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Chemical Equilibrium

Chemical Equilibrium. Chapter 13. Haber Process: N 2( g ) + 3H 2( g ) ⇌ 2NH 3( g ) . Chemical Equilibrium ⇌. a dynamic condition where two opposite changes occur at equal rates Chemical reactions: the forward rate of reaction = the reverse rate of rxn.

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Chemical Equilibrium

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  1. Chemical Equilibrium Chapter 13

  2. Haber Process: N2(g) + 3H2(g) ⇌ 2NH3(g)

  3. Chemical Equilibrium⇌ • a dynamic condition where two opposite changes occur at equal rates • Chemical reactions: the forward rate of reaction = the reverse rate of rxn. • Macroscopically – looks as if nothing is changing • Microscopically – both forward and reverse reactions continue

  4. What time represents the equilibrium point? a b c d

  5. Equilibrium lies towards the left:R E PReactants Equilibrium Products Equilibrium lies towards the right:R E PReactants Equilibrium Products

  6. Law of Mass Action (Chemical Equilibrium) • an equilibrium system at a given temperature has a constant ratio of products & reactants (K). Equilibrium Constant (K) (also written as: Kc or Keq) • equilibrium ratio of product to reactant concentrations.

  7. Equilibrium reaction aA + bB ⇌ cC + dD K = [C]c· [D]d [A]a· [B]b

  8. Problem: Write the equilibrium expression forH2O(g) + CO(g)⇌ H2(g) +CO2(g) K = [H2]∙[CO2] [H2O]·[CO]

  9. Problem: Write the equilibrium expression for N2(g) + 3 H2(g)⇌ 2 NH3(g) K = [NH3]2 [N2]·[H2]3 Write the equilibrium expression for ½ N2(g) + 3/2 H2(g)⇌ NH3(g) K = [NH3] [N2]½·[H2]3/2 =SQRT [NH3]2 [N2]·[H2]3

  10. Problem: • What is the K value for: N2(g)+3H2(g)⇌ 2 NH3(g) where [N2] =0.020 M, [H2] =0.15 M, [NH3] =0.10 M K = [0.1M]2 = 148 [0.02M]·[0.15]3

  11. Equilibrium Expression Summary • Equilibrium expression for the reverse reaction is the reciprocal of that for the forward reaction. • When a balanced equation for a reaction is multiplied by a factor of n, the new equilibrium expression is the original expression raised to the nth power; Knew = (Koriginal)n. • K values are usually written without units. • K is always the same value at a given temperature no matter what the initial quantities are of reactants or products.

  12. Equilibrium Expression Summary • At a given temperature for a reaction, K does NOT change • Sets of concentrations (“equilibrium positions”) can vary greatly, but their ratios when determining K will give the same result at equilibrium.

  13. Equilibrium Constant (K): • shows how far the reaction proceeds. • K >>1 favors products • K <<1 favors reactants

  14. Homogeneous Equilibria • Substances are in the same physical state: N2(g) + 3H2(g) ⇌ 2NH3(g) HCN(aq) ⇌ H+(aq) + CN−(aq) Heterogeneous Equilibria • Substances are in different physical states: 2KClO3(s) ⇌ 2KCl(s) + 3O2(g)

  15. Heterogeneous Equilibria • Concentration measurements (mol/L) apply only to gases and aqueous states. • A heterogeneous equilibrium does not depend on amounts of solids or liquids, so they are omitted from K equations. 2KClO3(s) ⇌ 2KCl(s) + 3O2(g) K = [O2]3

  16. K =? [CO2(g)] • CaCO3(s)⇌ CaO(s)+CO2(g) • H2O(l)⇌ H2O(g) • H2O(g)⇌ H2O(l) • C(s)+ H2O(g)⇌ CO(g) + H2 (g) • FeO(s)+CO(g)⇌Fe(s)+CO2(g) [H2O(g)] 1 . [H2O(g)] [H2(g)][CO(g)] [H2O(g)] [CO2(g)] [CO (g)]

  17. What is “PV=nRT”? Rearrange to isolate “P” Ideal gas law equation P = RT nV Units representing “n/V”? M = mol/L “n/V” = C (concentration) P = CRT

  18. C = Molar Concentration of a gas N2(g) + 3H2(g) ⇌ 2NH3(g) K = [NH3]2 = CNH32= Kc [N2]·[H2]3 (CN2)(CH23) Kp = (PNH3)2 . (PN2)(PH2)3

  19. Example: N2(g) + 3H2(g) ⇌ 2NH3(g) Find Kp if equilibrium pressures at a given temperature are:

  20. The Relationship Between K and Kp Kp = K(RT)Δn • Δn = sum of coefficients of gas products minus sum of coefficients of the gas reactants. • R = 0.08206 L·atm/mol·K • T = temperature (in K)

  21. Example N2(g) + 3H2(g) ⇌ 2NH3(g) Using the value of Kp (3.9 × 104) from the previous example, calculate the value of K at 35°C.

