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

1 CHEMICAL EQUILIBRIUM

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) + O2(g) Arrows going both directions ( ) indicates equilibrium in a chemical equation

3 2NO 2 (g) 2NO(g) + O 2 (g) Remember this from Chapter 12? Why was it so important to measure reaction rate at the start of the reaction (method of initial rates?)

4 2NO 2 (g) 2NO(g) + O 2 (g)

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

7 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

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

9 Conclusions about Equilibrium Expressions The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse 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)

10 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. 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 )

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

12 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

13 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

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

15 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 molecules of H 2 O and 6 molecules of CO. How many molecules of H 2 O, CO, H 2, and CO 2 are present at equilibrium? ICE Here, we learn about ICE – the most important problem solving technique in the second semester. You will use it for the next several chapters!

16 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:

17 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

18 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)

19 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

20 LeChateliers 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.

21 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

22 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

23 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

24 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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