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Thermodynamics 2

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Thermodynamics 2. Dr. Harris Suggested HW : Ch 23: 43, 59, 77, 81. Recap: Equilibrium Constants and Reaction Quotients. For any equilibrium reaction:. The equilibrium constant, K , is equal to the ratio of the concentrations/ pressures of products and reactants at equilibrium. . - PowerPoint PPT Presentation
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THERMODYNAMICS 2 Dr. Harris Suggested HW: Ch 23: 43, 59, 77, 81
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Page 1: Thermodynamics 2

THERMODYNAMICS 2

Dr. HarrisSuggested HW: Ch 23: 43, 59, 77, 81

Page 2: Thermodynamics 2

Recap: Equilibrium Constants and Reaction Quotients

β€’ For any equilibrium reaction:

π‘Žπ΄+𝑏𝐡 𝑐𝐢+𝑑𝐷

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 β€˜0’ denotes initial concentrations before shifting

toward equilibrium. Unlike K, Q is not constant.

Page 3: Thermodynamics 2

The Direction of Spontaneity is ALWAYS Toward Equilibrium

Q

Q

QKEquilibrium

Too much reactant

Too much product Q

Page 4: Thermodynamics 2

Recap: Thermodynamics of Equilibrium

β€’ When a system reaches equilibrium, the entropy is at a maximum, so the change in entropy is 0 (Ξ”Ssys = 0 at equilibrium)

β€’ Back reaction is required to maintain this disordered state.

Page 5: Thermodynamics 2

Recap: Spontaneity Depends on Enthalpy AND Entropy

βˆ†πΊ=βˆ†π»βˆ’π‘‡ βˆ†π‘†Dictates if a process is energetically favored

Dictates if a process is entropically favored

Page 6: Thermodynamics 2

Minimizing Ξ”G

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 TΞ”S term.

β€’ As the system approaches an entropically unfavorable composition, the back reaction occurs to prevent Ξ”G from becoming more positive. This is the basis of equilibrium.

β€’ Once equilibrium is reached, the free energy no longer changes

Page 7: Thermodynamics 2

ProductsReactants Equilibrium

Page 8: Thermodynamics 2

K=25K = 0.01 K = 1000

Page 9: Thermodynamics 2

When Ξ”G is Negative, the Value Tells Us the Maximum Portion of Ξ”U 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 = Ξ”G

Ξ”G = wmax

qmin

Page 10: Thermodynamics 2

Relating the Equilibrium Constant, Reaction Quotient, and Ξ”Go

rxn

β€’ The standard free energy change, Ξ”Go 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 StatePure element in most stable state

Ξ”Go is defined as ZERO

Gas 1 atm pressure, 25oCSolids and Liquids Pure state, 25oCSolutions 1M concentration

Page 11: Thermodynamics 2

Relating K, Q, and Ξ”Gorxn

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€’ Ξ”Go 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, Ξ”G, involves the Q term, and is given by:

βˆ†π†π«π±π§π¨ =βˆ’π‘π“ π₯𝐧𝐊

βˆ†π†π«π±π§=𝐑𝐓 π₯𝐧 𝐐𝐀

βˆ†π†π«π±π§π¨ =βˆ‘ π§βˆ†π†πŸ

𝐨 (𝐩𝐫𝐨𝐝 )βˆ’βˆ‘ 𝐧 βˆ†π†πŸπ¨(𝐫𝐱𝐭)

Page 12: Thermodynamics 2

Example πŸπ‘―π‘­ (π’ˆ ) π‘―πŸ (π’ˆ )+π‘­πŸ (π’ˆ )

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 Ξ”G. Which direction will the reaction proceed to reach equilibrium?

𝑄=( .150 )(.0425)

ΒΏΒΏ

β€’ The given pressures are NOT EQUILIBRIUM PRESSURES! (FIND Q) The conditions are NOT STANDARD!! (FIND Ξ”G)

βˆ†πΊπ‘Ÿπ‘₯𝑛=RT ln𝑄𝐾

βˆ†πΊπ‘Ÿπ‘₯𝑛=4.27 kJ /mol Reaction shifts left to reach equilibrium.

Page 13: Thermodynamics 2

The van’t 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:

βˆ†G rxno =βˆ† H rxn

o βˆ’T βˆ†Srxno

βˆ’RT ln K=βˆ† H oβˆ’T βˆ† Soβ€’ And relating this expression to the equilibrium constant, K, we obtain:

ln K=βˆ’βˆ†H rxn

o

RT+βˆ†Srxn

o

R

Page 14: Thermodynamics 2

Expanded Form of The van’t Hoff Equation

β€’ If you run the same reaction at different temperatures, T1 and T2:

ln K 1=βˆ’βˆ†H rxn

o

RT 1+βˆ† Srxn

o

Rln K 2=βˆ’

βˆ† H rxno

RT 2+βˆ† Srxn

o

R

β€’ Then subtraction yields:

ln K 2βˆ’ ln K1=βˆ† H rxn

o

R ( 1T 1 βˆ’1T 2 )

lnK2K1

=βˆ† H rxn

o

R ( 1T 1βˆ’1T 2 )

β€’ Which equals:

Expanded van’t 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.

Page 15: Thermodynamics 2

Example

β€’ CO(g) + 2H2(g) CH3OH(g) Ξ”Horxn= -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 LeChatlier’s Principle.

𝑙𝑛 𝐾 2

𝐾1=βˆ†π»π‘Ÿπ‘₯𝑛

π‘œ

𝑅 (𝑇2βˆ’π‘‡ 1

𝑇1𝑇 2)

lnK225000

=βˆ’90500 J

mol

(8.314 Jmol K ) (

1298K

βˆ’ 1598 K ) ln

K225000

=βˆ’18.32

eln K 2

25000=eβˆ’18.32

K225000

=1.1 x10βˆ’8 𝐊𝟐=𝟐 .πŸ•πŸ”π±πŸπŸŽβˆ’πŸ’

use ex to cancel ln term

Left. This is expected for an exothermic reaction at increased temperature.


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