Chapter 13 Chemical Equilibrium. Equilibrium is Dynamic. The Equilibrium Constant. Equilibrium and Catalysts. Characteristics of Chemical Equilibrium

Transcription

1 Characteristics of Chemical Equilibrium John W. Moore Conrad L. Stanitski Peter C. Jurs Chapter 13 Chemical Equilibrium Many reactions fail to go to completion. [Reactants] stop decreasing, and [Products] stop increasing. The reaction reaches equilibrium. If there are more: products than reactants = product favored. reactants than products = reactant favored. Stephen C. Foster Mississippi State University Equilibrium is Dynamic Equilibrium is Independent of Direction of Approach Reactants convert to products a A + b B c C + d D BUT, products can also react to make reactants. N 2 (g) (g) 2 N 3 (g) Equilibrium occurs when [A], [B], [C] and [D] stop changing - a dynamic equilibrium: Species do not stop forming OR being destroyed. Rate of formation = rate of removal. Concentrations are constant. or 2 mol N Start with either 3 1 mol N mol 2 and get the same equilibrium mixture. Equilibrium and Catalysts The Equilibrium Constant Catalysts do not affect equilibrium concentrations. Catalysts: speed up the forward reaction but also speed up the reverse reaction. decrease the time needed to reach equilibrium. 2-butene isomerization: C C 3 C C 3 At equilibrium: rate forward = rate reverse C 3 C C C 3 1

2 The Equilibrium Constant The isomerization is elementary, so: k forward [cis] = k reverse [trans] k or = [trans] forward [cis] k reverse This ratio is named the equilibrium constant, K c : k forward k reverse [trans] K c = = = 1.65 (at 500 K) [cis] The Equilibrium Constant For a general reaction: a A + b B k forward k reverse K c = = [C] c [D] d [A] a [B] b c C + d D Products raised to stoichiometric powers divided by reactants raised to their stoichiometric powers c for concentration based The Equilibrium Constant Equilibria Involving Pure Liquids & Solids Examples [Solid] is constant throughout a reaction. N 2 (g) + O 2 (g) 2 NO(g) K c = [NO] 2 [N 2 ] [O 2 ] pure solid concentration = density mol. wt g / L g / mol 1 S 8 (s) + O 2 (g) SO 2 (g) K = [SO2] 8 c [O 2 ] S 8 is ignored. All pure solids are omitted from K c. [S 8 ] = d S 8 / mol. wt S 8 d and mol. wt. are constants, so [S 8 ] is constant. This constant factor is absorbed into K c. Pure liquids are omitted for the same reason. Equilibria in Dilute Solutions Water is omitted from K c in dilute solution reactions: It is in large excess (pure water = 55.5 M). Even if consumed or produced, [ 2 O] constant. The constant factor is incorporated into K c. N 3 (aq) + 2 O(l) K c = [N 4 + ][O - ] [N 3 ] N 4+ (aq) + O - (aq) = 1.8 x 10-5 (at 25 C) (Units are customarily omitted from K c ) The Equilibrium Constant What is the equilibrium constant for: Si 4 (g) + 2 O 2 (g) SiO 2 (s) O(g)? Omit SiO 2 (a solid) Include 2 O (a gas) Write down K c for: 2 NaO(aq) + 2 SO 4 (aq) Na 2 SO 4 (aq) O(l) Omit 2 O (aq. solution) [ 2 O] 2 [Si 4 ][O 2 ] 2 [Na 2 SO 4 ] [ 2 SO 4 ][NaO] 2 2

3 K c for Related Reactions N 2 (g) (g) 2 N 3 (g) K c = [N 3] 2 [N 2 ][ 2 ] 3 K c for Related Reactions N 2 (g) (g) 2 N 3 (g) Change the stoichiometry change K c K c = [N 3] 2 [N 2 ][ 2 ] 3 ½ N 3 2 (g) (g) N 3 (g) K c = [N 3] [N 2 ] ½ [ 2 ] 3 2 reaction divided by 2 This new K c is the square root of the original K c. Multiply an equation by a factor Raise K c to the power of that factor. Reverse a reaction Invert K c : 2 N 3 (g) N 2 (g) (g) K c = [N 2][ 2 ] 3 [N 3 ] 2 K c for a Reaction that Combines Reactions If a reaction can be written as a series of steps: step 1 step 2 overall N 2 (g) + O 2 (g) 2 NO(g) + O 2 (g) 2 NO 2 (g) N 2 (g) + 2 O 2 (g) The overall K c is the product of the steps: K c (step1) x K c (step2) = 2NO(g) 2 NO 2 (g) [NO]2 [N 2 ][O 2 ] K c = [NO 2 ] 2 [NO] 2 [O 2 ] [NO]2 [N 2 ][O 2 ] K c = [NO 2] 2 [NO] 2 [O 2 ] K c = [NO 2] 2 [N 2 ][O 2 ] 2 = [NO 2] 2 [N 2 ][O 2 ] 2 Equilibrium Constants in Terms of P In a constant-v reaction, partial pressures change as concentrations change. At constant T, P is proportional to molar conc.: Ideal gas: and for gas A: PV = nrt P A V = n A RT n P A = A RT = [A]RT V P is easily measured, so another K is used Equilibrium Constants in Terms of P For a gas-phase reaction: a A(g) + b B(g) c C(g) + d D(g) Equilibrium Constants in Terms of P Calculate K p from K c for: N 2 (g) (g) 2 N 3 (g) K c = 5.8 x 10 5 at 25 C. In general: p for pressure based K p = P C c P D d P Aa P B b K p = K c (RT) Δngas K p T Δn gas = K c (RT) Δngas = = 298 K = 2 (3 + 1) = 2 Use R in L/mol units. [ ] has mol/l units. K p = 5.8 x L atm mol -1 K -1 (298 K) -2 with Δn gas = n gaseous products n gaseous reactants = (c + d) (a + b) = 9.7 x 10 2 mol 2 L -2 atm -2 K p = 9.7 x 10 2 atm -2 mol and L units are assumed to cancel (units omitted from K c ). 3

