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Chapter 9: Chemical Equilibrium The forward and reverse reaction are both taking place at the same rate.

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Presentation on theme: "Chapter 9: Chemical Equilibrium The forward and reverse reaction are both taking place at the same rate."— Presentation transcript:

1 Chapter 9: Chemical Equilibrium The forward and reverse reaction are both taking place at the same rate

2 Production and Decomposition of Ammonia Forward Reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g) Reverse Reaction: 2NH 3 (g)  N 2 (g) + 3H 2 (g) Equilibrium Reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g) Note the double headed arrow! The ammonia is decomposing as fast as it is being made at equilibrium

3 Equilibrium and the Law of Mass Action 2SO 2 (g) + O 2 (g)  2SO 3 (g) 5 mixtures of different initial compositions of gases were made and allowed to reach equilibrium at 1000K At first, you don’t see a trend in the data…

4 Equilibrium and the Law of Mass Action No trends, but if you calculate: You get the same value, regardless of initial concentration Note: K is unitless!

5 The Equilibrium Constant K is the equilibrium constant for the reaction

6 The Equilibrium Constant At equilibrium, the composition of the reaction mixture can be expressed in terms of an equilibrium constant where: For ideal gases, the concentrations are the partial pressures of the individual gases For solutions, the concentrations are the molar values of the individual atoms/ions/molecules

7 Examples of K setup aA (g) + bB (g)  cC (g) + dD (g) 

8 Units and Equilibrium Constants When working with equilibrium Constants, we’ll use the following unit conventions: 1.Gases: Units are bar 2.Aqueous Solutions: Unit is Molarity 3.Solids: The number 1 Solids have a single value (1) because the concentration of a solid doesn’t change.

9 Thermodynamic Origin of Equilibrium Constants The Free Energy changes as the composition of the reaction mixture changes All reactions will proceed towards equilibrium (by either forward or reverse reaction)  Gº is the free energy difference b/w the pure products and pure reactants

10 Thermodynamic Origins of Equilibrium Constants We can calculate the Free Energy change at any point along the reaction coordinate with the equation aA (g) + bB (g)  cC (g) + dD (g)  G r ° is the textbook Free Energy of reaction  G r is the Free Energy of value when the reactants and products are at particular concentrations 

11 Example: The standard free energy of reaction for: 2SO 2 (g) + O 2 (g)  2SO 3 (g) Is  G r °= -141.74 kJ/mole at 25°C. What is the Gibbs Free Energy of reaction when the partial pressure of each gas is 100.0 bar?

12 Example: The Standard Gibbs Free Energy of Reaction for N 2 O 4 (g) --> 2NO 2 (g) Is  G r ° = +4.73 kJ/mole at 298K. What is the value of  G r when the partial pressures are P N2O4 = 0.8 bar and P NO2 = 2.10 bar?

13 Free Energy of a Reaction at Equilibrium Q=K at equilibrium At equilibrium,  G=___ Therefore,  G =  G r º + RTlnK  G r º = -RTlnK (only at equilibrium) We can use this to compute equilibrium constants from  G r º values

14 K and the Extent of Reactions When K is very large, the reaction favors the products When K is very small, the reaction favors the reactants When K=1, the reaction is neither reactant nor product favored (Equilibrium)

15 The Direction of Reaction How can we tell if a reaction will continue towards the products or back towards the reactants at a given point along the reaction coordinate? When Q<K,  G is negative (product favored) When Q=K,  G = 0 When Q>K,  G is positive (reactant favored) Q = Reaction quotient used at any point in the coordinate K = Equilibrium constant

16 Equilibrium Calculations Toolbox 9.1: Know it. Love it. Use it. 

17 Example: Under certain conditions, nitrogen and oxygen react to form dinitrogen oxide, N 2 O. Suppose that 0.482 moles of N 2 and 0.933 moles of O 2 are transferred to a reaction vessel of volume 10.0L and allowed to form N 2 O @ 800K. At this temperature, K=3.2x10 -28 for the reaction: 2N 2 (g) + O 2 (g)  2N 2 O (g) What are the partial pressures of the gases at equilibrium?

18 Example: Chlorine and fluorine react at 2500K to produce ClF and reach the equilibrium: Cl 2 + F 2  2ClF With an equilibrium constant value of 20. If a gaseous mixture of 0.2 bar Cl 2, 0.1 bar F 2 and 0.1 bar ClF is allowed to reach equilibrium, what is the partial pressure of ClF in the mixture?

19 LeChatelier’s Principle When the equilibrium composition is perturbed by adding or removing a reactant of product, the reaction tends to proceed in the direction that brings Q closer to that of K. 

20 Consider the Equilibrium Reaction: 4NH 3 (g) + 3O 2 (g)  2N 2 (g) + 6H 2 O (g) What would result from the: a)Addition of N 2 b)Removal of NH 3 c)Removal of H 2 O

21 Effects of the Environment on Equilibria Compressing a Gas Phase Reaction –The reaction shifts so as to decrease the pressure Decrease the number of gas molecules Changing the Temperature of a Reaction –For exothermic reactions, lowering the temperature causes a shift towards the products –For endothermic reactions, increasing the temperature causes a shift towards the products


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