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 completelyto products. We will now learn that is often not thecase, at least not in a rigorous, quantitative sense.

We will now learn:1. to quantitatively describe how far a given

reaction proceeds toward completion.2. how altering conditions can shift equilibrium

concentrations in a reaction.

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I. The Equilibrium State

A. We will start by looking at a rxn. that clearlydoes not go to completion:

2 4 2 N O (g) W 2 NO (g) colorless reddish-brown

Aside on terms: 1. “Stuff” on the left is reactant, on the right is product.

22. The rxn. toward the right (forming NO ) is the

2 4forward rxn. The leftward (forming N O ) is thebackward rxn.

We can study this reaction by introducing a pure

2 4 2sample of either N O or NO into a reaction vessel.

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2 41. Starting with a pure sample of N O at 0.04 M (Fig.

2 4 213.1a), we see [N O ] decreases somewhat & [NO ]increases during early part of the time. Roughlyhalf-way through the observation time, the curveslevel off. (i.e., There is no further conc. change.)

2 4At “leveled off” point, [N O ] = .0337 M

2 [NO ] =.0125 M

22. Starting with a pure NO at 0.08 M (Fig. 13.1b):

2 2 4[NO ] decreases sharply & [N O ] increases duringthe early part of the observation time. Concentrations level off again. At this point,concentrations are same as 1 experiment.st

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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 havereached equilibrium. Do we ever really reach it, or are weapproaching it?

B. At equilibrium, the forward and backward ratesof the reaction are equal. (See Fig. 13.2)

1. Based on your work in Chap. 12, why is the rate ofthe forward rxn. decreasing as the rxn. proceeds?

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2. Likewise, why is the rate of the reverse rxn.increasing as the rxn. proceeds?

C. What would happen if you started with different

2 4 2initial [N O ] or [NO ]? (See Table 13.1)

2 4 2 2 4 0 2 01. [N O ] & [NO ] at equil. depend on [N O ] & [NO ]

2 2 42. The ratio [NO ] /[N O ] is a constant!!! (Table 13.1)2

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cII. The Equilibrium Constant K

eq(or K or just K) (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

cconstant expression: K = [C] [D] c d

(circa 1860's, Norway) [A] [B]a b

cNote: K = [product terms] [reactant terms]

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c1. K is constant for a specific rxn. at a specific temp.

2 42. For our rxn. with N O :

c 2At 25EC K = [NO ] = 4.64 x 102 !3

2 4 [N O ]

c B. Units of K

c1. If K is shown with units, they are determined by theconcentration units and the number of terms in the

cK expression. From above:

2[NO ] (M) 2 2

cK = = = M2 4 [N O ] (M)

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2. Your text says units are generally omitted when

cusing K , but this depends on the branch ofchemistry.

C. Equilibrium constant for the reverse rxn.

c 2 2 4What is K for: 2 NO (g) W N O (g) ?

c(The reverse rxn. equilibrium constant is called K N.)Try Probs. 13.1-.3, pp. 496-7. Key Concept Prob. 13.4

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pIII. The Equilibrium Constant K

When doing gas phase chemistry partial pressure unitsare sometimes used rather than molarity. The symbol

pused for the equilibrium constant in this case is K .

cHowever, these constants generally behave like K . The

cmain distinction from K is in B, below.

Comment re. respiratory physiology.

2 4(g) 2(g) A. Example, for N O W 2 NO :

p NO2 K = [P ] 2

X N O2 4(Define P ) [P ]

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B. Recall that there is a direct effect of temperatureon P: PV = nRT

p c1. Because of this, the numerical values of K and Kare not likely to be equal.

2. However, you can determine the relationship

p cbetween the 2 Ks: K = K x (RT)(c + d) - (a + b)

See text for derivation. (Can you do the derivation? Comment on derivations.)

On your own, Prob. 13.5 & 6.

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C. Biologists alert!!!

