Chapter 14 Chemial Equilibrium Chemial Equilibriu m Muh like water in a U-shaped tube, there is onstant mixing bak and forth through the lower portion of the tube. reatants produts It s as if the forward and reverse reations were ourring at the same rate. Nitrogen dioxide, NO (a reddish brown gas), is in equilibrium with dinitrogen tetroxide, N O 4 (a olorless gas). Nitrogen dioxide is present in many urban smogs, giving them a harateristi reddish brown olor. The system appears to stati (stationary) when, in reality, it is dynami (in onstant motion). See Figure 15. Chemial reations often seem to stop before they are omplete. Figure 14.: Catalyti Methanation Reation Approahes Equilibrium Atually, suh reations are reversible. That is, the original reatants form produts, but then the produts reat with themselves to give bak the original reatants. When these two reations forward and reverse our at the same rate, a hemial equilibrium exists. Chemial Equilibrium When ompounds reat, they eventually form a mixture of produts and unused reatants, in a dynami equilibrium. A dynami equilibrium onsists of a forward reation, in whih substanes reat to give produts, and a reverse reation, in whih produts reat to give the original reatants. Chemial equilibrium is the state reahed by a reation mixture when the rates of the forward and reverse reations have beome equal. Chemial Equilibrium For example, the Haber proess for produing ammonia from N and H does not go to ompletion. N (g) H(g) NH(g) It establishes an equilibrium state where all three speies are present. 1
A Problem to Consider Using the information given, set up the following table. N (g) H(g) NH(g) Starting 1.000 Change Equilibrium 1.000 - x.000 -x.000 - x 0 x x 0.080 mol The equilibrium amount of NH was given as 0.080 mol. Therefore, x 0.080 mol NH (x 0.040 mol). The Equilibrium Constant The equilibrium-onstant expression for a reation is obtained by multiplying the onentrations of produts, dividing by the onentrations of reatants, and raising eah onentration to a power equal to its oeffiient in the balaned hemial equation. aa bb C dd For the general equation above, the equilibrium-onstant expression would be: The molar onentration of a substane is denoted by writing its formula in square brakets. [C] [D] a [A] [B] d b A Problem to Consider Using the information given, set up the following table. N (g) H(g) NH(g) Starting 1.000 Change Equilibrium 1.000 - x.000 -x.000 - x 0 x x 0.080 mol Equilibrium amount of N 1.000-0.040 0.960 mol N Equilibrium amount of H.000 - ( x 0.040).880 mol H Equilibrium amount of NH x 0.080 mol NH The Equilibrium Constant The equilibrium onstant,, is the value obtained for the equilibrium-onstant expression when equilibrium onentrations are substituted. A large indiates large onentrations of produts at equilibrium. A small indiates large onentrations of unreated reatants at equilibrium. The Equilibrium Constant Every reversible system has its own position of equilibrium under any given set of onditions. The ratio of produts produed to unreated reatants for any given reversible reation remains onstant under onstant onditions of pressure and temperature. The Equilibrium Constant The law of mass ation states that the value of the equilibrium onstant expression is onstant for a partiular reation at a given temperature, whatever equilibrium onentrations are substituted. Consider the equilibrium established in the Haber proess. The numerial value of this ratio is alled the equilibrium onstant for the given reation. N ( g ) H (g ) NH (g)
The Equilibrium Constant The equilibrium-onstant expression would be [NH] [N ][H ] Note that the stoihiometri oeffiients in the balaned equation have beome the powers to whih the onentrations are raised. N (g) H(g) NH(g) NO4(g) NO(g) Ultimately, this reation reahes an equilibrium state where the rates of the forward and reverse reations are equal. Therefore, f [NO4 ] kr [NO ] k Combining the onstants you an identify the equilibrium onstant,, as the ratio of the forward and reverse rate onstants. k k f r [NO ] [N O ] 4 Equilibrium: A inetis Argument If the forward and reverse reation rates in a system at equilibrium are equal, then it follows that their rate laws would be equal. Consider the deomposition of N O 4, dinitrogen tetroxide. NO4(g) NO(g) If we start with some dinitrogen tetroxide and heat it, it begins to deompose to produe NO. However, one some NO is produed it an reombine to form N O 4. Figure 14.4: Temperature Effet on NO -N O 4 Equilibrium Do Now Page 585 Example 14. a. b.. d. Do exerise 14. See Problems 14.5 and 6 Equilibrium: A inetis Argument NO4(g) NO(g) Call the deomposition of N O 4 the forward reation and the formation of N O 4 the reverse reation. These are elementary reations, and you an immediately write the rate law for eah. k f k r Rate kf [NO4] Rate k [NO (forward) (reverse) r ] Here k f and k r represent the forward and reverse rate onstants. Obtaining Equilibrium Constants for Reations Equilibrium onentrations for a reation must be obtained experimentally and then substituted into the equilibrium-onstant expression in order to alulate. Consider the reation below (see Figure 15.5). H (g) CH (g) 4 HO(g) Suppose we started with initial onentrations of CO and H of 0.100 M and 0.00 M, respetively.
