Chapter 15 Chemical Equilibrium

Transcription

1 Chapter 5 Chemial Equilibrium 5. The Conept of Equilibrium Figure: 3. from Chemistry by MMurray & Fey Figure 3.(a) NO 4( g) NO( g) olorless brown we start with reatant, N O 4, so the solution is olorless as time progresses we generate produt, NO after a ertain length of time: the onentration of N O 4 stops dereasing the onentration of NO stops inreasing the solution goes bak and forth between lear and brown at this point the forward rate = reverse rate the prodution of NO and the onsumption of N O 4 is equal Figure 3. (b) NO( g) NO4( g) brown olorless this time we start with NO as time progresses we one again reah a point in whih: the onentration of NO stops dereasing the onentration of N O 4 stops inreasing the solution goes bak and forth between brown and lear at this point the forward rate = reverse rate the prodution of N O 4 and the onsumption of NO is equal this is an example of hemial equilibrium or dynami equilibrium: a phenomenon in whih the onentrations of reatants and produts remain onstant over time NOTE: this does not mean that the onentrations go to zero we an rewrite the above eqn as NO 4( g NO ) ( g ) 5. The Equilibrium Constant Evaluating [ C] [ D] aa bb C dd a b [ A] [ B] this is alled the equilibrium onstant expression or mass ation expression we all the equilibrium onstant sine there is a ertain set of onentrations for the speies in our general equation whih orresponds to equilibrium law of mass ation: the hemial equilibrium expression will give rise to a harateristi value for a given temperature d

2 Example: Write the equilibrium expression for the forward and the reverse reations for the following: a.) 4A 3B C D [C][D] [A][B] [A][B] [C][D] 4 3 ' 4 3 b.) 3H N NH3 3 [NH 3] ' [H ] [N ] 3 [H ] [N ] [NH 3] In both examples the equilibrium onstant in the forward diretion is the inverse of the equilibrium onstant in the reverse diretion: ' this means that when we reverse an eq reation we must take the inverse of it's eq onstant to get the reverse eq onstant Equilibrium Constants in Terms of p d ( PC) ( PD) if A, B, C, and D are gases then we would write: p a b ( PA) ( PB) Example: What is and p for the following reation? NO Cl NOCl ( g) ( g) ( g) ( PNOCl ) p NO Cl [ NOCl] [ NO] [ Cl ] ( P ) ( P ) How are and p related? NO NO O ( g) ( g) ( g) [ NO] [ O ] PNO PO p [ NO ] ( PNO ) ( ) ( ) n looking at the ideal gas law: P RT MRT M P V RT plugging this expression into gives: PNO PO RT O RT PNO PO P P NO NO [ NO] [ ] ( ) ( ) [ NO ] ( ) RT RT therefore the relationship btwn & p for this equation is: p p ( RT RT ) In general, ( RT) n where n = moles (g)produts moles (g)reatants p in this example n = when n = 0 then = p Example: For whih of the following reation will = p? SO O SO n 3 a. ( g) ( g) 3( g) b. Fe( ) CO( ) FeO( ) CO( ) n 0 s g s g. HO ( ) CO( ) H( ) CO( ) n0 g g g g therefore for b. &. Example: What is the p of the reation below at 35C given = 5.0?

3 n L Cl CO COCl ( g) ( g) ( g) Latm p ( RT) ( ) mol mol 0.0 p <5> 5.3 Understanding & Working with Equilibrium Constants The Magnitude of Equilibrium Constants > 0 3, produts predominate over reatants: reation proeeds to ompletion generating produt very little reatant is left [produts] < 0 3, reatants predominate over produts: [reatants] reation proeeds hardly at all very little produt is generated 0 3 > > 0 3, both reatants and produts are present at eq Handling Combined Equations () N( g) O( g) NO( g) () NO O NO ( g) ( g) ( g) N O NO ( g) ( g) ( g) Overall : N O NO the mass ation exp'n for the overall reation is: the mass ation exp'n for () is: NO( g) NO( g) ( g) ( g) ( g) [ NO] [ N ][ O ] [ NO ] NO O C [ NO ] [ N ][ O ] the mass ation exp'n for () is: [ ] [ ] we derive the mass ation exp'n from () & () for the overall reation thru x [ NO] [ NO ] [ NO ] [ N ][ O] [ NO] [ O [ ] N][ O] In general: overall = x x Example: Calulate the eq onst for D A Bgiven the info below. AB C 3.3 C D 0.04 we need to look from the reverse diretion for both of these reations C AB D C D AB How does multiplying a reation by a onstant affet the eq. onst? the eq. onstant is raised by the exponential value of the onstant