4 Determining Equilibrium Constants If all the equilibrium concentrations are known it s easy to calculate K c for a reaction. Example Consider: 2 A (aq) B (aq) At equilibrium [A] = 2.0 M & [B] = 4.0 M. What is K c? k K c = = [B] forward [A] 2 k reverse Determining Equilibrium Constants In other cases stoichiometry is used to find some concentrations. Example 2 (g) and I 2 (g) were added to a heated container. [ 2 ] initial = mol/l and [I 2 ] initial = mol/l. At equilibrium [I 2 ] = mol/l. Determine K c. 4.0 K c = = 1.0 (2.0) 2 2 (g) + I 2 (g) 2 I(g) Determining Equilibrium Constants Write a balanced equation; fill in [ ] initial : 2 (g) + I 2 (g) 2 I(g) [ ] initial Use stoichiometry to predict changes: [ ] change -x 1 2 2I so 2 lost = x Add [ ] initial and [ ] change to get [ ] equilib -x 2x 1I 2 2I so I 2 lost = x [ ] equilib x x 2x Let I made = 2x Determining Equilibrium Constants 2 (g) + I 2 (g) 2 I(g) [ ] equilib x x 2x Since [I 2 ] = M at equilibrium = x x = M [ 2 ] equilib = ( ) = M [I 2 ] equilib = ( ) = M [I] equilib = 2( ) = M [I] K c = 2 = [ 2 ][I 2 ] ( ) 2 ( )( ) = Meaning of the Equilibrium Constant When: K c >> 1 Reaction is strongly product favored. very little reactant remains. often written as a forward reaction only. assume reaction goes to completion. K c << 1 Reaction is strongly reactant favored. very little product forms. usually written as no reaction or NR. K c 1 Reactants & products present at equilibrium. use equilibrium methods discussed here. Predicting the Direction of a Reaction The reaction quotient, Q can predict the direction of a reaction: For the reaction: aa + bb cc + dd Q = [C]c [D] d [A] a [B] b Q looks identical to K c, BUT the concentrations are not equilibrium values. (K c = Q whenever a system is at equilibrium) 4

5 Predicting the Direction of a Reaction Predicting the Direction of a Reaction Q = [Products] [Reactants] Changes as a reaction moves to equilibrium K c = [Products] equilib [Reactants] equilib Constant! (If T is constant) If Q < K c, Q must increase to reach equilibrium. make more product (and less reactant). move forward. If Q > K c, Q must decrease to reach equilibrium. make less product (and more reactant). move back. Predicting the Direction of a Reaction 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) K c = 245 at 1000 K Predict the direction of the reaction if SO 2 (0.085 M), O 2 (0.100 M) and SO 3 (0.250 M) are mixed in a reactor at 1000 K. Q = [ SO 3 ] 2 [SO 2 ] 2 [O 2 ] (0.250) 2 = (0.085) 2 = 86.5 (0.100) Q < K c, Q must increase to reach equilibrium. Need more product (less reactant). Forward direction. If K c is known, [ ] equilib can be calculated. Problem 2 (g) + I 2 (g) 2 I(g) mol of I is placed in an empty 1.00 L flask at 600. K. What will be [I], [ 2 ] and [I 2 ] at equilibrium? K c = 76 at 600. K. 2 (g) + I 2 (g) 2 I(g) [ ] initial Change on reaching equilibrium: [ ] change +x 1 2 2I +x 1I 2 2I -2x [ ] equilib +x +x x Amount lost 76 x 2 = ( x) 2 Take the square root of both sides: x = ±( x) x = or x = [I] K c = 2 = [ 2 ][I 2 ] ( x) 2 (x)(x) x = 4.67 x 10-3 or x = x 10-3 Since K c = x 2 = ( x) 2 Ignore the negative root (x cannot be negative). 5

