Chapter 13, Chemical Equilibrium

Transcription

1 Chapter 13, Chemial Equilibrium You may have gotten the impression that when 2 reatants mix, the ensuing rxn goes to ompletion. In other words, reatants are onverted ompletely to produts. We will now learn that is often not the ase, at least not in a rigorous, quantitative sense. We will now learn: 1. to quantitatively desribe how far a given reation proeeds toward ompletion. 2. how altering onditions an shift equilibrium onentrations in a reation. 1

2 I. The Equilibrium State A. We will start by looking at a rxn. that learly does not go to ompletion: N O (g) 2 NO (g) olorless reddish-brown Aside on terms: 1. Stuff on the left is reatant, on the right is produt. 2. The rxn. toward the right (forming NO 2) is the forward rxn. The leftward (forming N2O 4) is the bakward rxn. We an study this reation by introduing a pure sample of either N O or NO into a reation vessel

3 1. Starting with a pure sample of N2O 4 at 0.04 M (Fig. 13.1a), we see [N2O 4] dereases somewhat & [NO 2] inreases during early part of the time. Roughly half-way through the observation time, the urves level off. (i.e., There is no further on. hange.) At leveled off point, [N2O 4] =.0337 M [NO ] =.0125 M 2 2. Starting with a pure NO 2 at 0.08 M (Fig. 13.1b): [NO 2] dereases sharply & [N2O 4] inreases during the early part of the observation time. Conentrations level off again. At this point, st onentrations are same as 1 experiment. 3

4 3. What happens at the leveled off point? a) Is the system frozen in plae? b) Are the reations (forward and bakward) still ourring, but at equal rates? Aside: At the leveled off point we say that we have reahed equilibrium. Do we ever really reah it, or are we approahing it? B. At equilibrium, the forward and bakward rates of the reation are equal. (See Fig. 13.2) 1. Based on your work in Chap. 12, why is the rate of the forward rxn. dereasing as the rxn. proeeds? 4

5 2. Likewise, why is the rate of the reverse rxn. inreasing as the rxn. proeeds? C. What would happen if you started with different initial [N O ] or [NO ]? (See Table 13.1) [N O ] & [NO ] at equil. depend on [N O ] & [NO ] 2. The ratio [NO ] /[N O ] is a onstant!!! (Table 13.1)

6 II. The Equilibrium Constant K (or K or just K) (Note: upper ase K) eq A. For any rxn. of the type: you an write an equilibrium onstant expression: a A + b B C + d D K = [C] [D] (ira 1860's, Norway) [A] [B] Note: K = [produt terms] [reatant terms] a d b 6

7 1. K is onstant for a speifi rxn. at a speifi temp. 2. For our rxn. with N2O 4: At 25 C K = [NO 2] = 4.64 x 10 [N O ] B. Units of K If K is shown with units, they are determined by the onentration units and the number of terms in the K expression. From above: [NO ] (M) 2 2 K 2 = [N = = M 2O 4] (M) 7

8 2. Your text says units are generally omitted when using K, but this depends on the branh of hemistry. C. Equilibrium onstant for the reverse rxn. What is K for: 2 NO (g) N O (g)? (The reverse rxn. equilibrium onstant is alled K.) Try Probs , pp Key Conept Prob

9 III. The Equilibrium Constant K p When doing gas phase hemistry partial pressure units are sometimes used rather than molarity. The symbol used for the equilibrium onstant in this ase is K p. However, these onstants generally behave like K. The main distintion from K is in B, below. Comment re. respiratory physiology. A. Example, for N O 2 NO : 2 4(g) 2(g) 2 NO2 N2O4 K = [P ] p (Define P ) [P ] X 9

10 B. Reall that there is a diret effet of temperature on P: PV = nrt 1. Beause of this, the numerial values of K p and K are not likely to be equal. 2. However, you an determine the relationship ( + d) - (a + b) between the 2 Ks: K = K x (RT) p See text for derivation. (Can you do the derivation? Comment on derivations.) On your own, Prob & 6. 10

