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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? Why was it so important to measure reaction rate at the start of the reaction (method of initial rates?)

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

5 Product Favored Equilibrium Large values for K signify the reaction is “product favored” k > 1 When equilibrium is achieved, most reactant has been converted to product

6 Reactant Favored Equilibrium Small values for K signify the reaction is “reactant favored” k < 1 When equilibrium is achieved, very little reactant has been converted to product

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

8 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)

9 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 )

10 Equilibrium Expressions Involving Pressure For the gas phase reaction: 3H 2 (g) + N 2 (g)  2NH 3 (g) K and Kp are NOT interchangeable!

11 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

12 CAN you….. Write an Keq expression?Write an Keq expression? Tell how Keq changes as the stoichiometry changes?Tell how Keq changes as the stoichiometry changes? Convert Keq in terms of Kc vs Kp?Convert Keq in terms of Kc vs Kp? Explain what K means for the rxn?Explain what K means for the rxn?

13 The Reaction Quotient When the reaction is NOT at equilibrium for a given time, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action. jA + kB  lC + mD Useful in determining what will happen under special conditions.

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? “RICE” Here, we learn about “RICE” – the most important problem solving technique in the second semester. You will use it for the next 4 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 “RICE” the problem, beginning with the R for balanced reaction x +x 8-x6-x x x Reaction

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 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.

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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