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Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9

Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9. The Keq is a constant - a number that does not change. Changing the volume , pressure , or any concentration , does not change the Keq. Only temperature changes the Keq.

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Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9

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  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

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