I. Introduction A. Chemical equilibrium – the state where the concentrations of all reactants and products remain constant with time. 13.1 The equilibrium condition A. Dynamic equilibrium – chemical changes are not “static” or unchanging once equilibrium has occurred. They are said to be dynamic, or a continually active process where the rate of the forward reaction is equal to the rate of the reverse reaction. B. Using figure 13.2 on page 603, when has equilibrium occurred in the reaction of CO + H2O CO2 + H2 . (When the concentrations of reactants & products are constant. C. How are reversible reactions illustrated in chemical equations? By usinga . D. The equilibrium can be shifted to the left (an increase in reactant concentrations)

or to the right (an increase in product concentrations) by changing the conditions of the reaction – a change in temperature, pressure, volume, or adding/removing a reactant or product (these changes are explained in 13.7). E. There are 2 possible reasons why a given chemical reaction appears to remain unchanged when mixed. 1. The system is at chemical equilibrium. 2. The forward and reverse reactions are so slow that the system moves toward equilibrium at a rate that can’t be detected. 13.2 The Equilibrium Constant A. The Law of Mass Action – describes the conditions of chemical equilibrium. 1. aA + bB cC + dD where the lower case letters represent the coefficients and the upper case letters represent the concentrations of reactants/products. 2. The equilibrium expression is then represented by: K is dependent on temp. K = [C]c[D]d by convention products always go on top. [A]a[B]b 3. What is the equilibrium expression for the following reaction? 4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g) K = [NO2]

4[H2O]6 [NH3]

4[O2]7

4. Calculate the value of K @ 127oC, in the Haber process when given at equilibrium: [NH3] = 3.1 x 10-2M, [N2] = 8.5 x 10-1 M, [H2] = 3.1 x 10-3M N2(g) + 3H2(g) 2NH3(g) K = [NH3]

2 = (3.1 x 10-2M)2 = 3.8 x 104 L2/ mol2 [N2][H2]

3 (8.5 x 10-1M)(3.1 x 10-3M)3 5. Any concentration can be determined by rearranging the equation.

5. If the reaction is reversed, then the new equilibrium expression becomes K' = [A]a[B]b = 1 [C]c[D]d K

6. If the balanced equation is multiplied by a factor of “n”, then K” = Kn 7. The law of mass action correctly describes behavior for gaseous and

aqueous solutions where the concentrations of reactants and products can change.

9. The equilibrium constant is specific to the temperature and concentrations of a particular reaction. 10.The value of the equilibrium constant determines the relative position of the equilibrium.

a) If K< 1……equilibrium shifted to the left (more reactants than products). b) If K> 1……equilibrium shifted to the right (more products than reactants). c) If K 1……equilibrium is about halfway between reactants and products).

13.3 Equilibrium Expressions Involving Pressures

1. What is the relationship between pressure (P) and concentration (n/V) from the ideal gas law? a. P = nRT/V = CRT …..C represents the molar concentration of a gas. 2. Given: N2(g) + 3H2(g) 2NH3(g) K = Kc = [NH3]

2 [N2][H2]

3 3. The equilibrium constant in terms of pressure would be Kp = PNH3

2

PN2 PH2 3

4. What is the relationship between the two equilibrium constants Kc to Kp ? a. Kp = Kc(RT)n b. Where R = 0.8206 L. atm/mole.K

c. T = Temperature of the reaction at equilibrium conditions d. n = The sum of the coefficients of the gaseous products minus

the sum of the coefficients of the gaseous reactants. e. Using the given equation n = -2 f. See Sample Exercises 13.4 & 13.5 to illustrate these processes.

13.4 Heterogeneous Equilibria 1. Many equilibria systems involving more than one phase are called heterogeneous equilibria. 2. How is equilibria affected by heterogeneous processes? a. If pure solids or pure liquids are involved in the chemical reaction, their concentrations are not included in the equilibrium expression for the reaction. b. Example Given: PCl5(s) PCl3(l) + Cl2(g) i. The equilibrium Kc = [Cl2] and Kp = PCl2

