Page 1 1 hemical Equilibrium EMIAL EQUILIBRIUM hapter 1 The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation. Properties of an Equilibrium Equilibrium systems are DYNAMI (in constant motion) REVERSIBLE can be approached from either direction hemical Equilibrium Fe + + SN - qe FeSN + + qe 4 Pink to blue o( O) 6 l o( O) 4 l + O o( O) 6 l qe o( O) 4 l + O Blue to pink o( O) 4 l + O ---> o( O) 6 l Fe( O) 6 + + SN - qe Fe(SN)( O) 5 + + O hemical Equilibrium Fe + + SN - qefesn + 5 Examples of hemical Equilibria 6 Phase changes such as O(s) q e O(liq) After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained.
Page Examples of hemical Equilibria 7 hemical Equilibria 8 Formation of stalactites and stalagmites ao (s) + O(liq) + O (g) q e a + (aq) + O - (aq) ao (s) + O(liq) + O (g) q e a + (aq) + O - (aq) At a given T and P of O, [a + ] and [O - ] can be found from the EQUILIBRIUM ONSTANT. Example 9 Remember we are looking at a net change. Reactions at equilibrium still react. It is just that the forward and reverse reactions are equal. O(g) + O(g) qe (g) + O (g) Figure 1.: (a) O and O are mixed in equal numbers and begin to react (b) to form O and. After time has passed, equilibrium is reached (c) and the numbers of reactant and product molecules then remain constant over time (d). Figure 1.: The changes in concentrations with time for the reaction O(g) + O(g) (g) + O (g) when equimolar quantities of O(g) and O(g) are mixed. 11 Figure 1.4: The changes with time in the rates of forward and reverse reactions for O(g) + O(g) qe(g) + O(g) when equimolar quantities of O(g) and O(g) are mixed. The rates do not change in the same way with time because the forward reaction has a much larger rate constant than the reverse reaction. haracteristics of hemical Equilibrium N (g) + (g) q en (g) When we mix N,, and N in a closed container at 5, we do no see a change in the concentrations over time regardless of the original amounts of gas. Two possibilities exist The system is at chemical equilibrium The forward and reverse reactions are so slow that the system moves towards equilibrium at a rate that cannot be detected. 1 The second reason applies. Remember N triple bond 941 kj/mol single bond 4 kj/mol ow can we speed up the reaction? atalyst
Page The Equilibrium onstant NOT TO BE ONFUED WIT TE RATE ONSTANT!!!!! Equilibrium onstant: The value obtained when equilibrium concentrations of the chemical species are substituted in the equilibrium expression. Equilibrium Expression: The expression (from the law of mass action) obtained by multiplying the product concentrations and dividing by the multiplied reactant concentrations, with each concentration raised to a power represented by the coefficient in the balanced equation. Law of Mass Action: A general description of the equilibrium condition; it defines the equilibrium constant expression. WAT TE *%#@! 1 The Law of Mass Action Two Norwegian chemist ato Maximilian Guldberg (186-190) Peter Waage (18-1900) Proposed in 1864 a general description of an equilibrium condition. Based upon observations of many chemical reactions. For some reaction ja + kb l + md The law of mass action is represented by the equilibrium expression & K is the constant: l m K D j k A B 14 Equilibrium Expression Example 4N (g) + 7O (g) qe 4NO (g) + 6 O(g) 4 6 K NO O N O 4 7 15 Example [N ].1 - M [N ] 8.5 - M [ ].1 - M N (g) + (g) q en (g) What is the value of K? [N ] (.1 ) K.8 ) 1 [N ][ ] (8.5 )(.1 16 4 Example [N ].1 - M [N ] 8.5 - M [ ].1 - M N (g)q en (g) + (g) Just the reverse of the previous example What is the value of K? [N ][ ] 1 K'.6 [N K ] 5 17 Example [N ].1 - M [N ] 8.5 - M [ ].1 - M ½ N (g) + / (g) q en (g) Just ½ of the original example What is the value of K? 1/ K' ' (K) 1.9 18
