1. 8–1 John A. Schreifels Chemistry 212 Chapter 15-1 Chapter 15 Chemical Equilibrium

2. 8–2 John A. Schreifels Chemistry 212 Chapter 15-2 Overview Describing Chemical Equilibrium –Chemical Equilibrium – A Dynamic Equilibrium (the link to Chemical Kinetics) –The Equilibrium Constant. –Heterogeneous Equilibria; solvents in homogeneous equilibria. Using the Equilibrium Constant –Qualitatively Interpreting the Equilibrium Constant –Predicting the direction of a Reaction –Calculating Equilibrium Concentrations Changing Reaction Conditions; Le Châtelier’s Principle. –Removing Products or Adding Reactants –Changing the Pressure or Temperature –Effect of a Catalyst.

3. 8–3 John A. Schreifels Chemistry 212 Chapter 15-3 Chemical Equilibrium – A Dynamic Equilibrium Upon addition of reactants and/or products, they react until a constant amount of reactants and products are present = equilibrium. Equilibrium is dynamic since product is constantly made (forward reaction), but at the same rate it is consumed (reverse reaction).

4. 8–4 John A. Schreifels Chemistry 212 Chapter 15-4 The Equilibrium State Not all reactants are completely converted to product. Reaction equilibria deal with the extent of reaction. Arrows between reactants and products separate them and qualitatively indicate the extent of reaction. –Single arrow points to dominant side: H 2 (g) + O 2 (g)  H 2 O(g) –Double arrow indicates both reactants and products present after equilibrium obtained: N 2 O 4 (g)  2NO 2 (g). Equilibrium exists when rates of forward and reverse reaction are the same. E.g. When rate of N 2 O 4 decomposition equal the rate of formation of N 2 O 4, reaction at equilibrium N 2 O 4 (g)  2NO 2 (g). Equilibrium can be obtained from any mixture of reactants and products.

5. 8–5 John A. Schreifels Chemistry 212 Chapter 15-5 The Link Between Chemical Equilibrium and Chemical Kinetics Reactions with one elementary step: At equilibrium: R f = R r. E.g. decomposition of N 2 O 4 : k f [N 2 O 4 ] = k r [NO 2 ] 2. or where K c is called the equilibrium constant. At equilibrium the ratio of concentrations equals a constant. A generalized form of this expression that describes the equilibrium condition: aA + bB + cC +...  mM + nN + oO....

6. 8–6 John A. Schreifels Chemistry 212 Chapter 15-6 The Link Between Chemical Equilibrium and Chemical Kinetics II Reactions with more than one elementary step Use the rate limiting step and “Steady State Approximation”: at equilibrium the rates of the forward and reverse reactions are equal. Eliminate intermediates. E.g.: Determine the equilibrium expression for the two step decomposition of ozone. Step 2 (rate limiting step)

7. 8–7 John A. Schreifels Chemistry 212 Chapter 15-7 Equilibrium Constant: Multi-step Rate equations lead to: Remove intermediate, O: Substitute into the first equation to get: Conclusion: Form of equilibrium expression is independent of mechanism.

8. 8–8 John A. Schreifels Chemistry 212 Chapter 15-8 The Equilibrium Constant K c Equilibrium constant of reverse reaction: aA + bB  cC + dD cC + dD  aA + bB E.g.1 Determine the equilibrium constant for the reaction: ½N 2 (g) + 3/2 H 2 (g)  NH 3 (g) given N 2 (g) + 3H 2 (g)  2NH 3 (g) K c = 1.7x10 2 E.g.2 determine the equilibrium constant for the formation of HI(g) if the equilibrium concentration of H 2, I 2 and HI are 0.0060 M, 0.106 M, and 0.189 M, respectively. H 2 (g) + I 2 (g)  2HI(g) K c = ? E.g. 3 Determine the equilibrium constant for the reaction below: ½ H 2 (g) + ½ I 2 (g)  HI(g) K c = ?

9. 8–9 John A. Schreifels Chemistry 212 Chapter 15-9 Equilibrium Constant K c, K p For gases K c and K p are used. K P same format as K c except pressures used instead of concentrations. E.g. Write out the equilibrium expression for K P using the reaction below: N 2 (g) + 3H 2 (g)  2NH 3 (g) K P = ? E.g.2 What is equilibrium expression (K P ) for the reaction below? ½ N 2 (g) + 3/2 H 2 (g)  NH 3 (g) K P = ? E.g.3 Determine the equilibrium constant (K P ) for the formation of one mole of ammonia if at 500K, P NH3 = 0.15 atm, P N2 = 1.2 atm. and P H2 = 0.81 atm.

