Post on 16-Dec-2015
THERMODYNAMICS 2
Dr. Harris
Suggested HW: Ch 23: 43, 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 to their respective orders at equilibrium.
โข The reaction quotient, Q, has the same form as K, but does not use equilibrium concentrations/pressures. Using the previous reaction:
Q=ยฟยฟโข The subscript โ0โ 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
โข When a system reaches equilibrium, the entropy is at a maximum, so the change in entropy is 0 (ฮSsys = 0 at equilibrium)
โข Back reaction is required to maintain this disordered state.
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 TฮS 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
ProductsReactants Equilibrium
K=25K = 0.01 K = 1000
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 portion of U converted
to work = ฮG
ฮG = wmax
qmin
Relating the Equilibrium Constant, Reaction Quotient, and ฮGo
rxn
โข The standard free energy change, ฮGo 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
ฮGo is defined as ZERO
Gas 1 atm pressure, 25oC
Solids and Liquids Pure state, 25oC
Solutions 1M concentration
Relating K, Q, and ฮGorxn
โข 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:
โข ฮGo can also be found using the free energies of formation (like Enthalpy) if the information is available:
โข Most processes, however, are non-standard. The non-standard free energy change, ฮG, involves the Q term, and is given by:
โ๐๐ซ๐ฑ๐ง๐จ =โ๐๐๐ฅ๐ง๐
โ๐๐ซ๐ฑ๐ง=๐๐๐ฅ๐ง๐๐ค
โ๐๐ซ๐ฑ๐ง๐จ =โ ๐งโ๐๐
๐จ (๐ฉ๐ซ๐จ๐ )โโ ๐ง โ๐๐๐จ (๐ซ๐ฑ๐ญ)
Example ๐๐ฏ๐ญ (๐ ) ๐ฏ๐ (๐ )+๐ญ๐ (๐ )
โข 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 ฮG. Which direction will the reaction proceed to reach equilibrium?
๐=( .150 )(.0425)
ยฟยฟ
โข The given pressures are NOT EQUILIBRIUM PRESSURES! (FIND Q) The conditions are NOT STANDARD!! (FIND ฮG)
โ๐บ๐๐ฅ๐=RT ln๐๐พ
โ๐บ๐๐ฅ๐=4.27 kJ /mol Reaction shifts left to reach equilibrium.
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:
โG rxno =โ H rxn
o โT โSrxno
โRT ln K=โ H oโT โ Soโข And relating this expression to the equilibrium constant, K, we obtain:
ln K=โโH rxn
o
RT+โSrxn
o
R
Expanded Form of The vanโt Hoff Equation
โข If you run the same reaction at different temperatures, T1 and T2:
ln K 1=โโH rxn
o
RT 1+โ Srxn
o
Rln K 2=โ
โH rxno
RT 2+โ Srxn
o
R
โข Then subtraction yields:
ln K 2โ ln K1=โ H rxn
o
R ( 1T 1โ1T 2 )
lnK2K1
=โ H rxn
o
R ( 1T 1โ1T 2 )
โข Which equals:
Expanded 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? Is this in line with LeChatlierโs Principle.
๐๐๐พ 2
๐พ1
=โ๐ป๐๐ฅ๐
๐
๐ (๐2โ๐ 1
๐1๐ 2)
lnK225000
=โ90500
Jmol
(8.314 Jmol K ) (
1298K
โ1
598K ) lnK225000
=โ18.32
eln
K 2
25000=eโ18.32
K225000
=1.1 x10โ8 ๐๐=๐ .๐๐๐ฑ๐๐โ๐
use ex to cancel ln term
Left. This is expected for an exothermic reaction at increased temperature.