4 D AB 3 ( ) [ A][ B] [ D] 3 6 [ A] [ B] 3 D AB6D 3A6B 6 [ D] Example: If the eq. onstant at a given temperature is.4 x 0 3 for SO O SO what is the eq. onstant for the reations below? ( g) ( g) 3( g) a.) SO O SO ( g) ( g) 3( g) SO SO O b.) 3( g) ( g) ( g).) SO SO O 3( g) ( g) ( g) -3 = (.4 x 0 ) = (.4 x 0 ) 47-3 = (.4 x 0 ) 0.4 Summary of manipulating the mass ation expression Ation New Eq Constant Reversing an equilibrium reation ' Multiplying an equilibrium reation by n ' n Adding several eq reations together ' 5.4 Heterogeneous Equilibria so far we have dealing with reations in whih everything is either in solution or is a gas homogeneous eq heterogeneous equilibrium: when the reatants/produts are in more than one phase the eq onstant is independent of solid speies solids have no pressure and the onentration is onstant (they are not in solution) [ CO][ H ] Ex: HO ( g) C( s) CO( g) H( g) [ HO ] the eq onstant is independent of H O (l) sine it is the solvent and therefore in exess [ HCO3] Ex: CO( g) HO ( l) HCO 3( aq) [ CO ] 5.5 Calulating Equilibrium Constants When Eq onentrations are known: it is a matter of plugging them into the equilibrium expression Ex: What is the for SO( g) O( g) SO3( g) if their equilibrium onentrations are [SO ]=0.5 M, [O ]=0.68, [SO 3 ]=.5. Write down the equilibrium expression symbolially: 3 SO SO O. Plug the give equilibrium values into the expression and solve: SO O SO When initial onentrations and some equilibrium onentrations are known: we use an "ICE" table, "I"nitial, "C"hange, "E"q & stoihiometry

5 Ex: What is the for SO( g) O( g) SO3( g) if the initial onentrations of reatants were [SO ]=0.50M and [O ]=0.680M and the equilibrium onentration of the produt is [SO 3 ]=0.050M?. Setup the ICE table SO O SO 3 Initial Change x x +x Eq?? Use the given equilibrium onentration to identify x x x Plug x into the table to get the missing equilibrium onentrations SO O SO 3 Initial Change (0.05) x Eq Write down the equilibrium expression symbolially and plug in values SO O SO When the initial onentrations and %dissoiation is known: Ex: What is the for SO( g) O( g) SO3( g) if 0.500M of both reatants will be 0.5% dissoiated in order to reah equilibrium?. Setup the ICE table SO O SO 3 Initial Change x x +x Eq???. Use the given %dissoiation to identify x % dissoiation x initial _ onentration x Plug x into the table to get the missing equilibrium onentrations SO O SO 3 Initial Change (0.005) x Eq Write down the equilibrium expression symbolially and plug in values SO O SO 5.6 Appliation of Equilibrium Constants Prediting the Diretion of Reation when a reation is not in equilibrium, the mass ation expression beomes:

6 [ C] [ D] aa bb C dd Q where Q is the reation quotient a b [ A] [ B] d Figure: 3.5 from Chemistry by MMurray & Fey when Q = then the reation is in equilibrium when Q < then more reatants need to be onsumed in order to reah eq therefore the reation will proeed in the forward diretion when Q < then more produts need to be onsumed in order to reah eq therefore the reation will proeed in the reverse diretion Example: Given the data below is the reation in equilibrium and if not in whih diretion will need to go in order to reah eq? A B =, [A] = 0.0 M, [B] =.0 M [ B].0 Q 0 < = therefore it will go in the forward diretion [ A] 0. Calulating Equilibrium Conentrations Example: The value of = at 98 for the reation below, determine the eq onentrations if initially [H O] = M and [Cl O] = M. HO ClO HOCl ( g) ( g) ( g). Write down the ICE table H O Cl O HOCl Initial Change x x +x Eq x x +x. Write down the equilibrium expression symbolially [ HOCl] [ HO] ClO 3. Fill in the expression and get the quadrati equation x 4x x x xx ( xx ) 4x x0.0900x 4x x x Use the quadrati equation to find x (hoose the positive value) Reall the quadrati equation:

