1 / 9

Temperature Dependence of Gibbs’ Free Energy

T, P initial state final state Temperature Dependence of Gibbs’ Free Energy Methods of evaluating the temperature dependence of the Gibbs’ free energy can be developed by beginning with (see the course notes on Maxwell’s Relations for a derivation of this equation):

jana
Télécharger la présentation

Temperature Dependence of Gibbs’ Free Energy

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. T, P initial state final state Temperature Dependence of Gibbs’ Free Energy Methods of evaluating the temperature dependence of the Gibbs’ free energy can be developed by beginning with (see the course notes on Maxwell’s Relations for a derivation of this equation): dG = V dP - S dT Dividing by dT, while holding the pressure constant, gives a partial derivative describing how the Gibbs’ free energy varies isobarically with temperature: (G / T)P = - S Applying this to a process or reaction: yields the equation: (DG / T)P = - DS Separating variables and integrating from the standard state at 298.2 K (a convenient choice of reference state, since data is readily available at this temperature): DGo298.2 KDGoT d(DGo) = - 298.2 KTDSo dT gives a starting point for calculating how the change in the Gibbs’ free energy varies with temperature: DGoT = DGo298.2 K - 298.2 KTDSo dT 33.1

  2. Tob, 1.00 atm C6H6 (l) C6H6 (g) If we assume that the change in entropy is not a function of temperature, then we can set DSoequal to DSo298.2 Kand further develop this equation along these lines: DGoT = DGo298.2 K - DSo298.2 K (T - 298.2 K) = (DHo298.2 K - 298.2 K DSo298.2 K) - DSo298.2 K (T - 298.2 K) = DHo298.2 K - T DSo298.2 K For example consider the vaporization of benzene at its normal boiling point: DHo298.2 K =DHof, 298.2 K [C6H6 (g)] - DHof, 298.2 K [C6H6 (l)] = (+ 82.93 kJ / mole) - (+ 49.0 kJ / mole) = + 33.9 kJ / mole DSo298.2 K =So298.2 K [C6H6 (g)] - So298.2 K [C6H6 (l)] = (+ 269.31 J / mole K) - (+ 173.3 J / mole K) = + 96.0 J / mole K Since at the normal boiling point, DGoTob equals zero (why?): Tob = DHo298.2 K /DSo298.2 K = (+ 33.9x103 J / mole) / (+ 96.0 J / mole K) = 353 K The experimental value is 353.3 K (Is this good or what?) 33.2

  3. If we assume that DS is a function of temperature then we have to start with the constant pressure temperature dependence of S: dS = Cp dT / T which when applied to a process or reaction becomes: d (DS) = DCp dT / T Integrating this expression from the standard state to T at constant pressure: DSo298.2 KDSoT d(DSo) = 298.2 KT (DCp / T) dT will give an expression for DSoT as a function of temperature. The actual form of this temperature dependence will depend on how we treat the temperature dependence of DCp. We will take the most general approach and assume that the temperature dependence of species i in the process or reaction is given by: Cp = ai + bi T + ci T2 + di T-2 DCp will then be of the form: DCp = S nproducts Cp, products - S nreactants Cp, reactants = A + B T + C T2 + D T-2 where: A = S nproducts aproducts - S nreactants areactants etc., for the constants B, C, and D. 33.3

  4. Substituting this temperature dependence into the integral for DSoT gives: DSoT = DSo298.2 K + 298.2 K T(A + B T + C T2 + D T-2) / T dT = DSo298.2 K + A ln (T / 298.2 K) + B [T - 298.2 K] + (C / 2) [T2 - (298.2 K)2] - (D / 2) [T-2 - (298.2 K)-2] = DSo298.2 K - A ln (298.2 K) - B (298.2 K) - (C / 2) (298.2 K)2 + (D / 2) (298.2 K)-2 + A ln (T) + B T + (C / 2) T2 - (D / 2) T-2 Defining the constant: E = DSo298.2 K - A ln (298.2 K) - B (298.2 K) - (C / 2) (298.2 K)2 + (D / 2) (298.2 K)-2 gives for the temperature dependence of DSoT: DSoT = E + A ln (T) + B T + (C / 2) T2 - (D / 2) T-2 33.4