  22. Consider the following equation: CO(g) + H2O(g) ⇌ CO2(g) + H2(g) Which of the following must be true at equilibrium? • [CO2] = [H2] because they are in a 1:1 mole ratio in the balanced equation. • The total concentration of reactants is equal to the total concentration of the products. • The total concentration of the reactants is greater than the total concentration of the products. • The total concentration of the products is greater than the total concentration of the reactants. • None of these is true.

  23. For which of the following reactions does K = Kp at the same temperature? • N2(g) + 3H2(g) ⇌ 2NH3(g) • CaCO3(s) ⇌ CaO(s) + CO2(g) • CO(g) + H2O(g) ⇌ CO2(g) + H2(g) • 3Fe(s) + H2O(g) ⇌ Fe3O4(s) + 4H2(g) Since Kp = K(RT)∆n then ∆n must = 0 to allow Kp = K.

  24. The equilibrium constant for A + 2B ⇌ 3C is 2.1  10–6. Determine the equilibrium constant for 2A + 4B ⇌ 6C. • 4.2  10–6 • 4.4  10–12 • 2.3  1011 • 1.8  10-11 • None of these K for the first reaction = [C]3/[A][B]2 and for the second reaction = [C]6/[A]2[B]4, the two differ by one being the square of the other. Consequently, the value for the equilibrium constant for the second reaction will be: (2.1  10–6)2 = 4.4  10–12

  25. The value of the equilibrium constant, K, is dependent on: • The temperature of the system. • The nature of the reactants and the products. • The concentrations of the reactants. • The concentrations of the products. • I, II • II, III • III, IV • I, II, III • I, II, III, IV K is influenced by the temperature and the phases of the reactants and products.

  26. Which of the following statements is false regarding chemical equilibrium? • A system that is disturbed from an equilibrium condition responds in a manner to restore equilibrium. • The value of the equilibrium constant for a given reaction mixture at constant temperature is the same regardless of the direction from which equilibrium is attained. • When two opposing processes are proceeding at identical rates, the system is at equilibrium. • A system moves spontaneously toward a state of equilibrium. • All of these statements are false. e) is the false statement because all of the above are true

  27. What is…? • k? Reaction rate constant • K? Equilibrium concentration expression constant • Kp ? Equilibrium pressure constant • When is K = Kp ? When Δn = 0 since Kp = K(RT)Δn

  28. Reaction Quotient (Q) Set up like K using current reaction concentrations. If Q is… • < Kc: reaction proceeds to right (forming more products) • = Kc: at equilibrium • > Kc: reaction proceeds to left (forming more reactants)

  29. Problem: • At 340°C, K = 0.064 for:Fe2O3(s) + 3H2(g)⇌ 2Fe(s)+ 3H2O(g)Determine which way the reaction will proceed if [H2]= 0.45 M and[H2O]= 0.37 M Q = [H2O]3 [H2]3 = (0.37)3 = 0.56. (0.45)3 Q>K, so rxn proceeds left

  30. Solving Equilibrium Problems: a) Write the balanced equation b) Write the equilibrium expression: Kp = … c) Determine the given concentrations/pressures d) Determine Q, if appropriate, to check position relative to the equilibrium e) Define the change needed to reach the equilibrium concentrations (ICE table) f) Substitute equilibrium concentrations into the equilibrium expression to solve for the unknown.

  31. Problem: • 3.00 mol H2 and 6.00 mol F2 are mixed in a 3.00 L flask. K at this temperature is 115. Find the equilibrium concentrations. • 1) Write the balanced equation: • H2(g) + F2(g)⇌ 2HF(g) • 2) Write the equilibrium expression: • 3) Determine the initial concentrations: K= [HF]2 [H2] [F2] [H2]0 = 3.00 mol/3.00 L = 1.00 M [F2]0 = 6.00 mol / 3.00 L = 2.00 M [HF]0 = 0

  32. Problem (cont.) • 4) Determine the value of Q: Q= [HF]2 = 0 so we are NOT at equilibrium [H2] [F2]

  33. Problem (cont.) • 5) Plug into your K expression: K= [2x]2 = 0 so we are NOT at equilibrium [1.00-x] [2.00-x]

  34. Problem: • At 35°C, K = 1.6 x 10-5 for the reaction below. Calculate the concentrations of all the species at equilibrium for the following initial concentrations: 2 NOCl (g) ⇌ 2 NO (g) + Cl2(g) a) 2.0 mol pure NOCl in a 2.0 liter flask K = [NO]2 [Cl2] [NOCl]2 Molar ratios show: 2 NOClare lost for every 2 NO’s formed and for every 1 Cl2formed.

