Chern 12 Notes II. 1 and 11.2 - Dynamic Equilibrium Goals: 1. 2. 3. 4. Realize that reactions can go both in forward and in reverse. Define equilibrium. Understand the concept of dynamic equilibrium. State the characteristics of a system at equilibrium We know that many reactions are REVERSIBLE! CReactants -;.. Products) or CReactants ~ Products) I Reactants form Products I Products form Reactants A system is said to be at IIEquilibrium lj when the ov vj C\("t\ (")(. r Cl \<...; EQUALSthe (LVt'f':::l{ '('1= ra1l! use, ~, If o(" 1::.;..--->,-\-O ShGMJ \ lv204 ~ 2 J.V02 \ Ex: Consider the reaction below in a CLOSEDsystem: CO\CUI(\ S\ b... 0v0Y\ )~ > 9a~,,3 Cl~. collide '-IHh each other, the bonds between the ;'N"s break 1Meach molecule splits up into h/o molecules of NOZ 1
- Imagine we are watching this reaction in the lab. We are measuring the rate of the forward reaction by measuring the il[reactant]. 11 c: N 2-0 lot1 At the beginning of the reaction (t= 0), there is a h~h [N 2 0 4 ] fj i,.,'~e, A graph of the forward reaction would.io?k like this:. '. _. '\ 5k~p.s\~ ( hljh C ~te'-l~'j... h \9h {C\t<- ) 0.6 [ J 0.4 0.3 ~-~) 0.2 0.1 0.0 (} 2 4 5 6 7 B 9 11} TIME (min) - As the reaction starts, the reactant is consumed so [N 2 0 4 ] de.lvzotu.s and the rate ~oes clo\.,0r\ (fewer reactant molecules, fewer collisions). - As slope of the line of the graph will get more gradual (less steep) Let's stop here and consider the REVERSEreaction! ~..-1Q\ ~ 2 molecules of N02 collide vitothe proper N'Ierg'J ~nd cotttsten geometry and form a molecule of N 0 2 4 2
If we graphed the rate of the reverse reaction, it would look like this: 0,6.,.--------------------, At t= 0, there is no N0 2 NO,l. \nq~ \\ \:MI' \,t \.Ilf' II. 0~.1 SO,S ra\z h'j\'\ fe.\{.e.v~ 0,4 0.3 0,2 ~ RA.TE (reverse) o 12345578 TIME (minl 9 10 As N0 2 accumulates, those molecules can collide and form N 2 0 4. Around t= 3, the rate begins to slow Let's compare the forward and reverse reaction rate graphs: 0.6 0.5 ~ RA.TE (forward) 0,- 0.4-0.3 0.2 0.1 R.=nE (reverse) 0.0 0 :2 4 6 8 10 12 TIME (mill) At t = 4 min, the rates are E.QU \ L Rate of forward reaction = rate of reverse reaction This system is at r.\\.i Y'\ C\ VY\ \ G equilibrium! ~[1 N02 is being used up at the SAME rate as it is being formed Ii -\-\{y,-l N204 is being formed at the SAME rate as it is being consumed 'o-e.c~u.~g 1V204 3 ~ "" 2 J.V02 C N 'J-0ltj'":: LenS \-UI\-tc:. NO..l. z. Lo" S-\"c;\.nI
Things you need to know about DYNAMIC EQUILIBRIUM:. \ ( t:\j-t n-yhou.j h S\Dpe = 0) The reaction has not.s+0 ppe.a Forward and reverse reaction continue at ~q.:\a C\ \ (" Q+e..s:. Changes are microscopic. No m C\ ( \f O~ Wp\ C changes occur (no visible, large scale changes so it appears as if nothing is happening). All observable properties are LJJ("lStCi(\\ (temp, [react], [prod], pressure, colour),\ " A system at equilibrium is a C, L...-OS ED II system. If we changed the temp, it would affect the equilibrium. -\-{.\'Y,p,S tch"\6tc\(\-t- 4 Characteristics ora SYstem at Equilibrium 1. The rate of the forward reaction = The rate of the reverse reaction 2. Microscopic processes. No macroscopic changes 3. The system is closed and the temperature is constant and uniform throughout. 4. The equilibrium can be approached from the left (starting with reactants) or from the right (starting with products). If sufficient Ea is available, systems not at equilibrium will tend to move towards equilibrium. http://www.absorblearning.com/media/item.action?guick=w8 STATIC EQUILIBRIUM: is the opposite of dynamic equilibrium. It will not move at all unless it is "pushed" in some way. Essentially, nothing is happening (system is balanced but particles are at rest).,~~rb\{s S1Ctt\Ofla~ rngt(b\q~ (O\'3-X- LASUC\.\\'1 ~ S L-U. So vel -"-""~-7- /) V------------- '. (' L..J Ii'::. - ". D r nv\ S 0») i L52. o ~ 0 J 4 ~ \-:J LS,. I <) '. \.l 0 )1 ~ SfA-l \L: ZI s» rv'a M\.L.
