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CHEMICAL EQUILIBRIUM Cato Maximilian Guldberg and his brother-in-law Peter Waage developed the Law of Mass Action.

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Presentation on theme: "CHEMICAL EQUILIBRIUM Cato Maximilian Guldberg and his brother-in-law Peter Waage developed the Law of Mass Action."— Presentation transcript:

1 CHEMICAL EQUILIBRIUM Cato Maximilian Guldberg and his brother-in-law Peter Waage developed the Law of Mass Action

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) 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 Example: Writing an Equilibrium Expression Write the equilibrium expression for the reaction:

10 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)

11 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 )

12 Example: Calculating the Value of K The following equilibrium concentrations were observed for the Haber process at C. a. Calculate the value of K at C for this reaction at C for this reaction b. Calculate the value of the equilibrium constant at C for reaction:

13 Example: Calculating the Value of K The following equilibrium concentrations were observed for the Haber process at C. c. Calculate the value of the equilibrium constant at C for reaction given by the equation:

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

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

16 Example: Equilibrium Expressions Involving Pressure The reaction for the formation of nitrosyl chloride: Was studied at 25 0 C. The pressures at equilibrium were found to be: a.Calculate the value of K p for this reaction at 25 0 C. b. Calculate the value of K for this reaction at 25 0 C.

17 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

18 Equilibrium Expressions for Heterogeneous Equilibria Write the expressions for K and Kp for the following process: ◦ a. Solid phosphorus pentachloride decomposes to liquid phosphorus trichloride & chlorine gas. ◦ b. Deep blue solid copper (II) sulfate pentahydrate is heated to drive off water vapor to from solid white copper (II) sulfate.

19 The Reaction Quotient Q K 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

20 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

21 Using the Reaction Quotient For the synthesis of ammonia at C, the equilibrium constant is 6.0x10 -2 Predict the direction in which the system will shift to reach equilibrium in each of the following cases: a. [NH 3 ] 0 = 1.0x10 -3 M; [N 2 ] 0 = 1.0x10 -5 M; [H 2 ] 0 = 2.0x10 -3 M b. [NH 3 ] 0 = 2.0x10 -4 M; [N 2 ] 0 = 1.5x10 -5 M; [H 2 ] 0 = 3.54x10 -1 M c. [NH 3 ] 0 = 1.0x10 -4 M; [N 2 ] 0 = 5.0 M; [H 2 ] 0 = 1.0x10 -2 M

22 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 4 chapters!

23 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:

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

25 Solving for Equilibrium Concentration 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) x = 4

26 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) x = 4

27 Calculating Equilibrium Pressure Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar landing for the NASA Apollo missions. In the gas phase it decomposes to gaseous nitrogen dioxide: Consider an experiment in which gaseous was placed in a flask & allowed to come to equilibrium at a temperature where Kp= At equilibrium, the pressure of was found to be at 2.71 atm. Calculate the equilibrium pressure of.

28 Calculating Equilibrium Concentrations Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen & fluorine has an equilibrium constant 0f 1.15x10 2 at a certain temperature. In an experiment, mol of each component was added to a L flask. Calculate the equilibrium concentrations of all species.

29 Solving Equilibrium Problems 1. Write a balanced equation. 2. Write the equilibrium expression using the law of mass action. 3. If given initial concentrations, use Q (reaction quotient) to predict the direction the reaction will proceed. 4. ICE the problem. ◦ I – initial concentration ◦ C – Change in concentration of reactants & products (remember stoichiometry here!) ◦ E – Equilibrium concentrations 5. Substitute the equilibrium concentrations into the equlibrium expression and solve for the unknown (x). 6. Check your calculated equilibrium [ ] by making sure they give the correct value of K.

30 Calculating Equilibrium Pressures Assume that the reaction for the formation of gaseous hydrogen iodide from hydrogen gas & iodine vapor has an equilibrium constant of 1.00 x 10 2 at a certain temperature. Suppose HI at x atm, H 2 at x atm, & I 2 at x atm are mixed in a L flask. Calculate the equilibrium concentrations of all species.

31 Calculating Equilibrium Pressures K=1.00 x 10 2 HI=5.000 x atm, H 2 =1.000 x atm, I 2 =5.000 x atm

32 Treating systems that have small equilibrium constants Treating systems that have small equilibrium constants We can make some approximations when the equilibrium constant is much smaller than the concentrations of the reactants & products. You must check to see if your approximation is valid using the “5% rule” You must note your approximation in the problem As a general rule of thumb – if the equilibrium constant is smaller than the concentrations by a magnitude of 10 3 or more – an approximation can be made.

33 Calculating Equilibrium Concentrations with small equilibrium constants Gaseous NOCl decomposes to form the gases NO & Cl 2. At 35 0 C K = 1.6x In an experiment in which 1.0 mol of NOCl is placed in an empty 2.0 L flask, what the equilibrium concentrations of all species?

34 Le Chatelier’s Principle

35 When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress and restore a state of equilibrium. Henry Le Chatelier

36 When you take something away from a system at equilibrium, the system shifts in such a way as to replace some of what you’ve taken away. Le Chatelier Translated: When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you’ve added.

37 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

38 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

39 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

40 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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