1. If you have an interest in respiratory physiology, you

2might use these constants. O levels (regardinghemoglobin binding, etc.) are usually expressed aspartial pressures. (Recovery room?)

22. Hemoglobin ~ half-saturated w/ PO at 26 mm Hg.

So far we have confined our comments to one phasesystems. Heterogeneous equilibria are also interesting.

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IV. Heterogeneous Equilibria (Qual. Scheme)

A. You should (will?) be familiar with thefollowing rxn. from your laboratory work:

(aq) (aq) (s)Ag + Cl W AgCl+ !

This system has components in 2 phases.

3 B. Similarly, in the decomposition of CaCO :

3 2(s) (s) (g)CaCO W CaO + CO

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c1. A K expression for this rxn. would be:

2 [CaO] [CO ]c 3“K ” = [CaCO ]

32. However, your text notes that as CaO and CaCO are

3solids, so [CaO] and [CaCO ] can’t change. (Thinkcollisions.) We can factor out these constant terms:

[CaO]c 2 3“K ” = [CO ] x �[CaCO ]�

3 [CaCO ]c 2“K ” x � [CaO] � = [CO ]

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3Because [CaO] and [CaCO ] are constant, we can

ccombine this term with the “K ” term:

3 [CaCO ]

c c“K ” x � [CaO] � = K

This is because: a constant × a constant = a constant.

c 23. Finally: K = [CO ]

4. Note that there may be additional reasons for leavingsolid or pure liquid component concentration termsout of equilibrium constant expressions.

Do Prob. 13.7, p. 502.

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c C. Differences in K values for heterogenoussystems are the basis for most of the cation partof the qual scheme.

V. Using the Equilibrium Constant Expression

A. Judging the extent of a reaction. How far(qualitatively) does it go?

c1. The K value tells us whether we will have largelyproducts or largely reactants at equilibrium.

2(g) 2(g) 2 (g)For example: 2 H + O W 2 H O

cK = (You fill it in!)15

cIf we go to the lab we can measure K . People have

cdone this. K = 2.4 x 10 at 500 K.47

2 2 2What does this mean re. [H O], [H ], [O ] at equil?

cLook at K expression above to figure this out.

2 2 2If [H O] = 5 M, what would the [H ] x [O ] be?2

Very small. Essentially all of the material in this

2system is present in the product, H O.

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

[HI] 2

c 2 2 cK = [H ] [I ] At 700 K, K = 57.0

2If we have an equilibrium condition where both [H ]2and [I ] are 0.10 M, what is [HI]?

c 2 2Solve for [HI]: [HI] = (K x [H ] x [I ]) 1/2

round

[HI] = (57.0 x 0.10 x 0.10) = 0.75498 ö 0.751/2

Here, a significant portion of material is present inboth the reactant and product components.

Problem 13.8, p.503.

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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 orreactants?

We can define the Reaction Quotient to answer thisquestion:

t [HI] 2

c 2 t 2 t Q = [H ] [I ]

a) t refers to some arbitrary time. b) This system is not necessarily at equilibrium!!!

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2 2 c2. Plug values for any set of [H ], [I ], and [HI] into Q .

c cThen, compare value of Q to value of K . Predictdirection the system goes (towards products or reactants)

2 2For ex., if [H ] = 0.07 M, [I ] = 0.2M & [HI] = 3.0 M,

2 2will this go toward product (HI) or reactants (H & I )?

2(g) 2(g) (g)H + I W 2 HI [3.0 M] 2

cQ = [0.07 M] [0.2 M] = 9.0 M /0.014 M = 6432 2

c cBecause Q is larger than K , & the system must gotoward the appropriate equilibrium ratios, [HI] must

2 2get smaller and [H ] & [I ] bigger. That is, the rxn.

2 2will go toward the reactants H & I .

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c c3. Summary: If Q < K , rxn. goes toward products.

c cIf Q > K , rxn. goes toward reactants.

c cIf Q = K , rxn. is already at equilibrium.

