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Spontaneity and Equilibrium

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## Spontaneity and Equilibrium

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**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. Given DHo and DSo of the the combustion of methane.**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.**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.**Fundamental equations of Thermodynamics**Maxwell Relations**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.**G dependence on n**H2O H2O Given a system consisting of two substances:**Chemical Equilibrium**a b Each subsystem is a mixture of substances. System at constant T and p dn1**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.**Pd membrane**a b • constant T & p Pure H2 N2 + H2 • Equilibrium never reached**Chemical reactions**CH4(g) +2O2(g)→ CO2(g) + 2H2O(g)**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)):**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)):**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**As the forward reaction proceeds:**x grows, dx positive, dx> 0**even though G of products larger than G of reactants, the**reaction proceeds!!!!!!!!!! Reason: DGmix**For a mixture:**• DG more negative if • DGpure is small • DGmix largely negative**Pure R**• Gpure • Gtotal • Pure P • DGmix**Equilibrium**constant R, P: Ideal gases**Kp relation to Kx**• ptotal in atm**Kp relation to Kc**• c in mol/L • R=0.0821 atmL/mol.K**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**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.**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**Vaporization Equilibria**Clausius-Clapeyron Equation Derive the above relations for the sublimation phase transition!**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**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**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**Ex. 4 Step 3. Solve for x**• Both sides are squared so we can take square root of both sides to simplify**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**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**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?**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