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Chemical Equilibrium

Chemical Equilibrium. Chapter 6. 6.0 Chemical Reaction. Chemical reaction may occur in two ways:- a) a non-reversible reaction - occurs in one direction b) a reversible reaction - occurs in both direction i.e (forward & reverse reaction). 6.1.1 A non-reversible reaction.

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Chemical Equilibrium

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  1. Chemical Equilibrium Chapter 6

  2. 6.0 Chemical Reaction Chemical reaction may occur in two ways:- a) a non-reversible reaction - occurs in one direction b) a reversible reaction - occurs in both direction i.e (forward & reverse reaction)

  3. 6.1.1 A non-reversible reaction • Example 1 • NaCl + AgNO3 AgCl + NO3 • One/both reactants will be completely consumed and amount of products formed depends on the limiting reactant. • A single arrow (→) is used to represent reaction.

  4. 6.1.2 Reversible Reaction • Example 2 • Occurs in both direction. • ⇌ shows that reactants react to form product, and product reacts to form reactants. • Amount of products formed does not depend on the amount of reactant used. • Reaction will reach equilibrium when the concentration of products & reactants remained constant. H2(g) + I2(g) ⇌ 2HI(g)

  5. 6.2 Equilibrium System • A system is at equilibrium when there is no observable change occurs. • Equilibrium system can be observed in: a. Physical Equilibrium b. Chemical Equilibrium

  6. ⇌ H2O (l) H2O (g) 6.2.1 Physical Equilibrium • Example 3 • vaporisation in a closed container. • Involves physical change of substance. • Level of H2O in the container remains constant at equilibrium because the rate of evaporation equals the rate of condensation.

  7. 6.2.2 Chemical Equilibrium • Example 4 • 2NO2(g) ⇌N2O4(g) • Initially NO2 molecules combine to form N2O4 (brown gas appears) • As soon as N2O4 is formed, it undergoes reverse reaction to form NO2. • At equilibrium, the system contains a constant amount of NO2 & N2O4.

  8. 6.3 Dynamic Equilibrium • A system reaches a dynamic equilibrium when • the rate of forward reaction equals the rate of reverse reaction. • the concentrations of both reactants and products remain constant. • no observable change occurs but the conversions of reactants to products and products to reactants continue.

  9. concentration t1 time After time t1, both concentrations remain constant. [N2O4] at equilibrium [NO2 ]at equilibrium The reaction does not stop but the rate of forward reaction equals the rate of reverse reaction.

  10. ⇌ aA (g) + bB (g)cC (g) + dD (g) [C]c[D]d Kc = [A]a[B]b 6.4 Equilibrium constant, K • Since concentrations at equilibrium remain constant, the equilibrium can be expressed by a constant. • Consider: • Kc is known as equilibrium constant Concentrations of species are expressed in molar.

  11. Table 6.1 ⇌ N2O4(g) 2NO2(g) [NO2]2 K = constant [N2O4] = 4.63 x 10-3

  12. 6.5 Equilibrium Law: Law of Mass Action • When a system has reached equilibrium, the ratio of multiplied concentrations of products to the multiplied concentrations of reactants (each raised to the power of the respective stoichiometric coefficient) is a constant at constant temperature. OR • Value of equilibrium constant, Kc is a constant at constant temperature.

  13. 6.6 Types of System in Chemical Equilibrium • a) Homogeneous equilibrium • Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. b) Heterogeneous equilibrium A system in which the reactants and products are not in the same phase.

  14. 6.7 Homogeneous Equilibrium System • 6.7.1 Liquid phase • Important variable: concentration • Equilibrium constant: Kc Example 5 CH3COOH (l) + CH3OH (l) ⇌ CH3COOCH3 (l) + H2O(l) Kc=

  15. Example 6 CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq) Kc=

  16. (PC)c (PD)d Kp = (PB)b (PA)a 6.7.2 Gas phase • The quantitative aspects to be considered are the concentration and pressure. • We can use Kcand Kp to represent the equilibrium constant. the pressure of gases are expressed in atm or other units of pressure

  17. Example 7 Write the equilibrium law in the form of Kp and Kc for the following reactions: a) N2(g) + 3H2(g) NH3(g) b) 2NO(g) + O2(g) N2O4(g) Answer: Kp = Kc =

  18. Kc = [SO2]2 [O2] [SO3]2 Kc1 = [SO2] [O2]1/2 [SO3] Kc2 = [SO3]2 [SO2]2 [O2] Expression of Kc Expression of Kc depends on the equilibrium equation given Example 8 2SO3 (g) ⇌ 2SO2 (g) + O2 (g) 1. SO3 (g) ⇌ SO2 (g) + ½ O2 (g) 2. 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)

