1 / 34

Equilibrium

Equilibrium. Chemical Equilibrium. Equilibrium occurs when opposing reactions are proceeding at equal rates Equilibrium can be reached from either direction Ratio of concentrations will become constant. Equilibrium Constant. Haber Process N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) aA + bB ↔ cC + dD

niabi
Download Presentation

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. Equilibrium

  2. Chemical Equilibrium • Equilibrium occurs when opposing reactions are proceeding at equal rates • Equilibrium can be reached from either direction • Ratio of concentrations will become constant

  3. Equilibrium Constant • Haber Process • N2(g) + 3H2(g) ↔ 2NH3(g) • aA + bB ↔ cC + dD • Ke q = (PC)c(PD)d (PA)a(PB)b • Ke q = [C]c[D]d [A]a[B]b • Equilibrium Expression • Ke q = equilibrium constant

  4. Equilibrium Constant • N2O4(g) ↔ 2NO2(g) • What is the equilibrium expression? • What do you notice about the equilibrium constant?

  5. Practice • Write the equilibrium expression for Ke q for the following reactions: • 2O3(g) ↔ 3O2(g) • 2NO(g) + Cl2(g) ↔ 2NOCl(g) • Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+(aq) • H2(g) + I2(g) ↔ 2HI(g) • Cd+2(aq) + 4Br-(aq) ↔ CdBr42-(aq)

  6. Equilibrium Constant • The magnitude of the constant can tell us what the make up equilibrium mixture is • If Ke q > 1 products predominate • If Ke q < 1 reactants predominate

  7. Practice • The reaction of N2 with O2 to form NO might be considered a means of “fixing” nitrogen. N2(g) + O2(g) ↔ 2NO(g) The value for the equilibrium constant for this reaction at 25°C is Ke q = 1x10- 3 0. Describe the feasibility of the reaction for nitrogen fixation. • The equilibrium constant for the reaction of H2(g) + I2(g) ↔ 2HI(g) varies with temperature as follows: Ke q = 794 at 298K; Ke q = 54 at 700K. Is the formation of HI favored more at high temperatures of low temperatures?

  8. Equilibrium Constant • Because equilibrium can be reached from either direction how we write equilibrium expressions is arbitrary. N2O4(g) ↔ 2NO2(g) • In terms of the forward reaction Keq = (PN O2)2 = 6.46 (@ 100°C) PN 2 O 4 • In terms of the reverse reaction Keq = PN 2 O 4 = 0.155 (@ 100°C) (PN O2)2 • The equilibrium constant expression for a reaction written in one direction is the reciprocal of the one for the reaction written in the reverse direction.

  9. Equilibrium Constant • N2O4(g) ↔ 2NO2(g) • Keq = (PN O2)2 = 6.46 (@ 100°C) PN 2 O 4 • 2N2O4(g) ↔ 4NO2(g) • Keq = (PN O 2)4 (PN 2 O 4)2 • The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to that power

  10. Equilibrium Constant • 2NOBr(g) ↔ 2NO(g) + Br2(g) Ke q= 0.42 • What is the equilibrium expression? • Br2(g) + Cl2(g) ↔ 2BrCl(g) Ke q= 7.2 • What is the equilibrium expression? • The sum of these two equations is 2NOBr(g) + Cl2(g) ↔ 2NO(g) + 2BrCl(g) Ke q= 3.0 • What is the equilibrium expression? • The equilibrium constant for a net equation made up of two or more steps is the product of the equilibrium constants for the individual steps.

  11. Practice • Given the following information, HF(aq) ↔ H+(aq) + F-(aq) Ke q= 6.8x10- 4 H2C2O4(aq) ↔ 2H+(aq) + C2O42 -(aq) Ke q= 3.8x10- 6 determine the value of the rate constant for the following reaction: 2HF(aq) + C2O42 -(aq) ↔ 2F-(aq) + H2C2O4(aq) • Answer: 0.12

  12. Units of Keq • Equilibrium constants have no units (are dimensionless) even though we put in concentrations and pressures. • What we actually put in are ratios of these quantities to a certain reference point. • For concentrations the reference is 1M • For pressures the reference is 1atm • Keq = (PN O 2/Pr e f)2 (PN 2 O 4/Pr e f) • This allows us to put both pressures and concentrations into the same expression since the units “disappear”