Page 4 Notes on Equilibrium Expressions (EE) The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse. When the equation for a reaction is multiplied by n, EE new (EE original ) n K new (K orginal ) n The units for K depend on the reaction being considered. ustomarily written without units. 19 Equilibrium with Pressure We have been talking about equilibrium of gases in terms of concentration. Why not pressure? You should know mols PV nrt & M Vol(L) n P RT V P P RT & RT 0 Equilibrium with Pressure [N ] K [N ][ ] K P N ( )( PN (P )(P N N ) K ) c 1 But There must be some sort of relationship between K and K P. So we have the following reaction. ja + kb qel + md We KNOW l [] [D] K j [A] [B] m k (P )(PD ) & KP j k (P )(P ) A l B m P RT A l j A K(RT) n B K P K(RT) n m D k B (P )(P KP (P )(P l m () (D) j k ( ) ( ) l m ) ( RT) (D RT) j k ) ( RT) ( RT) (RT) (RT) A l+ m j+ k B K(RT) (l+ m) (j+ k) Example K P 1.9 @ 5 NO(g) + l (g) qenol(g) alculate K from K P K P K(RT) n 1.9 K(RT) n n (+1) -1 1.9 K(RT) -1 1.9 (RT) K 1.9 (0.08057)(98) K 4.6 4 4
Page 5 eterogeneous Equilibria eterogeneous Equilibrium: an equilibrium involving reactants and/or products in more than one phase. Example: ao (s) qeao(s) + O (g) You would think that the equilibrium constant would be [O ][ao] K' [ao ] But heterogenous equilibrium does not depend on the amounts of pure solid or liquids present. Therefore K[O ] 5 Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O (g) qe SO (g) K [SO ] [O ] 6 Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. 7 The Reaction Quotient, Q In general, all reacting chemical systems are characterized by their REATION QUOTIENT, Q. 8 N (aq) + O(liq) q e N 4+ (aq) + O - (aq) K [N 4 + ][O - ] [N ] If Q K, then system is at equilibrium. If Q > K, then conc. of products too large shifts towards reactants. If Q < K, then conc. of reactants too large shifts towards products. The Meaning of K 1. an tell if a reaction is productfavored or reactant-favored. For N (g) + (g) qe N (g) K c [N ].5 x 8 [N ][ ] onc. of products is much greater than that of reactants at equilibrium. The reaction is strongly productfavored. 9 The Meaning of K For Agl(s) qe Ag + (aq) + l - (aq) K c [Ag + ] [l - ] 1.8 x -5 onc. of products is much less than that of reactants at equilibrium. The reaction is strongly reactant-favored. 0 Ag + (aq) + l - (aq) qe Agl(s) is product-favored.
Page 6 Product- or Reactant Favored 1 The Meaning of K K comes from thermodynamics. See hapter 19, page 81-81 G < 0: reaction is product favored G > 0: reaction is reactant-favored G o -RT ln K Product-favored Reactant-favored If K > 1, then G is negative If K < 1, then G is positive The Meaning of K The Meaning of K 4. an tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium. n-butane iso-butane [iso] K.5 [n] n-butane iso-butane [iso] K.5 [n] If [iso] 0.5 M and [n] 0.15 M, are you at equilibrium? If not, which way does the reaction shift to approach equilibrium? The Meaning of K All reacting chemical systems are characterized by their REATION QUOTIENT, Q. product concentrations Q reactant concentrations If Q K, then system is at equilibrium. Q conc. of iso conc. of n 0.5 0.15. Q (.) < K (.5) Reaction is NOT