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 to their respective orders at equilibrium.   1 = =   −1 • The reaction quotient, Q, has the same form as K, but does not use equilibrium concentrations/pressures. Using the previous reaction:   =   • The subscript ‘0’ denotes initial concentrations before shifting toward equilibrium. Unlike K, Q is not constant. 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. Determining the Direction of A Reaction Toward Equilibrium Too much product K Q Too much reactant Q Q Q 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. ∆ = ∆ + ∆ ∆ ≥ 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 ∆ = ∆ − ∆ 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 ΔG < 0 Reactants Equilibrium ΔG > 0 Products K = 0.01 K=25 K = 1000 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 qmin Work not accounted for by change in free energy must be lost as heat Maximum possible portion of U converted to work = ΔG 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. ∆ = ∆ − ∆ • In terms of the equilibrium constant of a particular reaction, the driving force to approach equilibrium under standard conditions is given by: ∆ = − • 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: ∆ = ∆ + ∆ = ∆ + Example #1 (No K value given) 2 + 32 23 () • 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. (0.50)2 = = .0277 (1.0)(3.0)3 • Now determine the standard free energy, ΔGo. If K is not given, you can calculate it from the standard table. • From appendix D in the back of the book: ∆ = ∆ = ∆ = −. ∆ = −. = −. • Solve for ΔG + ln ∆ = ∆ ∆ = −32800 + 8.314 ∆ = −41684.6 298 ln(.0277) Reaction moves to the right to reach equilibrium. Example #2 (Value of K given) + ln ∆ = ∆ + • 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? • Find Q .150 (.0425) = = 0.0255 (.500)2 • We have K, so we can determine ΔGrxn without using the standard table. ∆ = − + ln = RT ln ∆ = 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: o o o ∆Grxn = ∆Hrxn − T∆Srxn • And relating this expression to the equilibrium constant, K, we obtain: −RT ln K = ∆H o − T∆S o o o ∆Hrxn ∆Srxn ln K = − + RT R Deriving The van’t Hoff Equation o o ∆Hrxn ∆Srxn ln K = − + RT 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 Expanded Form of The van’t Hoff Equation • If you run the same reaction at different temperatures, T1 and T2: o o ∆Hrxn ∆Srxn ln K1 = − + RT1 R o o ∆Hrxn ∆Srxn ln K 2 = − + RT2 R • Then subtraction yields: o ∆Hrxn 1 1 ln K 2 − ln K1 = − R T1 T2 • Which equals: o K 2 ∆Hrxn 1 1 ln = − K1 R T1 T2 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. 2 ∆ 2 − 1 = 1 1 2 Example • CO(g) + 2H2(g) ΔHorxn= -90.5 kJ/mol CH3OH(g) 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. J −90500 K2 mol ln = J 25000 8.314 mol K use ex to cancel ln term 1 1 − 298K 598K e K2 ln25000 K2 = 1.1 x 10−8 25000 ln K2 = −18.32 25000 = e−18.32 = . − Left. This is expected for an exothermic reaction at increased temperature.