equation is slightly adjusted: DG° = DH° - TDS° Or from tabulated thermodynamic data: DG°rxn = S nDG°f(prod) - S mDG°f(react)

  6. - mDG0 (reactants) S S = f 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) DG0 DG0 DG0 - [ ] [ + ] = rxn rxn rxn [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ/mol = 12DG0 (CO2) 2DG0 (C6H6) f f 6DG0 (H2O) f nDG0 (products) f What is the standard free-energy change for the following reaction at 25 0C? Is the reaction spontaneous at 25 oC? DG0rxn = -6405 kJ/mol is < 0, therefore the reaction is spontaneous. Note that DGof (O2) = 0.0

  7. Calculate DGo for the reaction below where all components are at standard conditions, at 298K, provided that DGfo (NH3(g)) = -16.66 kJ/mol N2(g) + 3H2(g) D 2NH3(g) Solution Note that: DGfo(H2(g), 298K) = 0 (note: standard state of element); DGfo(N2(g), 298K) = 0 (standard state of element) DGorxn = SDGfo(products) - SDGfo(reactants) DGorxn = 2*(-16.66 kJ/mol) - (0 + (3*0)) DGorxn = -33.3 kJ/mol DGorxn is a negative value, which suggests that the reaction would proceed spontaneously to the right to produce more products. Note that DGof for both N2(g) and 3H2(g) is zero

  8. The standard enthalpy of formation of CO2 (g) is –394 kJ/mol. If So (CO2 (g)) = 213.6 J/mol K, So (O2 (g)) = 205.0 J/mol K, and So (C (s)) = 5.69 J/mol K. Find DSfo and DGfo for the formation of CO2 (g). Solution C(s) + O2 (g) g CO2 (g) DSfo = S nSo(products) - S mSo(reactants) DSfo = (So (CO2 (g)) - {So (O2 (g)) + So (C (s)} DSfo = 213.6 J/mol K - {205.0 J/mol K + 5.69 J/mol K} = 2.9 J/mol K DGfo = DHfo - TDSfo DGfo = = {-394 kJ/mol} – {298 K*(2.9*10-3 kJ/mol K)}= - 395 kJ/mol

  9. Consider the calculation for the following reaction at standard state: CH3OH (g) + O2(g) g CO2(g) + H2O (g) Determine DG°rxn 2 CH3OH (g) + 3 O2(g) g 2 CO2(g) + 4 H2O (g) DH°f - 201.2 0 -393.5 -241.82 S°f + 237.6 205.0 213.6 188.83 DG°f - 161.9 0 -394.4 -228.57

  10. DGorxn = {(4*(-228.57) + 2*(-294.4))} – {(3*0 + 2*(-161.9))} = -1379.3 kJ We should be able to get the same result from calculation of DHrxn and DSrxn DHorxn = {(4*(-241.82)+2*(-393.5)} – {(3*0 + 2*(-201.2))} = -1351.88 kJ DSorxn = {(4*188.83 + 2*213.6)} – {(3*205 + 2*237.6)} = + 92.32 J DSorxn = 92.32*10-3 kJ DGorxn = -1351 – (298*92.32*10-3) = -1379.4, which is the same as calculated from DGof values o o We can calculate DGorxn directly fromDGof

  11. Effect of temperature on Free Energy DG = DH - T DS

  12. Effect of temperature on Free Energy DG = DH - T DS both DH and DS are (+) Temperature will dictate outcome of DG. Tlow: Temperature small DH - TDS gDG (+) dominates negligible nonspontaneous Thigh: Temperature large DH - TDS gDG (-) negligible dominates spontaneous

  13. N2F4(g)g 2NF2(g) DH° - T DS° = DG° 85 kJ 198 J/K @ T Low a DH° dominates, since TDS will be small at low T DG° (+) a \ Nonspontaneous @ T High a -TDS° dominates DG° (-) a\ Spontaneous

  14. Effect of temperature on Free Energy DG = DH - T DS both DH and DS are (-) Temperature will dictate outcome of DG. Tlow: Temperature small DH - TDS gDG (-) dominates negligible spontaneous Thigh: Temperature large DH - TDS gDG (+) negligible dominates non-spontaneous

  15. Effect of temperature on Free Energy DG = DH - T DS DH is (-) and DS is(+) DG will always be negative and the reaction is always spontaneous DG = DH - T DS DH is (+) and DS is(-) DG will always be positive and the reaction is always non-spontaneous

  16. Can you predict the sign of DH and DS for each case?.

  17. 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° (-) 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

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