Presentation is loading. Please wait.

Presentation is loading. Please wait.

Reversible Reactions Reactions can be reversed depending on energy flow. For example if the battery in a car is dead, you can jump start it.

Similar presentations


Presentation on theme: "Reversible Reactions Reactions can be reversed depending on energy flow. For example if the battery in a car is dead, you can jump start it."— Presentation transcript:

1

2

3 Reversible Reactions Reactions can be reversed depending on energy flow. For example if the battery in a car is dead, you can jump start it.

4 We’ve already used the phrase “equilibrium” when talking about reactions. In principle, every chemical reaction is reversible... capable of moving in the forward or backward direction. 2 H 2 + O 2 2 H 2 O Some reactions are easily reversible... Some not so easy...

5 Equilibrium: the extent of a reaction In stoichiometry we talk about theoretical yields, and the many reasons actual yields may be lower. Another critical reason actual yields may be lower is the reversibility of chemical reactions: some reactions may produce only 70% of the product you may calculate they ought to produce. Equilibrium looks at the extent of a chemical reaction.

6

7

8 Equilibrium There is no net change in the amount of reactants and products from a chemical reaction. Products and reactants form at the same rate. Reactions can be pushed in either direction by adding or removing heat, reactants or products.

9 If Q < K eq, shift to right (toward product) If Q > K eq, shift to left (toward reactant)

10 The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

11 The Concept of Equilibrium As the reaction progresses –[A] decreases to a constant, –[B] increases from zero to a constant. –When [A] and [B] are constant, equilibrium is achieved.

12 A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

13 Dynamic Equilibrium The reaction can go both ways. You make something with two reactants and a product and you can reverse it to get what you started with. O 2 bonds with a free Oxygen to produce O 3 or ozone, O 3 breaks down to O 2 and a free O If the reaction can only go one way such as burning paper it is not dynamic.

14 Equilibrium is a case where the reaction does not go to completion, but wavers back and forth. Imagine a system like that to the left…people want to move where there is lots of land and food. Once there is an even number of people in each area they will move in and out of the areas at the same rate.

15 The Equilibrium Constant No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium. For a general reaction the equilibrium constant expression is where K eq is the equilibrium constant.

16 In an equilibrium reaction, the concentration of the products goes on the top and the concentrations of the reactants go on the bottom. Remember to include the coefficients.

17 What is the reaction here? 2 H 2 + CO  CH 3 OH

18 The Equilibrium Expression Write the equilibrium expression for the following reaction:

19 When K is large, the reaction will shift to the products (make more products).

20 When K is small, the reaction will favor the reactants (won’t react much).

21 The Equilibrium Constant The Equilibrium Constant in Terms of Pressure If K P is the equilibrium constant for reactions involving gases, we can write: K P is based on partial pressures measured in atmospheres.

22 The Equilibrium Constant Heterogeneous Equilibria When all reactants and products are in one phase, the equilibrium is homogeneous. If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. Consider: –experimentally, the amount of CO 2 does not seem to depend on the amounts of CaO and CaCO 3. Why?

23 The Equilibrium Constant Heterogeneous Equilibria Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant. (You can’t find the concentration of something that isn’t a solution!) We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions. The amount of CO 2 formed will not depend greatly on the amounts of CaO and CaCO 3 present. K eq = [CO 2 ]

24 Calculating Equilibrium Constants Steps to Solving Problems: 1.Write an equilibrium expression for the balanced reaction. 2.Write an ICE table. Fill in the given amounts. 3.Use stoichiometry (mole ratios) on the change in concentration line. 4.Deduce the equilibrium concentrations of all species. Usually, the initial concentration of products is zero. (This is not always the case.)

25 Applications of Equilibrium Constants Predicting the Direction of Reaction We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium as where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium.

26 The Reaction Quotient (Q) To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

27 Applications of Equilibrium Constants Predicting the Direction of Reaction If Q > K then the reverse reaction must occur to reach equilibrium (go left) If Q < K then the forward reaction must occur to reach equilibrium (go right)

28 If Q = K, the system is at equilibrium.

29 Example Problem: Calculate Concentration Note the moles into a L vessel stuff... calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = M 2 HI H 2 + I 2 Initial: Change: Equil: M00 -2x+x+x xxx What we are asked for here is the equilibrium concentration of H otherwise known as x. So, we need to solve this beast for x.

30 Example Problem: Calculate Concentration And yes, it’s a quadratic equation. Doing a bit of rearranging: x = or – Since we are using this to model a real, physical system, we reject the negative root. The [H 2 ] at equil. is M.

31 Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: x +2x 0.20-x 2x With an equilibrium constant that small, whatever x is, it’s near zero, and 0.20 minus zero is 0.20 (like a million dollars minus a nickel is still a million dollars) – x is the same as 0.20 x = 3.83 x M More than 3 orders of mag. between these numbers. The simplification will work here.

32 Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: x +2x 0.20-x 2x These are too close to each other x will not be trivially close to 0.20 here. Looks like this one has to proceed through the quadratic...

33 Le Chatlier’s Principle: Changing the environment of the reaction will shift the reaction in the direction favored by the environment. There are 3 ways to shift an equilibrium reaction: 1.Change the concentration of the reactants or products, 2.Change the temperature of the system. 3.Change the pressure of the system.

