Unit 7: General Equilibrium

1. Unit 7:General Equilibrium Ms. DiOrio Rm 109

2. Contents • Characteristics and Conditions of Chemical Equilibrium • Equilibrium Expression and the Equilibrium Constant • Applications of the Equilibrium Constant • Manipulations of Q and K • Solving Equilibrium Problems • Le Châtelier’s Principle • Free Energy and Equilibrium

3. Characteristics and Conditions of Chemical Equilibrium

4. Reversible Reactions • In our study of kinetics, we considered initial conditions in which we could ignore the rate of the reverse reaction. • However, many classes of reactions must consider both the forward and the reverse reactions • Dissolving and precipitation salts (chemical) • Binding and release of oxygen to hemoglobin (biochemical) • Attachment and release of molecules to receptor sites in the nose (biological) • Transfer of carbon and other dissolved substances between the atmosphere and biosphere (environmental) • Proton transfer in acid-base reactions (chemical) • Transfer of electrons in redox reactions (chemical)

5. Picture This • Imagine a container holds gases undergoing the following chemical reaction: • Over time, the reactant H2O and CO molecules decrease in concentration while the concentration of H2 and CO2 molecules increases. • However, since the reaction is reversible, the amount of H2O and CO never goes to zero

6. Picture This • As time goes on, what is happening to the rate of the forward reaction? • The forward reaction slows as the concentration of reactants decreases • What is happening to the rate of the reverse reaction? • The reverse reaction speeds up as more and more product molecules are made • Eventually, the rates of the forward and reverse reactions equal each other, and there is no further change to reaction rates. • This is the point where the reaction has reached equilibrium

7. Equilibrium • Chemical equilibrium is a dynamic situation where the rate of the forward reaction is equal to the rate of the reverse reaction • At any given point, the relative amounts of products and reactants are the same • However, there are constant forward and reverse reactions occurring

8. Equilibrium Expressions and the Equilibrium Constant

9. We can write two rate laws for the reaction assuming they are elementary steps • Forward: Ratefor = kfor[A]2[B] • Reverse: Raterev = krev[C]3[D] • At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This is known as an equilibrium expression

10. Law of Mass Action • The original assumption that the forward and reverse reactions happened in a single step led to the law of mass action to describe equilibrium • Keq or simply K is a unitlessvalue known as the equilibrium constant that describes the equilibrium ratio of reactants to products • The equilibrium constant only changes with temperature (based on the rate constant!) • It remains constant regardless of the amount of reactants/products or the presence of a catalyst

11. Example • The following equilibrium concentrations were observed for the Haber process at 127˚C. N2 + 3H2 2NH3 [NH3] = 3.1 x 10-2 mol/L [N2] = 8.5 x 10-1 mol/L [H2] = 3.1 x 10-3 mol/L Calculate the value of K at 127˚C for this reaction.

12. Equilibrium Position • K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially • For a reaction at a given temperature, there are infinitely many equilibrium positions but only onevalue for K. • Equilibrium position is a set of equilibrium concentrations

13. Equilibrium Expression with Pressure • K refers to the equilibrium constant involving concentration • However, equilibrium for gaseous reactions can also be described in terms of pressure. • Concentration and pressure are related through the ideal gas law • For a gas, concentration is directly related to the partial pressure of the gas *C = concentration

14. Equilibrium Constant with Pressure • Therefore, we can define a different equilibrium constant, Kp • We can relate Kp to the original K using P = (n/v)RT • Where ∆n is the difference in the sums of the coefficients for the gaseous products and reactants

15. Pure Solids and Pure Liquids • Look at our formulas for K and Kp • Notice that they only involve CONCENTRATION or PARTIAL PRESSURE • Equilibrium position does not depend on the amounts of pure solid or pure liquid present because their “concentration” is constant

16. Applications of the Equilibrium Constant

17. What can we learn from equilibrium constant? • Knowing the equilibrium constant for a reaction allows us to predict: • Tendency of the reaction to occur (but not rate) • Whether a given set of concentrations is at equilibrium • The equilibrium position that will be achieved given a set of initial concentrations

18. Consider this… • Assume the following reaction has an equilibrium constant of 16. • In a given experiment, the two types of molecules are mixed together in the following amounts:

19. Consider this… • From the reaction equation, we can write an equilibrium expression • Our starting conditions show us: • 9 molecules • 12 molecules

