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Thermodynamics, pt 2 Dr. Harris Lecture 22 (Ch 23) 11/19/12
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Transcript of Thermodynamics, pt 2 Dr. Harris Lecture 22 (Ch 23) 11/19/12
Thermodynamics, pt 2
Dr. HarrisLecture 22 (Ch 23)11/19/12HW: Ch 23: 27, 43, 57, 59, 77, 81
Recap: Equilibrium Constants and Reaction Quotients
โข For any equilibrium reaction:
๐๐ด+๐๐ต ๐๐ถ+๐๐ท
ยฟยฟโข The equilibrium constant, K, is equal to the ratio of the concentrations/
pressures of products and reactants at equilibrium.
โข The reaction quotient, Q, has the same form as K, but does not use equilibrium concentrations/pressures. Using the previous reaction:
๐ธ๐๐๐ ๐ธ๐=ยฟยฟโข The subscript โ0โ denotes arbitrary concentrations. Unlike K, Q is not
constant and depends on the starting concentrations.
Recap: Direction of Spontaneity
โข The direction of spontaneity is always toward equilibrium.
โข The value of Q/Kc tells us the direction in which a system not at equilibrium will proceed to reach equilibrium.
Recap: Entropy and the 2nd Law of Thermodynamics
โข 2nd Law of Thermodynamics:
Entropy is not conserved. The Entropy of the universe is continually Increasing.
โ๐๐ข๐๐๐ฃ=โ๐๐ ๐ฆ๐ +โ๐๐ ๐ข๐๐
โข The universe can never become more ordered after a process. Therefore, if a particular system becomes more ordered (ฮSsys<0), the surroundings must become even more disordered (ฮSsurr >0)
โข Entropy is a measure of the disorder of a system. Increasing disorder means that the change in entropy is positive.
โ๐๐ข๐๐๐ฃโฅ0
Recap: Thermodynamics of Equilibrium
โข When a system reaches equilibrium, the entropy is at a maximum, so the change in entropy is 0 (ฮSsys = 0 at equilibrium)
Recap: Spontaneity Depends on Enthalpy AND Entropy
โ๐บ=โ๐ปโ๐ โ๐Dictates if a process is energetically favored
Dictates if a process is entropically favored
Minimizing ฮG
โข In general, a system will change spontaneously in such a way that its Gibbs free energy is minimized.
โข The enthalpy term is independent of concentration and pressure. Entropy is not.
โข During a reaction, the composition of the system changes, which changes concentrations and pressures, leading to changes in the โTฮS term.
โข As the system approaches an entropically unfavorable composition, the back reaction occurs to prevent ฮG from becoming more positive. This is the basis of equilibrium.
โข Once equilibrium is reached, the free energy no longer changes
Recap: Correlation Between Gibbs Free Energy and Equilibrium
โข If ฮG is negative, the reaction is spontaneous
โข If ฮG is zero, the reaction is at equilibrium
โข If ฮG is positive, the reaction is spontaneous in the opposite direction
100% 0%
Decreasing G
Reactants
spontaneous
K > Q
Q > K
Q = K
ฮG = 0
When ฮG is Negative, the Value Tells Us the Maximum Portion of ฮU That Can Be Used to do Work
Gasoline with internal energy U
Work not accounted for by change in free energy must be lost as heat
Maximum possible fraction of U converted to work = -ฮG
ฮG = -wmax
Relating the Equilibrium Constant, Reaction Quotient, and ฮGo
rxn
โข Keep in mind that the standard free energy change, ฮGo, is not the same as the nonstandard free energy change, ฮG. ฮGo is determined under standard conditions. Those conditions are listed below.
State of Matter Standard StatePure element in most stable state ฮGo is defined as ZEROGas 1 atm pressure, 25oCSolids and Liquids Pure state, 25oCSolutions 1M concentration
โข For many elements, ฮGorxn
can be obtained from a table of values. ฮGorxn
can be calculated in the same manner as ฮHorxn using free energies of
formation:
โข In terms of the equilibrium constant of a particular reaction, the driving force to approach equilibrium under standard conditions is given by:
โ๐ฎ๐๐๐๐ =โ๐น๐ป ๐ฅ๐ง๐ฒ
โข When the reaction conditions are not standard, you must use the reaction quotient, Q.
โข The free energy change of a reaction (or the driving force to approach equilibrium) under non-standard conditions, ฮG, is given by:
โ๐ฎ๐๐๐=โ๐ฎ๐๐๐๐ +๐น๐ป ๐ฅ๐ง๐ธ
โ๐ฎ๐๐๐๐ =โ ๐โ๐ฎ ๐
๐ (๐๐๐๐ )โโ ๐โ๐ฎ ๐๐ (๐๐๐ )
Relating K, Q, and ฮGorxn
Example #1 (No K value given)
๐ 2 (๐ )+3๐ป2 (๐ )2๐๐ป3(๐)
โข Calculate ฮG at 298oK for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3. Which direction must the reaction shift to reach equilibrium?
โข We are finding the free energy change under non-standard conditions (ฮG). We must first Q.
โข Now determine the standard free energy, ฮGo. If K is not given, you can calculate it from the standard table.
