1. Chapter 15 Equilibrium

2. © 2009, Prentice-Hall, Inc. The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

3. © 2009, Prentice-Hall, Inc. A System at Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate. Once equilibrium is achieved, the amount of each reactant and product remains constant.

4. © 2009, Prentice-Hall, Inc. Depicting Equilibrium Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow. N 2 O 4 (g) 2 NO 2 (g)

5. © 2009, Prentice-Hall, Inc. The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate Law: Rate = k f [N 2 O 4 ] Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate Law: Rate = k r [NO 2 ] 2

6. © 2009, Prentice-Hall, Inc. The Equilibrium Constant Therefore, at equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

7. © 2009, Prentice-Hall, Inc. The Equilibrium Constant Consider the generalized reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD

8. The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression. K c = [Pb 2+ ] [Cl - ] 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq)

9. © 2009, Prentice-Hall, Inc. The Equilibrium Constant Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C c ) (P D d ) (P A a ) (P B b )

10. © 2009, Prentice-Hall, Inc. Relationship Between K c and K p From the Ideal Gas Law we know that Rearranging it, we get PV = nRT P = RT nVnV Since n/V = M =[ ]….

11. © 2009, Prentice-Hall, Inc. Relationship Between K c and K p Plugging this into the expression for K p for each substance, the relationship between K c and K p becomes where K p = K c (RT)  n  n = (moles of gaseous product) - (moles of gaseous reactant) In the synthesis of ammonia from nitrogen and hydrogen, K c = 9.60 at 300 °C. Calculate K p for this reaction at this temperature.

12. © 2009, Prentice-Hall, Inc. What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium. If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.

13. © 2009, Prentice-Hall, Inc. Manipulating Equilibrium Constants The K c of a reaction in the reverse reaction is the reciprocal of the K c of the forward reaction. The equilibrium constant for the reaction of N 2 with O 2 to form NO equals K c = 1 × 10 –30 at 25 °C: Using this information, write the equilibrium constant expression and calculate the equilibrium constant for the following reaction:

14. © 2009, Prentice-Hall, Inc. Manipulating Equilibrium Constants The K c of a reaction that has been multiplied by a number is the K c raised to a power that is equal to that number. K c = 0.212 at 100  C N 2 O 4(g) 2 NO 2(g) K c = 2 N 2 O 4(g) 4 NO 2(g)

15. © 2009, Prentice-Hall, Inc. Manipulating Equilibrium Constants The K c for a net reaction made up of two or more steps is the product of the K c for the individual steps. Given the following information, determine the value of K c for the reaction

16. © 2009, Prentice-Hall, Inc. An Equilibrium Problem A closed system initially containing 1.000 x 10 -3 M H 2 and 2.000 x 10 -3 M I 2 at 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 -3 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (s) 2 HI (g)

17. © 2009, Prentice-Hall, Inc. ICE Tables [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change At equilibrium1.87 x 10 -3

18. © 2009, Prentice-Hall, Inc. The Reaction Quotient (Q) Q gives the same ratio the equilibrium expression gives, but for a system that is not necessarily at equilibrium. To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.

19. What does the value of Q mean? If Q= K, the system is at equilibrium. IF Q> K, there is too much product present and the system must shift in reverse to achieve equilibrium If Q< K, there is too much reactant present and the system must shift forward to achieve equilbrium.

20. © 2009, Prentice-Hall, Inc. Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way:  G =  G  + RT lnQ (Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)

21. © 2009, Prentice-Hall, Inc. Free Energy and Equilibrium At equilibrium, Q = K, and  G = 0. The equation becomes 0 =  G  + RT lnK Rearranging, this becomes  G  =  RT lnK (note that  G  is in kJ and R is in J) If ΔG° is negative, then ln K must be positive…K must be greater than 1…more products are present at equilibrium…reaction is spontaneous in forward direction.

22. © 2009, Prentice-Hall, Inc. Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.” ****ONLY a CHANGE IN TEMPERATURE will affect the value of the equilibrium constant.

23. © 2009, Prentice-Hall, Inc. Catalysts Catalysts increase the rate of both the forward and reverse reactions. When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.