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Chemistry The Science in Context Chapter 15 Chemical Equilibrium

Chemistry The Science in Context Chapter 15 Chemical Equilibrium. Climate models are a system of differential equations derived form basic physics and Chemistry. These equations model kinetic processes but often we can approximate parts of the system by assuming that they reach equilibrium….

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Chemistry The Science in Context Chapter 15 Chemical Equilibrium

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  1. ChemistryThe Science in ContextChapter 15Chemical Equilibrium

  2. Climate models are a system of differential equations derived form basic physics and Chemistry. These equations model kinetic processes but often we can approximate parts of the system by assuming that they reach equilibrium…

  3. For example, climate models often employ the concept of radiative equilibrium incoming energy from the sun = out going energy from the earth One “simple model is (1 − a)Sπr2 = 4πr2εσT4

  4. We have observed that the oncentrations of O2, N2 and NO eventually stop changing. So the reaction can be written: N2+O2 = 2NO

  5. Equilibrium Equilibrium is reached when the forward and reverse rates of reaction are equal. R P kf[R] = kr[P]

  6. Equilibrium Equilibrium is reached when the forward and reverse rates of reaction are equal. R P kf/kr= [P]/[R] = Kc subscript “c” denotes conc.

  7. The Equilibrium Constant For the general reaction: a A + b B c C + d D Kc = ([C]c[D]d)/([A]a[B]b) The subscript “c” is used to indicate that the value of K is calculated based upon a ratio of concentrations. In pressure units.. Kp = (PC)c(PD)d)/(PA)a(PB)b

  8. The Equilibrium Constant The equilibrium expression for a chemical reaction is the ration of the concentration terms for products divided by that for reactant in accordance with the balanced chemical equation. The equilibrium constant (K) is the numerical value for the equilibrium expression for a chemical reaction The law of mass action states that the equilibrium constant will have a characteristic value at a given temperature.

  9. Properties of Equilibrium Constants • Keq applies only at equilibrium, i.e. when the rate forward = rate reverse • Keq is independent of initial conditions. The reaction can start with all products, all reactants, and any combination in between, as long as the temperature is constant, at equilibrium, the ratio of products to reactants is determined by the equilibrium constant. • Keqis related to the stoichiometry of the balanced (net) chemical equation. Based upon Keq the final state of the system can be calculated.

  10. Properties of Equilibrium Constants • Keq varies with temperature. It will increase with increasing temperature for an endothermic reaction. It will decrease with increasing temperature for an exothermic reaction. • Keq is reported as a unitless value. Although related to the ratio of the rate constants, K is a thermodynamic property. The previous thermodynamic properties; ΔH, ΔS, and ΔG, were all described relative to Standard State conditions:1 M for a solute and 1 atm for a gas. A more correct form of the equilibrium expression is also written relative to Standard State conditions.

  11. »PC version Explore the concept of dynamic equilibrium and learn to relate the equilibrium constant to molar concentrations and partial pressures of products and reactants. Includes practice exercises. Equilibrium Tutorial

  12. »PC version This unit describes the difference between Kc and Kp and explains how to convert between them. Includes practice exercises. Equilibrium in the Gas Phase Tutorial

  13. Problem 16 Kc = 8.7E6 at 700K for the reaction between NO and O2. The rate constant for the reverse reaction at this temperature is 0.54M-1s-1 What is the rate constant for the forward reaction at 700K? Answer: Keq = kf/kr kf = Keq * kr = 8.7E6 *0.54M-1s-1 kf = 4.7E6 M-2s-1

  14. Problem Write equilibrium constant expressions for the following reactions. a) 2 ClO(g) = Cl2(g) + O2(g) b) 2 O3(g) = 3 O2(g) What are the units for an equilibrium constant?

  15. Problem The value of the equilibrium constant (Kp) for the formation of ammonia N2(g) + 3 H2(g) = 2 NH3(g) Is 4.5E-5 at 450°C. What is the value of Kp for the following… NH3(g) = ½ N2(g) + 3/2 H2(g) Answer: 149

  16. Problem 35 Phosgene (COCl2) is used in manufacturing although it is highly poisonous. It is formed from the reaction of carbon monoxide and chlorine: CO(g) + Cl2(g) COCl2(g) The value of Kc for the reaction is 5.0 at 325°C. What is the value of Kp? Answer: 0.10

  17. Problem 28 A 100 mL reaction vessel initially contains 2.60  10–2 moles of NO and 1.30  10–2 moles of H2. At equilibrium, the concentration of NO in the vessel is 0.161 M. At equilibrium the vessel also contains N2, H2O, and H2. What is the value of the equilibrium constant Kc for the following reaction? 2H2(g) + 2NO(g) ↔ 2H2O(g) + N2(g) Answer: 19.5

  18. Problem 30 For the reaction: 2NO(g) + O2(g) 2NO2(g) Kc = 5E12. What is the value of Kc for each of the following: a) NO(g) + 1/2O2(g) NO2(g) b) 2NO2(g) 2NO(g) + O2(g) c) NO2(g) NO(g) + 1/2O2(g)

  19. The reaction quotient, Q, is defined in the same way as K, the equilibrium constant. Q is the numerical value for the mass action at some time point in the reaction. A ↔ B Q = [B]/[A] As reactions proceed toward equilibrium, Q increases or decreases until Q = K

