1. TOPIC 7.2 The Position of Equilibrium

2. Equilibrium Position  When the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance called chemical equilibrium.  The relative concentrations of the reactants and products at equilibrium constitute the equilibrium position of a reaction.

3. The Equilibrium Constant Equilibrium Constant (K c ) The equilibrium constant (Kc) is the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to a power equal to the number of moles of that substance in the balanced chemical equation The value of K c indicates the equilibrium position of a reaction

4. The Equilibrium Constant (K) Consider the following general reversible reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA (aq) + bB (aq) cC (aq) + dD (aq) where [ ] = concentration of each chemical in moles/ liter (M).

5. The Equilibrium Constant

7. Homogeneous vs. Heterogeneous Equilibria Homogeneous reactions – all reactants and products are in the same phase (gaseous, liquid, or aqueous). For example: 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) K c = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] C 2 H 5 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O (l) K c = [CH 3 COOC 2 H 5 ] [H 2 O ] [C 2 H 5 OH] [CH 3 COOH]

8. Homogeneous vs. Heterogeneous Equilibria Heterogeneous reactions- substances involved are in different phases (s, l, g, aq) of matter. The concentrations of solids and liquids do not appear in equilibrium expressions because their concentrations do not change. For example: K c = [CO 2 ] 2 [CO] 2 K c = [CO 2 ] K c = [Zn 2+ ]_ [Cu 2+ ]

9. Magnitude of K c Different reactions have different values of K c. What does this value tell us about a particular reaction? A value of K c greater than 1 means that products are favored over reactants (equilibrium lies to the right); a value of K c less than 1 means that reactants are favored over products (equilibrium lies to the left).

10. What Does the Value of K Mean? aA + bBcC + dD If K c >> 1, the reaction is product-favored; product predominates at equilibrium ( the reaction is considered to go almost to completion)

11. What Does the Value of K Mean? aA + bBcC + dD If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium (the reaction hardly proceeds).

12. Magnitude of K c The following three reaction and their K c values at 550 K: H 2 (g) + I 2 (g) ↔ 2 HI(g) K c = 2 H 2 (g) + Br 2 (g) ↔ 2 HBr(g) K c = 10 10 H 2 (g) + Cl 2 (g) ↔ 2 HCl(g) K c = 10 18 The large range in their K c values indicate the differing extents of these reactions. Here we can deduce that the reaction between H 2 and Cl 2 has reacted the most, while H 2 and I 2 have reacted the least.

13. Le Chatelier’s Principle Le Chatelier’s principle states that “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.” Simply put, this means that whatever we do to a system at equilibrium, the system will respond in the opposite way. If something is added, the system will react to remove it; if something is removed, the system will react to replace it. If something is added, the system will react to remove it something is removed, the system will react to replace it. Applying this principle enables us to predict the qualitative effect of typical changes that occur to systems at equilibrium.

14. Le Chatelier’s Principle Changes in Concentration Increasing the concentration of one of the reactants or decreasing the concentration of one of the products will shift the equilibrium to the right and more products will be produced. Similarly, decreasing the concentration of one of the reactants or increasing the concentration of one of the products will shift the equilibrium to the left. Equilibrium will later be re-established with new concentrations of all reactants and products. The value of K c will be unchanged. Often in industrial processes, the product will be removed as it forms. This ensures that the equilibrium continuously lies to the right, increasing the yield of product.

15. Le Chatelier’s Principle Change in pressure If a system at equilibrium is subject to an increase in pressure, the system responds to decrease the pressure by favoring the side with the smaller number of moles of gas (the side with less moles will have less pressure). Conversely, a decrease in pressure will cause a shift in the equilibrium to the side with the larger number of moles of gas (the side with more moles will have more pressure) A different equilibrium position will be achieved, but the value of Kc will be unchanged, as long as the temperature remains the same.

16. Le Chatelier’s Principle For example: CO(g) + 2 H 2 (g) ↔CH 3 OH(g) There are 3 moles of gas on the left side and 1 mole of gas on the right side. So, an increase in pressure will increase the production of methanol. Conversely, a decrease in pressure will increase the production of CO and H 2. CO(g) + 2 H 2 (g) ↔ CH 3 OH(g) + P ↑ ↓ → ←

17. Le Chatelier’s Principle Changes in temperature The change in the equilibrium with an increase or decrease in temperature depends on whether the reaction is exothermic or endothermic. If a reaction is exothermic, heat is released, so if the temperature is increased, the stress can be reduced by shifting the equilibrium to the left (reactants). However, if the temperature is decreased, the equilibrium will shift to the right and more products will be produced. If a reaction is endothermic, heat is absorbed, so if the temperature is increased, the stress can be reduced by shifting the equilibrium to make more products. However, if the temperature is decreased, the stress can be reduced by shifting equilibrium to left and favoring the reactants.

