1. Chemical Engineering Thermodynamics-II Topic: Excess Gibbs Free Energy, Data Reduction, Thermodynamic Consistency Prepatred by:- Patel Nirav (130110105032) Patel Parth(130110105033) Patel Prajesh (130110105034) Patel Rishi (130110105035) Patel Ronak (130110105036) PREPARED BY : DR. M S BHAKHAR PROF. HARESH K DAVE G H PATEL COLLEGE OF ENGINEERING AND TECHNOLOGY

2. Gibbs Free Energy  Gibbs free energy is a measure of chemical energy All chemical systems tend naturally toward states of minimum Gibbs free energy The gibbs free energy is defined as G = H - TS Where: G = Gibbs Free Energy H = Enthalpy (heat content) T = Temperature in Kelvins S = Entropy (can think of as randomness)

3. Products and reactants are in equilibrium when their Gibbs free energies are equal A chemical reaction will proceed in the direction of lower Gibbs free energy (i.e.,  G r < 0) The reaction won’t proceed if the reaction produces an increase in Gibbs free energy Gibbs Free Energy

4. Cont…  For a phase we can determine V, T, P, etc., but not G or H We can only determine changes in G or H as we change some other parameters of the system Example: measure  H for a reaction by calorimeter - the heat given off or absorbed as a reaction proceeds

5. Cont…  Arbitrary reference state and assign an equally arbitrary value of H to it: Choose 298.15 K/25°C and 0.1 MPa/1 atm/1 bar (lab conditions)...and assign H = 0 for pure elements (in their natural state - gas, liquid, solid) at that reference

6. Cont… In our calorimeter we can then determine  H for the reaction: Si (metal) + O 2 (gas) = SiO 2  H = -910,648 J/mol = molar enthalpy of formation of quartz (at 25°C, 1 atm) It serves quite well for a standard value of H for the phase Entropy has a more universal reference state: entropy of every substance = 0 at 0K, so we use that (and adjust for temperature) Then we can use G = H - TS to determine G for quartz = -856,288 J/mol

7. Cont… K=equilibrium constant at standard T T in kelvin 298.18K R=gas constant=1.987 cal/mol o

8. Example: What is the  G o R of calcite dissociation? Use data in appendix B for  G o f  G o R = [(-132.3)+(-126.17)] - [(-269.9)] = +11.43 kcal (+) means that the reaction goes from right to left so K must be small What is the value of K? K = 10 (-11.43/1.364) = 10 -8.3798 = 4.171 x 10 -9 CaCO 3 Ca 2+ + CO 3 2-

9. What if T  25 o C? Use the Van’t Hoff Equation Enthalpy of reaction R=1.987 cal/mol° T in Kelvin  G° r =  H° r -T  S° r and lnK T - lnK T° = (-  H° r /R)(1/T-1/T°) We can derive:

10. Example: What is K T of calcite dissociation at T=38°C? = [(-129.74)+(-161.8)] - [(-288.46)] = -3.08 When T increases, K decreases (K T° = 4.171 x 10 -9 )

11. Cont… G is a measure of relative chemical stability for a phase We can determine G for any phase by measuring H and S for the reaction creating the phase from the elements We can then determine G at any T and P mathematically Most accurate if know how V and S vary with P and T dV/dP is the coefficient of isothermal compressibility dS/dT is the heat capacity (Cp)

12. Excess properties The most important excess function is the excess Gibbs free energy G E Excess entropy can be calculated from the derivative of G E wrt T Excess volume can be calculated from the derivative of G E wrt P And we also define partial molar excess properties

13. Excess gibbs free enerrgy 

14. 

15. Activity coefficient  The activity coefficient of component i is found by differentiation of the excess Gibbs energy towards x i. This yields, when applied only to the first term and using the Gibbs–Duhem equation,Gibbs–Duhem equation In here A 12 and A 21 are constants which are equal to the logarithm of the limiting activit coefficients: and respectively. When, which implies molecules of same molecular size but different polarity, the equations reduce to the one-parameter Margules activity model:

16. Example of data reduction  The following is a set of VLE data for the system methanol(1)/water(2) at 333.15K P/kPax1y1P/kPax1y1 19.9530060.6140.52820.8085 39.2230.16860.571463.9980.60440.8383 42.9840.21670.626867.9240.68040.8733 48.8520.30390.694370.2290.72550.8922 52.7840.36810.734572.8320.77760.9141 56.6520.44610.774284.56211

17. Find parameter values for the Margules equation that give the best fit of G E /RT to the data, and prepare a P x y diagram that compares the experimental points with Curves determined from the correlation 1) Calculate EXPERIMENTAL values of activity coefficients  1 and  2 and G E

18. We have shown that:

19. Now we have our analytical model Lets calculate ln  1, ln  2, G E /x 1 x 2 RT, and:

20. RMS= SQRT (  (P i -P i calc ) 2 /n = 0.398 kPa

21. Thermodynamic consistency  We need to check that the experimentally obtained activity coefficients satisfy the Gibbs-Duhem equation of P-x ₁ -y ₁ data.  If the experimental data are inconsistent with the G-D equation, they are not correct

22. Consistency test

25. Solid lines are the result of data reduction adjusting G E /RT

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