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 H o and S o 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.
Special case: No work over and above pV-work w non p-V =0 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 temperature
Chemical potential
The value of G f o of Fe(g) is 370 kJ/mol at 298 K. If H f o of Fe(g) is 416 kJ/mol (assumed to be constant in the range K), calculate G f o of Fe(g) at 400 K. Transformation at constant pressure
G dependence on n H2OH2O H2OH2O Given a system consisting of two substances:
If there is no change in composition:
System at constant T and p dn 1 Each subsystem is a mixture of substances. Chemical Equilibrium
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.
Pure H 2 N 2 + H 2 Pd membrane Equilibrium never reached constant T & p
G and S of mixing of gases
Chemical reactions CH 4(g) +2O 2(g) → CO 2(g) + 2H 2 O (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 = H of formation reaction = F H Standard heat of formation = Hº of formation reaction = F Hº F Hº(NO (g) ): ½ N 2(g) +½ O 2(g) → NO (g) Hº F Hº(CO (g) ): C graphite(s) +½ O 2(g) → CO (g) Hº F Hº(O (g) ): ½ O 2(g) → O (g) Hº F Hº(C diamond(s) ): C graphite(s) → C diamond(s) Hº F Hº(O 2(g) ): O 2(g) → O 2(g) Hº=0 F Hº(C graphite(s) ): C graphite(s) → C graphite(s) Hº=0
G of Formation Gibbs energy of formation = G of formation reaction = F G Standard Gibbs energy of formation = Gº of formation reaction = F Gº F Gº(NO (g) ): ½ N 2(g) +½ O 2(g) → NO (g) Gº F Gº(CO (g) ): C graphite(s) +½ O 2(g) → CO (g) Gº F Gº(O (g) ): ½ O 2(g) → O (g) Gº F Gº(C diamond(s) ): C graphite(s) → C diamond(s) Gº F Gº(O 2(g) ): O 2(g) → O 2(g) Gº=0 F Gº(C graphite(s) ): C graphite(s) → C graphite(s) Gº=0
Applying Hess’s Law
Chemical reactions CH 4(g) +2O 2(g) → CO 2(g) + 2H 2 O (g) As the reaction proceeds: The number of moles of involved substances changes. G of system will change: Extent of reaction Reaction advancement Degree of reaction
As the forward reaction proceeds: grows, d positive, d > 0
even though G of products larger than G of reactants, the reaction proceeds!!!!!!!!!! Reason: G mix
For a mixture: G more negative if G pure is small G mix largely negative
Pure R Pure P G pure G mix G total
Equilibrium constant R, P: Ideal gases
K p relation to K x p total in atm
K p relation to K c c in mol/L R= atmL/mol.K
Consider the reaction N 2 O 4(g) → 2 NO 2(g) F Gº(NO 2(g) )=51.31 kJ/mol F Gº(N 2 O 4(g) )= kJ/mol Assume ideal behavior, calculate nono n eq nini x eq pipi N2O4N2O4 11-x 1+x (1-x)/(1+x)(1-x)/(1+x)*P tot NO 2 02x2x/(1+x)2x/(1+x)*P tot
Temperature dependence of K p
For the reaction N 2 O 4(g) → 2 NO 2(g) F Hº(NO 2(g) )=51.31 kJ/mol F Hº(N 2 O 4(g) )= kJ/mol K p (25 o C)=0.78 atm, calculate K p at 100 o C? For a given reaction, the equilibrium constant is 1.80x10 3 L/mol at 25 o C and 3.45x10 3 L/mol at 40 o C. Assuming H o to be independent of temperature, calculate H o and S o.
Heterogeneous Equilibria
F Gº kJ/mol F Hº kJ/mol CaCO 3(s) CaO (s) CO 2(g) Calculate The pressure of CO 2 at 25oC and at 827 o C? Gº =130.4 kJ º =178.3 kJ ln(K p )=ln(p CO2 )=-52.6 p CO2 =1.43x atm At 1100 K: ln(p CO2 )=0.17 p CO2 =0.84 atm
Vaporization Equilibria Clausius-Clapeyron Equation Derive the above relations for the sublimation phase transition!
Mass Action Expression (MAE) For reaction: aA + bB cC + 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
41 Calculate [X] equilibrium from [X] initial and K C Ex. 4 H 2 (g) + I 2 (g) 2HI (g) at 425 °C K C = If one mole each of H 2 and I 2 are placed in a L flask at 425 °C, what are the equilibrium concentrations of H 2, I 2 and HI? Step 1. Write Equilibrium Law
42 Ex. 4 Step 2. Concentration Table Conc (M)H 2 (g) +I2(g)I2(g) 2HI (g) Initial Change Equil’m Initial [H 2 ] = [I 2 ] = 1.00 mol/0.500L =2.00M Amt of H 2 consumed = Amt of I 2 consumed = x Amt of HI formed = 2x – x +2x – x– x 2.00 – x
43 Ex. 4 Step 3. Solve for x Both sides are squared so we can take square root of both sides to simplify
44 Ex. 4 Step 4. Equilibrium Concentrations Conc (M)H 2 (g) +I2(g)I2(g) 2HI (g) Initial Change Equil’m [H 2 ] equil = [I 2 ] equil = 2.00 – 1.58 = 0.42 M [HI] equil = 2x = 2(1.58) = 3.16 – –
45 Calculate [X] equilibrium from [X] initial and K C Ex. 5 H 2 (g) + I 2 (g) 2HI (g) at 425 °C K C = If one mole each of H 2, I 2 and HI are placed in a L flask at 425 °C, what are the equilibrium concentrations of H 2, I 2 and HI? Now have product as well as reactants initially Step 1. Write Equilibrium Law
46 Calculate [X] equilibrium from [X] initial and K C Ex. 6 CH 3 CO 2 H (aq) + C 2 H 5 OH (aq) CH 3 CO 2 C 2 H 5 (aq) + acetic acidethanol ethyl acetate H 2 O ( l ) K C = 0.11 An aqueous solution of ethanol and acetic acid, each with initial concentration of M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?
47 Calculating K C Given Initial Concentrations and One Final Concentration Ex. 2a H 2 (g) + I 2 (g) 2HI 450 °C Initially H 2 and I 2 concentrations are mol each in 2.00L (= 0.100M); no HI is present At equilibrium, HI concentration is M Calculate K C To do this we need to know 3 sets of concentrations: initial, change and equilibrium