18. Le Chatelier’s Principle For example: 2 NO 2 (g) ↔ N 2 O 4 (g) Δ H = -24 kJ mol -1 N 2 (g) + O 2 (g) ↔ 2 NO (g)Δ H = +181kJ mol - 1 + Heat ↑ in temp. ← → ↓ in temp. → ← + Heat

19. Le Chatelier’s Principle Unlike change the concentration or pressure, a change in temperature will also change the value of K c. For an exothermic reaction, the concentration of the products in the equilibrium mixture decreases as the temperature increases, so the value of K c will decrease. The opposite will be true for endothermic reactions.

20. Le Chatelier’s Principle Addition of a Catalyst The addition of a catalyst speeds up the rate of a reaction by lowering the activation energy and so increases the number of particles that have sufficient energy to react without raising the temperature. Increase the rate of both the forward and reverse reactions. Equilibrium is achieved faster, but the equilibrium composition remains unaltered So, a catalyst will have no effect on the position of the equilibrium or the value of K c.

21. a. Addition of product, shifts the equilibrium to the left, forming more PCl 5 b. The equation shows 2 mol of gaseous product and 1 mol of gaseous reactant. The increase in pressure is relieved if the equilibrium shifts to the left, because a decrease in the number of moles of gaseous substances produces a decrease in pressure. c. The removal of heat causes the equilibrium to shift to the left, because the reverse reaction is heat-producing. d. The removal of PCl 3 causes the equilibrium to shift to the right to produce more PCl 3 Le Chatelier’s Principle

22. Review Questions In a reaction at equilibrium, reactants and products a) decrease in concentration. b) form at equal rates. c) have equal concentrations. d) have stopped reacting. In the reaction 2NO 2 (g)  2NO(g) + O 2 (g), increasing the pressure on the reaction would cause a) the amount of NO to increase. b) the amount of NO 2 to increase. c) nothing to happen. d) the amount of O 2 to increase.

23. Industrial Applications In reactions involving the manufacture of a chemical, it is a goal to obtain as high a yield of product as possible. Applying Le Chatelier’s principle enables scientists to maximize the yield by choosing conditions that will cause the equilibrium to lie to the right. However, the yield of the reaction is only part of the consideration. The rate is also of great significance. For example, it would be of limited value if a process were able to yield 95% product, but would take several years. So, the economics of the process will depend not only on how much product, but how fast it can be produced. Sometimes these two factors work against each other and chemists must choose the best compromise between them.

24. It was not until the early 20th century that this method was developed to harness the atmospheric abundance of nitrogen to create ammonia, which can then be oxidized to make the nitrates and nitrites essential for the production of fertilizers and ammunitions.ammonia nitratesnitritesfertilizers ∆H = -92.4 kJ/n @ 25 0 C NH 3 + HNO 3  NH 4 NO 3 (explosives & fertilizers) Le Châtelier’s Principle & the Haber Process

25. The Haber Process- the production of ammonia The process is based on the reaction: N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) ∆H = -92.4 kJ/mol Applying Le Chatelier’s principle, we can consider the optimum conditions for this reaction: 1. Removing ammonia as it forms shifts the equilibrium to the right, increasing yield 2. Increasing the pressure will shift the equilibrium to the right, increasing yield. The usual process used in the Haber process is about 200 atmospheres 3. Lowering the temperature will shift the equilibrium to the right, but using a temperature that is too low would cause the reaction to be too slow. So, a moderate temperature of about 450 o is used. 4. The use of a catalyst will not increase the yield, however, it will speed up the rate of production and will help to compensate for the moderate temperature used. A catalyst of finely divided iron is used, with small amounts of aluminum and magnesium oxides added to improve its activity.

26. Le Châtelier’s Principle & the Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH 3 ) from the system as a liquid Removed by liquefaction Subsb. pt. NH 3 -33°C N2N2 -196°C H2H2 -253°C N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe Shift equilibrium to the right by increasing press & conc. of N 2 + H 2 and by removing NH 3 from system

27. The Contact Process- the production of sulfuric acid Sulfuric acid has the highest production of any chemical in the world. It is used in the production of fertilizers, detergents, dyes, explosives, drugs, plastics and in many other chemical industries. Its production, known as the Contact process, involves three simple reactions:  The combustion of sulfur S(s) + O 2 (g) → SO 2 (g)  The oxidation of sulfur dioxide 2SO 2 (g) + O 2 (g) ↔ 2SO 3 (g)  The combination of sulfur trioxide with water SO 3 (g) + H 2 SO 4 (l) → H 2 S 2 O 7 (l) +H 2 O(l) → 2H 2 SO 4 (l) It has been shown that the overall rate depends on step 2. So, applying Le Chatelier’s principle to this, we can predict the conditions that will most favor the formation of product.

28. The Contact process- the production of sulfuric acid 2SO 2 (g) + O 2 (g) ↔ 2SO 3 (g) ΔH = -196 kJ/mol Influence on reactionCondition used PressureForward reaction involves reduction in the number of moles of gas; high pressure will favor product 2 atm (this gives a very high yield, so a higher pressure is not needed. TemperatureForward reaction is exothermic; low temperature will increase yield, but decrease rate 450 o C CatalystIncreases the rate of the reaction Vanadium(V) oxide