at equilibrium, so [iso] must become and [n] must. 5 The Extent of a Reaction The inherent tendency for a reaction to occur it indicated by the magnitude of the equilibrium constant. A value of K much larger than 1 mean? At equilibrium the reaction system will consist of mostly products. Lies to the right. A very small value of K At equilibrium the reaction system will consist of mostly reactants. Lies to the left. Remember the size of K and the time required to reach equilibrium are NOT directed related. The time is dependent on the Activation Energy. 6
Page 7 MANY MANY EXAMPLES For the reaction N (g) + (g)q en (g) K6.0 - @ 500 Predict the direction in which the system will shift to reach equilibrium if [N ] 1.0-5 M [N ] 1.0-5 M [ ].0 - M [N ] (1.0 ) 7 1. [N ][ ] (1.0 )(.0 ) Q Therefore, since Q>K the direction goes towards the reactants 7 Apollo II lunar landing module at Tranquility Base, 1969. N O 4 (g)q e NO (g) 8 N O 4 used as one of the fuels on the lunar lander. losed container of N O 4 (g) allowed to reach equilibrium. K P 0.1 P NO4 at eq.71 atm P NO? PNO KP 0.1 P NO4 X 0.1.71atm X 0.60 0.600 9 40 Determining K NOl(g) qe NO(g) + l (g) Place.00 mol of NOl is a 1.00 L flask. At equilibrium you find 0.66 mol/l of NO. alculate K. Solution Set of an IE table of concentrations [NOl] [NO] [l ] Initial.00 0 0 hange Equilibrium 0.66 Determining K NOl(g) qe NO(g) + l (g) Place.00 mol of NOl is a 1.00 L flask. At equilibrium you find 0.66 mol/l of NO. alculate K. Solution Set of a table of concentrations [NOl] [NO] [l ] Initial.00 0 0 hange -0.66 +0.66 +0. Equilibrium 1.4 0.66 0. 41 Determining K NOl(g) qe NO(g) + l (g) [NOl] [NO] [l ] Initial.00 0 0 hange -0.66 +0.66 +0. Equilibrium 1.4 0.66 0. K [NO] [l ] [NOl] K [NO] [l ] [NOl] (0.66) (0.) (1.4) 0.080 4
Page 8 Typical alculations PROBLEM: Place 1.00 mol each of and I in a 1.00 L flask. alc. equilibrium concentrations. (g) + I (g) qe I(g) 4 (g) + I (g) q e I(g) K c 55. Step 1. Set up IE table to define EQUILIBRIUM concentrations. [ ] [I ] [I] Initial 1.00 1.00 0 44 hange K c [I] [ ][I ] 55. Equilib (g) + I (g) q e I(g) K c 55. 45 (g) + I (g) q e I(g) K c 55. 46 Step 1. Set up IE table to define EQUILIBRIUM concentrations. Step. Put equilibrium concentrations into K c expression. [ ] [I ] [I] Initial 1.00 1.00 0 hange -x -x +x K c [x] [1.00 - x][1.00 - x] 55. Equilib 1.00-x 1.00-x x where x is defined as am t of and I consumed on approaching equilibrium. K c (g) + I (g) q e I(g) K c 55. Step. Solve K c expression - take square root of both sides. [x] 55. 7.44 [1.00 - x][1.00 - x] x 0.79 Therefore, at equilibrium [ ] [I ] 1.00 - x 0.1 M [I] x 1.58 M x 1.00 - x 47 Nitrogen Dioxide Equilibrium N O 4 (g) q e NO (g) e 48
Page 9 49 50 Nitrogen Dioxide Equilibrium N O 4 (g) q e NO (g) K c [NO ] [N O 4 ] 0.0059 at 98 K Nitrogen Dioxide Equilibrium N O 4 (g) qe NO (g) K c [NO ] [N O 4 ] 0.0059 at 98 K If initial concentration of N O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an IE table [N O 4 ] [NO ] Initial 0.50 0 hange Equilib If initial concentration of N O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an IE table [N O 4 ] [NO ] Initial 0.50 0 hange -x +x Equilib 0.50 - x x Nitrogen Dioxide Equilibrium N O 4 (g) qe NO (g) Step. Substitute into K c expression and solve. K c 0.0059 [NO ] [N O 4 ] (x) (0.50 - x) Rearrange: 0.0059 (0.50 - x) 4x 0.009-0.0059x 4x 4x + 0.0059x - 0.009 0 This is a QUADRATI EQUATION ax + bx + c 0 a 4 b 0.0059 c -0.009 51 Nitrogen Dioxide