6 So: [ 2 ] equilib = x = M [I 2 ] equilib = x = M [I] equilib = x = M It s a good idea to check your work: Matches K c given in the problem Enough A is added to an empty container so that [A] initial = M. What are [A], [B] and [C] at equilibrium? A(g) B(g) + C(g) K c = [ ] Initial [ ] change -x x x [ ] equilib ( x) x x [I] K c = 2 ( M) = 2 = 76. [ 2 ][I 2 ] ( M)( M) [B][C] K c = = [A] (x)(x) ( x) x x = 0 x = 2 ( x) A quadratic equation ax 2 + bx + c = 0 has two roots (solutions): x = -b ± b 2-4ac 2a ere: x = 1x x = 0 a b c ± (2.500) 2 (4)(1)( ) 2(1) x = M or M Ignore the negative root (negative concentration!) Then x = M, and: [ A ] = x = M [ B ] = x = M [ C ] = x = M Most of the A is converted to products. [B][C] Check: = [A] (0.093)(0.093) = 2.5 = K c Le Chatelier s Principle Changing Concentrations a A + b B c C + d D If a system is at equilibrium and the conditions are changed so that it is no longer at equilibrium, the system will react to reach a new equilibrium in a way that partially counteracts the change A system at equilibrium resists change. If pushed, it pushes back Add A (or B) and the system adjusts to remove it. more product is created. the reaction shifts forward. Remove A (or B) the system adjusts to add more. more reactant is created. the reaction shifts backward. 6

7 Changing Concentrations 3 C C 3 C=C 3 C C=C C 3 Changing V or P in Gaseous Equilibria K does not change if P or V change. Equilibrium position may change. Consider: N 2 O 4 (g) 2 NO 2 (g) If V doubles, all concentrations divide by 2: 5 moles of NO 2 in 1 L, goes to 5 moles in 2 L. [NO 2 ] = 5 M becomes [NO 2 ] = 2.5 M. K c =Q= [ NO 2 ] 2 [ N 2 O 4 ] At equilibrium (½[ NO Q= 2 ]) 2 ¼[ NO = 2 ] 2 ½[ NO = 2 ] 2 = ½K ½[ N c 2 O 4 ] ½[ N 2 O 4 ] [ N 2 O 4 ] After V change, Q < K c. Equilibrium shifts to products Changing V or P in Gaseous Equilibria Consistent with Le Chatelier Changing V or P in Gaseous Equilibria P and V changes have no effect if Δn gas = 0: N 2 O 4 (g) 2 NO 2 (g) 2 (g) + I 2 (g) 2 I(g) V doubled = lower concentration. Minimize change by: Making more molecules (increase concentration). Shifting toward products; convert 1 molecule into 2. If V is doubled, the system cannot adjust (there are 2 gas molecules on each side). P doubled. Minimize the change by: Removing molecules (decrease P). Shifting toward reactants; convert 2 molecules into 1. K c =Q= [I] 2 [ 2 ][I 2 ] At equilibrium (½[I]) Q = 2 ¼[I] = 2 [I] = 2 ½[ = K c 2 ]½[I 2 ] ¼[ 2 ][I 2 ] [ 2 ][I 2 ] After V change, Q = K c still at equilibrium! Changing V or P in Gaseous Equilibria Adding an inert gas to change P has no effect. Total P increases V does not change, nor do n reactants or n products Changing Temperature A change in T changes K c. K c may increase or decrease if T is increased. N 2 (g) + O 2 (g) 2 NO(g) Δ = kj [reactant] and [product] do not change. P A = n A V RT = [A]RT T ( C) K c x x x 10-3 K c increases as T increases. Explained by LeChatelier 7

8 Changing Temperature N 2 (g) + O 2 (g) 2 NO(g) Δ = kj T increased? Minimize by removing heat. Forward reaction is endothermic. Equilibrium shifts forward, removing heat. T reduced? Add heat. Reverse reaction is exothermic. Equilibrium shifts in reverse, releasing heat. Changing Temperature Exothermic reactions: K c decreases as T increases. Are less product favored at higher T. Endothermic reactions: K c increases as T increases. Are more product favored at higher T. The aber-bosch Process The aber-bosch Process Production of N 3 for fertilizer from atmospheric N 2 N 2 (g) (g) 2 N 3 (g) Process developed by Fritz aber (chemist) and Carl Bosch (engineer). Perfected in Germany could not import guano from S. America and needed to make explosives N 2 (g) (g) 2 N 3 (g) Δ = kj It is: Exothermic = product favored at low-t, but very slow at room-t, and must be heated. Carried out at high-p (200 atm), to favor product. Uses a catalyst to speed the reaction. Run at 450 C; N 3 is continually liquefied and removed. The aber-bosch Process 8