7 b b 4a for ax bx 0 x a in our ase, a = 3.900, b = , = x x x x or 4 x Plug the x value into the E row of the table and find the onentrations: x must be greater than zero therefore x = 5.70x0 4 Using this value we an determine what the onentrations are at eq: [H O] = x = = M [Cl O] = x = M [HOCl] = x = M Example: The value of for the thermal deomposition of hydrogen sulfide is. x 0 6 at 400. HS H S ( g) ( g) ( g) A sample of gas in whih [H S] = 0.600M is heated to 400 in a sealed vessel. After hemial eq has been ahieved, what is the value of [H S]? Assume no H and S was present in the original sample.. Write down the ICE table H S H S Initial Change x +x +x Eq x +x +x. Write down the equilibrium expression symbolially: [ H][ S] HS 3. Fill in the expression x x 3 4x x x Here we annot use quadrati so instead we must use an assumption to find x: assume >> x0.600 x x 6 4x.0 x M x Verify assumption: %.94% 5% valid Use the E row to find the equilibrium onentrations: Using x = M we an determine what the onentrations are at eq: [H S] = x = * = M [H ] = x = 0.07 M and [S ] = M

8 Example: What are the eq onentrations of eah of the speies in the following reation, given the = 5. at 700 and the initial onentration of all speies is M? CO( g) H O( g) CO( g) H( g) CO H O CO H Initial Change x x +x +x Eq x x x x [ CO][ H] (0.050 x)(0.050 x) (0.050 x) 5. [ CO][ HO] (0.050 x)(0.050 x) (0.050 x) (0.050 x) (0.050 x) (0.050 x) (0.050 x).58(0.050 x) x x x x M using this value we determine the onentrations: [CO] = [H O] = ( )M = 0.03 M [CO ] = [H ] = ( )M = M 5.7 Le Châtelier s Priniple defn: when a stressor is applied to a system is at equilibrium, the system will adjust to ounterat the stressor in order to re establish eq stressors: adding or removing reatants or produts hanging the pressure/volume hanging the temperature we will talk about this muh later Change in reatant or produt onentration, A B [ B] if we inrease A then Q [ A] we will have to onsume A and produe more B to obtain eq the reation will go in the forward diretion [ B] if we inrease B then Q [ A] we will have to onsume B and produe more A to obtain eq the reation will go in the reverse diretion Change in pressure or volume, A( g ) B( g ) reall PV = nrt therefore P n and V n [ B] if we derease the volume then Q [ A] we inrease the number of ollisions btwn moleules: the pressure inreases then the no. of moles inreases the system will shift in the diretion that redues the no. of moles the reation will go in the forward diretion toward produt

9 [ B] if we inrease the volume then Q [ A] we derease the number of ollisions btwn moleules: the pressure dereases then the no. of moles dereases the system will shift in the diretion that inreases the no. of moles the reation will go in the reverse diretion toward reatant Example: For eah senario predit the diretion the reation goes to attain eq: CO(g) H(g) CH3 OH(g) a.) CO is added. reation goes toward produt (forward) b.) CH 3 OH is added. reation goes toward reatants (reverse).) Pressure is redued. n reatants = 3, n produts = reation goes toward reatants (reverse) d.) Volume is inreased. reation goes toward reatant (forward) Change in eq with T unlike our other hanges, the value of the eq onst hanges with T if the reation is exothermi and we add heat then we would predit thereation would go toward reatants: reatants produts + heat for an endothermi reation the addition leads toward more produt generation reatants+ heat produts in eah ase we are treating our heat as a reatant or produt Example: In what diretion will the eq shift when eah of the following hanges are made to the system at eq? NO NO H 58.0 kj 4g g (a) NO 4g is added. reation goes toward produt (b) NO g is removed. reation goes toward produt () the total pressure is inreased by adding N g. reation remains unhanged sine the partial pressures of the reating speies is onstant at onstant volume (d) the volume is dereased. reation goes toward reatant sine n reatants =, n produts = (e) the temperature is dereased. reation goes toward reatant sine the proess is endothermi or NO 58.0 kj NO 4 g g The Effet of Catalysts there is no impat on the eq onstant it just allows us to reah eq faster