Chapter 13, Chemical Equilibrium

Transcription

1 Chapter 13, Chemical Equilibrium You may have gotten the impression that when 2 reactants mix, the ensuing rxn goes to completion. In other words, reactants are converted completely to products. We will now learn that is often not the case, at least not in a rigorous, quantitative sense. We will now learn: 1. how to give quantitative descriptions of how far toward completion a given reaction proceeds. 2. how altering conditions can shift equilibrium concentrations in a reaction vessel. 1

2 Analogy: Escalator picture I. The Equilibrium State A. We will start by looking at a rxn. that clearly does not go to completion: N 2 O 4 (g) 2 NO 2 (g) colorless reddish-brown Aside on terms: 1. Stuff on the left is reactant, on the right is product. 2. Rxn. toward the right (forming NO 2 ) is the forward rxn. Leftward (forming N 2 O 4 ) is the backward rxn. 2

3 We can study this reaction by introducing a pure sample of either N 2 O 4 or NO 2 into a reaction vessel. 3

4 1. Starting with a pure sample of N 2 O 4 at 0.04 M (Fig. 13.1a), we see [N 2 O 4 ] decreases somewhat & [NO 2 ] increases during early part of the time. Roughly halfway through the observation time, the curves level off. (i.e., There is no further conc. change.) At leveled off point, [N 2 O 4 ] =.0337 M [NO 2 ] =.0125 M 2. Starting with a pure NO 2 at 0.08 M (Fig. 13.1b): [NO 2 ] decreases sharply & [N 2 O 4 ] increases during the early part of the observation time. Concentrations level off again. At this point, concentrations are same as 1 st experiment. 4

5 3. What happens at the leveled off point? a) Is the system frozen in place? b) Are the reactions (forward and backward) still occurring, but at equal rates? Aside: At the leveled off point we say that we have reached equilibrium. Do we ever really reach it, or are we approaching it? B. At equilibrium, the forward and backward rates of the reaction are equal. (See Fig. 13.2) 1. Based on your work in Chap. 12, why is the rate of the forward rxn. decreasing as the rxn. proceeds? 5

6 2. Likewise, why is the rate of the reverse rxn. increasing as the rxn. proceeds? C. What would happen if you started with different initial [N 2 O 4 ] or [NO 2 ]? (See Table 13.1) 1. [N 2 O 4 ] & [NO 2 ] at equil. depend on [N 2 O 4 ] 0 & [NO 2 ] 0 2. The ratio [NO 2 ] 2 /[N 2 O 4 ] is a constant!!! (Table 13.1) 6

7 II. The Equilibrium Constant K c (or K eq or just K) (Note: upper case K) A. For any rxn. of the type: a A + b B c C + d D you can write an equilibrium constant expression: K = [C] c [D] d c (circa 1860's, Norway) [A] a [B] b Note: K c = [product terms] [reactant terms] 7

8 1. K c is constant for a specific rxn. at a specific temp. 2. For our rxn. with N 2 O 4 : At 25 C K c = [NO 2 ] 2 = 4.64 x 10 3 [N 2 O 4 ] B. Units of K c 1. If K c is shown with units, they are determined by the concentration units and the number of terms in the K c expression. From above: [NO 2 ] 2 (M) 2 K c = = = M [N 2 O 4 ] (M) 8

9 2. Your text says is sometimes custom to omit units when using K c, but this depends on the branch of chemistry. C. Equilibrium constant for the reverse rxn. What is K c for: 2 NO 2 (g) N 2 O 4 (g)? (The reverse rxn. equilibrium constant is called K c.) Try Probs , p Conceptual Prob. 13.4, p (Next page) 9

10 III. The Link Between Kinetics & Equilibrium M + N Q + P Assume single step mechanism (works without this assumption, but it is a more complicated derivation). rate f = rate r = 10

11 What is true at equilibrium? Therefore; Rearrange to get: A constant a constant = another constant, so define: K = k f k r Note script, lower case!!! And substitute to complete the derivation! 11

12 V. The Equilibrium Constant K p (on your own) When doing gas phase chemistry partial pressure units are sometimes used rather than molarity. The symbol used for the equilibrium constant in this case is K p. However, these constants generally behave like K c. A. Example, for N 2 O 4(g) 2 NO 2(g) : K p = [P NO 2] 2 (Define P X ) [P N 2O4] B. Recall that there is a direct effect of temperature on P: PV = nrt 12

