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

• 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 at equilibrium.

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

𝑸𝒄𝒐𝒓 𝑸𝒑=¿¿• The subscript ‘0’ denotes arbitrary concentrations. Unlike K, Q is not

constant and depends on the starting concentrations.

Recap: Direction of Spontaneity

• The direction of spontaneity is always toward equilibrium.

• 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

• 2nd Law of Thermodynamics:

Entropy is not conserved. The Entropy of the universe is continually Increasing.

∆𝑆𝑢𝑛𝑖𝑣=∆𝑆𝑠𝑦𝑠+∆𝑆𝑠𝑢𝑟𝑟

• The universe can never become more ordered after a process. Therefore, if a particular system becomes more ordered (ΔSsys<0), the surroundings must become even more disordered (ΔSsurr >0)

• Entropy is a measure of the disorder of a system. Increasing disorder means that the change in entropy is positive.

∆𝑆𝑢𝑛𝑖𝑣≥0

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)

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

Recap: Correlation Between Gibbs Free Energy and Equilibrium

• If ΔG is negative, the reaction is spontaneous

• If ΔG is zero, the reaction is at equilibrium

• If ΔG is positive, the reaction is spontaneous in the opposite direction

100% 0%

Decreasing G

Reactants

spontaneous

K > Q

Q > K

Q = K

ΔG = 0

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 fraction of U converted to work = -ΔG

ΔG = -wmax

Relating the Equilibrium Constant, Reaction Quotient, and ΔGo

rxn

• Keep in mind that the standard free energy change, ΔGo, is not the same as the nonstandard free energy change, ΔG. ΔGo is determined under standard conditions. Those conditions are listed below.

State of Matter Standard StatePure element in most stable state ΔGo is defined as ZEROGas 1 atm pressure, 25oCSolids and Liquids Pure state, 25oCSolutions 1M concentration

• For many elements, ΔGorxn

can be obtained from a table of values. ΔGorxn

can be calculated in the same manner as ΔHorxn using free energies of

formation:

• In terms of the equilibrium constant of a particular reaction, the driving force to approach equilibrium under standard conditions is given by:

∆𝑮𝒓𝒙𝒏𝒐 =−𝑹𝑻 𝐥𝐧𝑲

• When the reaction conditions are not standard, you must use the reaction quotient, Q.

• The free energy change of a reaction (or the driving force to approach equilibrium) under non-standard conditions, ΔG, is given by:

∆𝑮𝒓𝒙𝒏=∆𝑮𝒓𝒙𝒏𝒐 +𝑹𝑻 𝐥𝐧𝑸

∆𝑮𝒓𝒙𝒏𝒐 =∑ 𝒏∆𝑮 𝒇

𝒐 (𝒑𝒓𝒐𝒅 )−∑ 𝒏∆𝑮 𝒇𝒐 (𝒓𝒙𝒕 )

Relating K, Q, and ΔGorxn

Example #1 (No K value given)

𝑁 2 (𝑔 )+3𝐻2 (𝑔 )2𝑁𝐻3(𝑔)

• Calculate ΔG 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?

• We are finding the free energy change under non-standard conditions (ΔG). We must first Q.

• Now determine the standard free energy, ΔGo. If K is not given, you can calculate it from the standard table.

𝑄=¿¿

∆𝑮𝒓𝒙𝒏=∆𝑮𝒓𝒙𝒏𝒐 +𝑹𝑻 𝐥𝐧𝑸

• From appendix D in the back of the book:

∆𝑮 𝒇𝒐𝑯𝟐=𝟎

∆𝑮 𝒇𝒐𝑵𝟐=𝟎

∆𝑮 𝒇𝒐𝑵𝑯𝟑=−𝟏𝟔 .𝟒

𝒌𝑱𝒎𝒐𝒍

∆𝑮𝒓𝒙𝒏𝒐 =𝟐(−𝟏𝟔 .𝟒 𝒌𝑱

𝒎𝒐𝒍 )=−𝟑𝟐 .𝟖 𝒌𝑱𝒎𝒐𝒍

𝑁 2 (𝑔 )+3𝐻2 (𝑔 )2𝑁𝐻3(𝑔)

• Solve for ΔG∆𝐺𝑟𝑥𝑛=∆𝐺𝑟𝑥𝑛

𝑜 +𝑅𝑇 ln𝑄

∆𝐺𝑟𝑥𝑛=−32800𝐽

𝑚𝑜𝑙+(8.314 𝐽𝑚𝑜𝑙𝐾 ) (298𝐾 ) ln(.0277)

∆𝐺𝑟𝑥𝑛=−32800𝐽

𝑚𝑜𝑙+(8.314 𝐽𝑚𝑜𝑙𝐾 )(298𝐾 )(−3.586)

∆𝐺𝑟𝑥𝑛=23915𝐽

𝑚𝑜𝑙Reaction moves to the left to reach equilibrium.

