1. Chemical Equilibrium

2. The Nature of chemical Equilibrium

3. Reversible Reactions • In theory, every reaction can continue in two directions, forward and reverse • Reversible reaction chemical reaction in which the products can react to re-form the reactants

4. Mercury (II) oxide • Decomposes when heated 2HgO(s)  2Hg(l) + O2(g) • Mercury and oxygen combine to form mercury (II) oxide when heated gently 2Hg(l) + O2(g)  2HgO(s)

5. Suppose HgO is heated in closed container where mercury and oxygen can’t escape • Once decomposition begins, mercury and oxygen released can recombine to form HgO again • Both reactions happen at same time

6. Rate of composition will eventually equal rate of decomposition • At equilibrium, mercury and oxygen will combine at same rate that mercury (II) oxide decomposes • Amounts of mercury(II) oxide, mercury and oxygen are expected to remain constant as long as conditions continue

7. At this point, a state of dynamic equilibrium has been reached between two reactions • No net change in composition of system

8. A reversible reaction is in chemical equilibriumwhen the rate of its forward reaction equals the rate of its reverse reaction and the concentrations of its products and reactants remain unchanged

9. Equation for chemical equilibrium written using double arrows 2HgO(s) ⇄ 2Hg(l) + O2(g)

10. Equilibrium, a Dynamic State • In some cases, the forward reaction is almost done before the rate of the reverse reaction is high enough for equilibrium

11. Here, products of forward reaction are favored (at equilibrium, there is higher concentration of products than reactants) HBr(aq) + H2O(l) H3O+(aq) + Br−(aq)

12. Other cases, reverse reaction is favored • More reactants than products H2CO3(aq) + H2O(l) H3O+(aq) + HCO3−(aq) • Sometimes forward and reverse happen almost same amount before equilibrium H2SO3(aq) + H2O(l) ⇄ H3O+(aq) + HSO3−(aq)

13. The Equilibrium Expression • Let’s say 2 substances, A and B, react to make products C and D • C and D react to make A and B nA + mB⇄xC + yD

14. At first, concentrations of C and D are zero and concentrations of A and B are max • Over time, rate of forward reaction decreases as A and B are used up • Rate of reverse increases as C and D are formed

15. When two reaction rates become equal, equilibrium is created • Individual concentrations of A, B, C, and D don’t change if conditions stay the same

16. After EQ, concentrations of products and reactants remain constant, so the ratio of their concentrations should also stay the same

17. Ratio of product [C]xx [D]y to product [A]n x [B]m for this reaction has definite value at given temp • Definite value is EQ constant of the reaction and is given letter K

18. Following equation describes equilibrium constant for theoretical equilibrium system • [C] = concentration of C in moles/liter

19. Concentrations of products in chemical reaction appear in numerator of equilibrium ratio • Each concentration to the power of the coefficient • These are products of forward reaction

20. Concentrations of reactants found in denominator (risen to power of coefficient) • These are reactants of forward reaction • K independent of original concentrations, but dependent on temperature

21. The Equilibrium Constant • Value of K for particular equilibrium system is found experimentally • Chemist analyzes equilibrium mixture and determines concentrations of all substances • Value of K for given reaction at given temp shows degree to which reactants converted to products

22. If K = 1, products of concentrations raised to appropriate power in numerator and denominator have the same value • Therefore, at equilibrium, there are equal concentrations of reactants and products

23. If K is low, forward reactions happens before equilibrium, and reactants are favored • If K is high, the original reactants are largely converted to products

24. Only concentrations of substances that can actually change are included in K • This means PURE solids and liquids are left out because their concentrations cannot change

25. Chemical-equilibrium Expression • In general, the equilibrium constant, K,  isthe ratio of the mathematical product of the concentrations of the products at equilibrium to the mathematical products of the concentrations of reactants

26. Each concentration raised to power equal to coefficient • Chemical-equilibrium expression the equation for K

27. The H2, I2, HI Equilibrium System • Consider reaction between H2 and I2 vapor in sealed flask at elevated temp • Rate of reaction followed by observing rate at which the violet color of iodine vapor lessens

