THERMODYNAMICS 2 Dr. Harris Suggested HW: Ch 23: 43, 59, 77, 81.

THERMODYNAMICS 2

Dr. Harris

Suggested HW: Ch 23: 43, 59, 77, 81

Recap: Equilibrium Constants and Reaction Quotients

• For any equilibrium reaction:

𝑎𝐴+𝑏𝐵 𝑐𝐶+𝑑𝐷

¿¿• The equilibrium constant, K, is equal to the ratio of the concentrations/

pressures of products and reactants to their respective orders at equilibrium.

• The reaction quotient, Q, has the same form as K, but does not use equilibrium concentrations/pressures. Using the previous reaction:

Q=¿¿• The subscript ‘0’ 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

• When a system reaches equilibrium, the entropy is at a maximum, so the change in entropy is 0 (ΔSsys = 0 at equilibrium)

• Back reaction is required to maintain this disordered state.

Recap: Spontaneity Depends on Enthalpy AND Entropy

∆𝐺=∆𝐻−𝑇 ∆𝑆Dictates if a process is energetically favored

Dictates if a process is entropically favored

Minimizing ΔG

• In general, a system will change spontaneously in such a way that its Gibbs free energy is minimized.

• The enthalpy term is independent of concentration and pressure. Entropy is not.

• During a reaction, the composition of the system changes, which changes concentrations and pressures, leading to changes in the TΔS term.

• As the system approaches an entropically unfavorable composition, the back reaction occurs to prevent ΔG from becoming more positive. This is the basis of equilibrium.

• Once equilibrium is reached, the free energy no longer changes

ProductsReactants Equilibrium

K=25K = 0.01 K = 1000

When ΔG is Negative, the Value Tells Us the Maximum Portion of ΔU 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 = ΔG

ΔG = wmax

qmin

Relating the Equilibrium Constant, Reaction Quotient, and ΔGo

rxn

• The standard free energy change, ΔGo 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

ΔGo is defined as ZERO

Gas 1 atm pressure, 25oC

Solids and Liquids Pure state, 25oC

Solutions 1M concentration

Relating K, Q, and ΔGorxn

• 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:

• ΔGo can also be found using the free energies of formation (like Enthalpy) if the information is available:

• Most processes, however, are non-standard. The non-standard free energy change, ΔG, involves the Q term, and is given by:

∆𝐆𝐫𝐱𝐧𝐨 =−𝐑𝐓𝐥𝐧𝐊

∆𝐆𝐫𝐱𝐧=𝐑𝐓𝐥𝐧𝐐𝐤

∆𝐆𝐫𝐱𝐧𝐨 =∑ 𝐧∆𝐆𝐟

𝐨 (𝐩𝐫𝐨𝐝 )−∑ 𝐧 ∆𝐆𝐟𝐨 (𝐫𝐱𝐭)

Example 𝟐𝑯𝑭 (𝒈 ) 𝑯𝟐 (𝒈 )+𝑭𝟐 (𝒈 )

• 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 ΔG. Which direction will the reaction proceed to reach equilibrium?

𝑄=( .150 )(.0425)

¿¿

• The given pressures are NOT EQUILIBRIUM PRESSURES! (FIND Q) The conditions are NOT STANDARD!! (FIND ΔG)

∆𝐺𝑟𝑥𝑛=RT ln𝑄𝐾

∆𝐺𝑟𝑥𝑛=4.27 kJ /mol Reaction shifts left to reach equilibrium.

The van’t Hoff Equation

• We know that rate constants vary with temperature.

• 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.

• Using our relationship of the standard free energy with standard enthalpy and entropy:

∆G rxno =∆ H rxn

o −T ∆Srxno

−RT ln K=∆ H o−T ∆ So• And relating this expression to the equilibrium constant, K, we obtain:

ln K=−∆H rxn

o

RT+∆Srxn

o

R

Expanded Form of The van’t Hoff Equation

• If you run the same reaction at different temperatures, T1 and T2:

ln K 1=−∆H rxn

o

RT 1+∆ Srxn

o

Rln K 2=−

∆H rxno

RT 2+∆ Srxn

o

R

• Then subtraction yields:

ln K 2− ln K1=∆ H rxn

o

R ( 1T 1−1T 2 )

lnK2K1

=∆ H rxn

o

R ( 1T 1−1T 2 )

• Which equals:

Expanded van’t Hoff equation

• 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

• CO(g) + 2H2(g) CH3OH(g) ΔHorxn= -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.

𝑙𝑛𝐾 2

𝐾1

=∆𝐻𝑟𝑥𝑛

𝑜

𝑅 (𝑇2−𝑇 1

𝑇1𝑇 2)

lnK225000

=−90500

Jmol

(8.314 Jmol K ) (

1298K

−1

598K ) lnK225000

=−18.32

eln

K 2

25000=e−18.32

K225000

=1.1 x10−8 𝐊𝟐=𝟐 .𝟕𝟔𝐱𝟏𝟎−𝟒

use ex to cancel ln term

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