General Equilibrium Chemial Equilibrium Most hemial reations that are enountered are reversible. In other words, they go fairly easily in either the forward or reverse diretions. The thing to remember about reations that have ome to equilibrium is that they have not stopped. The equilibrium is a dynami equilibrium. The reation is still proeeding as it was previously. However, the reverse reation is ourring at the same rate as the forward reation. This means that the onentration of the reatants and produts remaining the same after equilibrium has been reahed. A partiular reatant at any given time may onvert to a produt and vie versa. What happens to ause a reation to ome to equilibrium? Initially the reatant onentrations are large. Beause of this, the rate of the forward reation starts out rather high. As the reation proeeds the reatant onentrations derease ausing the forward rate to derease. At the same time, the reverse rate starts out at zero, beause the onentration of the produts is zero. As the produts are reated the reverse reation inreases. Eventually a point is reahed where the reatants are produed by the reverse reation just as quikly as they are onsumed by the forward reation. This is the state of hemial equilibrium. We need to remember that the reation mixture at equilibrium is ditated by both the forward and reverse rates AND the stoihiometry of the reation. The following reation is allowed to ome to equilibrium: NO + 7 H ¾ NH 3 + H O If the reation starts with 6.56 mol of Hydrogen gas and 8.95 mol of Nitrogen Dioxide in a 15 L ontainer and there are.6 mol of Ammonia at equilibrium, what is the omposition of the equilibrium mixture? [NO] [H ] [NH 3 ] [H O] Initial on. 0.597 1.77 0.000 0.000 Change -0.0755* -0.0755*7 +0.0755* +0.0755* Final on. 0.597-0.151 1.77-0.585 0.000+0.151 0.000+0.30 Final 0.6 1. 0.151 0.30
In this example, the onentration of eah ompound hanges by an amount that is a stoihiometri multiple of some given amount (0.0755 M). The hanges are given as positive or negative depending on whether that substane is being produed or onsumed. The final onentrations are dependent on the initial onentrations. If we start with different amounts of eah we will finish with different amounts. Equilibrium Constants At a given temperature we an define a onstant that indiates where along the reation path the equilibrium lies. This onstant is alled the equilibrium onstant,. If we onsider a general equilibrium reation: aa + bb ¾ C + dd the equilibrium onstant and expression is defined as C D [ ] [ ] a [ A] [ B] d b. where the subsript in denotes that this is derived from onentrations. This onstant is the same for a given reation at a given temperature regardless of the equilibrium onentrations. The expression for the equilibrium onstant is derived from the stoihiometri oeffiients in the hemial reation, unlike the exponents in the rate law. Looking at the above reation (NO and H ) what is the equilibrium expression? [ NH 3] [ H O] 7 [ NO] [ H O] Given that this is the equilibrium expression what is the equilibrium onstant? [ 0.151] [ 0.30] 7 [ 0.66] [ 1.] [ NH 3] [ H O] 1.9 10 7 [ NO] [ H O] Often the equilibrium onstants are given without units. The reason for this lies in thermodynamis where these onstants are defined in terms of the ativities of the speies present. The ativity is the onentration of the speies divided by 1 M. The result is unit-less so the equilibrium onstants are unit-less.
ineti argument for equilibrium onstants Look at the deomposition of N O. It proeeds aording to the following reation: N O ¾ NO This reation is at equilibrium when the forward and reverse rates are the same. It also proeeds as if the hemial equation is the only step in the mehanism, an elementary step. In that ase, the forward rate is given by: rate k 1 [N O ], and the reverse rate is given by: rate k -1 [NO ], and these rates are equal leading to the expression: k 1 [N O ] k -1 [NO ]. This an be rearranged to give: k k 1 1 [ NO ] [ N O ] k k 1 1 [ NO ] [ N O ] whih is just our equilibrium expression. If the reation ours by a multi-step mehanism, the equilibrium expression is a ombination of ratios of rate onstants from eah step in the mehanism. p The equilibrium onstants an be defined in terms of partial pressures when working with gases. Let s look at the relationship between and p. We an find the relationship by looking at the relationship between onentration and pressure via the Ideal Gas Law.