11 C. Biologists alert!!! 1. If you have an interest in respiratory physiology, you might use these onstants. O 2 levels (regarding hemoglobin binding, et.) are usually expressed as partial pressures. (Reovery room?) 2. Hemoglobin ~ half-saturated w/ PO 2 at 26 mm Hg. So far we have onfined our omments to one phase systems. Heterogeneous equilibria are also interesting. 11

12 IV. Heterogeneous Equilibria (Qual. Sheme) A. You should (will?) be familiar with the following rxn. from your laboratory work: Ag + + Cl AgCl (aq) (aq) (s) This system has omponents in 2 phases. B. Similarly, in the deomposition of CaCO : 3 CaCO CaO + CO 3(s) (s) 2(g) 12

13 1. A K expression for this rxn. would be: [CaO] [CO 2] K = [CaCO ] 3 2. However, your text notes that as CaO and CaCO 3 are solids, so [CaO] and [CaCO 3] an t hange. (Think ollisions.) We an fator out these onstant terms: [CaO] K = [CO ] x [CaCO ] 2 3 [CaCO 3] K x [CaO] = [CO ] 2 13

14 Beause [CaO] and [CaCO 3] are onstant, we an ombine this term with the K term: [CaCO 3] K x [CaO] = K This is beause: a onstant a onstant = a onstant. 3. Finally: K = [CO 2] 4. Note that there may be additional reasons for leaving solid or pure liquid omponent onentration terms out of equilibrium onstant expressions. Do Prob. 13.7, p

15 C. Differenes in K values for heterogenous systems are the basis for most of the ation part of the qual sheme. V. Using the Equilibrium Constant Expression A. Judging the extent of a reation. How far (qualitatively) does it go? 1. The K value tells us whether we will have largely produts or largely reatants at equilibrium. For example: 2 H + O 2 H O 2(g) 2(g) 2 (g) K = (You fill it in!) 15

16 If we go to the lab we an measure K. People have 47 done this. K = 2.4 x 10 at 500 K. What does this mean re. [H2O], [H 2], [O 2] at equil? Look at K expression above to figure this out. 2 If [H O] = 5 M, what would the [H ] x [O ] be? Very small. Essentially all of the material in this system is present in the produt, H O. 2 16

17 2. Another example: H 2(g) + I 2(g) 2 HI (g) 2 [HI] K = [H ] [I ] At 700 K, K = If we have an equilibrium ondition where both [H 2] and [I ] are 0.10 M, what is [HI]? 2 1/2 Solve for [HI]: [HI] = (K x [H 2] x [I 2]) round 1/2 [HI] = (57.0 x 0.10 x 0.10) = Here, a signifiant portion of material is present in both the reatant and produt omponents. Problem 13.8, p

18 B. Prediting Diretion of Rxn. Whih way does it go? 1. If we mix reatants & produts in speifi amounts, will the system proeed to form produts or reatants? We an define the Reation Quotient to answer this question: 2 t [HI] Q = [H ] [I ] 2 t 2 t a) t refers to some arbitrary time. b) This system is not neessarily at equilibrium!!! 18

19 2. Plug values for any set of [H 2], [I 2], and [HI] into Q. Then, ompare value of Q to value of K. Predit diretion the system goes (towards produts or reatants) For ex., if [H 2] = 0.07 M, [I 2] = 0.2M & [HI] = 3.0 M, will this go toward produt (HI) or reatants (H 2 & I 2)? H 2(g) + I 2(g) 2 HI (g) 2 [3.0 M] 2 2 Q = [0.07 M] [0.2 M] = 9.0 M /0.014 M = 643 Beause Q is larger than K, & the system must go toward the appropriate equilibrium ratios, [HI] must get smaller and [H 2] & [I 2] bigger. That is, the rxn. will go toward the reatants H & I

20 3. Summary: If Q < K, rxn. goes toward produts. If Q > K, rxn. goes toward reatants. If Q = K, rxn. is already at equilibrium. See Fig. 13.5, if it helps. Do prob. 13.9, p Do Key Conept Prob

