Chapter 15 Equilibrium. Reversible Reactions & Equilibrium. Reversible Reactions & Equilibrium. Reversible Reactions & Equilibrium 5/27/2014

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1 Amount of reatant/produt 5/7/01 quilibrium in Chemial Reations Lets look bak at our hypothetial reation from the kinetis hapter. A + B C Chapter 15 quilibrium [A] Why doesn t the onentration of A ever go to zero? What if we waited longer? What is happening here? time n reality, etter was to represent our reation would be A + B C quilibrium Some Reations are Reversible Tells us there is more than one reation taking plae. Reversible Reations & quilibrium 1 Fwd worker breaks reatant in two, sends piees down the line. + Reation 1 N O NO Reation Reatants Produts N O NO Forward reation and NO N O Reverse reation Will there be more reatants or more produts at equilibrium? Rev worker takes the two piees, puts bak together. Sends bak. However, it takes RV worker twie as long to put the piees together. How will the amounts of produts and reatants over time? Reversible Reations & quilibrium As the workday progresses.. Beause the fwd and rev workers are working at different rates, over time, there will be uild-up of produts Reversible Reations & quilibrium ventually 5 Beause there are now two RV workers, they together they work at the same rate as the FWD worker Reatants Produts Reatants Produts Sine the RV worker is being overwhelmed with more produts, the foremen sends in help!!! From this point on, the amounts of reatants and produts will stay the same they are in dynami equilibrium One dynami equilibrium is reahed, the amounts of reatants and produts DO NOT CHANG. The reation appears to have stopped, but the forward and reverse reation are taking plae at the same rate. 1

2 Reatant(s Produt(s Reatant(s Produt(s 5/7/01 xperimental videne of quilibrium quilibrium Constant NTAL FNAL Trial 1 Trial NTAL FNAL Notie that although eah trial started with a different ratios of reatant/produts, the ratio of onentrations of reatants/produts at equilibrium will be the same Do not onfuse the two kay s k lower-ase / small k Related to how fast a reation proeeds Determined based on the rate law UNTS!! Upper-ase / large Denotes how far are reation proeeds Determine by equilibrium amounts of reatants / produts NO UNTS!! quilibrium Constant The quantitative measure of how far a reation proeeds is represented by a onstant,. an be thought of as the result of two reations: the forward reation and the reverse reation. Whether we end up with more reatant and more produt in the end is determined by the rate of eah reation. quilibrium Constant What does the numerial value of atually mean? oeffiient [produt] atequilibriu m oeffiient [reatant] Let s assume the following reation is a one-step proess. We an determine the rate laws sine it is an elementary step: N O NO Forward reation Rate = k fwd [N O ] At equilibrium Reverse reation Rate = k rev [NO ] kfwd[no ] k [NO ] rev k k fwd rev [NO ] [N O ] oeffiient [produt] atequilibriu m oeffiient [reatant] Similar rxn rates Fwd faster than rev Rev faster than fwd Mostly reatants,, very small amount of produts Substantial amount of both reatants and produts Mostly produts, very small amount of reatants Reatant ~ Produt Produt Favored Reatant Favored <<<<1 <<1 1 >>1 >>>>1 Negligible Reatant Favored Produt Favored Complete quilibrium Constant and onentration For aqueous speies and gases we an use onentration in our onstant aa + bb C + dd d [D] atequilibriu m We an determine experimentally by measuring onentrations of reatants and produts at equilibrium N O NO nitial (M quilibrium (M xperiment [N O ] [NO ] [N O ] [NO ] Sine [NO ] (0.017M [N O ] ( values and equilibrium onentration pratie Oxygen gas an be onverted into ozone based on the following reation: O (g O (g f the equilibrium onentration of O is 0.5 and O is.x10-7 M, determine the value of the equilibrium onstant, f the equilibrium onentration of O is 0.1, what would be the equilibrium onentration of O? -8 M