See Fig. 13.5, if it helps. Do prob. 13.9, p. 505.

Do Key Concept Prob. 13.10

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C. Calculating Equilibrium Concentrations. Howfar (quantitatively) does it go?

1. It is useful to be able to predict how much of a givenreactant or product is present at equilibrium.

2. If you have equilibrium [values] for all variables butone, this is a straightforward problem. Solve for theone unknown, substitute the knowns, & crunchnumbers. (Try Problem 13.68 in rec.???, p. 530.)

3. More often, you will only have initial concentrationsfor the reactants, or some combination ofcomponents. Then you must be more creative. See

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Fig. 13.6, p. 506. Let’s apply this list to Prob. 13.13, p. 510:

Step 1: Write a balanced equation for the reaction:

(You fill it in:)

Step 2:Concentrations: Initial

Change

Final

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Step 3: Substitute into the equilibrium constant expressionand solve for x:

2 [NO ] 2

c 2 4 cK = [N O ] where K = 4.64 x 10!3

x = !b ± (b ! 4ac)2 ½

2a The quadratic formula.

Solve for x in the space below:

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Step 4: Once you have x, go back and calculate theequilibrium concentrations of the components.

2 4 2[N O ] = 0.0500 ! x [NO ] = 2 x

Your values here

Step 5: Check your work by substituting these values backinto the equilibrium constant expression.

2 [NO ] 2

2 44.64 x 10 = [N O ]!3

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Your check here:

VI. Factors That Alter the Composition of anEquilibrium Mixture

A. Once a system reaches equilibrium, it remainsthat 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.

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C. Can we predict how the system will adjust inresponse to a specific perturbation? (Recall LeChâtelier. The system responds in a way thatrelieves the “stress.”)

VII. 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].

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2. I personally find the collision theory approach moresatisfying. Adding more product increases collisionsbetween product molecules, and therefore thebackward rxn. rate. (Example?)

B. What if you remove of one of the products?

1. The stress of removing a product can be relieved byincreasing the amount of product. (Prob. 13.16)

2. Removing product decreases collisions betweenproduct molecules, and the backward rxn. rate.

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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 thatdecreases P. This can occur if the number ofreactant molecules in the gas phase is not equal tothe number of product molecules in the gas phase.

2. The reverse occurs if you increase the system V.Look at Prob. 13.17 & Key Concept 13.18, p. 517.

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IX. Altering Equilibrium Mixtures: T Changes

A. Outcome depends on whether the rxn. isexothermic or endothermic (!ÄH or +ÄH).We can obtain answers by treating heat as a product or reactant:

c1. For exothermic rxn., K decreases as T increases.

A + B + heat ö C + D How would this shift if Tgoes up?

Think upping T = adding heat.[C] [D]

cK =[A] [B]

c2. Likewise, endothermic, K increases as T increases.29

cReal: Haber process T dependence of K .

2 2 3N (g) + 3 H (g) º 2 NH (g) + 92.2 kJ (exothermic)

(ÄHE = !92.2 kJ)

3You need to make 10 metric tons of NH to sell; is 300 or 1000 K a better temp?

B. Proper explanation in section 16.11.

Try Key concept problem 13.21, p. 520.

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X. The Effect of a Catalyst on Equilibrium

A. A catalyst decreases t to reach equilibrium, but

c B. A catalyst doesn’t alter K . (Fig. 13.14, p. 521)

(Prob. 13.22, p. 521)This stuff gets way cool when you look at biological catalysts.

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XI. The Link Between Equilibrium & Kinetics

A. Let’s derive the equilibrium constant expressionfrom the definition of equilibrium.

1. At equilibrium, the forward rate = the reverse rate

f r2. Substitute/rearrange so k and k values are on the same side,

f with k . What assumptions, what definitions?r k (Look familiar?)

If we have time, Prob. 13.23, p. 523. If not, try on your own.

Why are you inhaling? See p.524-525 for related info.

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