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figure 15 3 some equilibrium compositions for the methanation reaction n.
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Figure 15.3: Some equilibrium compositions for the methanation reaction. PowerPoint Presentation
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Figure 15.3: Some equilibrium compositions for the methanation reaction.

Figure 15.3: Some equilibrium compositions for the methanation reaction.

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Figure 15.3: Some equilibrium compositions for the methanation reaction.

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  1. Figure 15.3: Some equilibrium compositions for the methanation reaction.

  2. It can be seen from the ideal gas equation that the partial pressure of a gas is proportional to its molarity. The Equilibrium Constant, Kp • In discussing gas-phase equilibria, it is often more convenient to express concentrations in terms of partial pressures rather than molarities (see Figure 15.4).

  3. Calculating Equilibrium Concentrations • Once you have determined the equilibrium constant for a reaction, you can use it to calculate the concentrations of substances in the equilibrium mixture.

  4. Heterogeneous Equilibria • A heterogeneous equilibrium is an equilibrium that involves reactants and products in more than one phase. • The equilibrium of a heterogeneous system is unaffected by the amounts of pure solids or liquids present, as long as some of each is present. • The concentrations of pure solids and liquids are always considered to be “1” and therefore, do not appear in the equilibrium expression.

  5. A Problem to Consider • Consider the following equilibrium. • A 50.0 L vessel contains 1.00 mol N2, 3.00 mol H2, and 0.500 mol NH3. In which direction (toward reactants or toward products) will the system shift to reestablish equilibrium at 400 oC? • Kc for the reaction at 400 oC is 0.500.

  6. 0.500 mol 1.00 mol 3.00 mol 50.0 L 50.0 L 50.0 L A Problem to Consider • First, calculate concentrations from moles of substances.

  7. The Qc expression for the system would be: A Problem to Consider • First, calculate concentrations from moles of substances. 0.0200 M 0.0600 M 0.0100 M

  8. A Problem to Consider • First, calculate concentrations from moles of substances. 0.0200 M 0.0600 M 0.0100 M • Substituting these concentrations into the reaction quotient gives:

  9. A Problem to Consider • First, calculate concentrations from moles of substances. 0.0200 M 0.0600 M 0.0100 M • Because Qc = 23.1 is greater than Kc = 0.500, the reaction will go to the left (toward reactants) as it approaches equilibrium.

  10. Le Chatelier’s Principle • Obtaining the maximum amount of product from a reaction depends on the proper set of reaction conditions. • Le Chatelier’s principle states that when a system in a chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the equilibrium will shift in a way that tends to counteract this change.

  11. “reactants” “products” Removing Products or Adding Reactants • Let’s refer back to the illustration of the U-tube in the first section of this chapter. • It’s a simple concept to see that if we were to remove products (analogous to dipping water out of the right side of the tube) the reaction would shift to the right until equilibrium was reestablished.

  12. “reactants” “products” Removing Products or Adding Reactants • Let’s refer back to the illustration of the U-tube in the first section of this chapter. • Likewise, if more reactant is added (analogous to pouring more water in the left side of the tube) the reaction would again shift to the right until equilibrium is reestablished.

  13. Forward WHY?

  14. x4

  15. Effects of Pressure Change • A pressure change caused by changing the volume of the reaction vessel can affect the yield of products in a gaseous reaction only if the reaction involves a change in the total moles of gas present (see Figure 15.9).

  16. Figure 15.9 A-C

  17. Effects of Pressure Change • If the products in a gaseous reaction contain fewer molesof gas than the reactants, it is logical that they would require less space. • So, reducing the volume of the reaction vessel would, therefore, favor the products. • Conversely, if the reactants require less volume (that is, fewer moles of gaseous reactant), then decreasing the volume of the reaction vessel would shift the equilibrium to the left (toward reactants).

  18. Effects of Pressure Change • Literally “squeezing” the reaction will cause a shift in the equilibrium toward the fewer moles of gas. • It’s a simple step to see that reducing the pressure in the reaction vessel by increasing its volume would have the opposite effect. • In the event that the number of moles of gaseous product equals the number of moles of gaseous reactant, vessel volume will have no effect on the position of the equilibrium.

  19. Conceptual Problem 15.18

  20. Effect of Temperature Change • Temperature has a significant effect on most reactions (see Figure 15.10). • Reaction rates generally increase with an increase in temperature. Consequently, equilibrium is established sooner. • In addition, the numerical value of the equilibrium constant Kc varies with temperature.

  21. Figure 15.10: The effect of changing the temperature on chemical equilibrium. Photo courtesy of American Color.

  22. Effect of Temperature Change • Let’s look at “heat” as if it were a product in exothermic reactions and a reactant in endothermic reactions. • We see that increasing the temperature is analogous to adding more product (in the case of exothermic reactions) or adding more reactant (in the case of endothermic reactions). • This ultimately has the same effect as if heat were a physical entity.

  23. Effect of Temperature Change • For example, consider the following generic exothermic reaction. • Increasing temperature would be analogous to adding more product, causing the equilibrium to shift left. • Since “heat” does not appear in the equilibrium-constant expression, this change would result in a smaller numerical value for Kc.

  24. Effect of Temperature Change • For an endothermic reaction, the opposite is true. • Increasing temperature would be analogous to adding more reactant, causing the equilibrium to shift right. • This change results in more product at equilibrium, amd a larger numerical value for Kc.