10. 8–10 John A. Schreifels Chemistry 212 Chapter 15-10 Relationship between K c and K p Relationship between concentration and pressure obtained from the ideal gas law. –Recall PV = nRT or –Substitute for P in equilibrium expression. Consider the reaction: aA + bB  cC + dD –Use this relationship to relate K P and K c E.g. Determine K p for the synthesis of ammonia at 25°C; N 2 (g) + 3H 2 (g)  2NH 3 (g) K c = 4.1x10 8 E.g. 2 Determine K p for the synthesis of HI at 458°C; ½ H 2 (g) + ½ I 2 (g)  HI(g) K c = 0.142.

11. 8–11 John A. Schreifels Chemistry 212 Chapter 15-11 Heterogeneous Equilbria Do not include a solvent or solid in the equilibrium expression. Their composition is constant and included in the equilibrium constant. –Water concentration is 55.5 M; very high compared with reactants and products. –The concentration of a solid such as CaCO 3 stays the same as long as some solid is present. E.g.: Determine the equilibrium expression for the reaction: CaCO 3 (s) + C(gr)  CaO(s) + 2CO(g). E.g.2 Determine the equilibrium expression for the reaction of acetic acid with water. CH 3 COOH(aq) + H 2 O(l)  CH 3 COO  (aq) + H 3 O + (aq)

12. 8–12 John A. Schreifels Chemistry 212 Chapter 15-12 Reaction Sequence As with  H in thermo. K c can be determined from a reaction sequence. Consider the reactions to the right. The third reaction is the result of the sum of the other two (called a reaction sequence.. E.g.: Determine the Kp for the reaction: CaCO 3 (s) + C(gr)  CaO(s) + 2CO(g). Given: CaCO 3 (s)  CaO(g) + CO 2 (g), K 1 = 0.039 atm. = P CO2, and C(g) + CO 2 (g)  2CO(g), K 2 = 1.9 atm.

13. 8–13 John A. Schreifels Chemistry 212 Chapter 15-13 Applications of the Equilibrium Constant Extent of reaction: The magnitude of the equilibrium constant allows us to predict the extent of the reaction. –Very large K (e.g. 10 10 )  mostly products. –Very small K (e.g. 10  10 )  mostly reactants. –When K is around 1, a significant amount of reactant and product present in the equilibrium mixture. E.g. For each of the following decide which species will predominate at equilibrium.

14. 8–14 John A. Schreifels Chemistry 212 Chapter 15-14 Using the Equilibrium Constant Direction of Reaction: the reaction quotient can be used to determine the direction of a reaction with certain initial concentrations. Reaction Quotient: For the general reaction, aA + bB  cC + dD,. where the t refers to the time that concentrations of the mixture are measured; not necessarily at equilibrium. Comparison of Q c with K c reveals direction of reaction. When only reactants Q c = 0; leads to –If Q c < K c, products will form. When only products present, Q c  . –If Q c > K c, reactants will form. When Q c = K c, no net reaction. E.g. Determine direction of reaction: H 2 (g) + I 2 (g)  2HI(g). Assume that [H 2 ] o = [I 2 ] o = [HI] o = 0.0020M at 490 o C for which K c = 46.

15. 8–15 John A. Schreifels Chemistry 212 Chapter 15-15 Calculating Equilibrium Concentrations Using initial concentrations, stoichiometry and K c, equilibrium concentrations of all components can be determined. E.g. Determine equilibrium amounts of hydrogen and iodine if you started with exactly 1.00 mol H 2 (g) and 2.00 mol I 2 in a 1.00 L flask; after equilibrium had been attained, [HI] = 1.86 M. H 2 (g) + I 2 (g)  2HI(g) E.g.2 [H 2 ] o = [I 2 ] o = 24 mM, were mixed and heated to 490 o C in a container. Calculate equil. composition. Given that K c = 46. Solution: set up an equilibrium table and solve for unknown after substitution into equilibrium expression. H 2 (g) +I 2 (g)  2HI(g)

