Thermodynamics, pt 2 Dr. Harris Lecture 22 (Ch 23) 11/19/12 HW: 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. [𝐶 ] 𝑐 [𝐷 ] 𝑑 [𝐴 ] 𝑎 [𝐵 ] 𝑏 = 𝑘 1 𝑘 −1 = 𝑲 𝒄 𝑜𝑟 𝑲 𝒑 The reaction quotient, Q, has the same form as K, but does not use equilibrium concentrations/pressures. Using the previous reaction: 𝑸 𝒄 𝒐𝒓 𝑸 𝒑 = [𝐶 ] 𝑜 𝑐 [𝐷 ] 𝑜 𝑑 [𝐴 ] 𝑜 𝑎 [𝐵 ] 𝑜 𝑏 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 Entropy is a measure of the disorder of a system. Increasing disorder means that the change in entropy is positive. 2nd Law of Thermodynamics: Entropy is not conserved. The Entropy of the universe is continually Increasing. ∆ 𝑆 𝑢𝑛𝑖𝑣 =∆ 𝑆 𝑠𝑦𝑠 +∆ 𝑆 𝑠𝑢𝑟𝑟 ∆ 𝑆 𝑢𝑛𝑖𝑣 ≥0 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)

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∆𝑆 Dictates 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 100% 0% Decreasing G Reactants 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 ΔG = 0 spontaneous Q > K K > Q Q = K

When ΔG is Negative, the Value Tells Us the Maximum Portion of ΔU That Can Be Used to do Work ΔG = -wmax Gasoline with internal energy U Maximum possible fraction of U converted to work = -ΔG Work not accounted for by change in free energy must be lost as heat

Relating the Equilibrium Constant, Reaction Quotient, and ΔGorxn 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 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 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: ∆ 𝑮 𝒓𝒙𝒏 𝒐 = 𝒏∆ 𝑮 𝒇 𝒐 𝒑𝒓𝒐𝒅 − 𝒏∆ 𝑮 𝒇 𝒐 𝒓𝒙𝒕 ∆ 𝑮 𝒓𝒙𝒏 𝒐 =−𝑹𝑻 𝐥𝐧 𝑲 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: ∆ 𝑮 𝒓𝒙𝒏 =∆ 𝑮 𝒓𝒙𝒏 𝒐 +𝑹𝑻 𝐥𝐧 𝑸

Example #1 (No K value given) ∆ 𝑮 𝒓𝒙𝒏 =∆ 𝑮 𝒓𝒙𝒏 𝒐 +𝑹𝑻 𝐥𝐧 𝑸 𝑁 2 𝑔 +3 𝐻 2 𝑔 2 𝑁𝐻 3 (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= (0.50 ) 2 (1.0)(3.0 ) 3 =.0277

From appendix D in the back of the book: 𝑁 2 𝑔 +3 𝐻 2 𝑔 2 𝑁𝐻 3 (𝑔) From appendix D in the back of the book: ∆ 𝑮 𝒇 𝒐 𝑯 𝟐 =𝟎 ∆ 𝑮 𝒓𝒙𝒏 𝒐 =𝟐 −𝟏𝟔.𝟒 𝒌𝑱 𝒎𝒐𝒍 =−𝟑𝟐.𝟖 𝒌𝑱 𝒎𝒐𝒍 ∆ 𝑮 𝒇 𝒐 𝑵 𝟐 =𝟎 ∆ 𝑮 𝒇 𝒐 𝑵𝑯 𝟑 =−𝟏𝟔.𝟒 𝒌𝑱 𝒎𝒐𝒍 Solve for ΔG ∆ 𝐺 𝑟𝑥𝑛 =∆ 𝐺 𝑟𝑥𝑛 𝑜 +𝑅𝑇 ln 𝑄 ∆ 𝐺 𝑟𝑥𝑛 =−32800 𝐽 𝑚𝑜𝑙 + 8.314 𝐽 𝑚𝑜𝑙 𝐾 298 𝐾 ln (.0277) ∆ 𝐺 𝑟𝑥𝑛 =−32800 𝐽 𝑚𝑜𝑙 + 8.314 𝐽 𝑚𝑜𝑙 𝐾 (298 𝐾)(−3.586) Reaction moves to the left to reach equilibrium. ∆ 𝐺 𝑟𝑥𝑛 =23915 𝐽 𝑚𝑜𝑙