3 5/7/01 Heterogeneous quilibria How do we deal with reations that have different states of matter within the same reation? For example: Fe(OH (s Fe + (aq + (aq - [Fe ][OH ] [Fe(OH ] Although a gas an be represented by or p, how do we represent solids? What is the onentration of a solid? Well, there is no need to alulate it sine it will not even if you the amount do we. So, we take it out of the expression. A solid or,liquid does not onentration quilibrium Constant and Pressure p Just as we an have an equilibrium onstant based on pressure of produts and reatants (, we an also use partial pressures when dealing with gases (Partial Pressureof Produts p atequilibriu m (Partial Pressureof Reatants aa + bb C + dd - '[Fe(OH ] [Fe ][OH ] - [Fe ][OH ] Pure liquids and solids are not represented in the equilibrium equation d ( PC ( PD p atequilibriu m ( P ( P A B P gas = the partial pressure the gas exerts in the reation mixture So for the following equation CO (g +C(s CO(g [CO] [CO ] solid arbon will not be a part of the equilibrium expression And for the following equation H + (aq + (aq H O(l 1 - [OH ][H ] liquid H O will not be a part of the equilibrium expression Based on onentrations [O ] [O ] So, s for the reation: O O Based on partial pressures ( PO p ( P O The Relationship between and p The relationship between and p is determined by the following equation (RT p Δn Latm mol Mol gas produts mol gas reatants from balaned equation Relationships between related s Remember that for the reation. C + dd aa + bb aa + bb C + dd But what about the reverse reation? d [D] 1 When we deal with reversible reations, we an write it any diretion we hoose. For example, the following reation: SO (g + O (g SO (g has a of.09x10 - at To determine p : (.09x10 [( ( (1000.].98x10 - Latm p mol Sine we have written the equation in reverse, produts and reatants are swithed. Thus, is different. d [D] The reation: H (g + Cl (g HCl(g has a p of.0x10 1 at 00.. What is? 1 (.0x10 ( [( (- (00.].0x10 1 Latm mol When the number of moles of gas are the same on both sides of the equation, = p Relationships between related s Let s say we have two reations that have a reatant/produt in ommon: aa + bb xx xx C x [X] 1 x [X] Reation Quotient and quilibrium Constant What do we do when we know the onentration of reatants before equilibrium (or before we mix them to perform a reation? Do we have an expression that an quantify this? d Q [D] atanytimenotatequilibriu m We an add the two reation together aa + bb C 1 How do we ombine expressions? x [X] 1 x [A] [B] [X] d [D] Q Q Q< Forward More Produts Formed Ratio of reatants higher than at equilibrium quilibrium Reverse More Reatants Formed d [D] Q Q Q> Ratio of produts higher than at equilibrium

4 5/7/01 N O + NO Dark olorless orange Le Chatelier s Priniple When an equilibrium is disturbed, the reation will proeed in the diretion that reestablishes an equilibrium. nrease the amount of a reatant Le Chatelier s Priniple What happens when a equilibrium is disturbed? One it is disturbed, it will shift toward reatants or produts to return to equilibrium Derease the amount of a produt Derease the amount of a reatant nrease the amount of a produt N O NO More NO added Wait N O? NO 10 N O NO minutes Q< Reation shifts toward produts A + B C + D Q< Reation shifts toward produts A + B C + D Q> Reation shifts toward reatants A + B C + D Q> Reation shifts toward reatants A + B C + D quilibrium mixture of NO and N O gives a yellowish olor. The addition of more NO turns the mixture orange (NO is orange in olor. Whih way did the reation shift to reah equilibrium? N O NO After 10 minutes, the olor returns as before, suggesting that the equilibrium mixture has returned. is reestablished is reestablished To make a reation shift toward produts Derease the amount of produt nrease the amount of reatant is reestablished is reestablished To make a reation shift toward reatants Derease the amount of reatant nrease the amount of produt Le Chatelier s Priniple Only reatants and/or produts that are part of the equilibrium equation, will affet equilibrium. Fe + CO CO Fe(OH CO CaCO Fe + CO CO CaO Fe(OH (s Fe + (aq + (aq CO Add more Fe(OH Conentration of Fe(OH ions does not CaCO (s CO (g + CaO(s Add more CaCO Fe + CO Conentration of CO does not CO Fe(OH CO CaCO Ca(CO Fe + CO CO CO - [Fe ][OH ] Adding more Fe(OH will not ause more ions to dissolve. No shift in equilibrium Fe(OH is not part of the equilibrium expression [CO ] Adding more CaCO will not ause more CO or CaO to be reated. No shift in equilibrium Le Chatelier s Priniple C in Temp n order to determine the effet of temperature on equilibria, it is onvenient to view heat either as a produt or reatant. A + B C + HAT Reation shifts toward reatants A + B C + HAT quilibrium is established w/ smaller A + B C + HAT xothermi reations (-ΔH A + B C + HAT Reation shifts toward produts A + B C + HAT quilibrium is established w/ larger A + B C + HAT HAT + A + B C Reation shifts toward produts HAT + A + B C quilibrium is established w/ larger HAT ndothermi reations (+ΔH + A + B C HAT + A + B C Reation shifts toward reatants HAT + A + B C quilibrium is established w/ smaller HAT + A + B C Derease in volume 1 atm Le Chatelier s Priniple An inrease in total pressure will shift a reation toward the side with the fewest moles of gas. atm atm The.C. Table A ertain reation has alaned equation of A B + C. After plaing 6.0 A in reation 1000., the reation begins. During the reation, the reatant dereased by1.m. What are the equilibrium onentrations and what is? N O NO N O NO N O NO quilibrium shifts to the side with the fewest moles of gas. More of that gas (produt or reatant depending on the balaned equation is produed atm atm CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CaCO Ca(CO CaCO CaCO CaCO (s CO (g + CaO(s CaCO (s CO (g + CaO(s CaCO (s CO (g + CaO(s Write the balaned equation We plug everything we know into the table. Sine A d -1.M, how should the onentration of B and C in omparison? Look at the balaned equation B should times as muh and C should times as muh a A. C nitial A B + C M +(1. (6.0 1.M (+.66M.67M.66M +(1. (+.99M.99M [B] (.66 (.99 [A]

