Equilibria and Perturbations (Stress)
Last Time
What happens to a system at equilibrium
if I change something like
The concentration of one of the
chemicals
The Pressure
Figure Copyright Houghton Mifflin
Company. All rights reserved
Each equilibrium has different concentrations,
but the same value for Kc
Principles of Chemistry II
© Vanden Bout
The Temperature
Principles of Chemistry II
© Vanden Bout
Qualitatively Understanding "stress"
-
+ H+
Le Chatlier's Principle
Yellow
Blue
In my demo last time, when the solution was a deep blue and at equilibrium. I then
perturb the system by adding a strong acid which had a high concentration of H+.
What did the reaction mixture do?
!
A.! !
the reaction shifted to a new equilibrium high in "reactant" concentration
!
B.! !
the reaction was unchanged as it was already at equilibrium
!
C.! !
the reaction generated even more "products"
Principles of Chemistry II
© Vanden Bout
If a chemical system at equilibrium
experiences a change,
then the equilibrium shifts to partially
counter-act the imposed change.
Principles of Chemistry II
© Vanden Bout
N2(g) + 3H2(g)
Stressing the concentrations
2NH3(g)
You find the system at equilibrium,
then you decide to add more H2 to the mixture
Add Reactants
Reaction Shifts
towards Product
Add Products
Reaction Shifts
towards Reactants
What happens as the reaction goes to a new equilibrium?
!
A.! !
!
the concentration
NH3 decreases by moving
The
system willof compensate
to "reduce" the stress.
C.! ! nothing when equilibrium is reached all the concentrations
will the same as You
beforeadded H2
The reaction will try to reduce the amount of H2
!
the concentration of N2 decreases
B.! !
Principles of Chemistry II
N2(g) + 3H2(g)
© Vanden Bout
Principles of Chemistry II
© Vanden Bout
Dealing with Stress from a Quantitative Perspective
2NH3(g)
You find the system at equilibrium at 1 atm,
then you decide to increase the pressure to 2 atm.
What happens as the reaction goes to a new equilibrium?
!
A.! !
moves towards the products as they have fewer molecules
!
B.! !
moves You
towards
the reactants
they have more molecules
increased
theaspressure
The
reaction will try to reduce the reduce the pressure
! C.! ! moves towards the products as they have a lower free energy
the only way to do this is to reduce the number of
molecules (move toward products)
Principles of Chemistry II
What if I increase the pressure?
© Vanden Bout
N2(g) + 3H2(g)
Equilibrium
[N2] = 0.921 M
[H2] = 0.763 M
[NH3] = 0.157 M
2NH3(g)
[NH3]2
Kc =
[N2][H2]3
[0.157]2
Kc =
= 0.06
[.921][.763]3
If I increase [N2] to 3 M the system will no longer be at equilibrium.
Which way will it shift to get back to equilibrium?
Principles of Chemistry II
© Vanden Bout
N2(g) + 3H2(g)
2NH3(g)
K is constant
not at equilibrium
Not at equilibrium
[N2] = 3 M
[H2] = 0.763 M
[NH3] = 0.157 M
Q=
[NH3]2
[N2][H2]3
K=
[0.157]2
Q=
=.0185
[3][.763]3
Constant!
So if products goes up
the reaction will shift to get
back to the same constant ratio
Q = 0.0185
K = 0.06
Q<K
therefore reaction needs to
increase products to get to equilibrium
Principles of Chemistry II
This can happen if
Product goes down slightly
and Reactant goes up slightly
© Vanden Bout
Principles of Chemistry II
Increasing Pressure
2NO2(g)
Kp =
PN2O4
PNO22
=
Products
Reactants
© Vanden Bout
Relating Kp and Kc
N2O4(g)
2NO2(g)
XN2O4
XN2O4 P
=
XNO22 P2
XNO22 P
Kc =
If you increase P
Then the mole fraction of NO2
must go down since K is constant
[N2O4]
[NO2]2
PN2O4 =
N2O4(g)
Kp =
PN2O4
PNO22
nN204RT
= [N2O4]RT
V
concentration
Principles of Chemistry II
© Vanden Bout
Principles of Chemistry II
© Vanden Bout
Relating Kp and Kc
2NO2(g)
Kp =
PN2O4
PNO22
=
Temperature Change
N2O4(g)
N2(g) + 3H2(g)
1
[N2O4]RT
=
K
c
[NO2]2(RT)2
RT
Kc =
In general
2NH3(g) + heat
this reaction is exothermic
[N2O4]
[NO2]2
If you increase T then to
"partially compensate" the reactions
shifts to the reactants (consuming heat)
KP = Kc(RT)!n
!n is the change in the number of moles of gas
Principles of Chemistry II
© Vanden Bout
How to change the pressure (constant T)
Principles of Chemistry II
Why does Temperature Change Equilibrium?
N2(g) + 3H2(g)
Increase P (decrease V)
Decrease P (increase V)
© Vanden Bout
2NH3(g) + heat
Shifts to side with fewer gas molecules
Shifts to side with more gas molecules
Kc =
[NH3]2
[N2][H2]3
Add an inert gas (one that doesn't react. Like He)
Constant P
Constant V
This is like diluting the system
This is like essentially doing nothing
increase in V
The partial pressures of all the
like lowering P
molecules that matter are unchanged
shift to side with more gas molecules
(the number of collisions is
unchanged)
the reaction is unchanged
Principles of Chemistry II
© Vanden Bout
K is a function of T!
Principles of Chemistry II
© Vanden Bout
lnK = -!RH°/RT + !RS°/R
!RG°(T) = -RT lnK
!RG°(T) = !RH° - T !RS°
exothermic
increasing T
smaller K
lnK
-RTlnK = !RG°(T) = !RH° - T !RS°
lnK = -!RH°/RT + !RS°/R
endothermic
increasing T
larger K
Temperature dependence of K depends on !RH°
Principles of Chemistry II
© Vanden Bout
Some Calculations
Principles of Chemistry II
© Vanden Bout
1/T
Principles of Chemistry II
T
y = mx + b
y is lnK
x is 1/T
m is -!RH°/R
b is !RS°/R
a different way to do
calorimetry
measure K to find
!RH°
© Vanden Bout