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

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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 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: 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 Q K Equilibrium Too much reactant Too much product Q

Recap: Thermodynamics of Equilibrium When a system reaches equilibrium, the entropy is at a maximum, so the change in entropy is 0 (ΔS sys = 0 at equilibrium) Back reaction is required to maintain this disordered state.

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

Products Reactants Equilibrium

K=25K = 0.01K = 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 = w max q min

Relating the Equilibrium Constant, Reaction Quotient, and ΔG o rxn The standard free energy change, ΔG o 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 MatterStandard State Pure element in most stable state ΔG o is defined as ZERO Gas1 atm pressure, 25 o C Solids and LiquidsPure state, 25 o C Solutions1M concentration

Relating K, Q, and ΔG o rxn 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: ΔG o 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:

Example At 298 o K, the initial partial pressures of H 2, F 2 and HF are 0.150 bar,.0425 bar, and 0.500 bar, respectively. Given that K p =.0108, determine ΔG. Which direction will the reaction proceed to reach equilibrium? The given pressures are NOT EQUILIBRIUM PRESSURES! (FIND Q) The conditions are NOT STANDARD!! (FIND ΔG) 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: And relating this expression to the equilibrium constant, K, we obtain:

Expanded Form of The van’t Hoff Equation If you run the same reaction at different temperatures, T 1 and T 2 : Then subtraction yields: 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) + 2H 2 (g) CH 3 OH(g) ΔH o rxn = -90.5 kJ/mol The equilibrium constant for the reaction above is 25000 at 25 o C. Calculate K at 325 o C. Which direction is the reaction favored at T 2 ? Is this in line with LeChatlier’s Principle. use e x to cancel ln term Left. This is expected for an exothermic reaction at increased temperature.

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