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15-1 15-2 15-3 15-4 15-5 15-6 Dynamic Equilibrium The Equilibrium Constant Expression Relationships Involving Equilibrium Constants The Magnitude of an Equilibrium Constant The Reaction Quotient, Q: Predicting the Direction of a Net Change Altering Equilibrium Conditions: Le Châtelier’ Principle s Equilibrium Calculations: Some Illustrative Examples Chapter 15: Principles of Chemical Equilibrium
Dr. Berk & Dr. Koel CHM 25 Spring 2007
Slide 1 of 46 CHM 25 SP07 15-7 Slide 2 of 46 CHM 25 SP07 Properties of an Equilibrium
Equilibrium systems are Equilibrium systems are DYNAMIC (in constant DYNAMIC (in constant motion) motion) REVERSIBLE REVERSIBLE can be approached from can be approached from either direction either direction
Pink to blue Co(H2O)6 + 4 Cl- ---> Co(Cl)4 + 6 H2O Blue to pink Co(Cl)4 + 6 H2O ---> Co(H2O)6 + 4 ClSlide 3 of 46 CHM 25 SP07 Dynamic Equilibrium Equilibrium –two opposing processes taking place at equal rates.
H2O(l) NaCl(s)
H2O H2O(g) NaCl(aq) CH3OH(g) I2(H2O) I2(CCl4) CO(g) + 2 H2(g)
Slide 4 of 46 CHM 25 SP07 Dynamic Equilibrium The Equilibrium Constant Expression Methanol synthesis is a reversible reaction.
CO(g) + 2 H2(g) CH3OH(g)
k-1 k1 CH3OH(g) CO(g) + 2 H2(g) CO(g) + 2 H2(g)
Slide 5 of 46 CHM 25 SP07 Slide 6 of 46 k1 k-1 CH3OH(g)
CHM 25 SP07 CHM 25 Sp07 1 Examples of Chemical Equilibria Chemical Equilibria
CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)
Formation of stalactites and stalagmites CaCO3(s) + H2O(liq) + CO2(g)
Slide 7 of 46 At a given T and P of CO2, [Ca2+] and [HCO3-] can be found from the Ca2+(aq) + 2 HCO3-(aq) EQUILIBRIUM CONSTANT.
CHM 25 SP07 Slide 8 of 46 CHM 25 SP07 Three Approaches to the Equilibrium Three Approaches to Equilibrium Slide 9 of 46 CHM 25 SP07 Slide 10 of 46 CHM 25 SP07 Three Approaches to Equilibrium
CO(g) + 2 H2(g)
k1 k-1 The Equilibrium Constant Expression
Forward: CO(g) + 2 H2(g) → CH3OH(g) Reverse: CH3OH(g) → CO(g) + 2 H2(g) At Equilibrium: Rfwrd = Rrvrs
k-1 k1 CH3OH(g) Rfwrd = k1[CO][H2]2 Rrvrs = k-1[CH3OH]
k1 k-1 CO(g) + 2 H2(g) CH3OH(g) k1[CO][H2]2 = k-1[CH3OH] k1 k-1
Slide 11 of 46 CHM 25 SP07 Slide 12 of 46 = [CH3OH] [CO][H2]2 = Kc
CHM 25 SP07 CHM 25 Sp07 2 The Reaction Quotient, Q
In general, ALL reacting chemical systems are characterized by their REACTION QUOTIENT, Q.
aA + bBcC + dD THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type aA + bBcC + dD the following is a CONSTANT (at a given T)
conc. of products K= [C]c [D]d [A]a [B]b conc. of reactants equilibrium constant If Q = K, then system is at equilibrium.
Slide 13 of 46 CHM 25 SP07 If K is known, then we can predict concs. of products or reactants.
