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Equilibrium

Equilibrium. Equilibrium. Systems at equilibrium are still dynamic (changing). However, no NET CHANGE will be observed. A system is at equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction. Changing concentration. 2 H 2 O  H 3 O + + OH -

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Equilibrium

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  1. Equilibrium

  2. Equilibrium • Systems at equilibrium are still dynamic (changing). However, no NET CHANGE will be observed. • A system is at equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction.

  3. Changing concentration • 2 H2OH3O+ + OH- • If I add more water • It will force the reaction to the right • Which means more hydronium and hydroxide will be produced • This is dilution (making the ratio of hydronium/hydroxide closer)

  4. Equilibrium • Add water • 2 H2OH3O+ + OH- • Stress + X 0 0 • Shift -2y +y +y • Final +X - 2y + y +y • Since the stress was added to the left we must take from the left and give to the right to relieve the stress • *where X is the amount of H2O added and larger than 2y

  5. That means… • Add water • 2 H2OH3O+ + OH- • Final + X - 2y + y +y • We increased water because X is always larger than y (with any coefficent). • We increased H3O+ because +y is an increase • We increased OH- because +y is an increase

  6. The only equilibrium calculation • That you will have to do with numbers is: • [OH-] [H3O+ ] = Kw • Kw is the equilibrium constant for water, it equals 1 x 10-14 M • We have already used the equation

  7. More Le Châtelier’s • If I add an acid to the equilibrium… • 2 H2OH3O+ + OH- • Stress 0 +X 0 • Shift +2y -y -y • Final + 2y +X- y -y •    • *Where X is larger than 2y • so adding acid will decrease the [OH-], only slightly increase the[H3O+ ], and increase water.

  8. 2 H2OH3O+ + OH- • If I remove hydroxide from the solution… • 2 H2OH3O+ + OH- • Stress 0 0 -X • Shift -2y +y +y • Final - 2y + y -X +y •    • *Where X is larger than 2y • So removing hydroxide increases [H3O+], only slightly decreases [OH-], and decrease the water

  9. Different equation • Adding hydrogen to the equilibrium • 2 NH3 3 H2 + N2 • Stress 0 +X 0 • Change +2y -3y -y • Final +2y +X-3y -y •    • *where X is larger than 3y • Increases the amount of NH3 decreases the amount of N2 and only slightly increases H2

  10. With heat • If I cool the following equilibrium • Heat+ Co2+ + 4 Cl- CoCl42- • stress -x 0 0 0 • Shift +y +4y -y • Final +y +4y -y •    • So cooling the solution will cause more Co2+ & Cl- and less CoCl42- to form

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