13 1. Because of this, the numerical values of K p and K c are not likely to be equal. 2. However, you can determine the relationship between (c + d) - (a + b) the 2 Ks: K p = K c x (RT) See text for derivation. On your own, Prob & 7. 13

14 C. Biologists alert 1. If you have an interest in respiratory physiology, you might use these constants. O 2 levels (regarding hemoglobin binding, etc.) are usually expressed as partial pressures. (Recovery room?) 2. Hemoglobin ~ half-saturated w/ PO 2 at 26 mm Hg. So far we have confined our comments to one phase systems. Heterogeneous equilibria are also interesting. See also: COPD, scoliosis, emphysema, etc. 14

15 V. Heterogeneous Equilibria (Qual. Scheme) A. You should (will?) be familiar with the following rxn. from your laboratory work: Ag + (aq) + Cl (aq) AgCl (s) This system has components in 2 phases. B. Similarly, in the decomposition of CaCO 3 : Important in manufacturing cement. CaCO 3(s) CaO (s) + CO 2(g) 15

16 1. A K c expression for this rxn. would be: [CaO] [CO 2 ] K c = [CaCO 3 ] 2. However, your text notes that as CaO and CaCO 3 are solids, [CaO] and [CaCO 3 ] are constant. We can factor out these constant terms: [CaO] K c = [CO 2 ] x [CaCO 3 ] [CaCO 3 ] K c x [CaO] = [CO 2 ] 16

17 Because [CaO] and [CaCO 3 ] are constant, we can combine this term with the K c term: [CaCO 3 ] K c x [CaO] = K c This is because: a constant a constant = a constant. 3. Finally: K c = [CO 2 ] 4. Note that there may be additional reasons for leaving solid or pure liquid component concentration terms out of equilibrium constant expressions. Do Prob. 13.8, p

18 C. Differences in K c values for heterogenous systems are the basis for most of the cation part of the qual scheme. VI. Using Equilibrium Constant Expressions A. Judging the extent of a reaction. How far (qualitatively) does it go? 1. The K c value tells us whether we will have largely products or largely reactants at equilibrium. For example: 2 H 2(g) + O 2(g) 2 H 2 O (g) K c = (You fill it in!) 18

19 If we go to the lab we can measure K c. People have done this. K c = 2.4 x at 500 K. What does this mean re. [H 2 O], [H 2 ], [O 2 ] at equil? Look at K c expression above to figure this out. If [H 2 O] = 5 M, what would the [H 2 ] 2 x [O 2 ] be? Very small. Essentially all of the material in this system is present in the product, H 2 O. 19

20 2. Another example: H 2(g) + I 2(g) 2 HI (g) [HI] 2 K c = [H 2 ] [I 2 ] At 700 K, K c = 57.0 If we have an equilibrium condition where both [H 2 ] and [I 2 ] are 0.1 M, what is [HI]? Solve for [HI]: [HI] = (K c x [H 2 ] x [I 2 ]) 1/2 [HI] = (57.0 x 0.1 x 0.1) 1/2 = 0.75 Here, a significant portion of material is present in both the reactant and product components. 20

21 B. Predicting Direction of Rxn. Which way does it go? 1. If we mix reactants & products in specific amounts, will the system proceed to form products or reactants? We can define the Reaction Quotient to answer this question: [HI] t 2 Q c = [H 2 ] t [I 2 ] t a) t refers to some arbitrary time. b) This system is not necessarily at equilibrium!!! 21

22 2. Plug values for any set of [H 2 ], [I 2 ], and [HI] into Q c. Then, compare value of Q c to value of K c. Predict direction the system goes (towards products or reactants) For ex., if [H 2 ] = 0.07 M, [I 2 ] = 0.2M & [HI] = 3.0 M, will this go toward product (HI) or reactants (H 2 & I 2 )? hidden in white H 2(g) + I 2(g) 2 HI (g) [3.0 M] 2 Q c = [0.07 M] [0.2 M] = 9.0 M 2 /0.014 M 2 = 643 Because Q c is larger than K c, & 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 reactants H 2 & I 2. 22

23 3. Summary: If Q c < K c, rxn. goes toward products. If Q c > K c, rxn. goes toward reactants. If Q c = K c, rxn. is already at equilibrium. See Fig. 13.5, if it helps. Do prob , p Do Conceptual Prob , p Same as Conceptual Prob in 2 nd edition. 23

