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Equilibrium

Equilibrium. and Reversible Reactions. Reversible reactions. Some reactions do not go to completion as we have assumed They may be reversible – a reaction in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously.

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Equilibrium

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  1. Equilibrium and Reversible Reactions

  2. Reversible reactions • Some reactions do not go to completion as we have assumed • They may be reversible – a reaction in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously

  3. NH4Cl(s)  NH3(g)  +  HCl(g) heat NH3(g)  +  HCl(g)    NH4Cl(s) But when the gases hit the cool test tube this happens

  4. for this reaction • The decomposition is endothermic  NH4Cl(s)  NH3(g)  +  HCl(g) • The synthesis of is exothermic NH3 (g)  +  HCl(g)     NH4Cl(s) + heat

  5. The two equations can be combined into one, by using a double arrow, which tells us that it is a reversible reaction: NH4Cl(s) ⇌ NH3(g)  +  HCl(g)

  6. You will see double arrows written several ways:

  7. Many reactions are not reversible and have the usual complete arrow  only pointing to the right.

  8. There is a balance between the rates of the concentrations of reactants and products.

  9. Things to note • If the direction of chemical change is reversed, the energy change is also reversed. Reversing the reaction conditions reverses the direction of chemical change.

  10. Things to note • The amounts of the reactants and products depend on the reaction conditions i.e. the temperature and pressure.

  11. The rate of reactants to products (forward reaction) is equal to the rate of products to reactants (reverse reaction).

  12. The concentrations of the reactants and products remain the same BUT the reactions don't stop!

  13. Even though the rates of the forward and reverse are equal, the concentrations of components on both sides may not be equal • An equlibrium position may be shown: A B or A B 1% 99% 99% 1% • It depends on which side is favored; almost all reactions are reversible to some extent

  14. Measuring the Rate • The rate of a reaction is measured by following the loss of a reactant or the formation of a product.

  15. Reactions can occur: • Very fast – such as a firecracker • Very slow – such as the time it took for dead plants to make coal

  16. The slope of the graph shows the rate of the reaction (how fast it is going).

  17. Dynamic equilibrium • is the point where the forward and reverse reactions take place at the same rate.

  18. End 1/4 Energy Level Diagrams. Endothermic Reaction witha Catalyst

  19. Exothermic Reaction with a Catalyst

  20. Equilibrium Constant • If at equilibrium there are significant amounts of both product and reactants, then the reaction can be described in terms of a K value, or equilibrium constant. For any reaction:

  21. Equilibrium Constant - Keq aA + bB → rR + sS A, B, R, S are the compounds a, b, r, s are the coefficient [R]r [S]s Keq= [A]a[B]b (brackets: [ ] = molarity concentration)

  22. Equilibrium Constant - Keq • The equilibrium constants provide valuable information, such as whether products or reactants are favored: Keq > 1, products favored at equilibrium Keq < 1, reactants favored at equilibrium Know this

  23. Equilibrium Constant - Keq Keq > 1, products favored at equilibrium – you will have a higher concentration of product at equilibrium Keq < 1, reactants favored at equilibrium – you will have a higher concentration of reactants at equilibrium

  24. Equilibrium Constant - Keq • K is a constant for the reaction at constant temperature. K has no units. • The concentrations at equilibrium always combine in the above manner to produce the K value, regardless of the initial concentrations of species

  25. Calculating Keq from Known Equilibrium Amounts • Write the equilibrium expression for the reaction: N2(g) + 3H2 (g)↔ 2NH3 (g) [NH3] Keq= [N2] [H2]

  26. [NH3] Keq= [N2] [H2] Calculating Keq from Known Equilibrium Amounts - Rules • When a reactant or product is preceded by a coefficient, its concentration is raised to the power of that coefficient in the Keq expression. N2(g) + 3H2 ↔ 2NH3(g) 2 3

  27. Calculating Keq from Known Equilibrium Amounts - Rules • Determine the molar concentrations or partial pressures of each species involved. [NH3] = 0.0100 M [N2] =0.0200 M [H2] = 0.0200M