  20. Problem 54 The equilibrium constant (Kc) for the reaction 2C D + E Is 3E-3. At a particular measurement time, the composition of the reaction mixture is: [C] = [D] = [E] = 5E-4. In which direction (forward or reverse) will the reaction proceed to reach equilibrium. Answer: Q = 1

  21. Keq = 3 E-3 [C] = [D] = [E] = 5E-4 Q = [D][E]/[C]2 = [5E-4][5E-4]/[5E-4]2

  22. »PC version Manipulate values for G and Kc to explore how Gibbs free energy is related to equilibrium constant and reaction spontaneity. Includes practice exercises. Equilibrium and Thermodynamics Tutorial

  23. K, Q, and DG There is a direct relationship between the equilibrium constant, K, and the free energy, DG. DG>O; K<1 “reactant favored” DG<0; K>1 “product favored” DG=0; K=1 DG = DG° + RT ln(Q) At equilibrium, DG = 0; Q=Keq DG° = -RTln(Keq)

  24. Problem 48 In glycolysis, the hydrolysis of ATP to ADP is used to drive the phosphorylation of glucose: glucose + ATP ↔ ADP + glucose 6-phophate DG° for the reaction equals, -16.7 kJ/mol What is Kc for the reaction at 298K? Answer: Kc=846

  25. Problem 72 Under the appropriate conditions, NO forms N2O and NO2: 3 NO(g) ↔ N2O(g) + NO2(g) Use the values of DG° for the following reactions to calculate the values of Kp for the above reaction at 500°C. 2 NO(g)+O2(g) ↔ 2NO2 (g) DG° = -69.7 kJ 2 N2O(g) ↔ 2NO(g) + N2(g) DG° = -33.8 kJ N2(g)+ O2(g) ↔ 2NO(g) DG° = 173.2 kJ

  26. Answer: Kc=1.16E7; Kp=1.83E5

  27. Disturbing an equilibrium and Le Châtelier’s Principle When a change is imposed on a system at equilibrium, the system will react in the direction (toward products or reactants) that reduces the effect and amount of change, (i.e. re-establishes equilibrium).

  28. When the total pressure is increased upon a gas-phase reaction, the reaction shifts to “relieve the stress” i.e. to the reactants or products side with smaller number of total moles.

  29. According to Le Châtelier’s principle, reactions will shift an equilibrium in an endothermic (exothermic) direction when the temperature is increased (decreased). Higher temperature Lower temperature

  30. »PC version In this tutorial you can subject a system at equilibrium to a stress and observe changes in the reaction quotient and how the system reacts according to Le Châtelier’s principle. Includes practice exercises. Le Châtelier’s Principle Tutorial

  31. Applications of Le Chatelier’s Principle Problem 86 How will the following changes affect the position of the equilibrium: 2NO2(g) ↔ NO(g) + NO3(g) a) The concentration of NO is increased b) The concentration of NO2 is increased c) The volume of the system is allowed to expand to 5-times its initial value. d) temperature increase or decrease

  32. Applications of Le Chatelier’s Principle Problem 79 Patients suffering from carbon monoxide poisoning are treated with pure oxygen (hyperbaric) to remove CO for the hemoglobin (Hb). The two relevant equilibria are: Hb + 4CO(g) ↔ Hb(CO)4(g) Hb + 4O2(g) ↔ Hb(O2)4(g) Kc for CO binding to hemoglobin is about 150 times that of O2 binding. How, then, does this treatment work?

  33. Reversing the first reaction…and summing, Hb(CO)4(g) ↔ Hb + 4CO(g) Hb + 4O2(g) ↔ Hb(O2)4(g) Net: Hb(CO)4(g) + 4O2(g) ↔ Hb(O2)4(g) + 4CO(g)

  34. How do we calculate how Keq changes with temperature? DG° = -RT ln(K) DG° = DH° - TDS° ln(K) = - DG°/RT ln(K) = - DH°/RT + TDS°/RT ln(K) = - (DH°/R)(T-1) + DS°/R

  35. Changing K with Temperature ln(K1) = - (DH°/R)(T1-1) + DS°/R ln(K2) = - (DH°/R)(T2-1) + DS°/R ln(K1) - ln(K2) = - (DH°/R)(T1-1) + (DH°/R)(T2-1) ln (K2/K1) = - (DH°/R)(T2-1 – T1-1) This is known as the van’t Hoff equation and can be used to calculate Kp or Kc at different temperatures. It assumes that the enthalpy value (DH) does not change over the temperature of interests. How/why does DH change with temperature?

  36. Changing K with Temperature Problem 86 The value of Kp for the reaction: N2(g) + 3H2(g) ↔ 2 NH3(g) Is 41 at 400K. If DH°r= -92.2 kJ/mol, Calculate the value of Kp at 700K. Answer:Kp(700K) = 2.84E-4

  37. Problem The equilibrium constant, Kp, for the thermal decomposition of NO2 2NO2(g) ↔ 2 NO(g) + O2(g) is 6.5E-6 at 450K. If a reaction vessel at this temperature initially contains 0.500 atm NO2, what will be the partial pressures of NO2, NO and O2 when equilibrium has been established?

  38. Problem 76 On a particularly smoggy day, the concentration of NO2 in the air over an urban area reaches 2.2x10-7M. If the temperature of the air is 25°C, what is the concentration of the NO2 dimer, N2O4, in the air? N2O4(g) ↔ 2 NO2(g) Kc = 6.1x10-3

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