16. 8–16 John A. Schreifels Chemistry 212 Chapter 15-16 Using the Equilibrium Constant (cont) E.g.2 For the reaction: PCl 5 (g)  PCl 3 (g) + Cl 2 (g), K c = 0.800 M at 340 o C. Find the equilibrium amounts of these compounds if they all start out at 0.120 M. Solution: Determine Q c so that you know which direction the reaction will proceed. Q c = 0.120 < K c  reaction proceeds to products. Equilibrium concentrations of each component are in last row. Substitute into equilibrium expression and solve for x. It may be necessary to rearrange so that quadratic equation can be used.. Rearrange to ax 2 + bx + c = 0; determine a, b, c and substitute into quadratic equation:

17. 8–17 John A. Schreifels Chemistry 212 Chapter 15-17 Factors that Alter the Composition of an Equilibrium Mixture A change to the system, which is initially at equilibrium, can cause a change in the equilibrium composition. Le Châtelier’s Principle: “If a stress is applied to a reaction mixture at equilibrium, reaction occurs in the direction that relieves the stress.” Types of stress on equilibrium: –Concentration of reactants or products. You can add or remove one or more components in a reaction mixture. –With gases changing the pressure or volume is a way of changing the concentrations of all components in the mixture. –Change temperature.

18. 8–18 John A. Schreifels Chemistry 212 Chapter 15-18 Changes in Concentrations Addition or removal of either reactant or product shifts the equilibrium to reduce the excess compound. Removal of product or addition of reactant has the same effect; they shift the equilibrium to the right. E.g. Consider the formation of NH 3 (g); the equilibrium conditions are listed in the first row; then half of the NH 3 (g) is removed. We calculate Q c and compare it with K c. Q c < K c  reaction goes to right to form more NH 3 (g). E.g.2 Instead lets add N 2 so that its concentration increases 10x. This produces a similar effect. Since hydrogen is a reactant (in denominator), Q c will become much smaller than K c. This means that product must form.

19. 8–19 John A. Schreifels Chemistry 212 Chapter 15-19 Changes in Concentrations 2 Removal of reactant or addition of product shifts the equilibrium in the opposite direction – to the left. E.g. Half of N 2 (g) is removed. We calculate Q c and compare it with K c. Notice that it is greater than the equilibrium constant. Therefore the reaction proceeds to the left to form more N 2 (g). E.g.2 Add NH 3 so that its concentration increases by a factor of 2. This produces a similar effect. Since ammonia is a product (in numerator), Q c > K c. Reaction proceeds to left since there is too much product.

20. 8–20 John A. Schreifels Chemistry 212 Chapter 15-20 Changes in Pressure and Volume Equilibria containing gases are sensitive to changes in pressure (volume). –Since pressure and volume are inversely proportional, an increase in pressure will give same result as a decrease in volume. –A decrease in the volume increases concentration of each of the components in the reaction mixture. –A decrease in V shifts equilibrium to side with least number of moles of gas. E.g. Determine the effect of a 2.00x increase in pressure (volume decreases by factor of 2;V f = ½ V I ). Concentrations of reactants and products increase by a factor of 2. Since Q c < K c, the reaction proceeds to the right to form more NH 3 (g).

21. 8–21 John A. Schreifels Chemistry 212 Chapter 15-21 Changes in Pressure and Volume 2 E.g.2 Determine the effect of a factor of 2.00 decrease in the pressure (volume increases by a factor of 2;V f = 2 V i ). We need to be vigilant so that we do not include any solids or solvents in this, since they do not affect the equilibrium. E.g. Determine the ratio of Q c /K c when the pressure increases by a factor of two and from this discuss the effect on the equilibrium: C(s) + CO 2 (g)  2CO(g).

22. 8–22 John A. Schreifels Chemistry 212 Chapter 15-22 Changes in Temperature  T   Q c.  T produces  K c Whether K c increases or decreases depends upon two factors,  T and  H rxn. –An increase in temperature causes the equilibrium to shift towards the endothermic side; –A decrease in temperature causes the equilibrium to shift in the opposite direction. E.g. Will an increase or a decrease in temperature increase the amount of CO produced from the following reaction 2CO 2 (g)  2 CO(g) +O 2 (g)  H = 566 kJ The Effect of a Catalyst It has no effect on the equilibrium. It speeds up attainment of equilibrium. K c related to the  H and not E a