  19. Kc1 = [SO2] [O2]1/2 [SO3] Kc2 = [SO3]2 [SO2]2 [O2] Example 9 Find the relationship between Kc1 and Kc2 for the following equilibrium equations. 1. SO3 (g) ⇌ SO2 (g) + ½ O2 (g) 2. 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)

  20. Example 10 Consider the following reaction: A(g) + B(g) ⇌ 2C(g) If 5 moles of A are allowed to mix and react with 3 moles of B in 1dm3 container, 2 moles of C are produced at equilibrium. What is the value of Kc for this reaction? (Ans: 0.5 )

  21. Example 11 Consider the reaction: CO(g) + 2H2(g) ⇌ CH3OH(g) The following equilibrium concentrations are achieved at 400K: [CO] = 1.03M [H2] = 0.332M [CH3OH] = 1.56M Determine the equilibrium constant at 400K (Ans: 13.74 M-2)

  22. Example 12 Nitrogen gas reacts with oxygen gas at high temperature to form nitrogen monoxide gas, NO. The equilibrium constant for the reaction is 4.1 x 10-4 at 200oC. If the concentration for both N2(g) and O2(g) at equilibrium are 0.40 moldm-3 and 1.3 moldm-3 respectively, calculate the concentration of NO gas at equilibrium? (Ans: 0.0146 M)

  23. Example 13 The equilibrium constant Kp for the reaction; 2NO2 (g) ⇌ 2NO (g) + O2 (g) is 158 at 1000K. What is the equilibrium pressure of O2 if the P NO2 = 0.400 atm and PNO = 0.270 atm?

  24. 2 PNO PO 2 Kp = 2 PNO 2 2 PNO PO 2 = Kp 2 2 PNO PO = 158 x (0.400)2 2 (0.270)2 Answer; = 347 atm

  25. ⇌ aA (g) + bB (g)cC (g) + dD (g) [C]c[D]d Kc = [A]a[B]b {PC /(RT)}c {PD / (RT)}d Kc = {PA /(RT)}a {PB /(RT)}b 6.7.3 Relationship between Kp and Kc Consider Since P = nRT V Thus, [ ] = P RT

  26. ⇌ aA (g) + bB (g)cC (g) + dD (g) PCc PDd (RT) (a + b) – (c+d) Kc = PAa PBb Kc = Kp (RT) (a + b) – (c+d) Kp = Kc (RT) (c + d) – (a+b) Kp = Kc (RT)Δn For the reaction

  27. The equilibrium concentrations for the reaction between carbon monoxide and chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. ⇌ CO (g) + Cl2 (g) COCl2 (g) 1.

  28. 0.14 [COCl2] = 0.012 x 0.054 [CO][Cl2] Answer; Kc= = 220 M -1 Kp = Kc(RT)n n = 1 – 2 = -1 Kp= 220 x (0.08206 x 347)–1 = 7.7 atm–1

  29. 6.8 Heterogeneous Equilibrium system • Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. • Kc and Kp can be used to represent the equilibrium constant. • The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

  30. ⇌ 1. CH3COOH (aq) + H2O (l) CH3COO-(aq) + H3O+ (aq) [CH3COO-][H3O+] Kc = [CH3COOH][H2O] [CH3COO-][H3O+] Kc = [CH3COOH] H2O a pure liquid, has a constant concentration and thus is not included in the expression. Therefore, the equilibrium constant is written as, Unit : M

  31. ⇌ 2. CH3COOH (aq) + C2H5OH (aq) CH3COOC2H5(aq) + H2O(l) [CH3COOC2H5] Kc = [CH3COOH] [C2H5OH] Alcohol Ester Carboxylic acid

  32. [CaO][CO2] ‘ Kc = [CaCO3] Kp = PCO 2 ⇌ CaCO3 (s) CaO (s) + CO2 (g) 3. [CaCO3] = constant [CaO] = constant Kc = [CO2]

  33. ⇌ CaCO3(s) CaO (s) + CO2(g) PCO 2 PCO does not depend on the amount of CaCO3 or CaO 2 = Kp

  34. 5. Consider the following equilibrium at 295 K: The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction? ⇌ NH4HS (s) NH3 (g) + H2S (g)

  35. P Kp = P H2S NH3 Kc = 0.0702 atm2 (0.08206 atm L mol–1 K–1 x 295 K )2 Answer; = 0.265 atm x 0.265 atm = 0.0702 atm2 Kp = Kc(RT)n n = 2 – 0 = 2 T = 295 K = 1.2 x 10–4 mol2 L–2 = 1.2 x 10–4 M2

  36. The value of Kc for the following equation is 1.0 x 10–3 M at 200°C. 2NOBr (g) ⇌ 2NO (g) + Br2 (g) Calculate Kc’ for the following equation at 200°C. NO (g) + ½ Br2 (g) ⇌ NOBr (g) 6.