Consider the reation below (see Figure 15.5). H (g) CH (g) 4 H O(g) Suppose we started with initial onentrations of CO and H of 0.100 M and 0.00 M, respetively. When the system finally settled into equilibrium we determined the equilibrium onentrations to be as follows. Reatants [CO] 0.061 M [H ] 0.189 M Produts [CH 4 ] 0.087 M [H O] 0.087 M H (g) CH (g) 4 HO(g) As an example, let s repeat the previous experiment, only this time starting with initial onentrations of produts: [CH 4 ] initial 0.1000 M and [H O] initial 0.1000 M We find that these initial onentrations result in the following equilibrium onentrations. Reatants [CO] 0.061 M [H ] 0.189 M Produts [CH 4 ] 0.087 M [H O] 0.087 M The equilibrium-onstant expression for this reation is: [CH4 ][HO] [CO][H ] If we substitute the equilibrium onentrations, we obtain: (0.087M)(0.087M) (0.061M)(0.189M).9 Regardless of the initial onentrations (whether they be reatants or produts), the law of mass ation ditates that the reation will always settle into an equilibrium where the equilibrium-onstant expression equals. Substituting these values into the equilibriumonstant expression, we obtain the same result. (0.087M)(0.087M) (0.061M)(0.189M).9 Whether we start with reatants or produts, the system establishes the same ratio. Do Now - Example 14. Do exerise 14. See problems 14.4-44 Figure 14.5: Some equilibrium ompositions for the methanation reation Figure 15.5: Some Equilibrium Composition for the Methanation Reation Do exerise 14.4 See problems 14.45-48 4
The Equilibrium Constant, p In disussing gas-phase equilibria, it is often more onvenient to express onentrations in terms of partial pressures rather than molarities (see Figure 14.6 It an be seen from the ideal gas equation that the partial pressure of a gas is proportional to its molarity. P ( V n ) RT MRT From the relationship n/vp/rt, we an show that n p (RT) where n is the sum of the moles of gaseous produts in a reation minus the sum of the moles of gaseous reatants. Consider the reation SO (g) O (g) SO (g) for the reation is.8 x 10 at 1000 o C. Calulate p for the reation at this temperature. Figure 14.6: The Conentration of a Gas at a Given Temperature is Proportional to the Pressure A Problem to Consider Consider the reation SO (g) O(g) SO(g) We know that n p (RT) From the equation we see that n -1. We an simply substitute the given reation temperature and the value of R (0.0806 L. atm/mol. ) to obtain p. If we express a gas-phase equilibria in terms of partial pressures, we obtain p. Consider the reation below. H (g) CH (g) 4 HO(g) The equilibrium-onstant expression in terms of partial pressures beomes: p P P CH 4 CO P P HO H In general, the numerial value of p differs from that of. Sine n p (RT) L atm -1.8 p 10 (0.0806 1000 ).4 mol Do exerise 14.5 See problems 14.51-54 5
1 Page 591 (1) () Equilibrium Constant for the Sum of Reations For example, nitrogen and oxygen an ombine to form either NO(g) or N O (g) aording to the following equilibria. N (g) O(g) NO(g) N (g) 1 O(g) NO(g) 4.1 x 10-1.4 x 10-18 Using these two equations, we an obtain for the formation of NO(g) from N O(g): with Either with Either 1 with A red () N O(g) 1 O(g) NO(g)? Equilibrium Constant for the Sum of Reations Similar to the method of ombining reations that we saw using Hess s law in Chapter 6, we an ombine equilibrium reations whose values are known to obtain for the overall reation. With Hess s law, when we reversed reations or multiplied them prior to adding them together, we had to manipulate the H s values to reflet what we had done. The rules are a bit different for manipulating. To ombine equations (1) and () to obtain equation (), we must first reverse equation (). When we do we must also take the reiproal of its value. (1) () () Equilibrium Constant for the Sum of Reations N (g) O (g) NO(g) 4.1 x 10-1 N 1 O(g) N(g) O (g) N O(g) O(g) 1 NO(g) 1.4 10 1.4 10 1 1 (overall) (4.1 10 ) ( 18) 1.7 10 Chek example page 591-18 Equilibrium Constant for the Sum of Reations 1. If you reverse a reation, invert the value of.. If you multiply eah of the oeffiients in an equation by the same fator (,, ), raise to the same power (,, ).. If you divide eah oeffiient in an equation by the same fator (,, ), take the orresponding root of (i.e., square root, ube root, ). 