1301 Dynamic Equilibrium, Keq,

Transcription

1 1301 Dynamic Equilibrium, Keq, and the Mass Action Expression The Equilibrium Process Dr. Fred Omega Garces Chemistry 111 Miramar College 1 Equilibrium

2 Concept of Equilibrium & Mass Action Expression Extent of a Chemical Reaction Many reactions do not convert 100% of reactants to products. There is often a point in a reaction when the products will back react to form reactants. The extent of the reaction i.e., 20% or 80%, can be determined by measuring the concentration of each component in solution. In general the extent of the reaction is a function of: Temperature, Concentration and degree of organization which is monitored by some constant value called the - The equilibrium constant (K eq ). 2 Equilibrium

3 (Dynamic) Equilibrium Chemical Equilibrium is a dynamic state in which the rates of the forward and the reverse reaction are equal. A teeter-totter is an analogy of a system in balance. The left is balanced with the right. In a chemical reaction this means that the Left (reactant) is changing at the same rate as the right (products). Reactant Product 3 Equilibrium

4 How Equilibrium is achieved Consider the reaction A B Forward A B rate = k f [A] Reverse B A rate = k r [B] Overall A D B rate forward (k f [A]) = rate reverse (k r [B]) Example: N 2 O 4 + E D 2 NO 2 Effect of temperature on the N 2 O 4 /NO 2 equilibrium. The tubes contain a mixture of NO 2 and N 2 O 4. As predicted by Le Chatelier s principle, the equilibrium favors colorless N 2 O 4 at lower temperatures, but shifts to the darker brown NO 2 at higher temperature. 4 Equilibrium

5 Equilibrium: Mass Action Expression When equilibrium is establish A D B That is rate forward = rate reverse or k f [A] = k r [B] Rearranging this equation can be rearrange to k f /k b = [B] /[A] The mass action expression defines the equilibrium constant (K eq ): K eq = [B] / [A] For any general chemical process at equilb. aa + bb D pp + qq A mass action expression can be written: K eq = [ P ] p [ Q ] q [ A ] a [ B ] b This is also referred to as the Law of Mass Action 5 Equilibrium

6 Heterogeneous Systems (1) Substances are in different phases at equilibrium i.e., solid and aqueous. Which solid is more concentrated? 100 g (1 cup) 200 g (2 cup) Concentration of a solid (and pure liquid) is always a constant; The concentration does not change! 6 Equilibrium

7 Heterogeneous Systems Consider the following reaction: CaCO 3 (s) D CaO (s) + CO 2 (g) K eq = [ CaO] CO 2 [ ] [ CaCO ] 3 but, [ CaO] = constant (solid) and [ CaCO ] 3 = constant (solid) K eq = [ Constant ] 1 CO 2 [ ] [ Constant ] 2 K eq = K p = [ CO ] 2 7 Equilibrium

8 Haber Process N 2(g) + 3H 2 (g) D 2NH 3 (g) (D signify that the reactants are not completely (100%) converted to products. As soon as NH 3 is form, some of it decomposes and form N 2 and H 2 the starting chemical. This takes place until the amount consumed is equal to the amount produced, i.e., equilibrium is reached Law of Mass Action: 2! NH # " 3 $ K eq =! N # " 2 $ H 3 = K p! # " 2$ 8 Equilibrium

9 Meaning of K eq Which is favored: Reactant or Product? K eq = [ P ] p [ Q ] q [ A ] a [ B ] b = [ Product ] x [ Reactant ] y For Keq > 1, the ratio [Product] x > [Reactant] y For Keq = 1, the ratio [Product] x = [Reactant] y For Keq < 1, the ratio [Product] x < [Reactant] y For Keq > 1, at equilibrium Product is favored. For Keq = 1, at equilibrium Product and Reactant are equal. For Keq < 1, at equilibrium Reactant is favored. 9 Equilibrium

10 Calculating Keq from reaction The mass action expression is equal to the equilibrium constant. Equilibrium constants are generally written without units. For Keq >> 1, at equilibrium Product is favored. For Keq < 1, at equilibrium Reactant is favored. 10 Equilibrium