At equilibrium... [reactants] -::/=- [products] (not equal in general) Analogy: if you have coins in your pockets (3 twoonies, 20 loonies) ~ -\-""<'1 CD\J\ \ 6\ b.u SOn'l0 CAS-e..S. *~b\;\ e XCh a"j0 1-T ~( 9.. L. * -~h-({\ '10lA e..)(cy\{l.('jl.+\t\-( O\-h-e.\{ uj Cl~ ~L~1-'T A mou(\-\-d~ Ch C\,Y'"\9 e. \ n -e,«c,\'"'\ POL"-e -\- \.s Uy\L\l C\V"j-{ J. bv\:\- C\.-M DV\ n t 0r ~ c.ctv'\ ~\ ~r. - Overtime, the back and forth rate is the same. The amount of money is each pocket has not changed. But the amount of money in each pocket is different! [two~nies] doesn't equal [loonies]. \~S. f0hr Ex: A < ---- > B (See page 40, Q 6), po (J(-er [J Rate forward = kforward[reactants] ~ 0,60 See the table of data. a) Plot the values of [A] vs time, and [B] vs time: \ \L \:D O:1J o Jo 0.'-\ O 'L 0 \ Rate reverse = kreverse [Products] '-V--,-l O~\ 0 C 5'1 : [3\,\:\I~ l.(f() J o 'L ~ '4 Co "1 ~ q \D +l (Y\ ~ l\'y\ ~(\ )
b) When are th,e RATESequal? (look at d~ta) ') L~au\tl~ @ '0 m \" ('00 \-\-\ ra-\-(.~ -:. D.. \ 00 What does this look like on the graph? Are the concentrations the same? @ 1) m;{\.}bd~r, \ '\("\-{S CAr c "\..e v.e.\' '. L A~ ~ Lt>,(, &\ C\ (\-t ) C \SJ e; co (\&-\ C4(y\=. _ c) When does [A] = [B]? o.-\- \ m,' (\ L AJ =F C Bj Isthe systems at equilibrium when [A] = [B]7_N 0. f G\ k.s oj e nd-t ecr-lata \ d) When is the forwa rd rate the greatest? ~. G\ I{\j in (~ (h;,,bh C (-\-'J ) e) What is the numerical value of the ratio [products]/ [reactants] at equilibrium? \' 000-0,.2-00 Ex 2) Hz + Brz < ---- > 2 HBr Will the ratio of [HBr]/ [H 2 ] be 2/ 1 \~ \}'0 O\i- ec\v\~v, [?m ~1 [ \{ ec\ l\j. N t n-l.(( e~~c\(\'~ (or 2:1) at equilibrium? \...-:.-:...::O~ ---"? 'D-e..~--tY\6's 0'(\ -\-\1\ 0 -.. L\-\B'f] O\~C\ [\-\;=J ~"'(\D\ \-\ 1- \Xt) c\ U (p~ ~ rrxo \ f'(\o \ \-\ 6y f etactl' 03. _ ('(\0 '\ \-\'- «fl.ct ch (1j \4Br.~ - \ But it is true to say that ratio of moles of HBr reacting to moles of H2 reacting is 2: 1 Do Questions p. 43 #8-13 in Hebden 6