Chapter 15: Equilibrium Part 2
Read:
HW:
BLB 15.4–5
BLB 15:27, 28, 29, 30, 32, 35, 37, 39, 46;
Supplemental 15:5–10
What does the size of K tell us?
Reactants
!If K ! 1
Know:
• homogeneous & heterogeneous equilibria
• equilibrium apps
• calculating equilibrium constants
products
[products] ! [reactants]
K!
[products]
[reac tan ts]
• math refresher in BLB Appendix A
Meaning of the magnitude of K
FINAL DEADLINE for credit on skill check
tests: _______________________________
Final Exam: __________________
Know your section's Exam location, Know
your section number, Bring your ID!
!If K >> 1
K!
[products]
big #
=
little # [reac tan ts]
!If K <<1
K!
Dr. Lori S. Van Der Sluys
Page 1
Chapter 15
products are favored
reactants are favored
little # [products]
=
big # [reac tan ts]
Dr. Lori S. Van Der Sluys
Page 2
Chapter 15
Demonstration
Is Keq large or small?
Complexation Reactions
!Very large Keq
!Related to Acid-Base Reactions
PbI2(s)
yellow
Pb2+(aq) + 2I- (aq)
clear
!Metal cation acts as an acid
clear
!Electron-rich molecule or anion (ligand)
acts as a base
Keq =
!Metal + ligand forms a complex (can be
an ion)
Cu2+(aq) + 4NH3(aq)
pale blue
clear
Cu(NH3)42+(aq)
deep blue
Keq =
[Cu(H2O)2(NH3)2]2+
Ni2+(aq) + 6NH3(aq)
pale green
clear
Ni(NH3)62+(aq)
lavender
[Cu(H2O)2(en)2]2+
!Ligands can bind to the metal in more
than one spot (chelating ligand or “claw”,
eg. EDTA, en)
Keq =
Dr. Lori S. Van Der Sluys
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Chapter 15
Dr. Lori S. Van Der Sluys
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Chapter 15
Example Complexation Reaction
Molecular Equation:
CuSO4(aq) + NH3(aq) "
Cu(NH3)4(SO4)2(aq)
Predicting the direction of a reaction
aA + bB
Reaction quotient # Q
Q=
Complete Ionic Equation:
Cu2+(aq) + SO42-(aq) + NH3(aq) "
[Cu(NH3)4]2+(aq) + SO42-(aq)
Net Ionic Equation:
cC + dD
[C ]c [D]d
[A]a [B ]b
Note: the concentrations used are NOT
equilibrium concentrations.
*
When Q = Kc
system IS at equilibrium
• When Q < Kc
reaction moves to right
(produces product)
• When Q > Kc
reaction moves to left
(produces reactant)
Dr. Lori S. Van Der Sluys
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Chapter 15
Dr. Lori S. Van Der Sluys
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Chapter 15
2HI(g)
Using Keq to obtain equilibrium
concentrations of reactants and products.
H2(g) + I2(g)
2 IBr(g)
KC = 1.25 x 10-3 =
Kc =
If we put 0.1 mole of HI in a 1L container, what
will happen?
[Br2 ][I2 ] = 2.5 "10#3
2
[ IBr]
Initially [IBr] = [I2] = [Br2] = 0.05M
!
1. reaction shifts to right "
2. reaction shifts to left $
3. no change occurs
Br2(g) + I2(g)
What are the final concentrations of
reactants and products?
If the concentrations of all three gases in the
vessel are 0.1 mol/L, what will happen?
1. reaction shifts to right "
2. reaction shifts to left $
3. no change occurs
Dr. Lori S. Van Der Sluys
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Chapter 15
Dr. Lori S. Van Der Sluys
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Chapter 15
Multistep Reactions:
How are the Equilibria Related?
Multistep Reactions:
Use Hess's Law: Use Keq of 2 or more
steps to get Keq for an overall process.
2
N2O4(g)
2NO2(g)
KC1= 0.212 @ 100°C
2NOBr(g)
2NO(g) + Br2(g)
K C1 =
[ NO ] [Br2 ] = 0.014
2
[ NOBr ]
2
Br2(g) + Cl2(g)
Equilibrium Expression?
KC2 =
2BrCl(g)
!
K C1=
2NOBr(g) + Cl2(g)
[BrCl ] = 7.2
Br
[ 2 ][Cl2 ]
2NO(g) + 2BrCl(g)
!
What is Keq for the overall reaction?
If we multiply by 2:
2N2O4(g)
4NO2(g)
Equilibrium Expression?
K C2=
K C2=
Dr. Lori S. Van Der Sluys
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Chapter 15
Dr. Lori S. Van Der Sluys
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Chapter 15
Manipulation of Equilibrium
Constants:
! Keq in the reverse direction is the
inverse of the forward reaction Keq.
! Keq of a reaction that has been
multiplied by an integer is the original
Keq raised to the integer power.
! If the net reaction is made up of 2 or
more steps, Keq is the product of the
individual Keq's.
Dr. Lori S. Van Der Sluys
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Chapter 15