  13. Heterogeneous Equilibria • Homogeneous equilibria – all substances are in the same state • Heterogeneous equilibria – substances are in different states PbCl2(s) ↔ Pb+ 2(aq) + 2Cl-(aq) Keq = [Pb2 +][Cl-]2 [PbCl2] • If a pure solid or liquid is involved in a heterogeneous equilibrium its concentration is not included in the expression

  14. Rules for Equilibrium Expressions • Partial pressures of gases are substituted into the equilibrium constant expression. • Molar concentrations of dissolved species are substituted into the equilibrium constant expression. • Pure solids, pure liquids, and solvents are not included in the equilibrium constant expression. • Must still be present at equilibrium • CaCO3(s) ↔ CaO(s) + CO2(g) • Ke q = PC O 2

  15. Practice • Write the equilibrium constant expressions for each of the following reactions: CO2(g) + H2(g) ↔ CO(g) + H2O(l) SnO2(s) + 2CO(g) ↔ Sn(s) + 2CO2(g) Sn(s) + 2H+(aq) ↔ Sn2 +(aq) + H2(g)

  16. Practice • Each of the following mixtures was placed in a closed container and allowed to stand. Which of the mixtures is capable of attaining the equilibrium expressed by this reaction - CaCO3(s) ↔ CaO(s) + CO2(g). a) pure CaCO3 b) CaO and a pressure of CO2 greater than the value of Ke q c) some CaCO3 and a pressure of CO2 greater than the value of Ke q d) CaCO3 and CaO • Answer: a,b,d – yes ; c – no

  17. Calculating Equilibrium Constants • To calculate the equilibrium constants for a reaction we need to know the equilibrium concentration for at least one of the substances. • We will do this using the following procedure • Tabulate the known initial and equilibrium concentrations of all species in the equilibrium constant expression. • For all species for which both the initial and equilibrium concentrations are known, calculate the change in concentration that occurs as the system reaches equilibrium. • Use the stoichiometry of the reaction (that is, use the coefficients in the balanced chemical equation) to calculate the changes for all the other substances in the equilibrium. • From the initial concentrations and the changes in concentration, calculate the equilibrium concentrations. Substitute these concentrations into the equilibrium expression.

  18. Calculating Equilibrium Constants • Enough ammonia is dissolved in 5.00 liters of water at 25°C to produce a solution that is 0.00124M in ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that the concentration of OH- is 4.64x10- 4 M. Calculate Ke q at 25°C for the reaction: NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq) Answer: 1.81x10- 5

  19. Practice • Sulfur trioxide decomposes at high temperatures in a sealed container: 2SO3(g) ↔ 2SO2(g) + O2(g). Initially the vessel is charged at 1000K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate the value of Ke q at 1000K. Answer: 0.338

  20. Predicting the Direction of Reaction • If you are given a set of concentrations or pressures and the equilibrium constant you can determine in which direction a reaction will proceed. • Substitute the given []'s or pressures in to the equilibrium expression. • The result is called the reaction quotient (Q) • If Q = K, the reaction is a equilibrium • If Q < K, the reaction will produce more products • If Q > K, the reaction will produce more reactants

  21. Practice • At 448°C the equilibrium constant, Keq, for the reaction H2(g) + I2(g) ↔ 2HI(g) is 51. Predict how the reaction will proceed to reach equilibrium at this temperature if we start with 2.0x10- 2 mol of HI, 1.0x10- 2 ofH2, and 3.0x10- 2 mol of I2 in a 2.00L container. Answer: The reaction will generate more products.

  22. Practice • At 1000K the value of Keq for the reaction 2SO3(g) ↔ 2SO2(g) + O2(g) is 0.338. Calculate the value of Q, and predict the direction in which the reaction will proceed toward to attain equilibrium if the initial partial pressures of reactants are PSO3 = 0.16 atm; PSO2 = 0.41atm; PO2 = 2.5atm. Answer: Q = 16, the reaction will proceed from right to left

  23. Calculating Equilibrium Concentrations • In a system where we know a few of the equilibrium pressures or concentrations and the equilibrium constant, we can solve for the ones we don't • In the Haber Process, N2(g) + 3H2(g) ↔ 2NH3(g), Ke q = 1.45x10- 5 at 500°C. In an equilibrium mixture of the three gases, the partial pressure of H2 is 0.928atm and that of N2 is 0.432atm. What is the partial pressure of NH3 in this equilibrium mixture? Answer: 1.45x10- 5