11 Equilibrium Constants: K eq, K c, K p, K a, K b, K sp, K f, K w K c versus K p : For reactions that undergoes reversibility, the efficiency of the reaction (extent to which the reaction forms product) is indicated by its equilibrium constant, K eq. If the equilibrium constant (K eq ) express through its mass action expression is based on molar concentration, then the equilibrium constant is written as Kc. If the equilibrium constant (K eq ) express through its mass action expression is based on partial pressure, then the equilibrium constant is written as Kp. K sp : If the rxn is a solid dissolving to ions, i.e., solubility of a salt, then K eq is written as K sp. K f : If the reaction is a formation of a transition complex, then K eq is written as K f. K a : If the reaction involves an acid losing a proton, then K eq is written as K a. K a2 : If the rxn involves an diprotic acid losing its second proton, then K eq is written as K a2. K a3 : If the reaction involves a polyprotic acid losing third proton, then K eq is written as K a3. Kb: If the reaction involves an base, then K eq is written as K b. Kw: If the reaction involves water, then K eq is written as K w. 11 Equilibrium

12 K eq : K c versus K p If the equilibrium constant (K eq ) express through its mass action expression is based on molar concentration, then the equilibrium constant is written as Kc. If the equilibrium constant (K eq ) express through its mass action expression is based on partial pressure, then the equilibrium constant is written as Kp. K c =K ( p RT) Δn K p =K ( c RT) +Δn [RT is positive exponent (+Δn)] where Δn = (gas moles) (gas moles) product reactant 12 Equilibrium

13 Derivation Between K c and K p Consider the Reaction: COCl 2 (g) D CO (g) + Cl 2 (g) K c = [CO] [Cl 2 ] [COCl 2 ] It can be shown that: or ( ) * ( p Cl ) 2 ( pcocl ) 2 K P = p CO K c =K ( p RT) Δn K p =K ( c RT) +Δn [RT is positive exponent (+Δn)] where Δn = (gas moles) (gas moles) product reactant 13 Equilibrium

14 1. Calculating K eq : K c - K p relationship Carbon dioxide is placed in a 5.00-L high-pressure reaction vessel at C. Given the following reaction, CO (g) + 1/2 O 2 (g) D CO 2 (g), calculate the equilibrium constant K c and K p after equilibrium is achieved with 4.95 mol of CO 2, mol of CO, and mol of O 2 found in the container. k c = T = 1400 C CO (g) O 2 (g)! CO 2 (g) V = 5.0 L Vol = 5.0 L [ CO 2 ] [ CO] [ O 2 ] = / M k c = i Δ mol e mol 0.025mol 4.95mol [ e] M M M [ -1 M] [ ] [ M] 1/2 14 Equilibrium

15 2. Calculating K eq ice - method At 300 C, mol of PCl 5 is placed in a 100mL reaction vessel. During the course of the reaction PCl 5 converts to PCl 3 and Cl 2 gas. At equilibrium the amount of PCl 3 formed is 0.200mol. What is the equilibrium constant for the following reaction. PCl 5(g) D PCl 3 (g) + Cl 2 (g) Note: 0.600mol/0.100L = 6.00 M and 0.200mol/0.100L = 2.00M PCl 5 (g) PCl 3 (g) + Cl 2 (g) i 6.00 M 0 0 V = 100 ml Δ -X +X +X = 0.100L " e # $ %? 2.00? 15 Equilibrium

16 Animation: PCl 5 D PCl 3 + Cl 2 Cl Cl C Cl P Cl C Cl P Cl Cl Cl Cl Cl PCl 5(g) D PCl 3 (g) + Cl 2 (g) Cl C C Cl Cl Cl P P Cl Cl Cl Cl Cl Cl C Cl P C C Cl Cl Cl P C P Cl Cl Cl Cl Cl P Cl Cl Cl PCl 5 (g) PCl 3 (g) + Cl 2 (g) i 6.00 M 0 0 V = 100 ml Δ -X +X +X = 0.100L " e # $ % 4.00? 2.00? Equilibrium

17 Summary: Equilibrium constant The mass action expression is equal to the equilibrium constant. Equilibrium constants are generally written without units. For Keq >> 1, at equilibrium Product is favored. For Keq < 1, at equilibrium Reactant is favored. 17 Equilibrium

18 LeChatelier s Principle Changing the concentration or pressure of a substance in a reaction at equilibrium is creating a stress. The removal of something creates a deficit stress, and adding something creates an excess stress. Reactions will always respond to stress by replacing the deficit (making more of what was removed) or by conversion of the excess (consuming some of what was added). This description works regardless of which side of the reaction is changed. Stress/Deficit will be in the form of: (1) Concentration, (2) Pressure and (3) Temperature variation 18 Equilibrium