Slide 14 of 46 CHM 25 SP07 Determining K
2 NOCl(g) 2 NO(g) + Cl2(g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of an “ ”table of concentrations ICE [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change Equilibrium 0.66
Slide 15 of 46 CHM 25 SP07 Determining K
2 NOCl(g) 2 NO(g) + Cl2(g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of a table of concentrations [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33
Slide 16 of 46 CHM 25 SP07 Determining K
2 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33
K K
Slide 17 of 46 The Significance of the Magnitude of the Equilibrium Constant. [NO]2 [ Cl2 ] [NOCl]2 = (0.66) 2 (0.33) (1.34)2 = 0.080
CHM 25 SP07 Slide 18 of 46 CHM 25 SP07 [NO]2 [ Cl2 ] [NOCl]2 CHM 25 Sp07 3 The Meaning of K
Can tell if a reaction is product-favored or reactant-favored. For N2(g) + 3 H2(g) 2 NH3(g)
Kc = [NH3 ]2 [N2 ][H2 ]3 = 3.5 x 10 8 The Meaning of K The Meaning of K
For AgCl(s) Ag+(aq) + Cl-(aq) Kc = [Ag+] [Cl-] = 1.8 x 10-5 Conc. of products is much than that of reactants at equilibrium. less Conc. of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored.
Slide 19 of 46 CHM 25 SP07 Ag++(aq) + Cl-(aq)) Ag (aq) + Cl-(aq AgCl(s)) AgCl(s The reaction with small K is strongly is product--favored. is product favored. reactant-favored. Slide 20 of 46 CHM 25 SP07 Product- or Reactant Favored Relationships Involving the Equilibrium Constant Reversing an equation causes inversion of K. Multiplying by coefficients by a common factor raises the equilibrium constant to the corresponding power. Dividing the coefficients by a common factor causes the equilibrium constant to be taken to that root.
Slide 22 of 46 CHM 25 SP07 Product-favored K>1
Slide 21 of 46 Reactant-favored K<1
CHM 25 SP07 Writing and Manipulating K Expressions
Adding equations for reactions
S(s) + O2(g) SO2(g)
Net equation
[SO2 ] K1 = [O2 ] Writing and Manipulating K Expressions
Changing coefficients
S(s) + 3/2 O2(g) SO3(g) 2 S(s) + 3 O2(g) 2 SO3(g)
K new Slide 24 of 46 K [SO 3 ] [O 2 ]3/2 SO2(g) + 1/2 O2(g) SO3(g)
S(s) + 3/2 O2(g) SO3(g)
Knet Slide 23 of 46 K2 = [SO3 ] [SO 2 ][O2 ]1/ 2 K new [SO3 ]2 [O2 ]3 [SO3 ]2 [O2 ]3 [SO3 ] [O2 ]3 / 2 Knet = K1 • K 2
CHM 25 SP07 = (K old )2
CHM 25 SP07 CHM 25 Sp07 4 Writing and Manipulating K Expressions
Changing direction S(s) + O2(g) SO2(g) SO2(g) S(s) + O2(g)
K new [O 2 ] 1 = [SO 2 ] K old
CHM 25 SP07 Writing and Manipulating K Expressions
Equilibrium constant expressions do not contain concentration terms for solid or liquid phases of a single component (that is, pure solids or liquids). S(s) + O2(g) SO2(g)
K
Slide 26 of 46 K [SO 2 ] [O 2 ] K new [O 2 ] [SO 2 ] Slide 25 of 46 [SO 2 ] [O 2 ] CHM 25 SP07 Writing and Manipulating K Expressions
Solids and liquids NEVER appear in equilibrium expressions. NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq)
K
Slide 27 of 46 Writing and Manipulating K Expressions
Concentration Units We have been writing K in terms of mol/L. These are designated by Kc But with gases, P = (n/V)• = conc •RT RT P is proportional to concentration, so we can write K in terms of P. These are designated by Kp. Kc and Kp may or may not be the same. [NH 4+ ][OH - ] [NH3 ] CHM 25 SP07 Slide 28 of 46 CHM 25 SP07 Burnt Lime
CaCO3(s) Kc = [CO2] CaO(s) + CO2(g) KP = PCO2(RT) The Reaction Quotient, Q: Predicting the Direction of Net Change.
CO(g) + 2 H2(g)
k1 k-1 CH3OH(g) Equilibrium can be approached various ways. Qualitative determination of change of initial conditions as equilibrium is approached is needed.
Qc = [G]tg[H]th [A]tm[B]tn At equilibrium Qc = Kc Slide 29 of 46 CHM 25 SP07 Slide 30 of 46 CHM 25 SP07 CHM 25 Sp07 5 Reaction Quotient Altering Equilibrium Conditions: Le Châtelier’ Principle s Temperature, catalysts, and changes in concentration affect equilibria. The outcome is governed by LE “ a system at equilibrium is disturbed, ...if the system tends to shift its equilibrium position to counter the effect of the disturbance.”