34 Disturbing an equilibrium will make the reaction start up again to regain its equilibrium. Reactants, products, or heat can be added or removed to disturb and equilibrium. CaCO 3  CaO + CO 2 If you remove CO 2, more CaCO 3 will break down to restore the equilibrium. Chemical engineers use this principle to produce huge amounts of different product.

35 Adding products to the system will cause the products to change back into the reactants. Changing the concentration of the products will shift the reaction to the left, making more reactants.

36 Adding reactants will cause the reaction to continue and make more products. Changing the concentration of the reactants will shift the reaction to the right (make more products).

37 Adding pressure will cause the reaction to go to the side with fewer molecules. In the case below, 4 reactant molecules make up 2 product molecules. Adding pressure to this system will cause the reaction to continue making products. In the case of sugar (C 6 H 12 O 6 ) and 6 O 2 reacting and changing into 6 CO 2 and 6 H 2 O, there are only 7 reactant molecules and 12 product molecules. Adding pressure to this system would cause the reactants to be formed, not the products. This happens cuz there is more space for fewer molecules. Rarely does the size of the molecule really matter.

38 As pressure is lessened, the reaction will reverse because there is more space. The 4 reactant molecules will be favored over the 2 product molecules, because the container wants to be full. In the case of sugar (C 6 H 12 O 6 ) and 6 O 2 reacting and changing into 6 CO 2 and 6 H 2 O, as pressure is released, the reaction will make more products, because there is enough space for them.

39 Temperature can change the direction of the reaction. If the reaction is exothermic, as in the case above, and heat is added to the system, the reaction will make more reactants than products. If the reaction is exothermic and heat is removed (the system is made cooler) as below, the reaction will continue to make products.

40

41 Solubility Product Principle Another equilibrium situation is slightly soluble products K sp is the solubility product constant K sp can be found on a chart at a specific temperature Since the product is solid on the left side, only the products (ions) are involved in the K sp expression

42 Solubility Equilibria Solubility Product Constant K sp for saturated solutions at equilibrium

43 Solubility Product (K sp ) = [products] x /[reactants] y but..... reactants are in solid form, so K sp =[products] x i.e. A 2 B 3 (s)  2A B 2– K sp =[A 3+ ] 2 [B 2– ] 3 Given: AgBr(s)  Ag + + Br – In a saturated o C, the [Ag + ] = [Br – ]= 5.7 x 10 –7 M. Determine the K sp value.

44 Problem: A saturated solution of silver chromate was to found contain g/L of Ag 2 CrO 4. Find K sp Eq. Expression: Ag 2 CrO 4 (s)  2Ag + + CrO 4 2– K sp = [Ag + ] 2 [CrO 4 2– ] So we must find the concentrations of each ion and then solve for K sp.

45 Problem: A saturated solution of silver chromate was to contain g/L of Ag 2 CrO 4. Find K sp Eq. Expression: Ag 2 CrO 4 (s)  2Ag + + CrO 4 2– K sp = [Ag + ] 2 [CrO 4 2– ] Ag + : g Ag 2 CrO 4 L 1.33 x 10 –4 CrO 4 –2 : 0.022g Ag 2 CrO 4 L 6.63 x 10 –5

46 Problems working from K sp values. Given: K sp for MgF 2 is 6.4 x o C Find: solubility in mol/L and in g/L MgF 2 (s)  Mg F – K sp = [Mg 2+ ][F – ] 2 I. C. E. N/A 0 0 N/A +x +2x K sp = [x][2x] 2 = 4x x 10 –9 = 4x 3 now for g/L: 7.3 x 10 –2

47 Solubility Product Principle

48 Example: Find the concentration of ions present in calcium fluoride (in water) and the molar solubility. CaF 2 (s) --> Ca F - K sp = [Ca +2 ] [F - ] 2 = 2 X If x = [Ca +2 ], then [F - ] = 2x [x] [2x] 2 = 2 X x 3 = 2 X x 3 = 5 X x = 3.68 X [Ca +2 ] = x = 3.68 X [F - ] = 2x = 7.37 X Solubility of CaF 2 = 3.68 X 10 -4

49 Practice: Write the equilibrium expression and calculate the equilibrium constant for each of the following reactions. 1.Zn + 2HCl  ZnCl 2 + H 2 Concentrations: Zn = 4M, HCl = 8M, ZnCl 2 = 3M, H 2 = 10M 2.C 3 H 8 + 5O 2  3CO 2 + 4H 2 O Concentrations: H 2 O = 7M, O 2 = 5M, C 3 H 8 = 1.5M, CO 2 = 2.6M 3.HCl + NaOH  NaCl + H 2 O Concentrations: H 2 O = 4.5M, HCl = 10.10M, NaCl = 0.5M, NaOH = 0.99M 4.2NO 2 + 2H 2  N 2 + 2H 2 O Concentrations: NO 2 = 0.56M, H 2 = 1.01M, N 2 = 0.05M, H 2 O = 0.9M.

50 Practice: Given the information below, write the chemical equilibrium reaction and figure out the chemical equilibrium constant. Will each reaction favor the reactants or products? 1.5M of Co reacts with 3M of Cl 2 to form 8M of CoCl M H 2 O and 0.5M NaCl are formed when 8M HCl and 4M of Na 2 O are reacted. 3.3M of K 2 Cr 2 O 7 and 4M of HCl react to form 1M KCl, 2M CrCl 2, 3M H 2 O, and 4M Cl 2.


Download ppt "Reversible Reactions Reactions can be reversed depending on energy flow. For example if the battery in a car is dead, you can jump start it."

Similar presentations


Ads by Google