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Ch. 13 Equilibrium

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Chemical Equilibrium Z The state where the concentrations of all reactants and products remain constant with time. Z On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation. Z (BDVD)

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Dynamic Equilibrium Z Reactions continue to take place. Z A + B C + D ( forward) Z C + D A + B (reverse) Z Initially there is only A and B so only the forward reaction is possible Z As C and D build up, the reverse reaction speeds up while the forward reaction slows down. Z Eventually the rates are equal.

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Reaction Rate Time Forward Reaction Reverse reaction Equilibrium

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What is equal at Equilibrium? Z Rates are equal. Z Concentrations are not. Z Rates are determined by concentrations and activation energy. Z The concentrations do not change at equilibrium.

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13.2 Law of Mass Action Z j A + k B l C + m D Z j, k, l, m are coefficients Z The law of mass action is represented by the equilibrium expression: Z K = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power Z K is called the equilibrium constant. Z is how we indicate a reversible reaction. [x] represents concentration.

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Playing with K Z If we write the reaction in reverse. Z l C + m D j A + k B Z Then the new equilibrium constant is Z K = [A] j [B] k = 1/K [C] l [D] m

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Playing with K Z If we multiply the equation by a constant Z nj A + nk B nl C + nm D Z Then the equilibrium constant is Z K = [A] nj [B] nk = ([A] j [B] k ) n = K n [C] nl [D] nm ( [C] l [D] m ) n

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K is CONSTANT Z At any temperature. Z Temperature affects rate. Z The equilibrium concentrations dont have to be the same only K. Z Equilibrium position is a set of concentrations at equilibrium. Z One value at each temperature, but there are an unlimited number of possibilities. Z Usually written without units.

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Equilibrium Expression 4NH 3(g) + 7O 2(g) 4NO 2(g) + 6H 2 O (g) What is the equilibrium expression for this reaction? #19a N 2 (g) + O 2 (g) 2NO(g) What is the equilibrium expression for this reaction? #19b N 2 O 4 (g) 2NO 2 (g)

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Calculate K Z N 2 (g) + 3H 2 (g) 2NH 3 (g) Z In the above reaction, at a given temperature. K = 0.013. Z Calculate the value of K in the following reactions. #21ab Z 1/2N 2 (g) + 3/2H 2 (g) NH 3 (g) Z 2NH 3 (g) N 2 (g) + 3H 2 (g)

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Calculate K Z N 2 (g) + 3H 2 (g) 2NH 3 (g) Z Initial At Equilibrium Z [N 2 ] 0 =1.000 M [N 2 ] = 0.921M Z [H 2 ] 0 =1.000 M [H 2 ] = 0.763M Z [NH 3 ] 0 =0 M [NH 3 ] = 0.157M

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Calculate K Z N 2 (g) + 3H 2 (g) 2NH 3 (g) Z Initial At Equilibrium Z [N 2 ] 0 = 0 M [N 2 ] = 0.399 M Z [H 2 ] 0 = 0 M [H 2 ] = 1.197 M Z [NH 3 ] 0 = 1.000 M [NH 3 ] = 0.203M Z K is the same no matter what the amount of starting materials (at the same temperature).

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13.3 Equilibrium and Pressure Z Some reactions are gaseous Z PV = nRT Z P = (n/V)RT Z P = CRT Z C is a concentration in moles/Liter Z C = P/RT

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Equilibrium and Pressure 2 SO 2 (g) + O 2 (g) 2SO 3 (g) Z In term of partial pressures: K p = (P SO3 ) 2 (P SO2 ) 2 (P O2 ) Z In terms of concentration: K c = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

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Practice Z N 2 (g) + 3H 2 (g) 2NH 3 (g) Z In the reaction above, the following equilibrium pressures were observed. Z P NH3 = 0.89 atm Z P N2 = 0.62 atm Z P H2 = 0.029 atm Z Calculate the value for the equilibrium constant K p at this temperature.

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K v. K p Z For j A + k B l C + m D K p = K(RT) n n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.