  35. Problem: [NO]2 [Cl2] [NOCl]2 NOCl (g) ⇌ 2 NO (g) + Cl2(g) K = Let’s use “x” to represent concentration of Cl2 formed. Initially we only have the reactant NOCl:K = [2x]2 [x] [1.0M −2x]2

  36. Lab Notes: Determination of Kc • Part I: • Write the reaction and Kc expression in your notebook. • Assume forward rxn due to large [Fe3+], giving you 100% FeSCN2+. • Plot graph of absorbance vs. [FeSCN2+]. You determined these concentrations in your pre-lab table. • Sketch or print your graph. Include the equation for your slope (shown on LabQuest) • Part II: - REVISED!! • Run Test Tube 3 only!! Read the absorbance. Locate this value on your graph (part I) to determine the equilibrium [FeSCN2+]. • Calculate initial [Fe3+]0 and [SCN− ]0 . • Use ICE table to determine equilibrium [Fe3+] and [SCN−]. Use the equilibrium concentrations to determine Kc.

  37. Le Châtelier’s Principle: • A stress imposed on a system at equilibrium will cause the equilibrium to shift in a direction that reduces the stress.

  38. Factors that alter chemical equilibrium are changes in • Concentrations (reactant and/or product) • Pressure/volume (for gases) • temperature

  39. 1. Change in Concentration: • Equilibrium shifts to reduce added substance or produce the removed substance

  40. 2. Pressure/ volume Changes • Increased P (decreased V) shifts equilibrium to side with fewest # of gas molecules • Less P shifts it to side with more gas molecules • Inert gases (non reactive) do not affect the equilibrium position.

  41. Examples • 2H2(g)+O2(g) ⇌ 2H2O(g) • NO(g)+O3(g) ⇌ O2(g) +NO2(g) • 2NO(g)+Br2(g) ⇌ 2NOBr(g)

  42. 3. Temperature Change • Heat is part of reaction: see which side it is on. • Exothermic rxns: Kc decreases with rising T • Endothermic rxns: Kc increases with rising T

  43. Add O2? → 2SO2(g) + O2(g) ⇌ 2SO3(g) → Remove SO3? ← Add SO3? → Increase pressure? What happens to Kc?

  44. Exo- or endotherm.? endo NH4Cl(s)+heat⇌NH3(g)+HCl(g) Increase T? → K at higher T? increases Increase P? ←

  45. 2 NO2 ↔ N2O4 + 57.2 kJ • Start with 2 moles of NO2(brown color) and 1 mole of N2O4 (clear) in each of these vessels. Change the temperatures. • After sufficient time for equilibrium to be established, which one of the four vessels will be the brownest? Which one will be the clearest? The one in hot water The one in ice water

  46. Suppose we have the following reaction at equilibrium: PCl5(g) ⇌ PCl3(g) + Cl2(g)Which of the following statements is false? • Adding PCl3 to the container shifts the equilibrium to form more PCl5. • Decreasing the volume of the container shifts the equilibrium to form more PCl5. • Removing PCl5 from the container shifts the equilibrium to form more PCl3.

  47. For a certain reaction at 25.0C, the value of K is 1.2  10–3. At 50.0C, the value of K is 3.4  10–1. This means that the reaction is: • exothermic. • endothermic. • More information is needed to answer the question. K is influenced by the temperature and the phases of the reactants and products.

  48. Problem Consider the reaction represented by the equation: Fe3+(aq) + SCN-(aq) ⇌ FeSCN2+(aq) Trial #1: 6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN2+(aq) is 4.00 M. What is the value for the equilibrium constant for this reaction?

  49. Trial 1 Solution:set up ICE Table Fe3+(aq) + SCN–(aq) ⇌ FeSCN2+(aq) Initial 6.00 10.00 0.00 Change – 4.00 – 4.00 +4.00 Equilibrium 2.00 6.00 4.00 K = 0.333

  50. Problem Consider the reaction represented by the equation: Fe3+(aq) + SCN-(aq) ⇌ FeSCN2+(aq) • Trial #2: Initial concentrations: 10.0 M Fe3+(aq) and 8.00 M SCN−(aq) (same temperature as Trial #1). K (from trial 1) = 0.333 Equilibrium: ? M FeSCN2+(aq)  5.00 M FeSCN2+

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