THERMODYNAMICS 2
Dr. Harris
Suggested HW: Ch 23: 43, 59, 77, 81
Recap: Equilibrium Constants and Reaction Quotients
β’ For any equilibrium reaction:
ππ΄+ππ΅ ππΆ+ππ·
ΒΏΒΏβ’ The equilibrium constant, K, is equal to the ratio of the concentrations/
pressures of products and reactants to their respective orders at equilibrium.
β’ The reaction quotient, Q, has the same form as K, but does not use equilibrium concentrations/pressures. Using the previous reaction:
Q=ΒΏΒΏβ’ The subscript β0β denotes initial concentrations before shifting
toward equilibrium. Unlike K, Q is not constant.
The Direction of Spontaneity is ALWAYS Toward Equilibrium
Q
Q
QKEquilibrium
Too much reactant
Too much productQ
Recap: Thermodynamics of Equilibrium
β’ When a system reaches equilibrium, the entropy is at a maximum, so the change in entropy is 0 (ΞSsys = 0 at equilibrium)
β’ Back reaction is required to maintain this disordered state.
Recap: Spontaneity Depends on Enthalpy AND Entropy
βπΊ=βπ»βπ βπDictates if a process is energetically favored
Dictates if a process is entropically favored
Minimizing ΞG
β’ In general, a system will change spontaneously in such a way that its Gibbs free energy is minimized.
β’ The enthalpy term is independent of concentration and pressure. Entropy is not.
β’ During a reaction, the composition of the system changes, which changes concentrations and pressures, leading to changes in the TΞS term.
β’ As the system approaches an entropically unfavorable composition, the back reaction occurs to prevent ΞG from becoming more positive. This is the basis of equilibrium.
β’ Once equilibrium is reached, the free energy no longer changes
ProductsReactants Equilibrium
K=25K = 0.01 K = 1000
When ΞG is Negative, the Value Tells Us the Maximum Portion of ΞU 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 portion of U converted
to work = ΞG
ΞG = wmax
qmin
Relating the Equilibrium Constant, Reaction Quotient, and ΞGo
rxn
β’ The standard free energy change, ΞGo is determined under standard conditions. It is the change in free energy that occurs when reactants in standard states are converted to products in standard states. Those conditions are listed below.
State of Matter Standard State
Pure element in most stable state
ΞGo is defined as ZERO
Gas 1 atm pressure, 25oC
Solids and Liquids Pure state, 25oC
Solutions 1M concentration
Relating K, Q, and ΞGorxn
β’ In terms of K of a particular reaction, we can describe the standard change in free energy as the driving force to approach equilibrium. This is expressed as:
β’ ΞGo can also be found using the free energies of formation (like Enthalpy) if the information is available:
β’ Most processes, however, are non-standard. The non-standard free energy change, ΞG, involves the Q term, and is given by:
βππ«π±π§π¨ =βπππ₯π§π
βππ«π±π§=πππ₯π§ππ€
βππ«π±π§π¨ =β π§βππ
π¨ (π©π«π¨π )ββ π§ βπππ¨ (π«π±π)
Example ππ―π (π ) π―π (π )+ππ (π )
β’ At 298oK, 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 ΞG. Which direction will the reaction proceed to reach equilibrium?
π=( .150 )(.0425)
ΒΏΒΏ
β’ The given pressures are NOT EQUILIBRIUM PRESSURES! (FIND Q) The conditions are NOT STANDARD!! (FIND ΞG)
βπΊππ₯π=RT lnππΎ
βπΊππ₯π=4.27 kJ /mol Reaction shifts left to reach equilibrium.
The vanβt Hoff Equation
β’ We know that rate constants vary with temperature.
β’ 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.
β’ Using our relationship of the standard free energy with standard enthalpy and entropy:
βG rxno =β H rxn
o βT βSrxno
βRT ln K=β H oβT β Soβ’ And relating this expression to the equilibrium constant, K, we obtain:
ln K=ββH rxn
o
RT+βSrxn
o
R
Expanded Form of The vanβt Hoff Equation
β’ If you run the same reaction at different temperatures, T1 and T2:
ln K 1=ββH rxn
o
RT 1+β Srxn
o
Rln K 2=β
βH rxno
RT 2+β Srxn
o
R
β’ Then subtraction yields:
ln K 2β ln K1=β H rxn
o
R ( 1T 1β1T 2 )
lnK2K1
=β H rxn
o
R ( 1T 1β1T 2 )
β’ Which equals:
Expanded vanβt Hoff equation
β’ 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
β’ CO(g) + 2H2(g) CH3OH(g) ΞHorxn= -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? Is this in line with LeChatlierβs Principle.
πππΎ 2
πΎ1
=βπ»ππ₯π
π
π (π2βπ 1
π1π 2)
lnK225000
=β90500
Jmol
(8.314 Jmol K ) (
1298K
β1
598K ) lnK225000
=β18.32
eln
K 2
25000=eβ18.32
K225000
=1.1 x10β8 ππ=π .πππ±ππβπ
use ex to cancel ln term
Left. This is expected for an exothermic reaction at increased temperature.
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