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Practice K p = K(RT) n Z At 25°C, K = 2.01 x 10 4 for the reaction Z CO (g) + Cl 2(g) COCl 2(g) Z What is the value of K p at this temperature?

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Practice #32 Z For which reaction in #31 is K p = K?

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Homogeneous Equilibria Z So far every example dealt with reactants and products where all were in the same phase. Z We can use K in terms of either concentration or pressure. Z Units depend on reaction.

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13.4 Heterogeneous Equilibria Z Are equilibria that involve more than one phase. Z If the reaction involves pure solids or pure liquids, the concentration of the solid or the liquid doesnt change. Z The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. Z As long as they are not used up we can leave them out of the equilibrium expression.

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For Example Z H 2 (g) + I 2 (s) 2HI(g) Z K = [HI] 2 [ H 2 ][ I 2 ] Z But the concentration of I 2 does not change, therefore: K = [HI] 2 [ H 2 ]

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13.5 Applying the Equilibrium Constant Reactions with large K (>>1), essentially to completion. Large negative E. Z Reactions with small K (<<1) consist mostly of reactants. Z Time to reach equilibrium is related to rate and AE. It is not related to size of K.

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The Reaction Quotient Z Tells you the directing the reaction will go to reach equilibrium Z Calculated the same as the equilibrium constant, but for a system not at equilibrium by using initial concentrations. Z Q = [Products] coefficient [Reactants] coefficient Z Compare value to equilibrium constant

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What Q tells us Z If Q

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Using the Reaction Quotient Z For the reaction Z 2NOCl(g) 2NO(g) + Cl 2 (g) Z K = 1.55 x 10 -5 M at 35ºC Z In an experiment 0.10 mol NOCl, 0.0010 mol NO(g) and 0.00010 mol Cl 2 are mixed in 2.0 L flask. Z Which direction will the reaction proceed to reach equilibrium?

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Using the Reaction Quotient For the synthesis of ammonia at 500°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. Ex. 13.7 a. [NH 3 ] 0 = 1.0 x 10 -3 M; [N 2 ] 0 = 1.0 x 10 -5 M; [H 2 ] 0 = 2.0 x 10 -3 M b. [NH 3 ] 0 = 1.0 x 10 -4 M; [N 2 ] 0 = 5.0 M; [H 2 ] 0 = 1.0 x 10 -2 M

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Problems Involving Pressure For the reaction N 2 O 4 (g) 2NO 2 (g) K P = 0.133 atm. At equilibrium, the pressure of N 2 O 4 was found to be 2.71 atm? Calculate the equilibrium pressure of NO 2 (g) EX.13.8

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Problems Involving Pressure At a certain temperature a 1.00 L flask initially contained 0.298 mol PCl 3 (g) and 8.70 x 10 -3 mol PCl 5 (g). After the system reached equilibrium, 2.00 x 10 -3 mol Cl 2 (g) was found in the flask. Calculate the equilibrium concentrations of all species and the value of K. EX.13.9 PCl 5(g) PCl 3(g) + Cl 2(g)

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No Equilibrium [x] Ex. 13.10 Z Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.00 mol of each component is mixed in a 1.00 L flask.

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Intro to ICE 2 SO 3 (g) 2SO 2 (g) + O 2 (g) At a certain temperature. 12.0 mol SO 3 is placed into a 3.0 L rigid container. The SO 3 dissociates during the reaction. At equilibrium, 3.0 mol SO 2 is present. What is the value of K for this reaction? (#43)

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Intro to ICE 2NH 3 N 2 + 3H 2 At a certain temperature. 4.0 mol NH 3 is introduced into a 2.0 L container. The NH 3 partially dissociates during the reaction. At equilibrium, 2.0 mol NH 3 remains. What is the value of K for this reaction? (#44) WAssign

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What if youre not given equilibrium concentration? H 2 + F 2 2HF The Equilibrium constant for the above reaction is 115 at a certain temperature. 3.000 mol of each component was added to a 1.500 L flask. Calculate the equilibrium concentrations of all species. (ex.13.11)

