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EQUILIBRIUM CHEMICAL. Chemical Equilibrium Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants Chemical.

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Presentation on theme: "EQUILIBRIUM CHEMICAL. Chemical Equilibrium Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants Chemical."— Presentation transcript:

1 EQUILIBRIUM CHEMICAL

2 Chemical Equilibrium Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants Chemical Equilibrium: When the rate of the forward reaction equals the rate of the reverse reaction and the concentration of products and reactants remains unchanged 2HgO(s)  2Hg(l) + O 2 (g) Arrows going both directions (  ) indicates equilibrium in a chemical equation

3 2NO 2 (g)  2NO(g) + O 2 (g)

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5 Law of Mass Action For the reaction For the reaction: Where K is the equilibrium constant, and is unitless jA + kB  lC + mD

6 Writing an Equilibrium Expression 2NO 2 (g)  2NO(g) + O 2 (g) K = ??? Write the equilibrium expression for the reaction:

7 Product Favored Equilibrium Large values for K signify the reaction is “product favored” When equilibrium is achieved, most reactant has been converted to product

8 Reactant Favored Equilibrium Small values for K signify the reaction is “reactant favored” When equilibrium is achieved, very little reactant has been converted to product

9 Equilibrium Expressions Involving Pressure For the gas phase reaction: 3H 2 (g) + N 2 (g)  2NH 3 (g)

10 Do Problems 21 a&b and 22 c&d 21 22

11 Do Problem 32

12 Conclusions about Equilibrium Expressions  The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse 2NO 2 (g)  2NO(g) + O 2 (g ) 2NO(g) + O 2 (g)  2NO 2 (g)

13 Conclusions about Equilibrium Expressions  When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power. 2NO 2 (g)  2NO(g) + O 2 (g ) NO 2 (g)  NO(g) + ½O 2 (g )

14 Do Problem 23

15 Heterogeneous Equilibria  The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present Write the equilibrium expression for the reaction: PCl 5 (s)  PCl 3 (l) + Cl 2 (g) Pure solid Pure liquid

16 Do Problem 34

17 Do Problem 36

18 The Reaction Quotient For some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action. jA + kB  lC + mD

19 Significance of the Reaction Quotient  If K = Q, the system is at equilibrium  If K < Q, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved  If K > Q, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

20 Do Problem 39

21 Have fun doing your HW problems: 21, 22, 23, 33, 35, 37.

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23 Do Problem 43

24 Solving for Equilibrium Concentration Consider this reaction at some temperature: H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) K = 2.0 Assume you start with 8 mol/L of H 2 O and 6 mol/L of CO. Wha is concentration of H 2 O, CO, H 2, and CO 2 at equilibrium? “ICE” Here, we learn about “ICE” – the most important problem solving technique.

25 Solving for Equilibrium Concentration H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) K = 2.0 Step #1: We write the law of mass action for the reaction:

26 Solving for Equilibrium Concentration H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) I Initial: Change: Equilibrium: Step #2: We “ICE” the problem, beginning with the Initial concentrations 8600 -x +x 8-x6-x x x

27 Solving for Equilibrium Concentration Equilibrium: 8-x6-xxx Step #3: We plug equilibrium concentrations into our equilibrium expression, and solve for x H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g)

28 Solving for Equilibrium Concentration Step #4: Substitute x into our equilibrium concentrations to find the actual concentrations H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) Equilibrium: 8-x6-xxx 8-4=46-4=244

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30 Do Problem 51

31 47.

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38 LeChatelier

39 LeChatelier’s Principle When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress. Translated: The system undergoes a temporary shift in order to restore equilibrium.

40 LeChatelier Example #1 A closed container of ice and water is at equilibrium. Then, the temperature is raised. Ice + Energy  Water The system temporarily shifts to the _______ to restore equilibrium. right

41 LeChatelier Example #2 A closed container of N 2 O 4 and NO 2 is at equilibrium. NO 2 is added to the container. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium. left

42 LeChatelier Example #3 A closed container of water and its vapor is at equilibrium. Vapor is removed from the system. water + Energy  vapor The system temporarily shifts to the _______ to restore equilibrium. right

43 LeChatelier Example #4 A closed container of N 2 O 4 and NO 2 is at equilibrium. The pressure is increased. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation. left


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