4. When you finally ombine (that is, add) the individual equations together, take the produt of the equilibrium onstants to obtain the overall. Heterogeneous Equilibria A heterogeneous equilibrium is an equilibrium that involves reatants and produts in more than one phase. The equilibrium of a heterogeneous system is unaffeted by the amounts of pure solids or liquids present, as long as some of eah is present. The onentrations of pure solids and liquids are always onsidered to be 1 and therefore, do not appear in the equilibrium expression. 6
Heterogeneous Equilibria Consider the reation below. C(s) H O(g) H (g) The equilibrium-onstant expression ontains terms for only those speies in the homogeneous gas phase H O, CO, and H. Do Example 14.4 Now! Page 59 [CO][H] [H O] Do exerise 14.6 See problems 14.55-56 Prediting the Diretion of Reation For the general reation aa bb C dd the Q expresssion would be: Q i a i d i b i [C] [D] [A] [B] Using the Equilibrium Constant 1. Qualitatively interpreting the equilibrium onstant Prediting the Diretion of Reation For the general reation aa bb C dd. Prediting the diretion of reation. Calulating equilibrium onentrations Do Exerise 14.7 Now See Problems 14. 56-60 If Q >, the reation will shift left toward reatants. If Q <, the reation will shift right toward produts. If Q, then the reation is at equilibrium. Prediting the Diretion of Reation How ould we predit the diretion in whih a reation at non-equilibrium onditions will shift to reestablish equilibrium? Figure 14.10: Diretion of reation To answer this question, substitute the urrent onentrations into the reation quotient expression and ompare it to. The reation quotient, Q, is an expression that has the same form as the equilibrium-onstant expression but whose onentrations are not neessarily at equilibrium. 7
A Problem to Consider Consider the following equilibrium. N (g) H(g) NH(g) A 50.0 L vessel ontains 1.00 mol N,.00 mol H, and 0.500 mol NH. In whih diretion (toward reatants or toward produts) will the system shift to reestablish equilibrium at 400 o C? for the reation at 400 o C is 0.500. First, alulate onentrations from moles of substanes. 1.00 mol 50.0 L 0.000 M.00 mol 50.0 L 0.0600 M 0.500 mol 50.0 L 0.0100 M Calulating Equilibrium Conentrations One you have determined the equilibrium onstant for a reation, you an use it to alulate the onentrations of substanes in the equilibrium mixture. For example, onsider the following equilibrium. H (g) CH 4(g) H O(g) Suppose a gaseous mixture ontained 0.0 mol CO, 0.10 mol H, 0.00 mol H O, and an unknown amount of CH 4 per liter. What is the onentration of CH 4 in this mixture? The equilibrium onstant equals.9. The Q expression for the system would be: [NH] Q [N][H] N (g) H (g) NH (g) 0.000 M 0.0600 M 0.0100 M Q (0.0100) (0.000)(0.0600).1 Beause Q.1 is greater than 0.500, the reation Will go to the left as it approahes equilibrium. Therefore, Ammonia will dissoiate Do exerise 14.8 See problems 14.61-6 Calulating Equilibrium Conentrations First, alulate onentrations from moles of substanes. H (g) CH 4(g) H O(g) 0.0 mol 1.0 L 0.10 mol 1.0 L?? 0.00 mol 1.0 L 0.0 M 0.10 M?? 0.00 M The equilibrium-onstant expression is: [CH4][H O] [CO][H ] Substituting the known onentrations and the value of gives: [CH4](0.00M).9 (0.0M)(0.10M) You an now solve for [CH 4 ]. 4 (.9)(0.0M)(0.10M) [CH ] (0.00M) 0.059 The onentration of CH 4 in the mixture is 0.059 mol/l. 8
