1 / 21

Chapter 9: Chemical Equilibrium

Chapter 9: Chemical Equilibrium. The forward and reverse reaction are both taking place at the same rate. Production and Decomposition of Ammonia. Forward Reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g). Reverse Reaction: 2NH 3 (g)  N 2 (g) + 3H 2 (g).

drea
Télécharger la présentation

Chapter 9: 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. Chapter 9: Chemical Equilibrium The forward and reverse reaction are both taking place at the same rate

  2. Production and Decomposition of Ammonia Forward Reaction: N2 (g) + 3H2 (g) 2NH3 (g) Reverse Reaction: 2NH3 (g)  N2 (g) + 3H2 (g) Equilibrium Reaction: N2 (g) + 3H2 (g) 2NH3 (g) Note the double headed arrow! The ammonia is decomposing as fast as it is being made at equilibrium

  3. Equilibrium and the Law of Mass Action 2SO2 (g) + O2 (g)  2SO3 (g) 5 mixtures of different initial compositions of gases were made and allowed to reach equilibrium at 1000K At first, you don’t see a trend in the data…

  4. Equilibrium and the Law of Mass Action No trends, but if you calculate: You get the same value, regardless of initial concentration Note: K is unitless!

  5. The Equilibrium Constant K is the equilibrium constant for the reaction

  6. The Equilibrium Constant At equilibrium, the composition of the reaction mixture can be expressed in terms of an equilibrium constant where: For ideal gases, the concentrations are the partial pressures of the individual gases For solutions, the concentrations are the molar values of the individual atoms/ions/molecules

  7. Examples of K setup aA (g) + bB (g)  cC (g) + dD (g) 

  8. Units and Equilibrium Constants When working with equilibrium Constants, we’ll use the following unit conventions: • Gases: Units are bar • Aqueous Solutions: Unit is Molarity • Solids: The number 1 Solids have a single value (1) because the concentration of a solid doesn’t change.

  9. Thermodynamic Origin of Equilibrium Constants The Free Energy changes as the composition of the reaction mixture changes • All reactions will proceed towards equilibrium (by either forward or reverse reaction) • Gº is the free energy difference b/w the pure products and pure reactants

  10. Thermodynamic Origins of Equilibrium Constants • We can calculate the Free Energy change at any point along the reaction coordinate with the equation aA (g) + bB (g)  cC (g) + dD (g) Gr° is the textbook Free Energy of reaction Gr is the Free Energy of value when the reactants and products are at particular concentrations 

  11. Example: The standard free energy of reaction for: 2SO2 (g) + O2 (g)  2SO3 (g) Is Gr°= -141.74 kJ/mole at 25°C. What is the Gibbs Free Energy of reaction when the partial pressure of each gas is 100.0 bar?

  12. Example: The Standard Gibbs Free Energy of Reaction for N2O4 (g) --> 2NO2 (g) Is Gr° = +4.73 kJ/mole at 298K. What is the value of Gr when the partial pressures are PN2O4 = 0.8 bar and PNO2 = 2.10 bar?

  13. Free Energy of a Reaction at Equilibrium • Q=K at equilibrium • At equilibrium, G=___ • Therefore, G = Grº + RTlnK Grº = -RTlnK (only at equilibrium) • We can use this to compute equilibrium constants from Grº values

  14. K and the Extent of Reactions • When K is very large, the reaction favors the products • When K is very small, the reaction favors the reactants • When K=1, the reaction is neither reactant nor product favored (Equilibrium)

  15. The Direction of Reaction How can we tell if a reaction will continue towards the products or back towards the reactants at a given point along the reaction coordinate? Q = Reaction quotient used at any point in the coordinate K = Equilibrium constant When Q<K, G is negative (product favored) When Q=K, G = 0 When Q>K, G is positive (reactant favored)

  16. Equilibrium Calculations Toolbox 9.1: Know it. Love it. Use it. 

  17. Example: Under certain conditions, nitrogen and oxygen react to form dinitrogen oxide, N2O. Suppose that 0.482 moles of N2 and 0.933 moles of O2 are transferred to a reaction vessel of volume 10.0L and allowed to form N2O @ 800K. At this temperature, K=3.2x10-28 for the reaction: 2N2 (g) + O2 (g)  2N2O (g) What are the partial pressures of the gases at equilibrium?

  18. Example: Chlorine and fluorine react at 2500K to produce ClF and reach the equilibrium: Cl2 + F2 2ClF With an equilibrium constant value of 20. If a gaseous mixture of 0.2 bar Cl2, 0.1 bar F2 and 0.1 bar ClF is allowed to reach equilibrium, what is the partial pressure of ClF in the mixture?

  19. LeChatelier’s Principle When the equilibrium composition is perturbed by adding or removing a reactant of product, the reaction tends to proceed in the direction that brings Q closer to that of K. 

  20. Consider the Equilibrium Reaction: 4NH3 (g) + 3O2 (g)  2N2 (g) + 6H2O (g) What would result from the: Addition of N2 Removal of NH3 Removal of H2O

  21. Effects of the Environment on Equilibria • Compressing a Gas Phase Reaction • The reaction shifts so as to decrease the pressure • Decrease the number of gas molecules • Changing the Temperature of a Reaction • For exothermic reactions, lowering the temperature causes a shift towards the products • For endothermic reactions, increasing the temperature causes a shift towards the products

More Related