Chemical Equilibrium
• Zumdahl Chapter 6 (6th Ed) Section 1, 2, 4, 6
• Define Reaction Equilibrium
– Contrast Equilibrium with Reaction Completion
• How we get to equilibrium (mass action)
• Follow how species change as we reach
equilibrium.
• The extent of a reaction in terms of activity
• Definition of Q and K
– Q The proper reaction quotient
– K: The equilibrium Constant a Number!!
• Problems: 6.2-4, 6.6, 6.9, 6.12-17, (in 6.15, the time is
moving left to right for the sequence of pictures), 6.19-29
Chemical Equilibrium
•
To this point we have assumed that all chemical
reaction go to completion. That is, a chemical reaction
is “complete” when all the reactions (up to the limting
reactant) to go to products.
•
Not all reactions go to completion. Even after an
infinitely long time, some reaction systems will remain
mixtures of reactants and products.
•
The reaction system includes …
– Concentrations/pressures of all species
– Temperature
Each reaction system has an equilibrium position at
which the average concentrations of all species are
constant for all time (assuming no perturbations).
A Chemical Reaction:
An internal Structural Rearrangement
• From Earth Moon Planets. 1997;78(1-3):29-35.
A model reaction occurs in asteroids on the Sun
side: H − C ≡ N ( g ) S H − N ≡ C ( g )
• If the bond strengths were equal, what ratio of
amounts (or concentrations) would you expect to
see?
An example of equilibrium:
The Great Crabapple War
vB
vM
cM
cB
The young boy can run much faster than the man; but fewer apples are scattered over his yard, so he has to run father to get the next apple. This gives the man an edge because apples are always handy, not far to go to find one. So the boy does not every get all the apples out of his yard. This is an example of dynamic equilibrium. Crabapple War
Expression for the Rate at which the two combatants remove
apples from the yard (a negative rate).
RateBoy = vBoy ⋅ C
1
2
BoyYard
and RateMan = vMan ⋅ C
1
2
ManYard
Number of Apples
C=
Area of Yard
Eventually, the boy’s superior speed will be matched by the
man’s larger concentration of crabapples, at which point:
RateBoy = RateMan
vBoy C
2
1
2
BoyYard
= vMan C
1
2
ManYard
⎛ vBoy ⎞
CManYard
= A Constant @ Equilibrium
⎜
⎟ =
C BoyYard
⎝ vMan ⎠
Watching a Chemical Reaction
• Demonstrate equilibrium with a dimerization
reaction: 2NO2 ( g ) R N 2O4 ( g )
• Or forming the monomers: N 2O4 ( g ) R 2 NO2 ( g )
• The monomer is brown, it is the main culprit in
smog, the dimer is colorless.
• Why does the color Change?
• What drives a chemical reaction?
Reaction Equilibrium Shifts (by Temperature)
Cold
N 2O4 ( g ) R 2 NO2 ( g )
colorless
brown
Hot
Molecular Representation of 2NO2→N2O4 in a
Closed Vessel; Reaction goes the other way.
FIGURE 6.1:
Time is not the only parameter controlling reactions and change.
The equilibrium is a dynamic process, the bond making and breaking
is going on at the molecular level, but the average number of different
types of molecules does not change.
How many steps of reaction separate these 4 different states?
When away from equilibrium; a system moves
(in time) to get to equilibrium.
N2O4 (g)
Colorless
2 NO2 (g)
Brown
N2O4 system moves (in time) to get to
equilibrium.
N2O4 (g)
Colorless
2 NO2 (g)
Brown
C
B
A
D
Characterize the Reaction
•
The two rates (forward and backward rates) compete
Rate forward
•
P ( N 2O4 )
⎛ P ( NO2 ) ⎞
= kf
and Ratebackward = k f ⎜
⎟
1Atm
⎝ 1Atm ⎠
At equilibrium the rates balance so:
Rate forward = Ratebackward and Po = 1Atm
kf
kf
kb
•
•
Peq ( N 2O4 )
Po
=
eq
⎛ Peq ( NO2 ) ⎞
= kb ⎜
⎟
Po
⎝
⎠
Peq 2 ( NO2 )
Po ⋅ Peq ( N 2O4 )
2
= A number : K P
eq
Regardless of the initial concentrations we have, we must end up
with the same ratio at equilibrium, although each species can have
quite different values, the ratio must be the same.