Calulating Equilibrium Conentrations Suppose we begin a reation with known amounts of starting materials and want to alulate the quantities at equilibrium. Consider the following equilibrium. Suppose you start with 1.000 mol eah of arbon monoxide and water in a 50.0 L ontainer. Calulate the molarity of eah substane in the equilibrium mixture at 1000 o C. for the reation is 0.58 at 1000 o C. Calulating Equilibrium Conentrations Solving for x. Starting 0.000 Change Equilibrium 0.000 Or: H O(g) CO (g) H(g) 0.000 0.000 x 0.58 (0.000 x) Taking the square root of both sides we get: x 0.76 (0.000 x) 0 0 x x x x First, alulate the initial molarities of CO and H O. H O(g) CO (g) H Starting Change Equilibrium 1.000 mol 50.0 L 1.000 mol 50.0 L 0.000 M 0.000 M 0 M 0 M (g) The starting onentrations of the produts are 0. We must now set up a table of onentrations (starting, hange, and equilibrium expressions in x). 0.000 0.000 0.000 0.000 0 0 x x x x Rearranging to solve for x gives: Solving for equilibrium onentrations. Starting 0.000 Change Equilibrium 0.000 0.000 0.76 x 0.0086 1.76 0.000 0.000 0 0 x x x x If you substitute for x in the last line of the table you obtain the following equilibrium onentrations. 0.0114 M CO 0.0086 M CO 0.0114 M H O 0.0086 M H Do Exerise 14.10 Look at Problems 14.67-68 The equilibrium-onstant expression is: Starting 0.000 Change Equilibrium 0.000 [CO][H] [CO][H O] H O(g) CO (g) H(g) 0.000 0.000 0 0 x x x x Substituting the values for equilibrium onentrations, we get: (x)(x) 0.58 (0.000 x)(0.000 x) Calulating Equilibrium Conentrations The preeding example illustrates the three steps in solving for equilibrium onentrations. 1. Set up a table of onentrations (starting, hange, and equilibrium expressions in x).. Substitute the expressions in x for the equilibrium onentrations into the equilibrium-onstant equation.. Solve the equilibrium-onstant equation for the values of the equilibrium onentrations. 9
Calulating Equilibrium Conentrations In some ases it is neessary to solve a quadrati equation to obtain equilibrium onentrations. The next example illustrates how to solve suh an equation. Substituting our equilibrium onentration expressions gives: (x) (1.00 x)(.00 x) Solving for x. Starting Change Equilibrium 1.00 1.00.00.00 0 x x Beause the right side of this equation is not a perfet square, you must solve the quadrati equation. Calulating Equilibrium Conentrations Consider the following equilibrium. H (g) I(g) HI(g) Solving for x. H (g) I(g) Starting 1.00.00 Change Equilibrium 1.00.00 HI(g) 0 x x Suppose 1.00 mol H and.00 mol I are plaed in a 1.00-L vessel. How many moles per liter of eah substane are in the gaseous mixture when it omes to equilibrium at 458 o C? at this temperature is 49.7. The equation rearranges to give: 0.90x.00x.00 0 The two possible solutions to the quadrati equation are: x. and x 0.9 Calulating Equilibrium Conentrations The onentrations of substanes are as follows. H (g) I(g) HI(g) Starting Change Equilibrium 1.00 1.00.00.00 0 x x The equilibrium-onstant expression is: [HI] [H ][I ] Solving for x. H (g) I(g) Starting Change Equilibrium 1.00 1.00.00.00 HI(g) 0 x x However, x. gives a negative value to 1.00 - x (the equilibrium onentration of H ), whih is not possible. Only x 0.9 remains. If you substitute 0.9 for x in the last line of the table you obtain the following equilibrium onentrations. 0.07 M H 1.07 M I 1.86 M HI Do Exerise 14.11 Look at Problems 14.69-70 10
Conept Chek Page 601 Removing Produts or Adding Reatants Let s refer bak to the illustration of the U-tube in the first setion of this hapter. reatants produts It s a simple onept to see that if we were to remove produts (analogous to dipping water out of the right side of the tube) the reation would shift to the right until equilibrium was reestablished. Likewise, if more reatant is added (analogous to pouring more water in the left side of the tube) the reation would again shift to the right until equilibrium is reestablished. Le Chatelier s Priniple Obtaining the maximum amount of produt from a reation depends on the proper set of reation onditions. Le Chatelier s priniple states that when a system in a hemial