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Chapter 14 -Chemical Equilibria

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A dynamic process with the forward and

backwards reactions both happening at thesame time - but in EQUILIBRIUMso no NETchange in reactants or products

H2O(l) H2O(g)

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E uilibrium Constant

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The equilibrium constant K c or K p

aA + bB cC + dD

Kc = [C] c [D]d /[A] a [B]b

Kp = P Cc PDd / P Aa PBb

! This is the version used for solution equilibria

! NO SOLIDS ORLIQUIDS

[A] concentration P A pressure

K values are a function of temperature andunitless

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Examples of Equilibria

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H2O(l) H2O(g)KP = P H2O

Vapor pressure of water = 0.031 atm at 25 oC

N2(g) + O 2(g) 2NO(g)

KP = (P NO)2 /(P N2 PO2 )

AgCl(s) Ag + (aq) + Cl - (aq)KC = [Ag + ] [Cl - ]

= 4.1 x 10 -31 at 25 oC

= 1.77 x 10 -10 at 25 oCK

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More Examples

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

KP = P CO2 KC = [CO 2]Note: CO 2 only 0.035% by volume in atmosphere

partial pressure = 3.5 x 10 -4 atm

Fe 2O3(s) + 3CO(g) 2Fe(l) + 2CO 2(g)

Kp = (P CO2 )2/(PCO )3 KC = [CO 2]2 /[CO] 3

Remember: Solids and liquids are not included!

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Rules for Combining or

Manipulating Keq

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1) Reverse the reaction get K = 1/K

2) Add two reactions: with K 1 and K2 getK = K 1K2

4) Multiply a reaction by a constant nK = K n

3) Subtract two reactions (1)-(2):

K = K 1 /K 2

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Reaction Quotient Q

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Same form as for K eq but using the [ ] or pressureyou have.

Q < K reaction will mover to the product

side (right) to achieve equilibrium

Q > K reaction will move to the reactantsside (left) to achieve equilibrium

Q = K reaction is at equilibrium

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Q Example

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H2O(g) + CO(g) H2(g) + CO 2(g)KP = 1.5 at a given T

KP =PH2PCO2 /P H2O PCO

If you have a gas mixture that is 0.60 atmH2O(g), 0.80 atm CO(g), 1.0 atm H 2(g) and 0.9atm CO 2 , which direction will the reaction go to

achieve equilibrium?

Q = (1.0)(0.9)/(0.6)(0.8) = 1.875

Q > Kreaction will go to the LEFT or REACTANTS

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Another Q Example

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If you mix two solutions of equal volume, onewith [Ag + ] = 0.110 ! M and another with [Cl - ]= 1.00 mM, will AgCl from a precipitate (ppt)

Q = [Ag + ][Cl- ] = {(0.110x10 -6 )/2}{(1.00x10 -3 )/2} = 2.75 x 10 -11

KC = 1.8x10 -10

Q < K so NO ppt

AgCl(s) Ag + (aq) + Cl - (aq)

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Calculating K eq fromReaction Data

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You place H 2(g) and I 2(g) in a container at 445 oC.What is K C for the following nal [ ]

Consider the reaction H 2(g) + I 2(g) 2HI(g)

Data from Table 14.1

[H2] = 0.11 M

[I2] = 0.11 M[HI] = 0.78 M

KC = [HI] 2 /[H 2][I2]KC = (0.78) 2 /(0.11)(0.11)

KC = 50.

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The ICE Table

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Example 14.6 Given: [CH 4] = 0.115 M;at equilibrium [C 2H2] = 0.035 M

2 CH 4(g) " C 2H2(g) + 3 H 2(g)

initial 0.115 0 0

change -0.070 +0.035 +0.105

nal 0.045 0.035 +0.105

KC = (0.035)(0.105) 3 /(0.045) 2 = 0.020

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Another Problem with ICE TableH2O(g) + CO(g) H2(g) + CO 2(g)

Given the following: 1.50x10 -2 moles H 2O and 1.50x10 -2 moles CO in a sealed 1 L vessel.At equilibrium there is 8.3x10 -3 molesof CO 2 .

What is the value of Kc?ICE table for the reaction

Initial 1.50x10 -2 1.50x10 -2 0 0

Reaction -.83x10 -2 -.83x10 -2 +.83x10 -2 +.83x10 -2

Final 0.67x10 -2 0.67x10 -2 0.83x10 -2 0.83x10 -2

H2O(g) + CO(g) H2(g) + CO 2(g)

KC =[H2][CO2]/[H 2O][CO]

= (.83x10-2

)2

/(.67x10-2

)2

KC = 1.5