Why is the pressure for the monomer squared?
2
Characterize the Reaction: Reaction Quotient, Q
• The relation of the equilibrium constant at amounts at
equilibrium lets us write at any point in the reaction.
P 2 ( NO2 )
QP =
Po ⋅ P ( N 2O4 )
therefore
QP
eq
= KP
0.42
QP ( start ) =
= 0.133
1.2
2.62
QP (equilb) =
= 67. = K P
0.1
Activity
• Recall, we are considering chemical reactions that occur in solutions with numerous species present.
• When discussing gases we saw that ideal behavior was recovered at low pressure.
• The same is true for species in solution. Ideal behavior is recovered at “infinite dilution”. (solute vs solvent)
• The activity of a substance is a measure of the non‐ideality of its properties. • When performing a proper thermodynamic description of equilibrium, one considers the activity of substances, not simply their concentrations.
Activity
• For our purposes, we are going to simply refer all of our concentrations (or pressures) to a standard state:
o
o
o
o
P0 = 1Atm
For gases: 1 atm
For solutions: 1 M
Co = 1M
For pure liquids: The state of the liquid itself
For solids: The state of the solid itself
• Given these definitions, we define activity (ai) for the species i in solution as follows:
CJ [ J ]
aJ =
=
Co 1 M
PJ
aJ =
Po
Solute in Solution Gas in Mixture
aJ ( pure solid ) = 1
aJ ( pure liquid ) = 1
aJ ( solvent ) = 1
Chapter 5: Pressure and Concentration for Gasses
• Given I.G. equation and a reference pressure and a
reference concentration:
Po = 1Atm and Co = 1M From PV = nRT ⇒
P = CRT
⎛ RTCo ⎞
P C
=
⋅τ where τ = ⎜
⎟
Po Co
⎝ Po ⎠
Work with P in Atm, and C in M, and the two are proportional
by the dimensionless temperature τ = 0.082 ⋅ T
For each species that is a gas, its activity in terms of partial
pressures (preferred) can be related to its activity in terms of
concentration.
a ( P ) = τ ⋅ a (C )
The Equilibrium Constant
• The equilibrium constant is then expressed as a ratio of activities of products versus reactants as follows:
bB + cC ↔
⎛ ( aD ) d ( aE )e
K =⎜
⎜ ( a )b ( a ) c
C
⎝ B
dD + eE
⎞
⎟
⎟
⎠ @ EQUILIBRIUM
• This definition of the equilibrium constant supersedes the definition provided by Zumdahl in Section 6.2. It is correct for ANY reaction.
• This definition results in K being unitless, as it should!
The Reaction Quotient
• The equilibrium constant is then expressed as a ratio of activities of products versus reactants as follows:
bB + cC ↔
⎛ ( aD ) d ( aE )e
Q= ⎜
⎜ ( a )b ( a ) c
C
⎝ B
dD + eE
⎞
⎟
⎟
⎠
ANY TIME
• This definition of the reaction quotient supersedes the definition provided by Zumdahl in Section 6.6. Valid for ANY reaction.
• This definition results in Q and K being unitless, as it should!
A Chemical Reaction:
An internal Structural Rearrangement
• From Earth Moon Planets. 1997;78(1-3):29-35.
A model reaction occurs in asteroids on the Sun
side: H − C ≡ N ( g ) S H − N ≡ C ( g )
• If the bond strengths were equal, what ratio of
amounts (or concentrations) would you expect to
see?
HNC ]
P ( HNC )
[
K=
=
P ( HCN ) @ Eq [ HCN ] @ Eq
K << 1 or K ≈ 1 or K >> 1
The Ammonia Reaction: Equilbrium
• Example from Zumdahl (done with activities)
N 2 ( g ) + 3H 2 ( g ) ←⎯
→ 2NH 3 ( g )
• At equilibrium [NH3] = 3.1 x 10‐2 M, [N2] = 8.5 x 10‐1 M, and [H2] = 3.1 x 10‐3 M. What is the equilibrium constant?
⎛ [ NH 3 ] ⎞
⎜
⎟
1
M
⎠
= ⎝
3
⎛ [ N2 ] ⎞ ⎛ [H2 ] ⎞
⎜
⎟⎜
⎟
⎝ 1 M ⎠⎝ 1 M ⎠
2
a )
(
=
( a )( a )
2
KC
NH 3
N2
H2
3
@ EQUILIBRIUM
• We write KC to indicate that we are using concentrations.