d [ C] [ D] a b [ A] [ B] n P [ C] V RT P Pd RT RT a Pa Pb RT RT ( RT) p n d b d P P d a b d ( RT) a b p( RT) P P ( + ( + )) n Where n is the number of moles of gaseous produts minus the number of moles of gaseous reatants. RT is a onstant at a given temperature. R must be in units of L-atm/mol- beause we are, in effet hanging between molarity (mol/l) to atmospheres (atm). What is the p for the reation between NO and H? p (RT) n (.1 10 - )((0.0806 L-atm/mol-) (98 )) (6-9) 1.5 10-8 Equilibrium Constants for the sum of reations If we add two more reations together, the equilibrium onstants are multiplied together. If we reverse a reation the equilibrium onstant gets inverted. Dividing a reation by a onstant, we take the orresponding root of the equilibrium onstant. Multiplying a reation by a onstant means we raise the equilibrium onstant to that power. Heterogeneous Equilibria This is equilibrium in whih the speies in the reation are in different phases. When we write the equilibrium expression for this equilibrium we omit the onentrations of pure solids and liquids. The reason for this is that the onentrations of these phases are onstant. As a result, the onentrations of these phases are absorbed into the equilibrium onstant. If the solvent takes plae in the reation its onentration is also effetively onstant so it is also absorbed into the equilibrium onstant. The onentration of the solvent is also not represented in the equilibrium expression. a b
Using the Equilibrium Constant We an use the equilibrium onstant in one of three ways: 1. Qualitative interpretation of the onstant. By looking at the onstant we an tell how far toward ompletion the reation goes.. Prediting the diretion of a reation. Given a mixture of produts and reatants, what diretion will the reation go? 3. Finding equilibrium onentrations. Given an initial mixture of produts and reatants, what will be the equilibrium onentrations? Qualitative interpretation Look at the magnitude of the equilibrium onstant. If the onstant is muh greater than 1, the reation lies toward the produts. If it is muh less than 1, the reation lies toward the reatants. If the equilibrium onstant is about 1, the reation mixture is about equal in produts and reatants. Prediting the diretion of a reation To predit the diretion of a reation we must first examine a quantity that is alled the reation quotient. The reation quotient is mathematially idential to the Equilibrium expression that gives rise to. However, the onentrations that are used in quotient are not equilibrium onentrations. [produts] [reatants] [produts] Q [reatants] x e y e x y The subsript e indiates an equilibrium value. We an predit the diretion of a reation by omparing Q and <Q >Q Q reation shifts to the left reation shifts to the right already at equilibrium so there is no hange Finding equilibrium onentrations of reatants We an use the equilibrium onstant and expression to alulate what the equilibrium onentrations (or partial pressures) will be. Sometimes the resulting
mathematial equation is too ompliated to solve diretly so we must resort to approximation methods. Example: For the reation: N O (g) + 3 O (g) ¾ NO (g) the equilibrium onstant ( ) is 1. 10. If.00 mol of Dinitrogen Monoxide and.00 mol of Oxygen are plaed into a 1.00 L ontainer what will the equilibrium onentrations of everything be? N O (g) + 3 O (g) ¾ NO (g) init.00 M.00 M 0.00M -x -3x +x final.00-x.00-3x x The equilibrium expression is: 1. 10 ( x) ( x) ( 3x) 3 This is a 5 th order equation that an be solved using an approximation method. (It an be solved exatly using a omputer.) To solve this we need to realize that the largest x an be is /3. If it were more than this the denominator would be negative whih it an t be. We an guess a value for x and solve for Q. We then adjust the value for x until we get Q to be lose to the value for. Beause is fairly large, we an guess that the x will be fairly lose to /3. We hoose 0.500 initially, whih gives: Initial guess: 0.500 Q 18 0.510 Q 17 0.505 Q 19 0.50 Q 1 0.50 Q 136 0.503 Q 10.19 (very lose) The value for x is then 0.503, whih means that the equilibrium onentration of N O is 0.99 M, O is 0.91 M, and NO is.01 M.
Using the quadrati equation or taking a root an solve most problems of this sort. Other problems may require other approximation methods. We ll examine the other approximation methods as we ome to them. Le Châtelier s Priniple Le Châtelier s priniple states that when a system at equilibrium is stressed or strained, the equilibrium will shift so as to relieve the stress or strain. If we add more reatants to a system at equilibrium, the system will respond by produing more produts, thus relieving the stress. If we take reatants away, the system will onsume produts to make more reatants. When we add anything to either side of the reation the equilibrium will shift to the opposite side. If we take anything away, the equilibrium will shift to the side that loses moleules. Effets of pressure hanges If we hange the pressure by adding an inert gas, suh as Argon, the equilibrium will not shift beause we are not hanging the partial pressures of the omponents of the reation. We an shift the equilibrium by hanging the pressure by other means though. If we redue the size of the ontainer the partial pressures of everything will inrease. The equilibrium will shift to the side of the reation with fewer moles of gas to relieve the inrease in total pressure. Conversely, if we inrease the size of the ontainer, whih dereases all of the partial pressures, the equilibrium will shift to the side of the reation with more moles of gas to bring the pressure bak up. The same argument above an be made for aqueous solutions. If we dilute a solution, the equilibrium will shift to the side of the reation with more moles of solute to inrease the overall onentration of the solution. Quantitative hanges of Q For the following reation A (g) + B (g) ¾ C (g) If we derease the size of the ontainer by a fator of the partial pressures of everything will double. This will ause Q to beome Q [ C] 1 [ A][ B]
So Q will beome ½ of. Beause is bigger than Q the reation will shift to the right. This is same onlusion we ame to above. The reation shifts to the side with fewer moles of gas. Effets of temperature hanges The effet on equilibria due to temperature hanges depends on if the reation is exothermi or endothermi. On a qualitative level we an think of heat as a produt (exothermi reation) or as a reatant (endothermi reation). Now Le Châtelier s priniple is applied. If the reation is exothermi and we add heat, we drive the reation to the reatant side. If we remove heat the reation is driven to the produt side. We must keep in mind; however, that what is hanging when we hange the temperature is not Q but. If the reation is exothermi and we add heat, inreases whih is what atually auses the shift in equilibrium. Effets of a atalyst on equilibrium Adding a atalyst to a system already at equilibrium does not hange the equilibrium. The atalyst will inrease the forward and reverse rates by the same amount. If the system is not yet at equilibrium, the atalyst auses the equilibrium to be reahed faster than it would have been without the atalyst.