5 5/7/01 Determining C from nitial and quilibrium Values Gas A is put in a reation flask and heated to 0.0 C until it s to gas B in a -to- ratio. Before the reation, gas A had a partial pressure of 1.00atm. At the end of the reation, after it has ooled gas A had a partial pressure of 0.0atm. Determine the equilibrium pressure of gas B and determine p for the reation. Write the balaned equation Sine we don t know the, we, an substitute with a variable x (temporarily. We an atually work bakward from equilibrium. C = nitial quilibrium = 1.00atm-0.0atm = 0.60atm and C = -x A B 1.00atm 0atm -x -0.60atm +0.0atm 0.0atm Therefore, we know that -x = x = 0.0 atm 0.0atm So x = 0.0atm Don t forget your oeffiients ( PB p ( PA (0.0 p.5 (0.0 Calulating quilibrium Conentration using A ertain reatant A rearranges into a produt B at room temperature in a 1-to-1 ration. f you started with.m A, what will be the onentration of both A and B at equilibrium if = 1.11x10-1 Write the balaned equation Sine we don t know the, we, an substitute with a variable x. quilibrium is just the initial plus the Substitute equilibrium onentrations A B.M -x.m - x + x [B] x x x 1.111x 0.70 [A].- x Double-hek by working bakward x 0. Sine [A] eq =.M - x [B] eq = x Then [A] eq =.0 [B] eq = 0.M Calulating quilibrium Conentration using Perfet Squares The reation between H and produes H. Assume 0.100mols of H and moles of are plaed into a 1.00L ontainer. The gas reat and are allowed to reah equilibrium. Calulate the equilibrium onentration of all the gases. for the reation at C is Write the balaned equation Sine we don t know the, we, an substitute with a variable x. Sine quilibrium is just the initial plus the Substitute equilibrium onentrations [H] 50.5 [H ][ ] 1 Perfet Square (x 50.5 ( x( x or x x H + H x -x x x + x x x Calulating quilibrium Conentration using Perfet Squares H + H x -x x x x = Sine [H ] eq and [ ] eq = x [H] eq = x + x [H ] eq and [ ] eq = 0. [H] eq = 0.156M x x x 7.11 ( x x x 9.11x Double-hek by working bakward ( (0.0(0.0 Calulating quilibrium Conentration using Quadrati Formula We an alulate equilibrium onentrations if we know initial onentrations and the equilibrium onstant,. f we start with an initial onentration of N O of 5. before deomposition, how muh NO will there be at equilibrium? for the reation is.6x10 -. Write the balaned equation 5. Sine we don t know the, we, an substitute with a variable x. -x quilibrium is just the initial plus the 5. - x x Substitute equilibrium onentrations N O NO Calulating quilibrium Conentration using Quadrati Formula N O NO 5. -x 5. - x x General form of a Quadrati ax + bx + = 0 x x = 0 (x.6x10 (5.0- x (x = (5.0-x(0.006 x = x x x = 0 Quadrati Formula b b a This is in the form of a quadrati a [NO ].6x10 [N O ] Sine (x.6x10 (5.0- x Solve for x (0.006 ( ((-0.0 5

6 5/7/01 Calulating quilibrium Conentration using Quadrati Formula Calulating quilibrium Conentration using Simplifying Assumptions [N O ] eq = 5.-x [N O ] eq =.9M (0.006 ((-0.0 ( This value for x does not make hemial sense beause you annot x = and x = have negative onentrations [NO ] eq = x [NO ] eq = 0.15M Double-hek by working bakward ( Consider the deomposition of H S, to give H and S. f = 1.67x10-7 at 800 C, determine the equilibrium onentrations for all gases if the initial amount of H S is At this point, you had better know whih steps to take!!! [H ] [S] 7 [HS] 1.67x10 (x (x 1.67x10 (0.050-x 7 H S H + S x x x This looks like it is going to be a ubi funtion; rd degree polynomial!!! That is: ax + bx + x +d = 0. don t feel like solving this!!! Although you an do this via a graphing alulator (whih you an t use on an exam, of ourse.. or using suessive approximations. Let s make some simplifying assumptions. For example, sine is so small, m going to assume that the in x will be small also. So.. (x (x 1.67x10 (0.050-x 7 Let s get rid of this..sine it should small (x (x 1.67x10 ( HHMMM.Better Calulating quilibrium Conentration using Simplifying Assumptions H S H + S x x x s our approximation appropriate? Calulate what perentage x is of your original [] s. f it is 5% or smaller, it is O..97x10 - x % Our approximation is appropriate (x (x 1.67x10 ( x x10-6.5x10 7 (1.67x10 (6.5x10 x x.61x10 11 x.97x10 Slightly off beause of approximation. But O.. [H S] eq = 0.05 x [H ] eq = x [S ] eq = x [H S] eq = 0. [H ] eq = 5.9x10 - M [S ] eq =.97x10 - M Double-hek by working bakward - - (5.9x10 (.97x x10 (