1. The Equilibrium Condition, the Equilibrium Constant and Equilibrium in Terms of Pressures Chemistry 142 B Autumn Quarter 2004 J. B. Callis, Instructor Lecture #17

2. Chemical Equilibrium Until Now we have assumed that a chemical reaction goes to completion as written. For example, we might suppose that when H 2 O is introduced into a flask with CO and sealed, it will all convert to H 2 and CO 2 : H 2 O(g) + CO (g) H 2 (g) + CO 2 (g) However, we note that the reaction does not go to completion, but rather forms a certain, predictable amount of products and does not proceed further. This new stable state of the system which includes both reactants and products is called the equilibrium state.

3. Molecular Picture of Equilibrium 2 NO 2 (g) = N 2 O 4 (g)

4. Concentration vs. Time CO(g) + H 2 O(g) = CO 2 (g) + H 2 (g)

5. Molecular Picture of Establishment of Equilibrium CO(g) + H 2 O(g) = CO 2 (g) + H 2 (g)

6. Kinetics of Approach to Equilibrium

7. The Haber Process N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

8. Characteristics of True Chemical Equilibria They show no macroscopic evidence of change. They are reached through spontaneous processes. A dynamic balance of forward and reverse processes exists within them. They are the same regardless of the direction from which they are approached.

9. The Equilibrium Constant - Definition For a reaction of the type jA + kB = lC+mD Where A, B, C and D represent chemical species and j, k, l, and m are coefficients of the balanced chemical equation, the law of mass action is represented in the following equilibrium expression: The square brackets indicate the concentrations of the species in equilibrium and K is a constant called the equilibrium constant.

10. Characteristics of the Equilibrium Expression The equilibrium expression for a reaction written in reverse is the reciprocal of that for the original expression. If the original reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power n. The apparent units for K are determined by the powers of the concentration terms.

11. Problem 17-1 Expressing the Equilibrium Constant At a certain temperature the value of the equilibrium constant for the reaction is CS 2 (g) + 3 O 2 (g) = CO 2 (g) + 2 SO 2 (g) is K 1. (a) Write an expression for the equilibrium constant, K 1. (b) Write an expression for the equilibrium constant, K 2, of the reverse reaction. (c) Write an expression for the related equilibrium constant, K 3 (1/3) CS 2 (g) + O 2 (g) = (1/3) CO 2 (g) + (2/3) SO 2 (g)

12. Problem 17-1 Calculation of the Equilibrium Constant Answers: (a) (b) (c)

13. Problem 17-2 Calculation of the Equilibrium Constant At 454 K, the following reaction takes place: 3 Al 2 Cl 6 (g) = 2 Al 3 Cl 9 (g) At this temperature, the equilibrium concentration of Al 2 Cl 6 (g) is 1.00 M and the equilibrium concentration of Al 3 Cl 9 (g) is 1.02 x 10 -2 M. Compute the equilibrium constant at 454 K. Answer:

14. Equilibrium Expressions Involving Pressures For a reaction of the type jA + kB = lC+mD It is sometimes convenient to write the equilibrium expression in terms of partial pressures, e.g. The the Ps indicate the partial pressures of the species in equilibrium and K P is a constant called the equilibrium constant in terms of partial pressures.

15. How is K P related to K? Answer: Through the use of the ideal gas law. Where  n = l+m-(j+k)

16. (Apparent) Units for K and K P At first sight, it would seem that the units for K would be in concentrations raised to a reaction- specific power, and those for K P in pressure units raised to a reaction-specific power. But these are ‘apparent’ units. For theoretical reasons, we will refer each concentration or pressure to a reference state, which always causes the units of concentration or pressure to cancel. Thus, K and K P are expressed without units.

17. Problem 17-3: Converting between K and K P Calculate K p for the following equilibrium: N 2 (g) + 3 H 2 (g) = 2 NH 3 (g), K = 2.4 x 10 -3 at 1000 K

18. Answers to Problems in Lecture 17 1.(a) (b) (c) 2. 3.