๐=ยฟยฟ
โ๐ฎ๐๐๐=โ๐ฎ๐๐๐๐ +๐น๐ป ๐ฅ๐ง๐ธ
โข From appendix D in the back of the book:
โ๐ฎ ๐๐๐ฏ๐=๐
โ๐ฎ ๐๐๐ต๐=๐
โ๐ฎ ๐๐๐ต๐ฏ๐=โ๐๐ .๐
๐๐ฑ๐๐๐
โ๐ฎ๐๐๐๐ =๐(โ๐๐ .๐ ๐๐ฑ
๐๐๐ )=โ๐๐ .๐ ๐๐ฑ๐๐๐
๐ 2 (๐ )+3๐ป2 (๐ )2๐๐ป3(๐)
โข Solve for ฮGโ๐บ๐๐ฅ๐=โ๐บ๐๐ฅ๐
๐ +๐ ๐ ln๐
โ๐บ๐๐ฅ๐=โ32800๐ฝ
๐๐๐+(8.314 ๐ฝ๐๐๐๐พ ) (298๐พ ) ln(.0277)
โ๐บ๐๐ฅ๐=โ32800๐ฝ
๐๐๐+(8.314 ๐ฝ๐๐๐๐พ )(298๐พ )(โ3.586)
โ๐บ๐๐ฅ๐=23915๐ฝ
๐๐๐Reaction moves to the left to reach equilibrium.
Example #2 (Value of K given)
๐๐ฏ๐ญ (๐ ) ๐ฏ๐ (๐ )+๐ญ๐ (๐ )
โข At 598oK, the initial partial pressures of H2, F2 and HF are 0.150 bar, .0425 bar, and 0.500 bar, respectively. Given that Kp = .0108, determine ฮG. Which direction will the reaction proceed to reach equilibrium?
โ๐บ๐๐ฅ๐=โ๐บ๐๐ฅ๐๐ +๐ ๐ ln๐
๐=( .150 )(.0425)
ยฟยฟโข Find Q
โข We have K, so we can determine ฮGorxn without using the standard table.
โ๐ฎ๐๐๐๐ =โ๐น๐ป ๐ฅ๐ง๐ฒ
โ๐บ๐๐ฅ๐=โ๐ ๐ ๐๐๐พ +๐ ๐ ln๐=RT ln ๐๐พ
โ๐บ๐๐ฅ๐=4.27 kJ /mol Reaction moves left to reach equilibrium.
Deriving The vanโt Hoff Equation
โข We know that rate constants vary with temperature.
โข Considering that equilibrium constants are ratios of rate constants of the forward and back reaction, we would also expect equilibrium constants to vary with temperature.
โข Using our relationship of the standard free energy with standard enthalpy and entropy:
โ๐บ๐๐ฅ๐๐ =โ๐ป๐๐ฅ๐
๐ โ๐ โ๐๐๐ฅ๐๐
โ๐ ๐ ln ๐พ=โ๐ป๐โ๐ โ๐๐
โข And relating this expression to the equilibrium constant, K, we obtain:
ln ๐พ=โโ๐ป๐๐ฅ๐
๐
๐ ๐ +โ๐๐๐ฅ๐
๐
๐
โข As we see in this expression, as we increase temperature, the enthalpy term becomes very small. The entropy term then becomes more important in determining K as T increases.
โข Thus, entropy is the dominant factor in determining equilibrium distributions at high temperatures, and enthalpy is the dominant factor at low temperatures.
โข A plot of ln K vs. 1/T will yield a linear plot with a slope of (โฮHorxn)/R
ln ๐พ=โโ๐ป๐๐ฅ๐
๐
๐ ๐ +โ๐๐๐ฅ๐
๐
๐
Deriving The vanโt Hoff Equation
Deriving The vanโt Hoff Equation
โข If you run the same reaction at different temperatures, T1 and T2:
ln ๐พ 1=โโ๐ป๐๐ฅ๐
๐
๐ ๐1+โ๐๐๐ฅ๐
๐
๐ ln ๐พ 2=โโ๐ป๐๐ฅ๐
๐
๐ ๐2+โ๐๐๐ฅ๐
๐
๐
โข Then subtraction yields:
ln๐พ 2โln ๐พ 1=โ๐ป๐๐ฅ๐
๐
๐ ( 1๐ 1โ 1๐ 2 )
๐๐ ๐พ 2
๐พ1=โ๐ป๐๐ฅ๐
๐
๐ (๐2โ๐ 1
๐1๐ 2)
โข Which equals:
vanโt Hoff equation
โข So if you know the equilibrium constant at any temperature, and the standard enthalpy of reaction, you can determine what K would be at any other temperature.
Example
โข CO(g) + 2H2(g) CH3OH(g) ฮHorxn= -90.5 kJ/mol
The equilibrium constant for the reaction above is 25000 at 25oC. Calculate K at 325oC. Which direction is the reaction favored at T2?
๐๐ ๐พ 2
๐พ1=โ๐ป๐๐ฅ๐
๐
๐ (๐2โ๐ 1
๐1๐ 2)
K1 = 25000, T1 = 298 K, T2 = 598 Kโข Find K2
๐๐๐พ 2
25000=โ90500 ๐ฝ
๐๐๐
(8.314 ๐ฝ๐๐๐ ๐พ )(
598๐พโ298๐พ178204๐พ 2 ) ๐๐
๐พ 2
25000=โ18.32
๐๐๐ ๐พ 2
25000=๐โ18.32
๐พ 2
25000=1.1 ๐ฅ10โ 8 ๐ฒ ๐=๐ .๐๐ ๐๐๐โ๐
use ex to cancel ln term
Example, contd.
โข Because the value of K2 is increasingly smaller as T increases, it is clear that the reaction is favored to the left.
CO(g) + 2H2(g) CH3OH(g) ฮHorxn= -90.5 kJ/mol
K1 = 25000T1 = 298 K
K2 = .000276T2 = 598 K