Chem_2_Ch_6

Chem_2_Ch_6 - Chapte 6 r Che ical e m quilibrium Please...

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Unformatted text preview: Chapte 6 r Che ical e m quilibrium Please note: Wewill not cove any of them rial in chapte 6 r ate r re d to re gase activitie or e late al s, s, quilibrium constants involving gas pre ssure ; re wedid s call, not cove C r hapte 5. r Equilibrium aA + bB PhaseC hange H2O (l) Gas Phase 2 NO2 (g) H O (g) 2 c C+ d D N O (g) (g H O (aq) + C H O (aq) 2 3 2 Aque ous HC H3O2(aq) + H2O(l) 2 He roge ous te ne Ag+ (aq) + C - (aq) l AgC (s) l Law of Mass Action aA + bB Re action Quotie nt Q< K Products c c C+ d D Products [C [D] ] Rea [B] b [A]actants d Q= = K(T) at Equ. Q> K Re actants Q= K EQUI LI BRI UM Pureliquids and solids do not appe in Q or K ar H2O(l) would not beinclude in Q or d K, but H2O(g) would be . AgC would not beinclude in K, l(s) d which is give thespe sym Ksp. n cial bol Changes in concentration with time for the reaction: H2O(g) + CO(g) H2(g) + CO2(g) Q K =Q Meaning of equilibrium constant aA + bB c c C+ d D Products [C [D] ] Rea [B] b [A]actants d at e qu., K= K > 1 m ans products pre inate re e dom , action goe s m ostly "forward" K < 1 m ans re e actants pre inate re dom , action goe s m ostly "re rse ve " Question Theorganic com pound 2-bute consists of two form ne s, cis-bute and trans-bute , that arein e ne ne quilibrium : cis-bute trans-bute ne ne S upposethee quilibriumconstant for this re action is K = 3.0. I f theconce ntration of cis-bute at e ne quilibriumis 1.0 M, what is theconce ntration of trans-bute ? ne Answer K= [trans-bute ] ne [cis-bute ] ne [trans-bute ] ne 1.0 M [trans-bute ] = 3.0 M ne = 3.0 = 3.0 Question 2NOCl(g) 2NO(g) + C 2(g) l At e quilibrium theconce , ntrations are : [NOCl] = 1.0 M [NO] = 0.025 M [C 2] = 0.050 M l What is thee quilibriumconstant for there action? Answer 2NOC 2NO(g) + C 2(g) l(g) l K = [NO] 2[C 2]/[NOCl] l Question 2NOC 2NO(g) + Cl2(g) l(g) Thee quilibriumconstant is K = 3.1 x 10-5. I f [NOC = 1.0 M and [Cl2] = 0.050 M at e l] quilibrium , what is [NO] at e quilibrium ? Answer 2NOC 2NO(g) + C 2(g) l(g) l I na na na R na na na E 1.0 x 0.050 Answer Cont. K = [NO] 2[C 2]/[NOCl] 2 = 3.1 x 10-5 l Re that [NO] = x call x2(0.050)/(1.0)2 = 3.1 x 10-5 x2 = 0.00062 x = [NO] = 0.025 M Question Conside a re r action A + B C+ D (all gase in a close containe s, d r) Le there t action com to e e quilibrium . The add som e A. n, e xtra What happe to theconce ns ntrations of B, C, and D (incre or de ase ase cre ?) K= [C ]eq [ D]eq [ A]eq [ B ]eq [C ][ D] Q= [ A][ B ] Answer K= [C ]eq [ D]eq [ A]eq [ B ]eq [C ][ D] Q= [ A][ B ] A + B C+ D Adding A incre s de ., m s Q < K. ase nom ake S eA m re with B to re stablish e om ust act -e quilibrium . Ne e w quilibriumhas highe [C and [D] r ] Ne e w quilibriumhas lowe [B]. r Question There action I (s) I (g) is at e quilibrium What will . 2) S e am 1) I ncre ase 3) De ase cre Answer K= K is I nde nde of solid pe nt I (s) I (g) Thecolor inte nsity of thegas will stay thesam . e conc. of I (g) 1) I ncre s ase 2) S e am 3) De ase cre s Question H2(g) + I 2(g) 2HI (g) C onside two e r xpts pe rform d in rigid containe of thesam size e rs e and at thesam te pe e m rature : A) Add 1 m of e re ole ach actant (H2 and I 2), allow syste to re m ach e quilibrium the add anothe 1 m of H2 and re e , n r ole ach quilibrium again B) Add 1 m of I 2 and 2 m s of H2 and allow syste to re ole ole m ach e quilibrium