1. 1 Chemical Equilibrium Chapter 14 Henri L. le Chatlier 1850-1936. Adapted thermodynamics to equilibria; formulated the principle known by his name.

2. 2 The Concept of Equilibrium Consider colorless N 2 O 4. At room temperature, it decomposes to brown NO 2 : N 2 O 4 (g)  2NO 2 (g). (colorless) (brown) The double arrow implies a dynamic equilibrium At some point, the forward and reverse rates become the same. This is equilibrium. Furthermore, it is a dynamic equilibrium. As NO 2 is produced, it reacts back to form N 2 O 4 : N 2 O 4 (g)  2NO 2 (g)

3. 3

4. 4 The Concept of Equilibrium Consider Forward reaction: A  B Rate = k f [A] Reverse reaction: B  A Rate = k r [B] At equilibrium k f [A] = k r [B]. For an equilibrium we write A B As the reaction progresses –[A] decreases to a constant, –[B] increases from zero to a constant. –When [A] and [B] are constant, equilibrium is achieved.

5. 5 The Concept of Equilibrium Consider reaction: A(g) B(g) At beginning: P A = P Ao P B = 0 As reaction proceeds, P A decreases and P B increases Equilibrium occurs when there is no further change in concentration as time progresses. PAPA PBPB

6. 6 The Concept of Equilibrium Alternatively: As reaction proceeds, until the two rates become the same. This is equilibrium. Or, when k f [A] = k r [B], equilibrium is achieved. forward rate, k f [A] decreases reverse rate, k r [B] increases

7. 7 The Equilibrium Constant Consider If we start with a mixture of N 2 and H 2 (in any proportions), the reaction will reach equilibrium with a constant set of pressures of N 2, H 2 and NH 3. If we start with just NH 3 and no N 2 or H 2, the reaction will proceed in reverse and N 2 and H 2 will be produced until equilibrium is achieved, with a constant set of pressures of N 2, H 2 and NH 3. When dealing with gases, partial pressures can conveniently be used.

8. 8 The Equilibrium Constant No matter the starting composition of reactants and products, the same ratio of concentrations (for solutions) or pressures (for gases) is achieved at equilibrium. For a general reaction where K eq is the equilibrium constant (in terms of concentration or pressure) the equilibrium constant expression is aA + bB pP + qQ

9. 9 The Magnitude of Equilibrium Constants The equilibrium constant, K eq, is the ratio of products to reactants. Therefore, the larger the K eq the more products are present at equilibrium. Conversely, the smaller the K eq the more reactants are present at equilibrium. If K eq >> 1, then products dominate at equilibrium and equilibrium lies to the right. If K eq << 1, then reactants dominate at equilibrium and the equilibrium lies to the left. K eq = k f /k b For A  B, K eq = [B] [A]

10. 10 Calculating the equilibrium constant for: [NO 2 ] 2 K c = [N 2 O 4 ]

11. 11 An equilibrium is established by placing 2.00 moles of frozen N 2 O 4 (g) in a 5.00 L and heating the flask to 407 K. It was determined that at equilibrium the amount of the NO 2 (g) is 1.31 mole. What is the value of the equilibrium constant? [NO 2 ] 2 K c = [N 2 O 4 ] N 2 O 4(g)  2 NO 2(g) [Initial] (mol/L) 0.400 M0 M [change] [Equilibrium] 0.262 + x = 0.262 -½ x = -0.131 0.269 K = (0.262) 2 /0.269 = 0.255 Set up a table. Let x = [NO 2 ] produced. given 2.00/5.00 = 0.400 M Given 1.31/5.00 = 0.262 M given Use molarities in table

12. 12 We just calculated that at 407 Kelvin, K c = 0.255 for For the reverse reaction, calculate the new K by taking the reciprocal: K c = 1/0.255 = 3.92 Proof: K c = [N 2 O 4 ]/[NO 2 ] 2 (0.269)/(0.262) 2 =3.92 (g () NO 2 g N 2 O 4 ) 2 For the “half” reaction, calculate the square root: K c = (0.255) 1/2 = 0.505 Proof: K c = [NO 2 ]/[N 2 O 4 ] 1/2 = (0.262)/(0.269) 1/2 = = 0.505 ½ N 2 O 4 ( g) NO 2 (g) Be careful! K c values must be evaluated for the reaction as written!