Example #2 (Value of K given)

𝟐𝑯𝑭 (𝒈 ) 𝑯𝟐 (𝒈 )+𝑭𝟐 (𝒈 )

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

∆𝐺𝑟𝑥𝑛=∆𝐺𝑟𝑥𝑛𝑜 +𝑅𝑇 ln𝑄

𝑄=( .150 )(.0425)

¿¿• Find Q

• We have K, so we can determine ΔGorxn without using the standard table.

∆𝑮𝒓𝒙𝒏𝒐 =−𝑹𝑻 𝐥𝐧𝑲

∆𝐺𝑟𝑥𝑛=−𝑅𝑇 𝑙𝑛𝐾 +𝑅𝑇 ln𝑄=RT ln 𝑄𝐾

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

Deriving 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:

∆𝐺𝑟𝑥𝑛𝑜 =∆𝐻𝑟𝑥𝑛

𝑜 −𝑇 ∆𝑆𝑟𝑥𝑛𝑜

−𝑅𝑇 ln 𝐾=∆𝐻𝑜−𝑇 ∆𝑆𝑜

• And relating this expression to the equilibrium constant, K, we obtain:

ln 𝐾=−∆𝐻𝑟𝑥𝑛

𝑜

𝑅𝑇 +∆𝑆𝑟𝑥𝑛

𝑜

𝑅

• 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.

• Thus, entropy is the dominant factor in determining equilibrium distributions at high temperatures, and enthalpy is the dominant factor at low temperatures.

• A plot of ln K vs. 1/T will yield a linear plot with a slope of (–ΔHorxn)/R

ln 𝐾=−∆𝐻𝑟𝑥𝑛

𝑜

𝑅𝑇 +∆𝑆𝑟𝑥𝑛

𝑜

𝑅

Deriving The van’t Hoff Equation

Deriving The van’t Hoff Equation

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

ln 𝐾 1=−∆𝐻𝑟𝑥𝑛

𝑜

𝑅𝑇1+∆𝑆𝑟𝑥𝑛

𝑜

𝑅 ln 𝐾 2=−∆𝐻𝑟𝑥𝑛

𝑜

𝑅𝑇2+∆𝑆𝑟𝑥𝑛

𝑜

𝑅

• Then subtraction yields:

ln𝐾 2−ln 𝐾 1=∆𝐻𝑟𝑥𝑛

𝑜

𝑅 ( 1𝑇 1− 1𝑇 2 )

𝑙𝑛 𝐾 2

𝐾1=∆𝐻𝑟𝑥𝑛

𝑜

𝑅 (𝑇2−𝑇 1

𝑇1𝑇 2)

• Which equals:

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?

𝑙𝑛 𝐾 2

𝐾1=∆𝐻𝑟𝑥𝑛

𝑜

𝑅 (𝑇2−𝑇 1

𝑇1𝑇 2)

K1 = 25000, T1 = 298 K, T2 = 598 K• Find K2

𝑙𝑛𝐾 2

25000=−90500 𝐽

𝑚𝑜𝑙

(8.314 𝐽𝑚𝑜𝑙 𝐾 )(

598𝐾−298𝐾178204𝐾 2 ) 𝑙𝑛

𝐾 2

25000=−18.32

𝑒𝑙𝑛 𝐾 2

25000=𝑒−18.32

𝐾 2

25000=1.1 𝑥10− 8 𝑲 𝟐=𝟐 .𝟕𝟔 𝒙𝟏𝟎−𝟒

use ex to cancel ln term

Example, contd.

• 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) ΔHorxn= -90.5 kJ/mol

K1 = 25000T1 = 298 K

K2 = .000276T2 = 598 K