equilibrium is disturbed by a hange of temperature, pressure, or onentration, the equilibrium will shift in a way that tends to ounterat this hange. Effets of Pressure Change A pressure hange aused by hanging the volume of the reation vessel an affet the yield of produts in a gaseous reation only if the reation involves a hange in the total moles of gas present See Figure 14.1 Changing the Reation Conditions Le Chatelier s Priniple 1. Change onentrations by adding or removing produts or reatants.. Changing the partial pressure of gaseous reatants or produts by hanging the volume.. Changing the temperature. 4. (Catalyst) 11
Effets of Pressure Change If the produts in a gaseous reation ontain fewer moles of gas than the reatants, it is logial that they would require less spae. So, reduing the volume of the reation vessel would, therefore, favor the produts. Conversely, if the reatants require less volume (that is, fewer moles of gaseous reatant), then dereasing the volume of the reation vessel would shift the equilibrium to the left (toward reatants). Literally squeezing the reation will ause a shift in the equilibrium toward the fewer moles of gas. It s a simple step to see that reduing the pressure in the reation vessel by inreasing its volume would have the opposite effet. In the event that the number of moles of gaseous produt equals the number of moles of gaseous reatant, vessel volume will have no effet on the position of the equilibrium. See Example 14.10 Do Exerise 14.1 Look at Problems 14.75-76 Effet of Temperature Change Let s look at heat as if it were a produt in exothermi reations and a reatant in endothermi reations. We see that inreasing the temperature is analogous to adding more produt (in the ase of exothermi reations) or adding more reatant (in the ase of endothermi reations). This ultimately has the same effet as if heat were a physial entity. Effet of Temperature Change Temperature has a signifiant effet on most reations (See Figure 14.1 Reation rates generally inrease with an inrease in temperature. Consequently, equilibrium is established sooner. In addition, the numerial value of the equilibrium onstant varies with temperature. Effet of Temperature Change For example, onsider the following generi exothermi reation. reatants produts "heat" ( H is negative) Inreasing temperature would be analogous to adding more produt, ausing the equilibrium to shift left. Sine heat does not appear in the equilibriumonstant expression, this hange would result in a smaller numerial value for. 1
Effet of Temperature Change For an endothermi reation, the opposite is true. " heat" reatants produts ( H is positive) Inreasing temperature would be analogous to adding more reatant, ausing the equilibrium to shift right. This hange results in more produt at equilibrium, amd a larger numerial value for. Effet of a Catalyst A atalyst is a substane that inreases the rate of a reation but is not onsumed by it. It is important to understand that a atalyst has no effet on the equilibrium omposition of a reation mixture (see Figure 15.15). A atalyst merely speeds up the attainment of equilibrium. Effet of Temperature Change In summary: For an endothermi reation ( H positive) the amounts of produts are inreased at equilibrium by an inrease in temperature ( is larger at higher temperatures). For an exothermi reation ( H is negative) the amounts of reatants are inreased at equilibrium by an inrease in temperature ( is smaller at higher temperatures). Oxidation of Ammonia Using a opper atalyst Results in N and H O Using a platinum atalyst Results in NO and H O Do Exerises 14.14 and 15 Look at Problems 14.77-78, 81-8 Operational Skills Applying stoihiometry to an equilibrium mixture Writing equilibrium-onstant expressions Obtaining the equilibrium onstant from reation omposition Using the reation quotient Obtaining one equilibrium onentration given the others Solving equilibrium problems Applying Le Chatelier s priniple 1