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Practical Example Z H 2 (g) + F 2 (g) 2HF(g) Z K = 1.15 x 10 2 at 25ºC Z Calculate the equilibrium concentrations if a 3.00 L container initially contains 3.00 mol of H 2 and 6.00 mol F 2. Z [ H 2 ] 0 = 3.00 mol/3.00 L = 1.00 M Z [F 2 ] 0 = 6.00 mol/3.00 L = 2.00 M Z [HF] 0 = 0

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Z Q= 0

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Z For H 2 and F 2 the change must be -X Z Using to stoichiometry HF must be +2X Z Equilibrium = initial + change H 2 (g) F 2 (g) 2HF(g) Initial1.00 M2.00 M 0 M Change Equilibrium

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Z Therefore, ice chart looks like this. Z Change in HF = twice change in H 2 H 2 (g) F 2 (g) 2HF(g) Initial1.00 M2.00 M 0 M Change-X -X +2X Equili 1.00 -X 2.00-X 2X

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Z Now plug these values into the equilibrium expression Z K = 1.15 x 10 2 = ( 2X ) 2 (1.00- x )(2.00-x) Z Solving this gives us a quadratic equation. Quadratic Calculator H 2 (g) F 2 (g) 2HF(g) Initial1.00 M2.00 M 0 M Change-X -X +2X Equili 1.00 -X 2.00-X 2X

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Z Now plug these values into the equilibrium expression Z K = 1.5 x 10 2 = ( 2X ) 2 (1.00- x )(2.00-x) Z Solving this gives us a quadratic equation. Z Quadratic gives us 2.14 mol/L and 0.968 mol/L. Only 0.968 is reasonable. H 2 (g) F 2 (g) 2HF(g) Initial1.00 M2.00 M 0 M Change-X -X +2X Equili 1.00 -X 2.00-X 2X

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Z [ H 2 ] = 1.00 M - 0.968 M = 3.2 x 10 -2 M Z [F 2 ] = 2.00 M - 0.968 M = 1.032 M Z [HF] = 2(0.968 M) = 1.936 M Z If substituted into the equilibrium expression we get 1.13 x 10 2 which is very close to given K.

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Practice H 2 O (g) + Cl 2 O (g) 2HClO (g) K = 0.090 Z In an experiment 1.0 g H 2 O (g) and 2.0 g Cl 2 O are mixed in a 1.00 L flask. Z Calculate the equilibrium concentrations. (#48)

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13.6 Solving Equilibrium Problems 1. Balance the equation. 2. Write the equilibrium expression. 3. List the initial concentrations. 4. Calculate Q and determine the shift to equilibrium. 5. Define equilibrium concentrations. 6. Substitute equilibrium concentrations into equilibrium expression and solve. 7. Check calculated concentrations by calculating K.

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Problems with small K K<.01

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Process is the same Z Set up table of initial, change, and equilibrium concentrations. Z Choose X to be small. Z For this case it will be a product. Z For a small K the product concentration is small.

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For example Z For the reaction 2NOCl 2NO +Cl 2 Z K= 1.6 x 10 -5 Z If 1.00 mol NOCl is put in a 2.0 L container, what are the equilibrium concentrations? Z K = [NO] 2 [Cl 2 ] = 1.6 x 10 -5 [NOCl] 2 Z [NOCl] 0 = 0.50M, [NO]& [Cl 2 ] =0

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K = [NO] 2 [Cl 2 ] = (2x) 2 (x) = 1.6 x 10 -5 [NOCl] 2 (0.50 -2x) 2 Z Since K is so small, we we can make an approximation that 0.50-2x = 0.50 Z This makes the math much easier. X = 1.0 x 10 -2 2NOCl 2NO+Cl 2 Initial 0.50 00 Change-2X +2X+X Equil0.50 -2X 2X X

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5% Rule Z Many of the systems we will deal with have very small equilibrium constants. Z When this is the case, there will be very little shift to the right to reach equilibrium. Since x is so small, we will ignore it. However, the final value must be checked against the initial concentration. If the difference is less than 5%, then our assumption is valid. Z In the previous problem X = 1.0 x 10 -2 Z 0.50 -2x = 0.50 - 2(1.0 x 10 -2 ) = 0.48 Z Is 1.0 x 10 -2 five percent or less than 0.50? Z 0.02/0.50 x 100 = 4%

