1 / 29

SECTION 1: THE CONCEPT OF EQUILIBRIUM

Chapter 16 Chemical Equilibrium. SECTION 1: THE CONCEPT OF EQUILIBRIUM. General Concepts of Equilibrium. Equilibrium only happens if a reaction is reversable The reaction must be a closed system at a constant temperature

maitland
Download Presentation

SECTION 1: THE CONCEPT OF EQUILIBRIUM

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 16 Chemical Equilibrium SECTION 1: THE CONCEPT OF EQUILIBRIUM

  2. General Concepts of Equilibrium • Equilibrium only happens if a reaction is reversable • The reaction must be a closed system at a constant temperature • At equilibrium, the forward and reverse reaction are proceeding at the same rate • rate of reaction = how fast reactants become products

  3. At equilibrium, the concentrations of all species appear to be constant ( unchanging) • dynamic (always changing) state • Equilibrium notation: • Reaches the same equilibrium concentrations regardless the rxn starts with reactants or products *Don’t think that all substances have the same amount or concentration at equilibrium!!

  4. Example: N2O4(g)2 NO2(g) colorless brown Equilibrium established

  5. The same equilibrium composition is reached from either the forward or reverse direction Example: N2O4(g)2 NO2(g) colorless brown Notice the different amounts!

  6. Do you the differences between this and the early graphs?

  7. Reaction Quotient (Q) aA + bB cC + dD *[ ] represents the molar concentration *The molarity of a pure solid or liquid is 1 **Notice that gas and aqueous are not included!!

  8. Q becomes Keq when... • the reaction is at equilibrium • Keq = equilibrium constant • Kc if molarities are used (c for concentration of a solution) • Kp if pressures are used (p for pressure of gas)

  9. Q vs. Keq *If Q > Kc, the rxn proceeds to reactants to increase the concentrations of reactants *If Q < Kc, the rxn proceeds to products to increase the concentrations of products *Kc and Q depend on temperature *Kc and Q are unitless

  10. The magnitude of Keq tells…. • Big Keq value = products dominant at equilibrium • Small Keq value = reactants dominant at equilibrium

  11. Example Equation: 2SO2(g) + O2(g)  2SO3(g) Keq = 85.0 • Products or reactants dominant at equilibrium? • If [SO2] = 1.0 M, [O2] = 2.0 M, and [SO3] = 3.0 M, the reaction at equilibrium?

  12. Practices Write the equilibrium law for each. • 2H2(g) + O2(g) 2H2O(g) • 2Hg(l) + Cl2(g) Hg2Cl2(s) • Na(s) + H2O(l) NaOH(aq) + H2(g) • NH3(g) + HCl(g) NH4Cl(s) • Ag2CrO4(s) 2Ag+(aq) + CrO42-(aq)

  13. Example #2: Analysis of an equilibrium mixture of nitrogen, hydrogen, and ammonia contained in a 2-L flask at 300 °C gives the following results: H2 = 0.15 mol; N2 = 0.25 mol: NH3 = 0.10 mol Calculate Keq for the reaction. 3H2 (g) + N2 (g)  2NH3 (g)

  14. Example #4: What is the equilibrium concentration of SO3 in the following reaction if the concentration of SO2 and O2 are each 0.0500M and Keq = 85.0? Equation: 2SO2(g) + O2(g)  2SO3(g)

  15. Various operations that can be performed on equilibrium expressions • When the direction of an equilibrium is reversed, the new equilibrium constant is the reciprocal of the original Switching the reactants and products,

  16. 2) When the coefficients in an equation are multiplied by a factor, the equilibrium constant is raised to a power equal to that factor Multiplying the equation by n,

  17. 3) When chemical equilibria are added, their equilibrium constants are multiplied

  18. Practices 8) At 25°C, Kc=7.0×1025 for the rxn 2SO2(g) + O2(g) 2SO3(g) What is the Kc value for the rxn SO2(g) + 1/2O2(g) SO3(g)? 9) At 25°C, the following reactions have the equilibrium constants. 2CO(g) + O2(g) 2CO2(g) Kc = 3.3x1091 2H2(g) + O2(g) 2H2O(g) Kc= 9.1x1080 Calculate Kc for the rxn H2O(g) + CO(g) CO2(g) + H2(g)

  19. Equilibrium Laws for Gaseous Rxns, Kp • Gas law: • Kp: Equilibrium constants written in terms of partial pressures

  20. Kc and Kp are related for gaseous rxns: aA + bB cC + dD

  21. 10) Methanol, CH3OH, is a promising fuel that can be synthesized from carbon monoxide and hydrogen according to the rxn CO(g) + 2H2(g) CH3OH(g) For this rxn at 200°C, Kp=3.8x10-2. Calculate Kc. 11) Nitrous oxide, N2O, is a gas used as an anesthetic. The compound has a strong tendency to decompose into nitrogen and oxygen following the equation 2N2O(g) 2N2(g) + O2(g) The decomposition reaction has Kc=7.3x1034 at 25°C. What is the value of Kp for this reaction at 25°C.

  22. The size of the equilibrium constant gives a measure of how the reaction proceeds • General statements can be made about the equilibrium constant (either Kc or KP)

  23. K>>1 (large K): A large amount of product and very little reactant at equilibrium K≈1: Approximately equal amounts of reactant and product are present at equilibrium. K<<1: Mostly reactant and very little product are present at equilibrium.

  24. Le Châtelier’s principle • Khan Academy (15 min) • If equilibrium is disturbed, the reaction tries to remove or minimize the disturbance, so it can reestablish equilibrium. • The disturbance is called a stress • A disturbance changes the molarity of a reactant or product

  25. Some common “stresses” • Adding or removing a product or reactant • Changing the volume of container • Changing the pressure by changing the container’s volume • Changing the temperature • Adding a catalyst • Adding inert gas

  26. Consider the equilibrium PCl3(g) + Cl2(g) PCl5(g) ΔH°=-88kJ About this reaction, must know:

  27. Consider the equilibrium PCl3(g) + Cl2(g) PCl5(g) ΔH°=-88kJ How will the reaction respond? • PCl5 is added. • Cl2 is removed. • The volume of the container is increased

  28. Consider the equilibrium PCl3(g) + Cl2(g) PCl5(g) ΔH°=-88kJ How will the reaction respond? (4) The temperature is lowered (5) He gas is added (6) A catalyst is added

  29. Haber Process N2(g) + 3 H2(g) <-> 2 NH3(g) + heat How can the production of NH3 be increased?

More Related