28. Suppose hydrogen gas is present in excess and reaction continues until all iodine is used up • Color (from iodine) gradually disappears because the product, HI and hydrogen gas are both colorless

29. In reality, color fades to constant intensity but does not disappear completely • The reaction is reversible • HI decomposes to form hydrogen and iodine

30. Rate of reverse reaction increases as concentration of HI increases • Rate of forward reaction decreases accordingly

31. Concentrations of hydrogen and iodine decrease as they are used up • As rates of opposing reactions become equal, equilibrium is established

32. Constant color indicates that equilibrium exists • Net chemical equation: H2(g) + I2(g) ⇄ 2HI(g)

33. H2(g) + I2(g) ⇄ 2HI(g) • From this chemical equation we can write chemical-equilibrium expression

34. Sample Problem • An equilibrium mixture of N2, O2 , and NO gases at 1500 K is determined to consist of 6.4 x 10–3 mol/L of N2, 1.7 x 10–3 mol/L of O2, and 1.1 x 10–5 mol/L of NO. What is the equilibrium constant for the system at this temperature?

35. 1. Analyze • Given: • [N2] = 6.4 × 10−3 mol/L • [O2] = 1.7 × 10−3 mol/L • [NO] = 1.1 × 10−5 mol/L • Unknown: K

36. 2. Plan • The balanced chemical equation is • N2(g) + O2(g) ⇄ 2NO(g) • The chemical equilibrium expression is

37. 3. Compute • = 1.1 x 10-5

38. Practice Problems • At equilibrium a mixture of N2, H2, and NH3 gas at 500°C is determined to consist of 0.602 mol/L of N2, 0.420 mol/L of H2, and 0.113 mol/L of NH3.What is the equilibrium constant for the reaction N2(g) + 3H2(g) ⇄ 2NH3(g) at this temperature? • 0.286

39. The reaction AB2C(g) ⇄ B2(g) + AC(g) reached equilibrium at 900 K in a 5.00 L vessel. At equilibrium 0.084 mol of AB2C, 0.035 mol of B2, and 0.059 mol of AC were detected. What is the equilibrium constant at this temperature for this system? • 4.9 × 10−3

40. At equilibrium a 1.0 L vessel contains 20.0 mol of H2, 18.0 mol of CO2, 12.0 mol of H2O, and 5.9 mol of CO at 427°C. What is the value of K at this temperature for the following reaction? CO2(g) + H2(g) ⇄ CO(g) + H2O(g) • 0.20

41. A reaction between gaseous sulfur dioxide and oxygen gas to produce gaseous sulfur trioxide takes place at 600°C. At that temperature, the concentration of SO2 is found to be 1.50 mol/L, the concentration of O2 is 1.25 mol/L, and the concentration of SO3 is 3.50 mol/L. Using the balanced chemical equation, calculate the equilibrium constant for this system. • 4.36

42. Shifting Equilibrium

43. In chemical equilibrium, forward and reverse reactions happen at same rate • Any change alters the rates and disturbs equilibrium

44. System will try to find a new equilibrium state • By shifting equilibrium in desired direction, chemists can improve the yield of product they want

45. Predicting the Direction of Shift • Le Chatelier’s principle provides a way of predicting the influence of stress factors on equilibrium systems

46. Le Chatelier’s principle if a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that relieves the stress

47. Principle true for all dynamic equilibria • Changes in pressure, concentration, and temperature demonstrate the application of Le Chatelier’s principle to chemical equilibrium

48. Changes in Pressure • Change in pressure only affects equilibrium systems where gases are involved • For changes to affect system, the total number of moles of gas on left side must be different from total moles on right side CaCO3(s) ⇄CaO(s) + CO2(g) 1 mol1 mol + 1 mol

49. CaCO3(s) ⇄CaO(s) + CO2(g) • High pressure favors reverse reaction b/c fewer CO2 molecules are produced • Lower pressure favors increased production of CO2

50. Haber process • For synthesis of ammonia N2(g) + 3H2(g) ⇄ 2NH3(g) • Consider an increase in pressure which causes increase in concentration of all species • System reduces number of molecules (total pressure) by shifting equilibrium to right