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Practice Problem Z For the reaction N 2 O 4(g) 2NO 2 (g) Z K = 4.0 x 10 -7 Z 1.0 mol N 2 O 4(g) is placed in a 10.0L vessel. Calculate the concentrations of all species at equilibrium. Z (#52)

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Practice Problem Z For the reaction COCl 2(g) CO (g) + Cl 2 (g) Z K p = 6.8 x 10 -9 Z If COCl 2(g) at an initial pressure of 1.00 atm decomposes, calculate the equilibrium pressures of all species? Z (#54)

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Practice Problem Z At 25°C, K p = 2.9 x 10 -3 Z NH 4 OCONH 2(s) 2NH 3g) + CO 2 (g) Z In an experiment, a certain amount of NH 4 OCONH 2(s) is placed in an evacuated rigid container and allowed to come to equilibrium. Calculated the total pressure in the container at equilibrium. Z (#55)

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Practice Problem #56WA Z 2AsH 3(g) 2As (s) + 3H 2(g) Z In an experiment pure AsH 3 was placed in a rigid, sealed flask at a pressure of 392.0 torr. After 48 hours the pressure was observed to be constant at 488.0 torr. Z Calculate the Equilibrium pressure of H 2(g). Z Calculate K p for this reaction. (#56)

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Le Chateliers Principle Z If a change is applied to a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce the change. Z 3 Types of change: Concentration (adding or reducing reactant or product), Pressure, Temperature.

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Change amounts of reactants and/or products Z Adding product makes Q>K Z Removing reactant makes Q>K Z Adding reactant makes Q

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Change Pressure Z By changing volume. Z When volume is reduced, the system will move in the direction that has the least moles of gas. Z COCl 2(g) CO (g) + Cl 2 (g) Z When volume is increased, the system will move in the direction that has the greatest moles of gas. Z Because partial pressures (and conc.) change a new equilibrium must be reached. Z System tries to minimize the moles of gas.

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Change in Pressure Z By adding an inert gas. Z Partial pressures of reactants and product are not changed. Z No effect on equilibrium position.

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Change in Temperature Z Affects the rates of both the forward and reverse reactions. Z Doesnt just change the equilibrium position, changes the equilibrium constant. Z The direction of the shift depends on whether it is exo - or endothermic.

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Exothermic H < 0 Z Releases heat. Z Think of heat as a product. Z N 2 (g) + 3H 2 (g) 2NH 3 (g) + Heat Z Raising temperature push toward reactants. Z Shifts to left.

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Endothermic H > 0 Z Absorbs heat. Z Think of heat as a reactant. Z Heat + COCl 2(g) CO (g) + Cl 2 (g) Z Raising temperature push toward products. Z Shifts to right.

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Free Energy and Equilibrium G tells us spontaneity at current conditions. When will it stop? It will go to the lowest possible free energy which may be an equilibrium. At equilibrium G = 0, Q = K Gº = -RTlnK from [ G = Gº + RTlnK] Gº = 0 < 0 > 0 K = 1 > 1 < 1

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Free Energy and Equilibrium The overall reaction for the corrosion of iron by oxygen is 4Fe(s) + 3O 2 (g)2Fe 2 O 3 (s) Use the following data, calculate the equilibrium constant for this reaction at 25°C. Substance H° f (kj/mol)S°(J/Kmol) Fe 2 O 3 (s)-82690 Fe(s) 027 O 2 (g) 0205 Ex. 16.15 Gº = -RTlnK

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Practice Calculate K for the following reaction (at 298 K) H 2 (g) + Cl 2 (g)2HCl H° f (kj/mol) = -184S°(J/Kmol) = 20. If each gas is placed in a flask such that the pressure of each gas is 1 atm, in which direction will the system shift to reach equilibrium at 25°C? Gº = -RTlnK

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