1 / 22

Chemical Equilibrium

Chemical Equilibrium. Brown, LeMay Ch 15 AP Chemistry. 15.1: Chemical Equilibrium. Occurs when opposing reactions are proceeding at the same rate Forward rate = reverse rate of reaction Ex: Vapor pressure: rate of vaporization = rate of condensation

ronald
Download Presentation

Chemical 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. Chemical Equilibrium Brown, LeMay Ch 15 AP Chemistry

  2. 15.1: Chemical Equilibrium • Occurs when opposing reactions are proceeding at the same rate • Forward rate = reverse rate of reaction Ex: • Vapor pressure:rate of vaporization = rate of condensation • Saturated solution:rate of dissociation = rate of crystallization • Expressing concentrations: • Gases: partial pressures, PX • Solutes in liquids: molarity, [X]

  3. Forward reaction: A → B Rate = kforward [A] • Reverse reaction: B → A Rate = kreverse [B] or R = 0.0821 L•atm mol•K Forward reaction: Reverse reaction:

  4. If equilibrium: A ↔ B forward rate =reverse rate or

  5. [A]0or PA0/ RT [A] or PA / RT PX or [X] [B] or PB / RT Time → Figure 1: Reversible reactions Equilibrium is established

  6. Reaction Rate Time Reversible Reactions and Rate Forward rate Equilibrium is established: Forward rate = Backward rate Backward rate When equilibrium is achieved: [A] ≠ [B] and kf/kr = Keq

  7. 15.2: Law of Mass Action • Derived from rate laws by Guldberg andWaage (1864) • For a balanced chemical reactionin equilibrium: a A + b B ↔ c C + d D • Equilibrium constant expression (Keq): Cato Guldberg Peter Waage (1836-1902) (1833-1900) or • Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism). • Units: Keq is considered dimensionless (no units)

  8. Relating Kc and Kp • Convert [A] into PA: where Dx == change in coefficents of products – reactants (gases only!)= (c+d) - (a+b)

  9. Magnitude of Keq • Since Keq a [products]/[reactants], the magnitude of Keq predicts which reaction direction is favored: • If Keq > 1 then [products] > [reactants] and equilibrium “lies to the right” • If Keq < 1 then [products] < [reactants] and equilibrium “lies to the left”

  10. 15.3: Types of Equilibria • Homogeneous: all components in same phase (usually g or aq) N2 (g) + H2 (g) ↔ NH3 (g) 1 3 2 Fritz Haber(1868 – 1934)

  11. Heterogeneous:different phases CaCO3 (s) ↔ CaO (s) + CO2 (g) Definition: What we use: • Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq (see p. 548). • Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium.

  12. 15.4: Calculating Equilibrium Constants Steps to use “ICE” table: • “I” = Tabulate known initial and equilibrium concentrations of all species in equilibrium expression • “C” = Determine the concentration change for the species where initial and equilibrium are known • Use stoichiometry to calculate concentration changes for all other species involved in equilibrium • “E” = Calculate the equilibrium concentrations

  13. Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is 0.0124 M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction: NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

  14. NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq) X 0.0124 M 0 M 0 M X - x + x + x X 0.0119 M 4.64 x 10-4 M 4.64 x 10-4 M x = 4.64 x 10-4 M

  15. Ex: A 5.000-L flask is filled with 5.000 x 10-3 mol of H2 and 1.000 x 10-2 mol of I2 at 448ºC. The value of Keq is 1.33. What are the concentrations of each substance at equilibrium? H2 (g) + I2 (g) ↔ 2 HI (g)

  16. H2 (g) + I2 (g) ↔ 2 HI (g) 1.000x10-3 M 2.000x10-3 M 0 M - x M - x M + 2x M (1.000x10-3 – x) M (2.000x10-3 – x) M 2x M 4x2 = 1.33[x2 + (-3.000x10-3)x + 2.000x10-6] 0 = -2.67x2 – 3.99x10-3x + 2.66x10-6 Using quadratic eq’n: x = 5.00x10-4or –1.99x10-3; x = 5.00x10-4 Then [H2]=5.00x10-4 M; [I2]=1.50x10-3 M; [HI]=1.00x10-3 M

  17. 15.6: Le Châtelier’s Principle • If a system at equilibrium is disturbed by a change in: • Concentration of one of the components, • Pressure, or • Temperature …the system will shift its equilibrium position to counteract the effect of the disturbance. Henri Le Châtelier(1850 – 1936)

  18. 4 Changes that do not affect Keq: • Concentration • Upon addition of a reactant or product, equilibrium shifts to re-establish equilibrium by consuming part of the added substance. • Upon removal of reactant or product, equilibrium shifts to re-establish equilibrium by producing more of the removed substance. • Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l) • Add HCl, temporarily inc forward rate • Add H2O, temporarily inc reverse rate

  19. 2. Volume, with a gas present (T is constant) • Upon a decrease in V (thereby increasing P),equilibrium shifts to reduce the number of moles of gas. • Upon an increase in V (thereby decreasing P),equilibrium shifts to produce more moles of gas. Ex: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) • If V of container is decreased, equilibrium shifts right. • XN2 and XH2dec • XNH3 inc Since PT also inc, KP remains constant.

  20. 3. Pressure, but not Volume • Usually addition of a noble gas, p. 560 • Avogadro’s law: adding more non-reacting particles “fills in” the empty space between particles. • In the mixture of red and blue gas particles, below, adding green particles does not stress the system, so there is no Le Châtelier shift.

  21. 4. Catalysts • Lower the activation energy of both forward and reverse rxns, therefore increases both forward and reverse rxn rates. • Increase the rate at which equilibrium is achieved, but does not change the ratio of components of the equilibrium mixture (does not change the Keq) Ea, uncatalyzed Ea, catalyzed Energy Rxn coordinate

  22. 1 Change that does affect Keq: • Temperature:consider “heat” as a part of the reaction • Upon an increase in T, endothermic reaction is favored (equilibrium shifts to “consume the extra heat”) • Upon a decrease in T, equilibrium shifts to produce more heat. Effect on Keq • Exothermic equilibria: Reactants ↔ Products + heat • Inc T increases reverse reaction rate which decreases Keq • Endothermic equilibria: Reactants + heat ↔ Products • Inc T increases forward reaction rate increases Keq • Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l); DH=+? • Inc T temporarily inc forward rate • Dec T temporarily inc reverse rate

More Related