21 C. Calulating Equilibrium Conentrations. How far (quantitatively) does it go? 1. It is useful to be able to predit how muh of a given reatant or produt is present at equilibrium. 2. If you have equilibrium [values] for all variables but one, this is a straightforward problem. Solve for the one unknown, substitute the knowns, & runh numbers. (Try Problem in re.???, p. 530.) 3. More often, you will only have initial onentrations for the reatants, or some ombination of omponents. Then you must be more reative. See 21

22 Fig. 13.6, p Let s apply this list to Prob , p. 510: Step 1: Write a balaned equation for the reation: (You fill it in:) Step 2: Conentrations: Initial Change Final 22

23 Step 3: Substitute into the equilibrium onstant expression and solve for x: 2 [NO 2] K = [N O ] where K = 4.64 x ½ x = b ± (b 4a) 2a The quadrati formula. Solve for x in the spae below: 23

24 Step 4: One you have x, go bak and alulate the equilibrium onentrations of the omponents. [N2O 4] = x [NO 2] = 2 x Your values here Step 5: Chek your work by substituting these values bak into the equilibrium onstant expression. 2 [NO 2] x 10 = [N O ]

25 Your hek here: VI. Fators That Alter the Composition of an Equilibrium Mixture A. One a system reahes equilibrium, it remains that way til the system is perturbed. B. A variety of perturbations an our. 1. Conentration of a produt or reatant an be altered. 2. Pressure and/or volume an be hanged. 3. Temperature an be hanged. 25

26 C. Can we predit how the system will adjust in response to a speifi perturbation? (Reall Le Châtelier. The system responds in a way that relieves the stress. ) VII. Altering an Equilibrium Mixture: Changes in Conentration A. What if you add more of 1 of the produts? 1. Stress of adding produt eased by reduing [produt]. 26

27 2. I personally find the ollision theory approah more satisfying. Adding more produt inreases ollisions between produt moleules, and therefore the bakward rxn. rate. (Example?) B. What if you remove of one of the produts? 1. The stress of removing a produt an be relieved by inreasing the amount of produt. (Prob ) 2. Removing produt dereases ollisions between produt moleules, and the bakward rxn. rate. 27

28 VIII. Altering an Equilibrium Mixture: Changes in P and V A. What happens if you inrease the pressure (P) by dereasing the system volume (V)? 1. The system will respond (if possible) in a way that dereases P. This an our if the number of reatant moleules in the gas phase is not equal to the number of produt moleules in the gas phase. 2. The reverse ours if you inrease the system V. Look at Prob & Key Conept 13.18, p

29 IX. Altering Equilibrium Mixtures: T Changes A. Outome depends on whether the rxn. is exothermi or endothermi ( ÄH or +ÄH). We an obtain answers by treating heat as a produt or reatant: 1. For exothermi rxn., K dereases as T inreases. A + B + heat C + D How would this shift if T goes up? Think upping T = adding heat. [C] [D] K = [A] [B] 2. Likewise, endothermi, K inreases as T inreases. 29

30 Real: Haber proess T dependene of K. N (g) + 3 H (g) 2 NH (g) kj (exothermi) (ÄH = 92.2 kj) You need to make 10 metri tons of NH 3 to sell; is 300 or 1000 K a better temp? B. Proper explanation in setion Try Key onept problem 13.21, p

31 X. The Effet of a Catalyst on Equilibrium A. A atalyst dereases t to reah equilibrium, but B. A atalyst doesn t alter K. (Fig , p. 521) (Prob , p. 521) This stuff gets way ool when you look at biologial atalysts. 31

32 XI. The Link Between Equilibrium & Kinetis A. Let s derive the equilibrium onstant expression from the definition of equilibrium. 1. At equilibrium, the forward rate = the reverse rate 2. Substitute/rearrange so k and k values are on the same side, f r with k f. k r What assumptions, what definitions? (Look familiar?) If we have time, Prob , p If not, try on your own. Why are you inhaling? See p for related info. 32