13.5 Applications of the Equilibrium Constant 1. Knowing the value of the equilibrium constant describes the tendency of a reaction, but not the speed of the reaction. a. If K is much larger than 1, at equilibrium the system is mostly products. i. It is said that equilibrium lies to the right. ii. Reactions with very large K values go essentially to completion. b. If K is much smaller than 1, at equilibrium the system is mostly reactants. i. It is said that equilibrium lies to the left. ii. These reactions do not occur to any significant extent. c. Note: The size of K and the time it takes to reach equilibrium are not directly related. 2. Reaction Quotient. a. When reactants are mixed it is helpful to compare the initial concentrations of your reactants and products to the equilibrium ratios provided by K. b. Reaction Quotient (Q) i. If one of the reactants or products is absent – the system will shift in in the direction of the missing component. ii. If there are initial concentrations of each reactant and product, we use the law of mass action to determine the value of Q using the initial concentrations in the expression.

iii. Given: N2(g) + 3H2(g) 2NH3(g) Q = [NH3]o2

[N2]o[H2]o3

c. Comparing Q to K i. When Q is equal to K. The system is at equilibrium, no change occurs. ii. When Q > K. The initial ratios of product concentrations to initial reactant concentrations is too high. To reach equilibrium some of the products have to change into reactants – the system shifts to the left until equilibrium is reached. iii. When Q < K. The initial ratios of product concentrations to initial reactant concentrations is too low. To reach equilibrium some of the reactants have to change into products – the system will shift to the right until equilibrium is reached. 3. Sample problems – study the processes shown in exercises 13.7 – 13.11. 4. You need to have the ability to use the following concepts in order to solve equilibrium problems a. Once Q has been compared to K, you know what direction the reaction proceeds. Many problems require the development of a data table. i. Given: N2(g) + 3H2(g) 2NH3(g) If the initial concentrations of each reactant and product are 1.0 M, and Q > K (by calculations) what are the net changes of each reactant & product? ii. N2 = 1.0 + x H2 = 1.0 + 3x NH3 = 1.0 - 2x K = (1.0 - 2x)2

(1.0 + x)(1.0 + 3x)3 iii. You need to be able to solve for "x" using the concepts of perfect

squares and the quadratic equation to calculate each concentration.

13.6 Solving Equilibrium Problems – This will be taught from book. 13.7 Le Chatelier's Principle 1. Definition Statement – When a stress (change) is imposed on a System at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce the stress. 2. What are the stresses that can affect the equilibrium of a system? a. Changes in pressure, volume, temperature, and the addition or removal of a reactant or product. 3. Change in Concentration. a. If a reactant or product increases in concentration, the equilibrium will shift away (to the other side of the reaction). If a reactant or product decreases in concentration, the equilibrium will shift away (to the other side of the reaction). b. Given: N2(g) + 3H2(g) 2NH3(g) What affect does ↑ [H2]? i. Shifts the reaction to the right. ii. Decreases the [N2], increases the [NH3]. 4. Changes in Pressure a. Increasing the pressure favors the formation of fewer gaseous molecules (this reduces the overall pressure). Decreasing the pressure favors the formation of more gaseous molecules. b. Given: N2(g) + 3H2(g) 2NH3(g) Reducing the pressure. i. Reactants have 4 gaseous molecules, products have 2. ii. Reaction proceeds to the left. iii. [N2] & [H2] increase, [NH3] decrease. c. Given: H2(g) + I2(g) 2HI(g) Increasing the pressure. i. Reactants have 2 gaseous molecules, products have 2. ii. There is no net change in number of gaseous molecules, equilibrium does not change. 5. Reducing the volume has the same affect as increasing the pressure, and vice versa for increasing the volume. 6. Temperature Change a. Increasing the temperature favors the endothermic reaction. Increasing the temperature increases the average kinetic energy of both reactions. Only the endothermic reaction requires additional energy. Reducing the temperature favors the exothermic reaction. b. Given: N2(g) + O2(g) 2NO(g) Ho = +181 kJ i. Since Ho is positive, forward reaction is endothermic. ii. Reaction proceeds to the right. iii. [NO] increases, while [N2] & [O2] decrease. iv. The reverse reaction is exothermic in nature....... Ho = -181kJ

Figure 13.8 a-c (a) The Initial Equilibrium Mixture of N2, H2, and NH3 (b) Addition of N2. (c.)

The New Equilibrium Position for the System Containing More N2 (due to Less H2, and More

NH3 than in (a)

Figure 13.9 a-c (a) A Mixture of NH3(g), N2(g), and H2(g) at Equilibrium (b) The Volume is Suddenly Decreased (c) The New Equilibrium Position for

the System Containing More NH3 and Less N2 and H2

Anhydrous Ammonia is Injected into the Solid to Act as a Fertilizer

That’s all folks.