Chapter 13, Chemical Equilibrium

Transcription

1 Chapter 13, Chemical Equilibrium You may have gotten the impression that when you mix two reactants together, the ensuing reaction goes to completion. In other words, the reactants are converted completely to products. In this chapter we will learn that is frequently 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. 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) W 2 NO 2 (g) colorless reddish-brown Aside on terms: 1. The stuff on the left is reactant, on the right is product. 2. The rxn. toward the right (forming NO 2 ) is the forward rxn. The leftward (forming N 2 O 4 ) is the backward rxn. We can study this reaction by introducing a pure sample of either N 2 O 4 or NO 2 into a reaction vessel. 1

2 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 concentration change.) At the leveled off point, [N 2 O 4 ] =.0337 M [NO 2 ] =.0125 M 2. Starting with a pure sample of NO 2 at 0.08 M (Fig. 13.1b): [NO 2 ] decreases sharply and [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. 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? 2. Likewise, why is the rate of the reverse rxn. increasing as the rxn. proceeds? 2

3 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) II. The Equilibrium Constant K c (Note: upper case K) A. For any rxn. of the type: a A + b B W c C + d D you can write an equilibrium constant expression: K c = [C] c [D] d (circa 1860's, Norway) [A] a [B] b Note: K c = [product terms] [reactant terms] 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: K c = [NO 2 ] 2 = (M) 2 = M [N 2 O 4 ] (M) 2. Your text says is sometimes custom to omit units when using K c, but this depends on the branch of chemistry. 3

4 C. Equilibrium constant for the reverse rxn. What is K c for: 2 NO 2 (g) W N 2 O 4 (g)? (The reverse rxn. equilibrium constant is called K c N.) Try Probs , p Key Concept Prob III. The Equilibrium Constant K p 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) W 2 NO 2 (g): K p = [P NO2 ] 2 (Define P X ) [P N 2O4] B. Recall that there is a direct effect of temperature on P: PV = nrt 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 & 6. 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?) 4

5 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. IV. Heterogeneous Equilibria (re. Qual. Scheme) A. You should be familiar with the following rxn. from your laboratory work: Ag + (aq) + Cl! (aq) W AgCl (s) This equilibrium has components in 2 different phases. B. Similarly, in the decomposition of CaCO 3 : CaCO 3(s) W CaO (s) + CO 2(g) 1. A K c expression for this rxn. would be: [CaO] [CO2 ] 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 ] 5

6 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 X 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. 13.7, p C. Differences in K c values for heterogenous systems are the basis for most of the cation part of the qual scheme. V. Using the Equilibrium Constant Expression 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) W 2 H 2 O (g) K c = (You fill it in!) 6

7 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. 2. Another example: H 2(g) + I 2(g) W 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. 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? 7

8 We can define the Reaction Quotient to answer this question: 2 [HI] t Q c = [H 2 ] t [I 2 ] t 1. t refers to some arbitrary time. 2. This system is not necessarily at equilibrium!!! 2. We can plug values for any set of [H 2 ], [I 2 ], and [HI] into the Q c. Then, by comparing the value of the Q c to the value of K c, we can predict the direction the system will go. (i.e., 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 )? H 2(g) + I 2(g) W 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 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. 13.9, p Do Key Concept Prob

9 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. 537.) 3. More often, you will only have initial concentrations for the reactants, or some combination of components. Then you must be more creative. See Fig. 13.6, p Let s apply this list to Prob , p. 516: Step 1: Write a balanced equation for the reaction: (You fill it in:) Step 2: Concentrations: Initial Change Final 9

10 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: 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. Your check here: [NO 2 ] x 10!3 = [N 2 O 4 ] 10

11 VI. Factors That Alter the Composition of an Equilibrium Mixture A. Once a system reaches equilibrium, it remains that way until 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. 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. ) VII. Altering an Equilibrium Mixture: Changes in Concentration A. What if you add more of one 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?) 11

12 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. VIII. 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 & Key Conc , p IX. Altering an Equilibrium Mixture: T Changes A. The outcome depends on whether the rxn. in question is exothermic or endothermic (!)H or +)H). 1. For exothermic rxn., K c decreases as T increases. 2. For endothermic rxn., K c increases as T increases. 12

13 B. We can explain why when we reach section X. The Effect of a Catalyst on Equilibrium A. A catalyst does decrease the time to reach equilibrium, but B. A catalyst does not alter the value of K c. (Fig , p. 527, Prob , p ) XI. The Link Between Equilibrium and Kinetics A. Derive the equilibrium constant expression from the definition of equilibrium. 1. At equilibrium, the forward rate = the reverse rate 2. Substitute/rearrange so k f and k r values are on the same side, with k f. (Look familiar?) k r If we have time, Prob , p If not, try on your own. Why are you inhaling? See p for related info. 13