  25. Effect of Temperature Change • In summary: • For an endothermic reaction (DH positive) the amounts of products are increased at equilibrium by an increase in temperature (Kc is larger at higher temperatures). • For an exothermic reaction (DH is negative) the amounts of reactants are increased at equilibrium by an increase in temperature (Kc is smaller at higher temperatures).

  26. Effect of a Catalyst • A catalyst is a substance that increases the rate of a reaction but is not consumed by it. • It is important to understand that a catalyst has no effect on the equilibrium composition of a reaction mixture (see Figure 15.12). • A catalyst merely speeds up the attainment of equilibrium.

  27. Figure 15.12: Oxidation of ammonia using a copper catalyst.Photo courtesy of James Scherer. 4 NH3+3 O22N2 + 6H2O Cu Pt 4NH3 + 5 O24NO+ 6H2O

  28. Conceptual Problem 15.19

  29. Suggested problems for Ch 21 20, 23, 25, 27, 29, 33, 35, 37, 47, 49, 51, 88, 63, 69 NO ELECTRONIC HOMEWORK FOR Ch 21!!

  30. Proton or Neutron Electron or Positron or Gamma photon Nuclear Equations • A nuclear equation is a symbolic representation of a nuclear reaction using nuclide symbols. • Other particles are given the following symbols.

  31. Nuclear Equations • A nuclear equation is a symbolic representation of a nuclear reaction using nuclide symbols. • Note that if all reactants and products but one are known in a nuclear equation, the identity of the missing nucleus (or particle) is easily obtained. • This is illustrated in the next example.

  32. The nuclear equation is • From the superscripts, you can write A Problem To Consider • Technetium-99 is a long-lived radioactive isotope of technetium. Each nucleus decays by emitting one beta particle. What is the product nucleus?

  33. The nuclear equation is A Problem To Consider • Technetium-99 is a long-lived radioactive isotope of technetium. Each nucleus decays by emitting one beta particle. What is the product nucleus? • Similarly, from the subscripts, you get

  34. The nuclear equation is A Problem To Consider • Technetium-99 is a long-lived radioactive isotope of technetium. Each nucleus decays by emitting one beta particle. What is the product nucleus? • Hence A = 99 and Z = 44, so the product is

  35. Nuclear Stability • The existence of stable nuclei with more than one proton is due to the nuclear force. • The nuclear force is a strong force of attraction between nucleons that acts only at very short distances (about 10-15 m). • This force can more than compensate for the repulsion of electrical charges and thereby give a stable nucleus.

  36. Nuclear Stability • Several factors appear to contribute the stability of a nucleus. • The shell model of the nucleus is a nuclear model in which protons and neutrons exist in levels, or shells, analogous to the shell structure exhibited in electron configurations. • Experimentally, note that nuclei with certain numbers of protons and neutrons appear to be very stable.

  37. Nuclear Stability • Several factors appear to contribute the stability of a nucleus. • These numbers, called magic numbers, are the numbers of nuclear particles in a completed shell of protons or neutrons. • Because nuclear forces differ from electrical forces, these numbers are not the same as those for electrons is atoms.

  38. Nuclear Stability • Several factors appear to contribute the stability of a nucleus. • For protons, the magic numbers are • 2, 8, 20, 28, 50, and 82 • For neutrons, the magic numbers are • 2, 8, 20, 28, 50, 82, and 126

  39. Nuclear Stability • Several factors appear to contribute the stability of a nucleus. • Evidence also points to the special stability of pairs of protons and pairs of neutrons.

  40. Nuclear Stability • Several factors appear to contribute the stability of a nucleus. • Finally, when you plot each stable nuclide on a graph of protons vs. neutrons, these stable nuclei fall in a certain region, or band. • The band of stability is the region in which stable nuclides lie in a plot of number of protons against number of neutrons. (see Figure 21.3)

  41. Figure 21.3: Band of stability.

  42. Nuclear Stability • Several factors appear to contribute the stability of a nucleus. • No stable nuclides are known with atomic numbers greater than 83. • On the other hand, all elements with Z equal to 83 or less have one or more stable nuclides.

  43. Alpha emission (abbreviated a): emission of a nucleus, or alpha particle, from an unstable nucleus. • An example is the radioactive decay of radium-226. Types of Radioactive Decay • There are six common types of radioactive decay.

  44. This is equivalent to the conversion of a neutron to a proton. Types of Radioactive Decay • There are six common types of radioactive decay. (see Table 21.2) • Beta emission (abbreviated b or b-): emission of a high speed electron from a stable nucleus.

  45. An example is the radioactive decay of carbon-14. Types of Radioactive Decay • There are six common types of radioactive decay. (see Table 21.2) • Beta emission (abbreviated b or b-): emission of a high speed electron from a stable nucleus.

  46. This is equivalent to the conversion of a proton to a neutron. Types of Radioactive Decay • There are six common types of radioactive decay. • Positron emission (abbreviated b+): emission of a positron from an unstable nucleus.

  47. The radioactive decay of techencium-95 is an example of positron emission. Types of Radioactive Decay • There are six common types of radioactive decay. • Positron emission (abbreviated b+): emission of a positron from an unstable nucleus.

  48. In effect, a proton is changed to a neutron, as in positron emission. Types of Radioactive Decay • There are six common types of radioactive decay. • Electron capture (abbreviated EC): the decay of an unstable nucleus by capturing, or picking up, an electron from an inner orbital of an atom.