Slide 31 of 46 CHM 25 SP07 Slide 32 of 46 CHM 25 SP07 CHATELIER’ PRINCIPLE S Altering Equilibrium Conditions: Le Châtelier’ Principle s
Henri Le Chatelier 1850-1936 Studied mining engineering. Interested in glass and Le Châtelier’ Principle s
2 SO2(g) + O2(g)
k1 k-1 2 SO3(g) Kc = 2.8102 at 1000K ceramics.
Q= What happens if we add SO3 to this equilibrium?
Slide 33 of 46 CHM 25 SP07 Slide 34 of 46 [SO3]2 = Kc [SO2]2[O2] Q > Kc
CHM 25 SP07 Effect of Condition Changes Adding a gaseous reactant or product changes Pgas. Adding an inert gas changes the total pressure. Relative partial pressures are unchanged. Effect of Change in Volume
Kc = [G]g[H]h [C]c[D]d = nG nH
g h a a nA nB V(a+b)-(g+h) Changing the volume of the system causes a change in the equilibrium position.
nSO3 Kc = [SO3] [SO2]2[O2] = V
Slide 35 of 46 = nG nH g h a a nA nB V-Δn 2 2 nSO3 V nSO2
2 = nO2 V nSO2 nO2 2 V When the volume of an equilibrium mixture of gases is reduced, a net change occurs in the direction that produces fewer moles of gas. When the volume is increased, a net change occurs in the direction that produces more moles of gas.
CHM 25 SP07 Slide 36 of 46 CHM 25 SP07 CHM 25 Sp07 6 Effect of the Change of Volume Effect of Temperature on Equilibrium Temperature change ---> change in K Consider the fizz in a soft drink CO2(aq) + HEAT CO2(g) + H2O(liq) K = P (CO2) / [CO2] Increase T. What happens to equilibrium position? To value of K? K increases as T goes up because P(CO2) increases and [CO2] decreases.
CHM 25 SP07 Slide 38 of 46 CHM 25 SP07 ntot = KP =
Slide 37 of 46 1.16 mol gas 415 1.085 mol gas 338 Effect of Temperature on Equilibrium Decrease T. Now what? Equilibrium shifts left and K decreases. Effect of a Catalyst on Equilibrium A catalyst changes the mechanism of a reaction to one with a lower activation energy. A catalyst has no effect on the condition of equilibrium. But does affect the rate at which equilibrium is attained.
Slide 40 of 46 CHM 25 SP07 Overall we see: Raising the temperature of an equilibrium mixture shifts the equilibrium condition in the direction of the endothermic reaction. Lowering the temperature causes a shift in the direction of the exothermic reaction.
Slide 39 of 46 CHM 25 SP07 EQUILIBRIUM AND EXTERNAL EFFECTS butane ButaneIsobutane Equilibrium
K= [isobutane] 2.5 [butane] Catalytic exhaust system Add catalyst ---> no change in K A catalyst only affects the RATE of approach to equilibrium.
Slide 41 of 46 CHM 25 SP07 Slide 42 of 46 isobutane
CHM 25 SP07 CHM 25 Sp07 7 Butane Isobutane Butane Isobutane
Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5 Solution Calculate Q immediately after adding more butane and compare with K. At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5. Add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]?
Slide 43 of 46 CHM 25 SP07 Q= [isobutane] 1.25 = 0.63 [butane] 0.50 + 1.50 Q is LESS THAN K. Therefore, the Q is LESS THAN K. Therefore, the reaction will shift to the ____________. reaction will shift to the ____________.
Slide 44 of 46 CHM 25 SP07 Butane Isobutane
You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Butane Isobutane
You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution Solution Q is less than K, so equilibrium shifts right — away from butane and toward isobutane. Set up ICE table [butane] Initial Change Equilibrium
Slide 45 of 46 K = 2.50 = [isobutane] 1.25 + x [butane] 2.00 - x [isobutane]
1.25 +x 1.25 + x
CHM 25 SP07 0.50 + 1.50 -x 2.00 - x x = 1.07 M At the new equilibrium position, [butane] = 0.93 M and [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane.
Slide 46 of 46 CHM 25 SP07 CHM 25 Sp07 8 ...

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