Equilibrium N O 4 (g) qe NO (g) Solve the quadratic equation for x. ax + bx + c 0 a 4 b 0.0059 c -0.009 x -b ± b - 4ac a x -0.0059 ± (0.0059) - 4(4)(-0.009) (4) x -0.00074 ± 1/8(0.046) 1/ -0.00074 ± 0.07 5 Nitrogen Dioxide Equilibrium N O 4 (g) qe NO (g) x -0.0059 ± (0.0059) - 4(4)(-0.009) (4) x -0.00074 ± 1/8(0.046) 1/ -0.00074 ± 0.07 x 0.06 or -0.08 But a negative value is not reasonable. onclusion: x 0.06 M [N O 4 ] 0.050 - x 0.47 M [NO ] x 0.05 M 5 Solving Quadratic Equations Recommend you solve the equation exactly on a calculator or use the method of successive approximations 54
Page Solving Equilibrium Problems 55 Example 56 1. Balance the equation.. Write the equilibrium expression.. List the initial concentrations. 4. alculate Q and determine the shift to equilibrium. 5. Define equilibrium concentrations. 6. Substitute equilibrium concentrations into equilibrium expression and solve. 7. heck calculated concentrations by calculating K. (g) + F (g) q ef(g) K 1.15 @ some temperature..00 mols of each component is added to a 1.50 L flask. alculate the equilibrium concentrations of all species. I E (g) +.00 M F (g) qe.00 M F(g).00 M 57 But which direction does the reaction go? I E (g) +.00 M F (g) qe.00 M F(g).00 M [F] (.000) Q 1.000 [ ][F ] (.000)(.000) Q < K, so the reaction go towards the products Keep Going (g) + F (g) qe F(g) I.00 M.00 M.00 M -x -x +x E.00 - x.00 -x.00 + x [F] (.00 + x) K 1.15 [ ][F ] (.00 x)(.00 x).00 + x 1.15.00 x x 1.58 58 Finally x 1.58 So, [ ].00 1.58 0.47 M [F ].00 1.58 0.47 M [F].00 + (1.58) 5.056 M 59 Example (g) + F (g) q ef(g) K 1.15 @ some temperature..00 mols of in a.000 L flask 6.00 mols F in a.000 L flask alculate the equilibrium concentrations of all species. (g) + F (g) qe F(g) 60 I 1.00 M.00 M 0.00 M -x -x +x E 1.00 - x.00 - x x
Page 11 [F] (x) K 1.15 [ ][F ] (1.00 x)(.00 x) 1.11 (x ).45 (x) +.0 0 b ± b 4ac x a x.14 M or 0.968 M annot be.14 M Why? (Look at [ ]) x 0.968 61 Another #$%& Example! (g) + I (g) qe I(g) K P 1.00 The following gasses are mixed in a 5.000 L Flask P I 5.000-1 atm P 1.000 - atm P I 5.000 - atm heck Q 6 6 64 Q (P ) 1 I (P )(PI ) (1.000 atm)(5.000 Q 5.000 (5.00 atm) atm) Q>K, therefore the system will shift toward the reactants (g) + I (g) qe I(g) I 1.00-5.00-5.00-1 K P 1 (PI ) (5.00 - x) (P )(P ) (1.000 + x)(5.000 I Reduces to 1 (9.60 )x +.5x (.45 1 ) 0 + x) +x +x -x x.55 - E 1.00 - + x 5.00 - + x 5.00-1 -x I 1.00 - M +x (g) + 5.00 - M +x I (g) qe I(g) 5.00-1 M -x 65 ONE MORE NOl(g) q eno(g) + l (g) In an experiment in which 1.0 mol NOl is placed in a.0 L flask, what are the equilibrium concentrations? 66 E 1.00 - + x x.55 - atm Therefore at eq. P I 4.9-1 atm P 4.55 - atm P I 4.05 - atm 5.00 - + x 5.00-1 -x I [NO] [l K [NOl] NOl(g)q e 0.50 -x 0 +x ] 1.6 NO(g) + 0 +x 5 l (g) E 0.5 - x x x
Page 1 [NO] [l K [NOl] ] 1.6 5 (x) (x) (0.50 x) The problem is you end up with an x component, which are not fun to solve for But, since K is so small the reaction will not proceed that far to the right. Which means that x is a relatively small number. So x does nothing to 0.50 67 [NO] [l K [NOl] [NO] [l ] [NOl] ] 1.6 5 (x) (x) (0.50 x) (x) (x) (0.50) 5 K 1.6 Therefore, x 1.0-68 4x (0.50) I NOl(g)q e 0.50 0 NO(g) + 0 l (g) 69 EQUILIBRIUM AND EXTERNAL EFFETS 70 E -x 0.5 - x x 1.0 - Therefore at eq. [NOl] I 0.50 M [NO].0 - M [l] 1.0 - M +x x +x x Temperature, catalysts, and changes in concentration affect equilibria. The outcome is governed by LE ATELIER S PRINIPLE...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance. EQUILIBRIUM AND EXTERNAL EFFETS enri Le hatelier 1850-196 Studied mining engineering. Interested in glass and ceramics. 71 EQUILIBRIUM AND EXTERNAL EFFETS Temperature change ---> change in K onsider the fizz in a soft drink O (aq) + EAT qe O (g) + O(liq) K P (O ) / [O ] Increase T. What happens to equilibrium position? To value of K? K increases as T goes up because P(O ) increases and [O ] decreases. Decrease T. Now what? Equilibrium shifts left and K decreases. 7
