1 / 21

Equilibrium

Learn about the concept of equilibrium in reversible chemical reactions and how the rates of the forward and reverse reactions balance each other. Discover how to calculate equilibrium concentrations using Keq.

eleonore
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. equilibrium • Many chemical reactions are reversible: • Reactants → Products or Reactants ← Products Reactants form ProductsProducts form Reactants • For example, under certain conditions, one mole of the colourlessgas N2O4 will decompose to form two moles of brown NO2 gas: • N2O42 NO2 • Under other conditions, you can take 2 moles of brown NO2 gas and change it into one mole of N2O4 gas: N2O42 NO2

  3. equilibrium • In other words, this reaction is as written can go forward or in reverse depending on the conditions If we were to put some N2O4 in a flask, the N2O4 molecules would collide with each other some of them would break apart to form NO2 This is called a FORWARD reaction N2O4 2 NO2

  4. equilibrium • Once this has happened for awhile, there is a build up of NO2 molecules in the same flask Once in awhile, two NO2 molecules will collide with each other and join to form a molecule of N2O4 ! This is called a REVERSE reaction: N2O4 2 NO2

  5. Equilibrium • 2 things to notice: • 1. as long as N2O4 present, the forward reaction will keep on happening • 2. as long as there is NO2 present, the reverse reaction will keep on happening! • At one particular time a molecule of N2O4 might be breaking up, and at the same time two molecules of NO2 might be joining to form another molecule of N2O4

  6. Equilibrium • In any reversible reaction, the forward reaction and the reverse reaction are going on at the same time N2O4 2 NO2 The double arrow means that both the forward and reverse reaction are happening at the same time.

  7. Rate of reaction • Rate of the forward reaction Rate will slow down Rate of forward reaction will be fast at first

  8. Rate of reaction • Rate of reverse reaction

  9. Rate of reaction

  10. Rate of reaction

  11. Rate of reaction • The rate of the forward reaction = the rate of the reverse reaction • N02 is being used up at the same rate that it is being formed • N2O4 2 NO2 • Rate of reaction does not change because both N202 and NO2 are constant

  12. Equilibrium • When this happens we say the system is at EQUILIBRIUM • We have reached a state of DYNAMIC EQUILIBRIUM

  13. Key points • The reaction has not stopped • The forward and reverse reactions are still taking place but their rates are equal so there are no changes in concentrations of the reactants and products • It is happening at a microscopic level so we don’t see the individual molecules reacting (macroscopic)

  14. Key points • All observable properties are constant • include the reactants, products, total pressure, temperature • If no changes were made to the condition a system would stay at equilibrium forever

  15. Key points • Temperature must remain constant • A system at equilibrium is a closed system • Equilibrium can be approaches form the left (reactants) or from the right (products) https://www.youtube.com/watch?v=dUMmoPdwBy4 https://www.youtube.com/watch?v=wlD_ImYQAgQ

  16. Calculations • Equilibrium calculations • We can figure out what the concentrations are for the products and reactants when they are at equilibrium • We use: • Keq • It tells us the ratio of products: reactants when the reaction is at equilibrium

  17. Calculations • to figure out the concentration of the products and reactants at equilibrium we use: • Keq = [products] • [reactants] • A + B C + D • Keq = [C] [D] • [A] [B]

  18. Calculations • Given the equilibrium system: • PCl5(g) PCl3(g) + Cl2(g) • The system is analyzed at a certain temperature and the equilibrium concentrations are as follows: • [PCl5] = 0.32 M, [PCl3] = 0.40 M and the [Cl2] = 0.40 M. • Calculate the Keq for this reaction at the temperature this was carried out.

  19. Calculations • SOLUTION: • Step 1 - Use the balanced equation to write the Keq expression: • PCl5(g) PCl3(g) + Cl2(g) • Keq = [PCl3] [Cl2] • [PCl5] • Step 2 - "Plug in" the values for the equilibrium • Keq = [0.4] [0.4] • [0.32] • Step 3 - Calculate the value of Keq : • Keq = 0.5

  20. Calculations • Given the equilibrium system: • 3 PCl5(g) 2PCl3(g) + Cl2(g) • [PCl5] = 0.32 M, [PCl3] = 0.40 M and the [Cl2] = 0.40 M. • We write it as: • Keq = [PCl3]2 [Cl2] • [PCl5]3 • Raise to the power of the coefficient • Keq = [0.4]2 [0.4] • [0.32]3

  21. Calculations • At 200°C, the Keq for the reaction: N2(g) + 3H2(g) 2NH3(g) is 625 • If the [N2] = 0.030 M, and the [NH3] = 0.12 M, at equilibrium, calculate the equilibrium [H2]. • Solution: All concentrations given are at equilibrium so: Write out the Keq expression: • Keq = [NH3]2 • [N2] [H2]3 • 625 = [0.12]2 • [0.30] [H2]3 • (625) [H2] 3 (0.030) = (0.12) 2

More Related