  24. Practice • At 500K the reaction PCl5(g) ↔ PCl3(g) + Cl2(g) has a Ke q = 0.497. In an equilibrium mixture at 500K, the partial pressure of PCl5 is 0.860atm and that of PCl3 is 0.350atm. What is the partial pressure of chlorine in the equilibrium mixture? Answer: 1.22atm

  25. Calculating Equilibrium Concentrations • In systems where all we know is the equilibrium constant and the initial concentrations, finding the equilibrium concentration will require a bit more algebra. • A 1.000L flask is filled with 1.000mol of H2 and 2.000mol of I2 at 448°C. The value of the equilibrium constant, Ke q, for the reaction at 448°C is 50.5. What are the partial pressures of H2, I2 and HI in the flask at equilibrium? • Answer: PH 2 = 3.9atm, PI 2 = 63.1atm, PH I = 110.6atm

  26. Practice • For the equilibrium, PCl5(g) ↔ PCl3(g) + Cl2(g), the equilibrium constant, Ke q, has the value 0.497 at 500K. A gas cylinder at 500K is charged with PCl5(g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl5, PCl3, and Cl2 at this temperature? • Answer: PPCL5 = 0.967 atm, PPCl3 = PCl2 = 0.693 atm

  27. Le Châtelier's Principle • If a system at equilibrium is disturbed by an change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance. • There are 3 main ways to change equilibrium • Adding or removing a reactant or product • Changing the pressure • Changing the temperature

  28. Changes in Reactant or Product [ ]'s • If a chemical system is at equilibrium and we add a substance (reactant or product) the reaction will shift so as to reestablish equilibrium by consuming part of the added substance. • Removing a substance will cause the reaction to move in the direction that forms more of that substance.

  29. Changes in Pressure and Volume • At a constant temperature, reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas. • Increasing the volume causes a shift in the direction that produces more gas molecules. • Video • What would happen to equilibrium if a gas that was not involved in the reaction such as Argon was added to the following system? • N2(g) + 3H2(g) ↔ 2NH3(g)

  30. Changes in Temperature • Changes in [ ] and volume shift equilibrium without changing the equilibrium constant. • Changes in temperature do cause the equilibrium constant to change. • Endothermic: reactants + heat ↔ products • Exothermic: reactants ↔ products + heat • When the temperature is increased, it is as if we have added a reactant, or a product, to the system at equilibrium. The equilibrium shifts in the direction that consumes the excess reactant or product, namely heat.

  31. Changes in Temperature • Endothermic • ↑ T causes ↑ in Ke q • Exothermic • ↓ T causes ↓ in Ke q • Co(H2O)62 +(aq) + 4Cl-(aq) ↔ CoCl42 -(aq) + 6H2O(l) ∆H > 0

  32. Practice • Consider the following equilibrium: • N2O4(g) ↔ 2NO2(g) ∆H° = 58.0 kJ • In what direction will the equilibrium shift when each of the following changes is made to a system at equilibrium: • Add N2O4 • Remove NO2 • Increase the total pressure by adding N2(g) • Increase the volume • Decrease the temperature

  33. Effect of Catalysts • A catalyst increases the rate at which an equilibrium is achieved, but does not change the composition of the equilibrium mixture.

  34. Integrative Practice At temperatures near 800°C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form CO and H2: C(s) + H2O(g) ↔ CO(g) + H2(g) The mixture of gases that results is an important industrial fuel called water gas. • At 800°C the equilibrium constant for this reaction is 14.1. What are the equilibrium partial pressures of H2O, CO, and H2 in the equilibrium mixture at this temperature if we start with solid carbon and 0.100 mol of H2O in a 1.00L vessel? • What is the minimum amount of carbon required to achieve equilibrium under these conditions? • What is the total pressure in the vessel at equilibrium? • At 25°C the value of Ke q for this reaction is 1.7x10- 2 1. Is this reaction exothermic or endothermic? • To produce the maximum amount of CO and H2 at equilibrium, should the pressure of the system be increased or decreased?

More Related