Will final com position of sam s (in A and B) bediffe nt? ple re Answer H2(g) + I 2(g) 2HI(g) I n e casetheam ach ounts of I 2 and H2 adde arethe d sam . K is a constant , so thesam e e e quilibrium com position will bere d re ache gardle of how m ss any ste it took to add there ps actants. Question Thevalueof thee quilibriumconstant, K, de nds on which pe of thefollowing? A) I nitial conce ntrations of products B) I nitial conce ntrations of re actants C) Te pe m rature D) C m ide he ical ntity of re actants & products Answer Thevalueof thee quilibriumconstant, K, de nds on which pe of thefollowing? A) I nitial conce ntrations of products B) I nitial conce ntrations of re actants C) Te pe m rature D) C m ide he ical ntity of re actants & products Question Theconce ntrations of products at e quilibriumde nd on pe which of thefollowing? A) I nitial conce ntration of products B) I nitial conce ntration of re actants C) Te pe m rature D) C m ide he ical ntity of re actants & products Answer Theconce ntrations of products at e quilibriumde nd on pe which of thefollowing? A) I nitial conce ntration of products B) I nitial conce ntration of re actants C) Te pe m rature D) C m ide he ical ntity of re actants & products Question A(g) + B(g) C (g) K = [C]/[A][B] = 2.0 At e quilibriumin a 1.0 L containe [A] = 2.0 M, [B] = r, 1.0 M, [C = 4.0 M ] Now, add anothe 3.0 m s of B. r ole What will thene e w quilibriumconce ntrations of A, B and Cbe ? Answer A(g) + B(g) C (g) I 2.0 1.0 +4.0 4.0 2.0 3.0 4.0 R -x -x +x E 2.0 x 4.0 - x 4.0 + x K = [C ]/[A][B] = 2.0 Answer-cont. K = [C ]/[A][B] 2 = (4+x)/(2-x)(4-x) 2 = (4+x)/(8-6x+x2) 16 12x + 2x2 = 4 + x 2x2 13x + 12 = 0 solvequadratic e quation x = (13 8.54)/4; only sign give positiveconce s ntrations of [A], [B] and [C so ], x = 1.12; now calculatetheconce ntrations: [A] =2.0-x, [B]=4.0-x and [C ]=4.0+x [A] = 0.9 M, [B] = 2.9 M, [C] = 5.1 M Question A(g) + 3B(g) 2C (g) I n a 1.0 L containe theinitial conce r, ntration of [A] = 4.0 M, [B] = 8.0 M and [C = 0 M. ] At e quilibrium[C = 4.0. ] C alculateK? Answer A(g) + 3B(g) 2C (g) I 4.0 8.0 R -2.0 -6.0 ? E 2.0 2.0 ? 0.0 +4.0 4.0 4.0 K = [C /[A][B] = (4) /(2)(2) = 1.0 ] 2 3 2 3 Question S 5(g) S l3(g) + C 2(g) bCl bC l Whe 89.7 g of S l5 is place in a 15.0-L containe at n bC d r 180C 29.2%of theS 5 has de pose whe thesyste , bCl com d n m has re d e ache quilibrium . CalculateK at this te pe m rature . S 5 MW= 299.00 g/m ; [S l5] o = 0.0200 bCl ole bC Answer S l5(g) S l3(g) + Cl2(g) bC bC I 0.0200 0 0 R -0.00584 +0.00584 +0.00584 E 0.0142(0.02-29.2%) ?0.00584 0.00584 ? K = [S l ][C ]/[S l ] = bC l bC 3 2 5 Finding equilibrium constants for "new" reactions Rxn 1: aA + bB cC+ dD Rxn 2: cC+ dD aA + bB [C ] [ D] K1 = a b [ A] [ B ] a b c d [ A] [ B ] 1 K2 = = c d [C ] [ D] K1 Re rsere ve action inve Keq rt Keq for new reactions--cont. Rxn 1: A+BC Rxn 2: A + C D Rxn 3: 2A + B D [C ] K1 = [ A][ B] [ D] K2 = [ A][C ] [ D] K3 = = K1 K 2 2 [ A] [ B ] S re um actions m ultiply Keq Keq for new reactions--cont. Rxn 1: 2A + B C Rxn 2: 4A + 2B 2C [C ] K1 = 2 [ A] [ B ] [C ] 2 K2 = = K1 4 2 [ A] [ B ] 2 Multiply re action by constant raiseKeq to powe (Notee r xpone in K ). nts Le Chtelier's Principle: I f som stre is brought to be on a syste in e ss ar m e quilibrium thee , quilibriumis displace in the d dire ction which te to undo thee ct of the nds ffe stre ss. Equilibrium and Le Chtelier's Principle applied to Gases 2NO (g) N O (g) Q= [N O ] =K at Equilibrium(T) Le