20. Consider this… • We can use reaction stoichiometry to identify some number of molecules (x) that be used up by the reactants to form products • We can solve for x by plugging into the equilibrium expression

21. Consider this.. • If x = 8, then our equilibrium position for this reaction occurs when: = 9 – 8 = 1 molecule = 12 – 8 = 4 molecules = 8 molecules = 8 molecules

22. ICE Tables • We use this concept to form what are called ICE tables in equilibrium problems x = 8

23. Example CO + H2O ⇋ CO2 + H2 A mixture of 1.0 mole carbon dioxide and 1.0 mole carbon monoxide are contained in a 1 liter vessel. Later 2.0 moles of water vapor is then introduced into the vessel. This reaction has an equilibrium constant of 0.64. How many moles of the different molecules will be present after equilibrium is obtained? 1 mol 2 mol 1 mol 0 mol -x -x +x +x 1 - x 2 - x 1 + x x 1.58 mol 1.42 mol 0.42 mol 0.58 mol Since the reaction is occurring in a 1 liter vessel, we can use moles as a substitute for concentration since moles = concentration

24. Example 2HI(g) ⇋ I2(g) + H2(g) Consider the reaction for the decomposition of HI at 448oC. The initial concentration of HI(g) was 1.00 M. Once at equilibrium, the concentration of HI(g) was measured to be 0.078 M. Calculate the equilibrium constant, K. 1.00 M 0 M 0 M -2x +x +x 1 - 2x x x 0.461 M 0.461 M 0.078 M

25. The Extent of a Reaction • The magnitude of the equilibrium constant represents the tendency of a reaction to occur • A value of K much larger than 1 means that the equilibrium position consists of mostly products • In other words, the reaction has essentially gone to completion • A very small value of K means that many reactants remain in the reaction mixture, so the reaction is not very likely to occur • Note: the size of K and the time required to reach equilibrium are not directly related

26. The Reaction Quotient • It is useful to know if a set of conditions is at equilibrium and, if not, which direction the system will shift to reach equilibrium. • The reaction quotient (Q) is obtained by applying the law of mass action using initial concentrations instead of equilibrium concentrations • We can determine the direction the reaction will proceed by comparing Q to K K = Q System at equilibrium K < Q Shifts to the left K > Q Shifts to the right Using concentrations NOT at equilibrium!

27. Manipulations of Q and K

28. When a reaction is reversed… • The value of of Q and K for the reverse reaction are the inverse of the forward reaction.

29. When multiplying by a coefficient... • The value of Q or K must be raised to the coefficient the balanced equation is multiplied by               A + B ⇋ C                 K              2A + 2B ⇋ 2C         K2              ½ A + ½ B ⇋ ½ C     K1/2 or

30. When reactions are performed in steps… • When reactions are added together through the presence of a common intermediate, Q and K of the resulting reaction is a product of the values of Q or for the original reactions. • Example: 2NO(g) ⇋ N2(g) + O2(g) K1= 1 x 1030 N2(g) + Br2(g) + O2(g) ⇌ 2NOBr(g) K2= 2 x 10-27 K for the reaction Br2(g) + 2NO(g) ⇌ 2NOBr(g) is calculated as…

31. Solving Equilibrium Problems

32. General Steps • Write the balanced equation for the reaction • Write the equilibrium expression using the law of mass action • List the initial concentrations • Calculate Q, and determine the direction of the shift to equilibrium • Define the change needed to reach equilibrium, and define equilibrium concentrations by applying to ICE tables • Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown • Check your calculated equilibrium concentrations by making sure they give the correct value of K

33. Example • Given the initial concentrations shown below, find the equilibrium concentrations for A, B, and C. A(g) + B(g)  2C(g) K = 9.0 x 10-3 0.500 M 0.500 M 0.500 M +x +x -2x 0.5 + x 0.5 + x 0.5 - 2x so the reaction shifts left

34. Example • Given the initial concentrations shown below, find the equilibrium concentrations for A, B, and C. A(g) + B(g)  2C(g) K = 9.0 x 10-3 0.500 M 0.500 M 0.500 M +x +x -2x 0.5 + x 0.5 + x 0.5 - 2x 0.716 M 0.068 M 0.716 M I cannot take away 0.574 M from C!

35. Systems with Small Equilibrium Constants • When K is small, we can assume that the reaction does not proceed very far to the right • There is a large amount of reactants relative to the amounts of products • We can use this knowledge to simplify our math by assuming that any change in equilibrium for our reactants will be a relatively small number. • In other words, we are ignoring the change in x for the reactants • The 5% Rule of Thumb • Once you solve for x, check to see if the assumption is justified • If the error is less than 5%, then the assumption was justified!

