1 / 47

Spontaneity and Equilibrium

Spontaneity and Equilibrium. isolated system : . Isothermal process. Maximum work obtained in a process at constant temperature is equal to the decrease in Helmholtz energy of the system. Calculate the maximum work that can be obtained from the combustion of 1 mole of methane at 298 K.

layne
Download Presentation

Spontaneity and Equilibrium

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. Spontaneity and Equilibrium isolated system: Isothermal process • Maximum work obtained in a process at constant temperature is equal to the decrease in Helmholtz energy of the system.

  2. Calculate the maximum work that can be obtained from the combustion of 1 mole of methane at 298 K. Given DHo and DSo of the the combustion of methane.

  3. Transformation at constant temperature and pressure • Maximum work, over and above pV-work, obtained in a process at constant temperature and pressure is equal to the decrease in the Gibbs energy of the system.

  4. No work over and above pV-work wnon p-V=0 Special case: Calculate the maximum non-pV work that can be obtained from the combustion of 1 mole of methane at 298 K.

  5. Fundamental equations of Thermodynamics Maxwell Relations

  6. Transformation at constant temperature

  7. Chemical potential m

  8. Transformation at constant pressure The value of DGfo of Fe(g) is 370 kJ/mol at 298 K. If DHfo of Fe(g) is 416 kJ/mol (assumed to be constant in the range 250-400 K), calculate DGfo of Fe(g) at 400 K.

  9. G dependence on n H2O H2O Given a system consisting of two substances:

  10. If there is no change in composition:

  11. Chemical Equilibrium a b Each subsystem is a mixture of substances. System at constant T and p dn1

  12. Equilibrium is established if chemical potential of all substances in the system is equal in all parts of the system. • Matter flows from the part of system of higher chemical potential to that of lower chemical potential.

  13. Pd membrane a b • constant T & p Pure H2 N2 + H2 • Equilibrium never reached

  14. DG and DS of mixing of gases

  15. Chemical reactions CH4(g) +2O2(g)→ CO2(g) + 2H2O(g)

  16. Heat of Formation Formation reaction: reaction of forming 1 mole of product from the elements in their stable form at 25ºC and 1 atm. Heat of formation = DH of formation reaction = DFH Standard heat of formation = DHº of formation reaction = DFHº ½N2(g)+½O2(g)→NO(g)DHº DFHº(NO(g)): Cgraphite(s)+½O2(g)→CO(g)DHº DFHº(CO(g)): ½O2(g)→O(g)DHº DFHº(O(g)): Cgraphite(s)→Cdiamond(s)DHº DFHº(Cdiamond(s)): O2(g)→O2(g)DHº=0 DFHº(O2(g)): Cgraphite(s)→Cgraphite(s)DHº=0 DFHº(Cgraphite(s)):

  17. DG of Formation Gibbs energy of formation = DG of formation reaction = DFG Standard Gibbs energy of formation = DGº of formation reaction = DFGº ½N2(g)+½O2(g)→NO(g)DGº DFGº(NO(g)): Cgraphite(s)+½O2(g)→CO(g)DGº DFGº(CO(g)): ½O2(g)→O(g)DGº DFGº(O(g)): Cgraphite(s)→Cdiamond(s)DGº DFGº(Cdiamond(s)): O2(g)→O2(g)DGº=0 DFGº(O2(g)): Cgraphite(s)→Cgraphite(s)DGº=0 DFGº(Cgraphite(s)):

  18. Applying Hess’s Law

  19. Chemical reactions CH4(g) +2O2(g)→ CO2(g) + 2H2O(g) • As the reaction proceeds: • The number of moles of involved substances changes. • G of system will change: • x • Extent of reaction • Reaction advancement • Degree of reaction

  20. As the forward reaction proceeds: x grows, dx positive, dx> 0

  21. even though G of products larger than G of reactants, the reaction proceeds!!!!!!!!!! Reason: DGmix

  22. For a mixture: • DG more negative if • DGpure is small • DGmix largely negative

  23. Pure R • Gpure • Gtotal • Pure P • DGmix

  24. Equilibrium constant R, P: Ideal gases

  25. Kp relation to Kx • ptotal in atm

  26. Kp relation to Kc • c in mol/L • R=0.0821 atmL/mol.K

  27. Consider the reaction N2O4(g)→ 2 NO2(g) DFGº(N2O4(g))=102.00 kJ/mol DFGº(NO2(g))=51.31 kJ/mol Assume ideal behavior, calculate

  28. Temperature dependence of Kp

  29. For the reaction N2O4(g)→ 2 NO2(g) DFHº(N2O4(g))=102.00 kJ/mol DFHº(NO2(g))=51.31 kJ/mol Kp(25oC)=0.78 atm, calculate Kp at 100oC? • For a given reaction, the equilibrium constant is 1.80x103 L/mol at 25oC and 3.45x103 L/mol at 40oC. Assuming DHo to be independent of temperature, calculate DHo and DSo.

  30. Heterogeneous Equilibria

  31. Calculate The pressure of CO2 at 25oC and at 827oC? DGº =130.4 kJ DHº =178.3 kJ ln(Kp)=ln(pCO2)=-52.6 pCO2=1.43x10-23atm At 1100 K: ln(pCO2)=0.17 pCO2=0.84 atm

  32. Vaporization Equilibria Clausius-Clapeyron Equation Derive the above relations for the sublimation phase transition!

  33. Mass Action Expression (MAE) • For reaction:aA + bBcC + dD Reaction quotient • Numerical value of mass action expression • Equals “Q” at any time, and • Equals “K” only when reaction is known to be at equilibrium

  34. Calculate [X]equilibrium from [X]initial and KC Ex. 4H2(g) + I2(g) 2HI(g) at 425 °C KC = 55.64 • If one mole each of H2 and I2are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI? • Step 1. Write Equilibrium Law

  35. Ex. 4 Step 2. Concentration Table – x –x +2x 2.00 – x 2.00 – x +2x • Initial [H2] = [I2] = 1.00 mol/0.500L =2.00M • Amt of H2 consumed = Amt of I2 consumed = x • Amt of HIformed = 2x

  36. Ex. 4 Step 3. Solve for x • Both sides are squared so we can take square root of both sides to simplify

  37. Ex. 4 Step 4. Equilibrium Concentrations – 1.58 –1.58 +3.16 0.42 0.42 +3.16 • [H2]equil = [I2]equil = 2.00 – 1.58 = 0.42 M • [HI]equil = 2x = 2(1.58) = 3.16

  38. Calculate [X]equilibrium from [X]initial and KC Ex. 5H2(g) + I2(g) 2HI(g) at 425 °C KC = 55.64 • If one mole each of H2, I2and HI are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI? • Now have product as well as reactants initially • Step 1. Write Equilibrium Law

  39. Calculate [X]equilibrium from [X]initial and KC Ex. 6 CH3CO2H(aq) + C2H5OH(aq)CH3CO2C2H5(aq) + acetic acid ethanolethyl acetate H2O(l) KC = 0.11 • An aqueous solution of ethanol and acetic acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?

  40. Calculating KC Given Initial Concentrations and One Final Concentration Ex. 2a H2(g) + I2(g) 2HI(g)@ 450 °C • Initially H2 and I2 concentrations are 0.200 mol each in 2.00L (= 0.100M); no HI is present • At equilibrium, HI concentration is 0.160 M • Calculate KC • To do this we need to know 3 sets of concentrations: initial, change and equilibrium

More Related