Page 1 Temperature Effects on Equilibrium N O 4 (colorless) + heat qe NO (brown) 7 EQUILIBRIUM AND EXTERNAL EFFETS Add catalyst ---> no change in K A catalyst only affects the RATE of approach to equilibrium. 74 o + 57. kj K c [NO ] [N O 4 ] K c (7 K) 0.00077 K c (98 K) 0.0059 atalytic exhaust system 75 76 aber-bosch Ammonia Synthesis N Production N (g) + (g) qe N (g) + heat K.5 x 8 at 98 K Fritz aber 1868-194 Nobel Prize, 1918 arl Bosch 1874-1940 Nobel Prize, 191 77 EQUILIBRIUM AND EXTERNAL EFFETS oncentration changes no change in K only the position of equilibrium changes. Le hatelier s Principle 78 Adding a reactant to a chemical system. 1_slide78.mov
Page 14 79 80 Le hatelier s Principle Le hatelier s Principle Removing a reactant from a chemical system. Adding a product to a chemical system. 1_slide79.mov 1_slide80.mov 81 8 Le hatelier s Principle Removing a product from a chemical system. butane isobutane Butane- Isobutane Equilibrium K [isobutane] [butane].5 1_slide81.mov Butane Isobutane butane 8 Butane qe Isobutane 84 isobutane Assume you are at equilibrium with [iso] 1.5 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.5 Solution alculate Q immediately after adding more butane and compare with K. At equilibrium with [iso] 1.5 M and [butane] 0.50 M. K.5. Add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? 1_slide8.mov Q [isobutane] [butane] 1.5 0.50 + 1.50 0.6 Q is LESS TAN K. Therefore, the reaction will shift to the.
Page 15 Butane qe Isobutane You are at equilibrium with [iso] 1.5 M and [butane] 0.50 M. Now add 1.50 M butane. Solution Q is less than K, so equilibrium shifts right away from butane and toward isobutane. Set up IE table [butane] [isobutane] Initial 0.50 + 1.50 1.5 hange -x + x Equilibrium.00 - x 1.5 + x 85 Butane e Isobutane You are at equilibrium with [iso] 1.5 M and [butane] 0.50 M. Now add 1.50 M butane. Solution K.50 [isobutane] [butane] 1.5 + x.00 - x x 1.07 M At the new equilibrium position, [butane] 0.9 M and [isobutane]. M. Equilibrium has shifted toward isobutane. 86 EQUILIBRIUM AND EXTERNAL EFFETS 87 Le hatelier s Principle 88 Temperature, catalysts, and changes in concentration affect equilibria. The outcome is governed by LE ATELIER S PRINIPLE...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance. hange T change in K therefore change in P or concentrations at equilibrium Use a catalyst: reaction comes more quickly to equilibrium. K not changed. Add or take away reactant or product: K does not change Reaction adjusts to new equilibrium position Nitrogen Dioxide Equilibrium N O 4 (g) q e NO (g) 89 Nitrogen Dioxide Equilibrium N O 4 (g) qe NO (g) 90 e K c [NO ] [N O 4 ] 0.0059 at 98 K K c [NO ] [N O 4 ] 0.0059 at 98 K Increase P in the system by reducing the volume (at constant T). Increase P in the system by reducing the volume. In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). Therefore, reaction shifts LEFT and P of NO decreases and P of N O 4 increases. 1_slide89.mov