  37. Kc = [NO]2 [Br2] Kc’ = [NOBr] [Br2]1/2 [NO] [NOBr]2 1 1 Kc [NOBr]2 = = [NO]2 [Br2] [NO]2 [Br2] [NOBr]2 ½ 1 Kc [NOBr] = ½ [NO] [Br2] Answer; = Kc’

  38. ½ 1 Kc Kc’ = Kc’ = 1 ½ 1.0 x 10–3 M = 31.6 M-½ = 31.6 dm3/2 mol–½

  39. ⇌ H2(g) + I2(g) 2HI(g) At 440C, the equilibrium constant Kc for reaction, 7. has a value of 49.5. If 0.200 mole of H2 and 0.200 mole of I2 are placed into a 10.0 L vessel and permitted to react at this temperature, what will be the concentration of each substance at equilibrium?

  40. ⇌ H2 (g) + I2 (g) 2HI (g) Answer; [H2]0 = 0.200 /10.0 = 0.0200 M [I2]0 = 0.200 /10.0 = 0.0200 M [ ]0 / M 0.0200 0.0200 0 Δ[ ] / M -x -x +2x 0.0200 - x 0.0200 - x +2x [ ]⇌/ M

  41. Kc = [HI]2 [H2] [I2] 49.5 = (2x)2 (0.0200 – x) (0.0200 – x) 7.036 = 2x 0.0200 - x x = 0.0156

  42. ⇌ H2 (g) + I2 (g) 2HI (g) [ ]0 / M 0.0200 0.0200 0 Δ[ ] / M -x -x +2x 0.0200 - x 0.0200 - x +2x [ ]⇌/ M = 0.0044 = 0.0044 = 0.0312 Therefore concentration of each substance at equilibrium: [H2] = 0.0044 M [I2] = 0.0044 M [HI] = 0.0312 M

  43. A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium constant, Kp for the reaction, 8. H2 (g) + I2 (g) ⇌ 2HI (g) at 448°C is 50.5. What are the partial pressures of H2 , I2 and HI in the flask at equilibrium.

  44. PH = = (1.000)(0.08206)(721) n RT H2 2 V 1.000 = (2.000)(0.08206)(721) PI = n RT I2 2 V 1.000 Answer; Initial pressures of H2 and I2. = 59.19 atm = 118.4 atm

  45. ⇌ H2 (g) + I2 (g) 2HI (g) 50.5 = (2x)2 2 Kp = PHI ⇒ (59.19 – x) (118.4 – x) P P H2 I2 PHI = 110.6 atm P = 3.89 atm ; P = 63.1 atm ; H2 I2 P0 59.19 118.4 0 ΔP -x -x +2x P⇌ 59.19 - x 118.4 - x +2x 46.5x2 – 8969x + 353904 = 0 Partial pressure of gases at equilibrium:

  46. ΔC α = C0 6.2.6 Degree of Dissociation, α The fraction of a molecule dissociated. ΔC = changes in concentration C0 = initial concentration A complete dissociation occurs if α = 1

  47. x α = a AB ⇌ A+ + B– a 0 0 C0 / M +x -x ΔC / M +x a-x +x +x C⇌ / M

  48. x 0.0089 α = = a 0.5 9. The concentration of H+ ion measured for HCOOH 0.5 M is 0.0089 M. What is the degree of dissociation of the compound? x = 0.0089 a = 0.5 HCOOH ⇌ H+ + HCOO– C0 / M 0.5 0 0 ΔC / M +0.0089 -0.0089 +0.0089 C⇌/ M 0.5-0.0089 0.0089 0.0089 = 0.4911 = 0.0178 = 1.78%

  49. 6.2.7 Predicting the direction of reaction • For any reversible reaction, we can determine the direction of a reaction (whether is moving forward or reverse) by comparing the value of Q with Kp or Kc. • Q is the reaction quotient • Q has the same expression of Kc and Kp but the numerical value gained is NOT at equilibrium.

  50. ⇌ aA (g) + bB (g)cC (g) + dD (g) [C]c[D]d Qc = [A]a[B]b Predicting the direction of reaction:

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