19 LeChatelier Principle Methods of disturbing Equilibrium and LeChatelier response. Stress on Reactant Stress on Product Relief on Reactant Relief on Product Reactant D Product Equilibrium Revisited: Equilibrium - Situation in which changes occur at equal rates so no net change is apparent. LeChatelier Principle: A stress on the a system at equilibrium causes changes to the system to reduce the stress until a new equilibrium is established. 19 Equilibrium

20 Reactant Change: Stress added Consider the affect on Equilibrium as a result of increasing the concentration of the reactant. Stress on the reactant Which direction does the reaction shift towards? i) A system at equilibrium. ii) A stress is added to the reactant. The reaction shifts to the right or left? 20 Equilibrium

21 Reactant Change: Stress added Consider the affect on Equilibrium as a result of increasing the concentration of the reactant. Stress on the reactant Which direction does the reaction shift towards? i) A system at equilibrium. ii) A stress is added to the reactant. iii) The system self-adjust to reduce the stress. iv) Until a new equilibrium is reestablished The reaction shifts to the right. The reaction favors the product. 21 Equilibrium

22 Product Changes: Stress added Consider the affect on Equilibrium as a result of increasing the concentration of the product. Stress on the product Which direction does the reaction shift towards? i) A system at equilibrium. ii) A stress is added to the product. The reaction shifts to the right or left? 22 Equilibrium

23 Product Changes: Stress added Consider the affect on Equilibrium as a result of increasing the concentration of the product. Stress on the product Which direction does the reaction shift towards? i) A system at equilibrium. ii) A stress is added to the product. iii) The system self-adjust itself to reduce the stress. iv) Until a new equilibrium is reestablished. The reaction shifts to the left. The reaction favor the reactant. 23 Equilibrium

24 Reactant Change: Stress Relief Consider the affect on Equilibrium as a result of decreasing the concentration of the reactant. What shift in direction occurs on the reaction? Deficit on the reactant i) A system at equilibrium. ii) A de-stress is place to the reactant. The reaction shifts to the right or left? 24 Equilibrium

25 Reactant Change: Stress Relief Consider the affect on Equilibrium as a result of decreasing the concentration of the reactant. What shift in direction occurs on the reaction? Deficit on the reactant i) A system at equilibrium. ii) A de-stress is place to the reactant. iii) The system self-adjust itself to replace the missing reactants. iv) Until a new equilibrium is reestablished. The reaction shifts to the left. The reaction favors the reactant. 25 Equilibrium

26 Product Change: Stress Relief Consider the affect on Equilibrium as a result of decreasing the concentration of the Product. What shift in direction occurs on the reaction? Deficit on the reactant i) A system at equilibrium. ii) A de-stress is place to the product. The reaction shifts to the right or left? 26 Equilibrium

27 Product Change: Stress Relief Consider the affect on Equilibrium as a result of decreasing the concentration of the Product. What shift in direction occurs on the reaction? Deficit on the product i) A system at equilibrium. ii) A de-stress is place to the product. iii) The system selfadjust itself to replace the displaced products. iv) Until a new equilibrium is reestablished. The reaction shifts to the right. The reaction favors to the product. 27 Equilibrium

28 LeChatelier s Principle (1): Concentration Effect If a chemical system is at equilibrium and then a substance is added (either a reactant or product), the reaction will shift so as to reestablish equilibrium by subtracting part of the added substance. Conversely, removal of a substance will result in the reaction moving in the direction that forms more of the substance. Consider the Haber reaction at equilibrium: N 2 + 3H 2 2NH 3 If some H 2 is added to the reaction which was at equilibrium, the system self-adjust to remove the excess H 2 by converting it to NH 3 until equilibrium is reestablish; in the process some N 2 is also consumed. 28 Equilibrium

29 LeChatelier s Principle (2): Volume / Pressure Effect Increasing the pressure (by reducing the volume) of a gaseous mixture causes the system to shift in the direction that reduces the number of moles of gas.* Consider the reaction : N 2 O 4 (g) D 2NO 2 (g) + P Suppose the volume of the container is reduced? The equilibrium shifts to the side that reduces the total number of moles of gas involved in the reaction. In this example, 2 moles of product (NO 2 ) will change to 1 mol of reactant (N 2 O 4 ). The total moles in the reaction mixture is reduced to compensate for the increase in pressure. Note: Adding an inert gas will not shift the equilibrium because the inert gas is not involved in the overall reaction. 29 Equilibrium

30 LeChatelier s Principle (2): Volume / Pressure Effect Examples: Which side is sensitive to pressure? 1) 2SO 2 (g) + 2H 2 O (g) D 2H 2 S (g) + 3O 2 (g) + P What happens to pressure if the volume is decrease? Pressure increase Which direction the reaction shift if the volume is decrease? R.S.* Reactant (Left) What happens to pressure if the volume is increase? Which direction the reaction shift if the volume is increase? Pressure decrease R.S. Product (Right) i) A system at equilibrium. ii) A stress is added to the product. 30 Equilibrium iii) The system self-adjust itself to reduce the stress. iv) Until a new equilibrium is reestablished.