  28. 2 [0.0100]2 [0.0200] [0.0200]3 [NH3] = Keq= 3 [N2] [H2] Calculating Keq from Known Equilibrium Amounts - Rules • Substitute into the equilibrium expression and solve for K. [NH3] = 0.0100 M [N2] =0.0200 M [H2] = 0.0200M

  29. 0.0001 [0.0200] [.000008] [0.0100]2 [0.0200] [0.0200]3 = Calculating Keq from Known Equilibrium Amounts - Rules • Substitute into the equilibrium expression and solve for K. Keq = = 625

  30. Calculating Keq from Known Equilibrium Amounts - Rules • Pure solids do not appear in the equilibrium expression. • Pure liquids do not appear in the equilibrium expression. • Water, either as a liquid or solid, does not appear in the equilibrium expression.

  31. Calculating Keq from Known Equilibrium Amounts • Calculate the Keq of the following equilibrium if the concentration of Ag2CrO4 (s)  2Ag + (aq) + CrO42- (aq) [Ag+] = 1.30* 10 -5, [CrO42-]= 6.5*10-5.

  32. Calculating Keq from Known Equilibrium Amounts • Write the Keq Ag2CrO4 (s)  2Ag + (aq) + CrO42- (aq) [Ag+]2 [CrO42-] Keq= [Ag2CrO4 (s) ]

  33. Calculating Keq from Known Equilibrium Amounts • Fill in the concentrations • Remember pure solids do not appear in the equilibrium expression. [Ag+] = 1.30* 10 -5, [CrO42-]= 6.5*10-5 [1.30*10 -5 ]2 [6.5*10-5] 1.1*10-14 Keq= = [Ag2CrO4 (s) ] (Ignore)

  34. Calculate an unknown concentration using Keq • At 200°C, the Keq for the reaction: N2(g) + 3H2 ↔ 2NH3(g) is 625 • If the [N2] = 0.030 M, and the [NH3] = 0.12 M, at equilibrium, calculate the equilibrium [H2].

  35. Write out the Keq expression: N2(g) + 3H2 ↔ 2NH3(g) [NH3]2 Keq= [N2] [H2]3

  36. Fill in everything you know [N2] = 0.030 M, and the [NH3] = 0.12 M, Keq =625 [0.12]2 625= [0.030] [H2]3

  37. Cross-multiply: (625) [H2]3 (0.030) = (0.12)2 Solve for H2 [H2]3 = (0.12)2 (625)(0.030)

  38. 3 3 Solve for H2 by taking the cube root of both sides [H2]3 = 0.0144 = 18.75 0.000768

  39. Take the cube root of both sides: [H ] = 9.2x10 2M = 0.092M

  40. Calculating Keq from Known Equilibrium Amounts - Rules The Keq of a reaction occurring in the reverse direction is simply the inverse of the Keq of the reaction occurring in the forward direction (1/Keq).

  41. The Equilibrium Constant Kp = PcCPdD PaAPbB Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written as a pressure-dependent constant:

  42. The Equilibrium Constant Since, from the Ideal Gas Law: PV =nRT Pi =(ni/V)RT=ciRT

  43. Relationship between Kc and Kp Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes: Kp =Kc(RT)Δn Where: Δ n = (moles of product) – (moles of reactant)

  44. Equilibrium A chemical system can be thought of as being either: 1. At Equilibrium 2. Not At Equilibrium A system which is not at equilibrium will move spontaneously to a position of being at equilibrium.

  45. The Reaction Quotient (Q) [C]c [D]d Q= [A]a[B]b To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.

  46. The Reaction Quotient (Q) Q gives the same ratio the equilibrium expression gives, but for a system that is notat equilibrium.

  47. The Reaction Quotient (Q) aA +bB ↔ cC +dD If Q = K, then the system is at equilibrium. If Q > K, there is too much product and the equilibrium will shift to the left. If Q < K, there is too much reactant and the equilibrium will shift to theright.

  48. Equilibrium Constant - Keq

  49. Le Châtelier’s principle

  50. Stresses to equilibria • Changes in reactant or product concentrations is one type of “stress” on an equilibrium • Other stresses are temperature, and pressure. • The response of equilibria to these stresses is explained by Le Chatelier’s principle: If an equilibrium in a system is upset, the system will tend to react in a direction that will reestablish equilibrium

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