  5. Substituting this result into: DGoT = DGo298.2 K - 298.2 KTDSo dT gives: DGoT = DGo298.2 K - 298.2 KT [E + A ln (T) + B T + (C / 2) T2 - (D / 2) T-2 ]dT = DGo298.2 K - E (T - 298.2 K) - A [T ln (T) - T - (298.2K) ln (298.2 K) + (298.2 K)] - (B / 2) [T2 - (298.2 K)2] - (C / 6) [T3 - (298.2 K)3] - (D / 2) [1/T - 1/(298.2 K)] = DGo298.2 K + E (298.2 K) + A (298.2 K) ln (298.2 K) - A (298.2 K) + (B / 2) (298.2 K)2 + (C / 6) (298.2 K)3 + (D / 2) / (298.2 K) - E T - A T ln (T) + A T - (B / 2 ) T2 - (C / 6 ) T3 - (D / 2) (1 / T) Defining the yet another constant F as: F = DGo298.2 K + E (298.2 K) + A (298.2 K) ln (298.2 K) - A (298.2 K) + (B / 2) (298.2 K)2 + (C / 6) (298.2 K)3 + (D / 2) / (298.2 K) gives for the temperature dependence of DGoT: DGoT = F - E T - A T ln (T) + A T - (B / 2 ) T2 - (C / 6 ) T3 - (D / 2) (1 / T) 33.5

  6. Tob, 1.00 atm C6H6 (l) C6H6 (g) Consider again the vaporization of benzene at the normal boiling point: DSo298.2 K and DGo298.2 K for the vaporization were calculated in the last example. The constant pressure heat capacities of benzene liquid and vapor are: Cp [C6H6 (l)] = 136.1 J / mole K Cp [C6H6 (g)] = 81.67 J / mole K For this vaporization what are the values of the constants A, B, C, D, E, and F in the temperature dependent expression for DGoT? Use this temperature dependent expression for the vaporization of benzene that you just developed to calculate the normal boiling point of benzene? Starting with: (G / T)P = - S and the definition of Gibbs’ free energy: G = H - T S could you derive either or both of these forms of theGibbs - Helmholtz equation: [ (G / T) / T]P = - H / T2or[ (G / T) /  ln (T)]P = - H These equations are often used as starting points for calculating how the Gibbs’ free energy varies with temperature. 33.6

  7. Gibbs’ free energy functions: (GoT - Ho298 K) / T vary only slowly with temperature and hence tables of these functions allow values of the Gibbs free energy to be accurately interpolated at different temperatures. Taking the difference (products minus reactants) of Gibbs’ free energy functions for a reaction or phase change leads to an expression for DGoT: DGoT = + T S nprod [(GoT - Ho298 K)/T]prod - T S nreact [(GoT - Ho298 K)/T]react + DHo298 K Could you derive this equation, starting with the definition of the Gibbs’ free energy function? 33.7

  8. Consider the sublimation of cesium at 1 bar and 1000 K: 1000 K Cs (s) ------------> Cs (g) 1.00 bar A portion of a table of Gibbs free energy functions that apply in this case is shown below: (GoT - Ho298 K) / T (cal / mole K) DHo298 K 298.2 K 500 K 1000 K 1500 K (kcal/mole) Cs (s) - 20.16 - 21.70 - 25.23 ------ Cs (g) - 41.94 - 42.51 - 44.48 - 45.99 + 18.67 Manipulation of the free energy function definition leads to an expression that can be used to calculate DGoT: DGoT = T ([(GoT - Ho298 K) / T]gas - [(GoT - Ho298 K) / T]solid) + DHo298 K For the vaporization of cesium at 1000 K and 1.00 bar: DGo1000 K = 1000 K [(- 0.04448 kcal / K) - (- 0.02523 kcal / K)] + 18.67 kcal = - 0.58 kcal What is DGo1000 K assuming that DSoT  f(T)? What thermodynamic quantity is the Gibbs’ free energy function at 298.2 K equal to? 33.8

  9. Use the following Gibbs free energy functions : (GoT - Ho298 K) / T (cal / mole K) DHo298 K 298.2 K 500 K 1000 K 1500 K (kcal/mole) UO2 (s) - 18.63 - 20.52 - 27.69 - 33.62 - 259.2 UO3 (s) - 23.58 - 25.9 - 34.6 ----- - 291.0 O2 (g) - 49.01 - 49.83 - 52.78 - 55.19 to calculate the Gibbs’ free energy change for the oxidation of urinite to urinate at 1 bar and 750 K: 1000 K UO2 (s) + 1/2 O2 (g) ------------> UO3 (s) 1.00 bar 33.9

More Related