13. 13 Heterogeneous Equilibria When all reactants and products are in one phase, the equilibrium is homogeneous. If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. Consider: –experimentally, the amount of CO 2 does not depend on the amounts of CaO and CaCO 3. Solids and pure liquids are not considered in the equilibrium constant expression. Thus, solids K c = [CO 2 ]

14. 14 Calculating Equilibrium Constants For 2 NO(g) + Cl 2 (g)  2NOCl(g), calculate K eq given that: P NO =0.095 atm; P Cl2 =0.171 atm and P NOCl = 0.28 atm at equilibrium. = (0.28) 2 /(0.095) 2 (0.171) = (0.0784)/(0.009025)(0.171) = (0.0784)/(0.00154) = 50.8

15. 15 Calculating Equilibrium Constants 1.374 g H 2 and 70.31 g Br 2 heated in 2-L vessel at 700 K are allowed to react. At equilibrium, the vessel contains 0.566 g H 2. Calculate K c. The reaction is: H 2 (g) + Br 2 (g)  2 HBr (g) What is known? start: 1.374 g 70.31 g 0 at equil 0.566 g ? ? Start by changing everything from grams to moles: H 2 : 1.374 g = 0.687 mol; at equil H 2 : 0.566 g = 0.283 mol Br 2 : 70.31 g = 0.440 mol Now divide by 2 to give molarities (the vessel is 2-L) Initial [H 2 ]= 0.344 mol; Final [H 2 ] = 0.142 Initial [Br 2 ] = 0.220

16. 16 Set up table (values are molarities): H 2 (g) + Br 2 (g)  2 HBr (g) Initial 0.344 0.220 0 Change Final 0.142 Now fill in changes: H 2 (g) + Br 2 (g)  2 HBr (g) Initial 0.344 0.220 0 Change -0.202 -0.202 +0.404 Final 0.142 Now calculate final values: H 2 (g) + Br 2 (g)  2 HBr (g) Initial 0.344 0.220 0 Change -0.202 -0.202 +0.404 Final 0.142 0.018 0.404 K c = [HBr] 2 /[H 2 ][Br 2 ] = (0.404) 2 /(0.142)(0.018) = 63.8 4 th step 1 st step2 nd step3 rd step

17. 17 Applications of Equilibrium Constants Predicting the Direction of Reaction We define Q, the reaction quotient, for a general reaction where [A], [B], [P], and [Q] are molarities at any time. Q = K eq only at equilibrium. as

18. 18 Applications of Equilibrium Constants Predicting the Direction of Reaction If Q > K eq then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K eq ). If Q < K eq then the forward reaction must occur to reach equilibrium.

19. 19 Le Châtelier’s Principle Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will shift to remove the disturbance. Three types of disturbances: (1)Amount of material. If you add a reactant or remove a product, the equilibrium shifts to the right. If you remove a reactant or add a product, the equilibrium shifts to the left. (2)Pressures. By increasing the pressure you shift the equilibrium in the direction of the lesser number of gas moles; by decreasing the pressure you shift the equilibrium in the direction of the greater number of gas moles. (3)Temperature. By increasing the temperature, you shift an exothermic reaction to the left, and an endothermic reaction to the right. By decreasing the temperature, you shift an exothermic reaction to the right, and an endothermic reaction to the left.

20. 20 Le Châtelier’s Principle Consider the production of ammonia (the Haber process) As the temperature decreases, the amount of ammonia at equilibrium increases (reaction goes to right). As the pressure increases, the amount of ammonia present at equilibrium increases (reaction goes to right). The following have been observed for this exothermic reaction: + heat If ammonia is removed, the equilibrium shifts to the right. If hydrogen is added, the equilibrium shifts to the right. If hydrogen is removed, the equilibrium shifts to the left.

21. 21 Le Châtelier’s Principle The Effect of Catalysts A catalyst lowers the activation energy barrier for the reaction. Therefore, a catalyst will decrease the time taken to reach equilibrium. A catalyst does not effect the composition of the equilibrium mixture. Catalysis is a kinetics matter, not one of equilibrium. Catalysis speeds up a reaction, but the reaction arrives at the same equilibrium point (it just gets there faster).

22. 22 A catalyst changes the rate of a chemical reaction, by lowering the activation energy E a., but the ΔH is not changed. ΔH