36. Example • A sample of COCl2 is allowed to decompose: COCl2(g) ⇋ CO(g) + Cl2(g) • The value of K is 2.2 x 10-10 at 100˚C. If the initial concentration of COCl2 is 0.095 M, what will be the equilibrium concentrations for each of the species involved?

37. Example • A sample of COCl2 is allowed to decompose: COCl2(g) ⇋ CO(g) + Cl2(g) • The value of K is 2.2 x 10-10 at 100˚C. If the initial concentration of COCl2 is 0.095 M, what will be the equilibrium concentrations for each of the species involved? (Assuming x is negligible)

38. Example • A sample of COCl2 is allowed to decompose: COCl2(g) ⇋ CO(g) + Cl2(g) • The value of K is 2.2 x 10-10 at 100˚C. If the initial concentration of COCl2 is 0.095 M, what will be the equilibrium concentrations for each of the species involved? Our assumption was valid.

39. Types of Equilibrium Problems • Writing Equilibrium Expressions given Chemical Equation • Manipulating Values of K (or Q) • Determining the Direction of Equilibrium Shift given Initial Concentrations and K • Solving for K given Equilibrium Concentrations • Solving for an Equilibrium Concentration given K and All Other Equilibrium Concentrations • Solving for K given Initial Concentrations and At Least One Equilibrium Concentration • Solving for Equilibrium Concentrations given Initial Concentrations and K

40. Le Châtelier’s Principle

41. Le Châtelier’s Principle If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. • Changes to a system at equilibrium will either change the value of Q or the value of K, therefore pushing the reaction to a new equilibrium position. • Equilibrium position is effected by • Concentration • Pressure • Temperature • Note: Equilibrium constant only changes with temperature

42. The Effect of a Change in Concentration • If a component (reactant or product) is added to a reaction system at equilibrium at constant temperature and pressure/volume, the equilibrium position will shift in the direction that lowers the concentration of that component. • If a component is removed, the opposite is true. • This can be explained in terms of reaction rate: • Increasing the number of reactant molecules in a system increases the frequency of collisions increasing the rate of the forward reaction *Remember that pure solids and pure liquids do NOT affect equilibrium

43. The Effect of a Change in Pressure • There are three ways to change pressure of a reaction involving gases: • Add or remove a gaseous reactant or product • Add an inert gas (one not involved in the reaction) • Change the volume of the container • Adding or removing product is effectively the same as changing concentration • The addition of an inert gas increases the total pressure but has no effect on the partial pressures of the reactants or products. • There is no change to the number of reactant collisions with sufficient energy, so there is no change in reaction rate to change equilibrium position

44. The Effect of a Change in Pressure • A change in the volume of a container changes the concentration of both gaseous products and reactants • When a container holding a gaseous system is reduced, the system responds by reducing its own volume by decreasing the total number of gaseous molecules in the system.

45. The Effect of a Change in Temperature • A change in temperature is the only perturbation that changes the equilibrium constant (K) rather than the reaction quotient (Q). • Changes in heat will effect exothermic and endothermic reactions differently • Treat heat as a reactant (endothermic) or product (exothermic) and consider changes in temperature as changes in concentration to predict shifts

46. Free Energy and Equilibrium

47. When ∆G = 0 • Equilibrium position occurs at the lowest value of free energy available to the reaction system (∆G = 0) • When the value of free energy reaches zero, there is no longer any driving force to push a reaction in either direction At equilibrium:

48. Free Energy of an Ideal Gas • Recall that we related the free energy of an ideal gas on its pressure: • Go = the free energy of the gas at a pressure of 1 atm • G = the free energy of the gas at a pressure of P • R = the universal gas constant in the same units of pressure • T = the Kelvin temperature

49. N2(g) + 3H2(g) ⇌ 2NH3(g) Where Using the Q with partial pressures since these are gases!

50. ∆G= ∆Gowhen all components are in the standard state (1 atm for gases) • ∆G = 0 when the free energies of reactants and products are equal and the system is at equilibrium • ∆G< 0 (exergonic) when the free energies of the reactants is less than the free energy of the products • The system will adjust to the right to reach equilibrium • ∆G< 0 (endergonic) when the free energies of the reactants is more than the free energy of the products • The system will adjust to the left to reach equilibrium