24 C. Calculating Equilibrium Concentrations. How far (quantitatively) does it go? 1. It is useful to be able to predict how much of a given reactant or product 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, & crunch numbers. (Try Problem in rec.???, p. 544.) 3. More often, you will only have initial concentrations for the reactants, or some combination of components. Then you must be more creative. See 24

25 Fig. 13.6, p. 522, if it helps. Worked examples???? Let s apply this list to Prob , p. 526: Step 1: Write a balanced equation for the reaction: (You fill it in:) Step 2: Concentrations: Initial Change Final 25

26 Step 3: Substitute into the equilibrium constant expression and solve for x: [NO 2 ] 2 K c = [N 2 O 4 ] where K c = 4.64 x 10 3 x = b ± (b 2 4ac) ½ 2a The quadratic formula. Solve for x in the space below: 26

27 Step 4: Once you have x, go back and calculate the equilibrium concentrations of the components. [N 2 O 4 ] = x [NO 2 ] = 2 x Your values here Step 5: Check your work by substituting these values back into the equilibrium constant expression. [NO 2 ] x 10 3 = [N 2 O 4 ] 27

28 Your check here: 28

29 VII. Factors That Alter the Composition of an Equilibrium Mixture A. Once a system reaches equilibrium, it remains that way til the system is perturbed. B. A variety of perturbations can occur. 1. Concentration of a product or reactant can be altered. 2. Pressure and/or volume can be changed. 3. Temperature can be changed. 29

30 C. Can we predict how the system will adjust in response to a specific perturbation? (Recall Le Châtelier. The system responds in a way that relieves the stress. ) 30

31 VIII. Altering an Equilibrium Mixture: Changes in Concentration A. What if you add more of 1 of the products? 1. Stress of adding product eased by reducing [product]. 2. I personally find the collision theory approach more satisfying. Adding more product increases collisions between product molecules, and therefore the backward rxn. rate. (Example?) 31

32 B. What if you remove of one of the products? 1. The stress of removing a product can be relieved by increasing the amount of product. (Prob ) 2. Removing product decreases collisions between product molecules, and the backward rxn. rate. 32

33 IX. Altering an Equilibrium Mixture: Changes in P and V A. What happens if you increase the pressure (P) by decreasing the system volume (V)? 1. The system will respond (if possible) in a way that decreases P. This can occur if the number of reactant molecules in the gas phase is not equal to the number of product molecules in the gas phase. 2. The reverse occurs if you increase the system V. Look at Prob & Conceptual Prob , p

34 Conceptual Prob 13.19, p

35 X. Altering Equilibrium Mixtures: T Changes A. Outcome depends on whether the rxn. is exothermic or endothermic ( ÄH or +ÄH). We can obtain answers by treating heat as a product or reactant: 1. For exothermic rxn., K c decreases as T increases. A + B + heat C + D [C] [D] K c = [A] [B] How would this shift if T goes up? Think upping T = adding heat. 2. Likewise, endothermic, K c increases as T increases. 35

36 Real: Haber process T dependence of K c. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) kj (exothermic) (ÄH = 92.2 kj) You need to make 10 metric tons of NH 3 to sell; is 300 or 1000 K a better temp? B. Proper explanation in section Try Conceptual problem 13.22, p

37 XI. The Effect of a Catalyst on Equilibrium A. Catalyst decreases t to reach equilibrium, but B. Catalyst doesn t alter K c. (Fig , p. 536) 37

38 (Prob , p. 537) Gets way cool w/ biological catalysts. FYI. Equilibrium and O 2 (& CO 2?)Transport A. Are you an obligate aerobe? Y or N Consequences? B. Is O 2 very soluble in water? C. Is how far/fast you can run dependent on how efficiently you can deliver O 2 to your cells? D. Selection for effective hemoglobin (Hb)? Biology: rbc s flowing though capillaries: Chemistry: See next page re. Hb-O 2 binding properties. Biochemistry/Structure: See pdbsum: 38

39 Hill Analysis of Hb-O 2 Interactions Vertical axis? is è, fractional saturation. Red = arterial PO 2, blue = venous PO 2 Hb shifts its binding curve roughly between n = 2 and n = 4 depending on the environment: arterial vs. venous [base vs. acid], O 2, CO 2, CO, BPG, etc. Never challenge an alligator to a breath holding contest!!! 39