31 LeChatelier s Principle (2): Volume / Pressure Effect Examples: Which side is sensitive to pressure? 1) 2SO 2 (g) + 2H 2 O (g) D 2H 2 S (g) + 3O 2 (g) + P What happens to pressure if the volume is decrease? Pressure increase Which direction the reaction shift if the volume is decrease? R.S.* Reactant (Left) What happens to pressure if the volume is increase? Which direction the reaction shift if the volume is increase? Pressure decrease R.S. Product (Right) i) A system at equilibrium. ii) A de-stress is place to the product. 31 Equilibrium iii) The system selfadjust itself to replace the displaced products. iv) Until a new equilibrium is reestablished.

32 LeChatelier s Principle (2): Volume / Pressure Effect Examples: Which side is sensitive to pressure? 2) N 2 (g) + O 2 (g) D 2 NO (g) What happens to pressure if the volume is decrease? Pressure increase Which direction the reaction shift if the volume is decrease? No shift What happens to pressure if the volume is increase? Which direction the reaction shift if the volume is increase? No shift Pressure decrease Add stress to reaction Create a deficit to system at equilibrium. System has no change 32 Equilibrium

33 LeChatelier s Principle (2): Volume / Pressure Effect Examples: Which side is sensitive to pressure? P + 3) 2CO (g) D C (s) + CO 2 (g) What happens to pressure if the volume is decrease? Pressure increase Which direction the reaction shift if the volume is decrease? R.S. Product (Right) What happens to pressure if the volume is increase? Pressure decrease Which direction the reaction shift if the volume is increase? R.S. Reactant (Left) i) A system at equilibrium. ii) A stress is added to the reactant. iii) The system self-adjust to reduce the stress. iv) Until a new equilibrium is reestablished 33 Equilibrium

34 LeChatelier s Principle (2): Volume / Pressure Effect Examples: Which side is sensitive to pressure? P + 3) 2CO (g) D C (s) + CO 2 (g) What happens to pressure if the volume is decrease? Pressure increase Which direction the reaction shift if the volume is decrease? R.S. Product (Right) What happens to pressure if the volume is increase? Pressure decrease Which direction the reaction shift if the volume is increase? R.S. Reactant (Left) i) A system at equilibrium. ii) A deficient is place to the reactant. iii) The system self-adjust itself to replace the missing reactants. iv) Until a new equilibrium is reestablished. 34 Equilibrium

35 The Math Behind Pressure Effects due to Volume changes Consider: C (s) + CO 2 (g) D 2CO (g) 1 mol of CO is in equilibrium with 1 mol of CO 2 in 1 L container. Keq = [1M] 2 / [1 M] Now suppose the volume is reduced from 1L to 0.5 L, What happens to the concentration? [CO] = 1 mol /.5 L = 2 M, [CO 2 ] = 1 mol /.5 L = 2 M Q (not at equilb) = [2M] 2 / [2 M] = 2 2 > 1, but reaction needs to go back to equilb. Q reduces to become K eq again. Reaction shifts to the left. K eq = 1 Q = 2 35 Equilibrium

36 Energy of Reaction: Review The energy of a reaction depends on the energies of the reactant relative that of the product. If the reaction is downhill (a) then energy is a product (Exothermic Reaction). If the reaction is uphill (b), then energy is a reactant (Endothermic) (a) Heat product (-) Exothermic Reactant Product + E (b) Heat reactant (+) Endothermic Reactant + E Product 36 Equilibrium

37 Temperature Effect (3): Endothermic Consider the affect on Equilibrium as a result increasing the temperature of an endothermic reaction. What shift in direction occurs on the reaction? Recall that an endothermic reaction treats heat (energy) as a reactant. Increase in temperature, increase in stress of reactant. E E E i) A system at equilibrium. ii) An increase in temperature is a stress to the reactant. iii) The system self-adjust itself to reduce the stress.. iv) Until a new equilibrium is reestablished The reaction shifts to the right. 37 Equilibrium