Chtelier's Principle Q = [N2O4]/[NO2] 2 Q= K Q> K 2NO V (g) (m cule ole s le crowde ss d) Q= K N O (g) Rxn runs in re rse ve (m s m ake ore gas particle s) Question A(g) + B(g) C + D(g) (g) Am ixtureof A, B, C, and D is allowe to re d ach e quilibriumin a close containe d r. Now, add m B and allow thesyste to re a ne ore m ach w e quilibrium . What happe to theconce ns ntration of C? A) Incre ase B) S thesam C) De ase tay e cre Answer A(g) + B(g) C + D(g) (g) Adding m of a re ore actant cause thesyste 's s m e quilibriumto shift so as to formm product. ore A) Incre ase B) S thesam C) De ase tay e cre Question A(g) + B(g) C + D(g) (g) Am ixtureof A, B, C, and D is allowe to re d ach e quilibriumin a close containe d r. Now, re ovesom of theD and allow thesyste to m e m re its ne e ach w quilibrium . What happe to theconce ns ntration of C? A) Incre ase B) S thesam C) De ase tay e cre Answer A(g) + B(g) C + D(g) (g) Re oving som of oneproduct cause thesyste 's m e s m e quilibriumto shift so as to formm products (i.e to ore . re placeD). A) Incre ase B) S thesam C) De ase tay e cre Question--Le Chtelier's Principle Hydrofluoric acid is a we acid: ak HF(aq) H+(aq) + F-(aq) Bariumfluoride BaF2, is a solid with ve low , ry solubility. What will happe to theH+ conce n ntration if Ba(NO3)2 is adde to an aque solution of HF? d ous A) Incre ase B) S thesam C) De ase tay e cre Answer HF(aq) H+(aq) + F-(aq) Thefollowing re action re ove F-, a product of this m s re action, fromsolution: Ba2+(aq) + 2F-(aq) BaF2(s) the forethee re quilibriumshifts to m m product ake ore (m H+). ore A) Incre ase B) S thesam C) De ase tay e cre Equilibrium/Heat Most re actions e r producehe (e ithe at xothe ic; he is a rm at "product") or absorb he (e at ndothe ic; he is a rm at "re actant"). Exothe ic: com rm bustion of gasoline 2C H18(g) + 25O2(g) 16C 2(g) + 18H2O(g) + he O at 8 Endothe ic: vaporization of wate rm r H2O(l) + he H2O(g) at Question Thete pe m raturefor 2NO (g) N O (g) is raise What happe to the d. ns color inte nsity at e quilibrium ? (Rxn is e xothe ic, produce rm s he at) 1) I ncre ase 2) S e am 3) De ase cre Answer 2 NO (g) K N O (g) + he at K e xothe ic rm (produce he s at) e ndothe ic rm (consum s he e at) T T 1) I ncre ase 2) S e am 3) De ase cre Ammonia Synthesis (Habe r-Bosch 1910) N2 (g) + 3 H2 (g) 2 NH3 (g) + heat Que stion: Rxn is carrie out at low T, but it's too slow at d low T. How do weim provetheyie ld? Answe usehigh P to driveRxn to product (product sidehas r: fe r m cule we ole s). Le Chtelier's summary actant / product shifts e quilibriumtoward Adding re products/ re actants m actant / product shifts e quilibriumtoward Re oving re re actants/ products asing/ de asing volum shifts e cre e quilibriumtoward side I ncre with m / fe r gas m cule ore we ole s m ratureshifts e quilibriumtoward Raising te pe re actants/ products for e xothe ic/ e rm ndothe ic re rm actions Le Chtelier's summary--cont. Position of e quilibriumis not affe d by: cte m m cie s Adding or re oving a che ical spe s that doe not re with spe s involve in thee act cie d quilibrium(e .g. spe ctator ions) e I ntroducing a catalyst (spe ds up both forward and re rsere ve actions, so it doe changeposition of sn't e quilibrium ) ...
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This note was uploaded on 04/27/2009 for the course CHEM chem1A taught by Professor Hooker during the Spring '09 term at UCSB.

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