Example #2 (Value of K given) ∆ 𝐺 𝑟𝑥𝑛 =∆ 𝐺 𝑟𝑥𝑛 𝑜 +𝑅𝑇 ln 𝑄 𝟐𝑯𝑭 𝒈 𝑯 𝟐 𝒈 + 𝑭 𝟐 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? Find Q 𝑄= .150 (.0425) (.500 ) 2 =0.0255 We have K, so we can determine ΔGorxn without using the standard table. ∆ 𝑮 𝒓𝒙𝒏 𝒐 =−𝑹𝑻 𝐥𝐧 𝑲 ∆ 𝐺 𝑟𝑥𝑛 =−𝑅𝑇 𝑙𝑛 𝐾+𝑅𝑇 ln 𝑄 =RT ln 𝑄 𝐾 Reaction moves left to reach equilibrium. ∆ 𝐺 𝑟𝑥𝑛 =4.27 kJ/mol

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: ∆ 𝐺 𝑟𝑥𝑛 𝑜 = ∆ 𝐻 𝑟𝑥𝑛 𝑜 −𝑇∆ 𝑆 𝑟𝑥𝑛 𝑜 And relating this expression to the equilibrium constant, K, we obtain: −𝑅𝑇 ln 𝐾 =∆ 𝐻 𝑜 −𝑇∆ 𝑆 𝑜 ln 𝐾=− ∆ 𝐻 𝑟𝑥𝑛 𝑜 𝑅𝑇 + ∆𝑆 𝑟𝑥𝑛 𝑜 𝑅

Deriving The van’t Hoff Equation ln 𝐾=− ∆ 𝐻 𝑟𝑥𝑛 𝑜 𝑅𝑇 + ∆𝑆 𝑟𝑥𝑛 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

Deriving The van’t Hoff Equation If you run the same reaction at different temperatures, T1 and T2: ln 𝐾 1 =− ∆ 𝐻 𝑟𝑥𝑛 𝑜 𝑅 𝑇 1 + ∆𝑆 𝑟𝑥𝑛 𝑜 𝑅 ln 𝐾 2 =− ∆ 𝐻 𝑟𝑥𝑛 𝑜 𝑅 𝑇 2 + ∆𝑆 𝑟𝑥𝑛 𝑜 𝑅 Then subtraction yields: ln 𝐾 2 − ln 𝐾 1 = ∆ 𝐻 𝑟𝑥𝑛 𝑜 𝑅 1 𝑇 1 − 1 𝑇 2 Which equals: 𝑙𝑛 𝐾 2 𝐾 1 = ∆ 𝐻 𝑟𝑥𝑛 𝑜 𝑅 𝑇 2 − 𝑇 1 𝑇 1 𝑇 2 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 𝑙𝑛 𝐾 2 𝐾 1 = ∆ 𝐻 𝑟𝑥𝑛 𝑜 𝑅 𝑇 2 − 𝑇 1 𝑇 1 𝑇 2 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? K1 = 25000, T1 = 298 K, T2 = 598 K Find K2 𝑙𝑛 𝐾 2 25000 = −90500 𝐽 𝑚𝑜𝑙 8.314 𝐽 𝑚𝑜𝑙 𝐾 598 𝐾 −298 𝐾 178204 𝐾 2 𝑙𝑛 𝐾 2 25000 =−18.32 𝑒 𝑙𝑛 𝐾 2 25000 = 𝑒 −18.32 use ex to cancel ln term 𝐾 2 25000 =1.1 𝑥 1 0 −8 𝑲 𝟐 =𝟐.𝟕𝟔 𝒙 𝟏 𝟎 −𝟒

Example, contd. CO(g) + 2H2(g) CH3OH(g) ΔHorxn= -90.5 kJ/mol T1 = 298 K K2 = .000276 T2 = 598 K Because the value of K2 is increasingly smaller as T increases, it is clear that the reaction is favored to the left.