38 LeChatelier s Principle (3): Temperature Effect Examples: Which side is sensitive to pressure? Question 1) 2SO 2 (g) + 2H 2 O (g) D 2H 2 S (g) + 3O 2 (g) + E Which direction the reaction shift if the temperature decreases? Which direction the reaction shift if the temperature is increases? 2) N 2 (g) + O 2 (g) D 2NO (g), ( ΔH = kj ) Which direction the reaction shift if the temperature decreases? Which direction the reaction shift if the temperature is increases? 3) 2CO (g) D C (s) + CO 2 (g) (Exothermic reaction ) Which direction the reaction shift if the temperature decreases? Which direction the reaction shift if the temperature is increases? 38 Equilibrium

39 LeChatelier s Principle (3): Temperature Effect Examples: Which side contains Energy? 1) 2SO 2 (g) + 2H 2 O (g) D 2H 2 S (g) + 3O 2 (g) + E Which direction the reaction shift if the temperature decreases? Which direction the reaction shift if the temperature is increases? R.S. Product (Right) R.S. Reactant (Left) 2) N 2 (g) + O 2 (g) D 2NO (g), ( ΔH = kj ) Which direction the reaction shift if the temperature decreases? Which direction the reaction shift if the temperature is increases? R.S. Reactant (Left) R.S. Product (Right) 3) 2CO (g) D C (s) + CO 2 (g) (Exothermic reaction ) Which direction the reaction shift if the temperature decreases? Which direction the reaction shift if the temperature is increases? R.S. Product (Right) R.S. Reactant (Left) 39 Equilibrium

40 LeChatelier s Principle (3): Temperature increase An increase in the temperature of a system at equilibrium will shift the reaction so that it will absorb the heat. The equilibrium constant changes with temperature. It will either increase or decrease depending on the exothermicity or endothermicity of the reaction. Endothermic: Reactant + E D products Energy = Reactant Temp increase Shift rxn to Right, (K eq increases) T increase K eq (old) K eq (new) Temp increase Shift rxn to Right, (K eq increases) Exothermic: Reactant D products + E Energy = Product Temp increase Shift rxn to Left, (K eq decrease) T increase K eq (new) K eq (old) Temp increase Shift rxn to Left, (K eq decreases) 40 Equilibrium

41 LeChatelier s Principle (3): Temperature decrease An decrease in the temperature of a system at equilibrium will shift the reaction so that it will compensate for the energy deficit. The equilibrium constant changes with temperature. It will either decrease or increase depending on the exothermicity or endothermicity of the reaction. Endothermic: Reactant + E D products Energy = Reactant Temp decrease Shift rxn to Left, (K eq decreases) T decrease K eq (new) K eq (old) Temp decrease Shift rxn to left, (K eq decreases) Exothermic: Reactant D products + E Energy = Product Temp decrease Shift rxn to Right, (K eq increase) T decrease K eq (old) K eq (new) Temp decrease Shift rxn to Right, (K eq increases) 41 Equilibrium

42 Solubility of Gas: Temperature Effect What is the solubility of a Gas? (Lake Nyos) CO 2 (g) D CO 2 (aq) 42 Equilibrium

43 Solubility of Gas: Temperature Effect What is the solubility of a Gas? (Lake Nyos) CO 2 (g) D CO 2 (aq) 43 Equilibrium

44 LeChatelier s Principle (4): Catalyst Catalyst added to a system at equilibrium lowers the activation energy of a reaction and therefore accelerates the rate of the reaction in both direction. A catalyst does not change the value of the equilibrium constant. A catalyst decrease the activation barrier between the reactant and product. This result in increasing the rate of the forward and the reverse reaction to the same extent. k f = k r, the K eq is unaffected: K eq = k f = 2k f = 100k f k r 2k r 100k r 44 Equilibrium

45 Summary of Temperature & Pressure Effects on Solubility ΔH (s, l or g) Temp Direction Solubility (+) Endothermic h g Prod h increase (+) Endothermic i f React i decrease (-) Exothermic h f React i decrease (-) Exothermic i g Prod h increase *Pressure Direction Solubility Gas solute h g Prod h increase Gas solute i f React i decrease * An inert gas will not change solubility. 45 Equilibrium

46 9.60 Exercises Equilibrium