1 / 12

Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9

Learn how to predict if a reaction is at equilibrium, calculate trial Keq, and shift equilibrium. Understand the factors influencing Keq and equilibrium shifts. Practice solving equilibrium problems with detailed examples.

nelsonl
Download Presentation

Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9

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. Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9

  2. The Keqis a constant- a number that does not change Changing the volume, pressure, or any concentration, does notchange the Keq. Onlytemperature changes the Keq Increasing a [Reactant] shifts right and maintains the Keq Increasing the temperature of an endothermic equilibrium shifts right andincreases the Keq

  3. Ktrial How can you tell if a system is in equilibrium or not? Calculatea trial Keq. Use initial concentrations in the equilibrium expression- evaluate.

  4. How can you tell if a system is in equilibrium or not? Calculatea trial Keq. Put initial concentrations into the equilibrium expression and evaluate. If Kt < Keq If Kt = Keq If Kt > Keq Shifts right equilibrium Shifts left Kt Kt Kt Keq 5 20 35

  5. 1. 10.0 moles of NH3, 15.0 moles of N2, and 10.0 moles of H2 are put in a 5.0 L container. Is the system in equilibrium and how will it shift if it is not? 2NH3(g)⇄ N2(g) + 3H2(g) Keq = 10 2.0 M 3.0 M 2.0 M [N2][H2]3 Kt = = (3)(2)3 = 6 (2)2 [NH3]2 Not in equilibrium Kt < Keq Shifts right!

  6. 2. 4.56 x 10-5 moles of NH3, 5.62 x 10-4 moles of N2, and 2.66 x 10-2 moles of H2 are put in a 500.0 mL container. Is the system in equilibrium and how will it shift if it is not? 2NH3(g)⇄ N2(g) + 3H2(g) Keq = 10 9.12 x 10-5 M 1.124 x 10-5 M 5.32 x 10-2 M [N2][H2]3 Kt = = (1.124 x 10-3 )(5.32 x 10-2 )3 = 20.3 (9.12 x 10-5)2 [NH3]2 Not in equilibrium Kt > Keq Shifts left!

  7. 4. If 4.00 moles of CO, 4.00 moles H2O, 6.00 moles CO2, and 6.00 moles H2 are placed in a 2.00 L container at 670 oC. CO(g) + H2O(g) ⇄ CO2(g) + H2(g) Keq = 1.0 2.00 M 2.00 M 3.00 M 3.00 M Is the system at equilibrium? +x +x -x -x (3)(3) Kt = = 2.25 (2)(2) 2.00 + x 2.00 + x 3.00 - x 3.00 - x Not in equilibrium Shifts left! [CO2][O2] Keq = [CO][H2O] Calculate all equilibrium concentrations.

  8. (3 - x)2 = 1.0 (2 + x)2 3 - x = 1.0 2 + x 3 - x = 2 + x 1 = 2x x = 0.50 M [CO2] = [H2] = 3.00 - 0.50 = 2.50 M [CO] = [H2O] = 2.00 + 0.50 = 2.50 M

  9. Size of the Keq

  10. Big Keq products Keq = reactants Keq = 10

  11. Little Keq products Keq = reactants Keq = 0.1 Note that the keq cannot be a negative number!

  12. Keq= 1 products Keq = reactants

More Related