FIGURE 6.5:
Concentration Profile for Ammonia Reaction
Equilibrium of Ammonia
• Table from Zumdahl:
KC =
(
aNH3
)
2
( a )( a )
N2
H2
3
@ EQUILIBRIUM
• Always the same equilibrium constant value no matter what the initial concentrations are.
• Determine K and Q from the initial concentrations (Warning!!!!)
The Equilibrium Constant (cont.)
• Back to our example reaction:
• What if the initial concentrations are different?
The Equilibrium Constant (cont.)
• Continuing on
⎛ [ NH 3 ] ⎞
⎜
⎟
1M ⎠
⎝
=
3
⎛ [ N2 ] ⎞ ⎛ [H2 ] ⎞
⎜
⎟⎜
⎟
1
M
1
M
⎝
⎠⎝
⎠
2
a )
(
=
( a )( a )
2
KC
NH 3
N2
H2
3
2
⎛ 3.1× 10 M ⎞
⎜
⎟
1
M
⎝
⎠
4
3.8
10
=
=
×
3
−1
−3
⎛ 8.5 × 10 M ⎞ ⎛ 3.1× 10 M ⎞
⎜
⎟⎜
⎟
1
M
1
M
⎝
⎠⎝
⎠
−2
• Notice the effect of dividing by a standard state is to render each component of K unitless.
We must be consistent with how we wrote the
reaction.
• What if we consider the opposite reaction:
2NH 3 ( g ) R N 2 ( g ) + 3H 2 ( g )
• The equilibrium concentrations remain the same with [NH3] = 3.1 x 10‐2 M, [N2] = 8.5 x 10‐3 M, and [H2] = 3.1 x 10‐3 M. What is the equilibrium constant?
KC
a )( a )
(
'=
(a )
N2
H2
2
NH 3
3
⎛ [ N2 ] ⎞ ⎛ [H2 ] ⎞
⎜
⎟⎜
⎟
1 M ⎠⎝ 1 M ⎠
⎝
=
2
⎛ [ NH 3 ] ⎞
⎜
⎟
⎝ 1M ⎠
3
The Equilibrium Constant (cont.)
• Continuing on
KC
a )( a )
(
'=
(a )
N2
H2
2
NH 3
3
⎛ [ N2 ] ⎞ ⎛ [H2 ] ⎞
⎜
⎟⎜
⎟
1 M ⎠⎝ 1 M ⎠
⎝
=
2
⎛ [ NH 3 ] ⎞
⎜
⎟
1
M
⎝
⎠
3
3
⎛ 8.5 × 10 M ⎞ ⎛ 3.1× 10 M ⎞
⎜
⎟⎜
⎟
1
M
1
M
1
1
⎠⎝
⎠ =
=⎝
=
2
4
−2
KC
3.8
10
×
⎛ 3.1× 10 M ⎞
⎜
⎟
1
M
⎝
⎠
−1
−3
• Reverse the reaction, equilibrium constant is inverted.
The Equilibrium Constant (cont.)
• What if we multiply all of the stoichiometric coefficients by 2?
2N 2 ( g ) + 6H 2 ( g ) ←⎯
→ 4NH 3 ( g )
• Still at equilibrium [NH3] = 3.1 x 10‐2 M, [N2] = 8.5 x 10‐1 M, and [H2] = 3.1 x 10‐3 M. What is the equilibrium constant?
K C '' =
(
aNH 3
)
(a ) (a )
2
N2
(
4
H2
6
)
⎛
aNH 3
=⎜
⎜ a
a
N
H2
2
⎝
2
( )( )
2
⎞
⎟ =K 2
C
3
⎟
⎠
• Multiply the coefficients by a constant, you raise the equilibrium constant to the power of the coefficient.
The Equilibrium Constant (cont.)
• Properties of the equilibrium constant:
K for a reaction written in reverse is the reciprocal of the original reaction (Knew = (Koriginal)‐1).
If you multiply a chemical equation by a factor n, the new K is simply the old K raised to the power of the factor (Knew = (Koriginal)n). n=‐1 reverses the direction of the written reaction.
• The equilibrium constant, and reaction quotient, are always unit‐less!