Download Equilibrium - AP Chemistry

Chemical Equilibrium
Chapter 14
Equilibrium: the extent of a reaction
In stoichiometry we talk about theoretical
yields, and the many reasons actual yields
may be lower.
Another critical reason actual yields may be
lower is the reversibility of chemical
reactions: some reactions may produce only
70% of the product you may calculate they
ought to produce.
Equilibrium looks at the extent of a chemical
reaction.
Equilibrium is a state in which there are no observable
changes as time goes by.
Chemical equilibrium is achieved when:
•
the rates of the forward and reverse reactions are equal and
•
the concentrations of the reactants and products remain
constant
Physical equilibrium
H2O (l)
H2O (g)
Chemical equilibrium
N2O4 (g)
2NO2 (g)
14.1
Rate of sale of
cookies
=
Rate of replacing
cookies
The Concept of Equilibrium
• Consider colorless frozen N2O4. At room temperature, it
decomposes to brown NO2:
N2O4(g)  2NO2(g).
• At some time, the color stops changing and we have a mixture of
N2O4 and NO2.
• Chemical equilibrium is the point at which the rate of the forward
reaction is equal to the rate of the reverse reaction.
At that point, the concentrations of all species are constant.
• Using the collision model:
– as the amount of NO2 builds up, there is a chance that two NO2
molecules will collide to form N2O4.
– At the beginning of the reaction, there is no NO2 so the reverse
reaction (2NO2(g)  N2O4(g)) does not occur.
The Concept of Equilibrium
•
•
•
•
As the substance warms it begins to decompose:
N2O4(g)  2NO2(g)
When enough NO2 is formed, it can react to form N2O4:
2NO2(g)  N2O4(g).
At equilibrium, as much N2O4 reacts to form NO2 as
NO2 reacts to re-form N2O4
The double arrow implies the process is dynamic.
N2O4(g)
2NO2(g)
The Concept of Equilibrium
As the reaction progresses
– [A] decreases to a constant,
– [B] increases from zero to a constant.
– When [A] and [B] are constant,
equilibrium is achieved.
A
B
N2O4 (g)
2NO2 (g)
equilibrium
equilibrium
equilibrium
Start with NO2
Start with N2O4
Start with NO2 & N2O4
14.1
constant
14.1
The Equilibrium Constant
• No matter the starting composition of reactants and
products, the same ratio of concentrations is achieved at
equilibrium.
• For a general reaction
aA + bB(g)
pP + qQ
the equilibrium constant expression is
Kc 
p
q
P Q
a
b
A B
where Kc is the equilibrium constant.
N2O4 (g)
K=
2NO2 (g)
[NO2]2
[N2O4]
= 4.63 x 10-3
aA + bB
K=
cC + dD
[C]c[D]d
Law of Mass Action
[A]a[B]b
Equilibrium Will
K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants
14.1
Homogenous equilibrium applies to reactions in which all
reacting species are in the same phase.
N2O4 (g)
Kc =
[NO2
2NO2 (g)
]2
Kp =
[N2O4]
2
PNO
2
PN2O4
In most cases
Kc  Kp
aA (g) + bB (g)
cC (g) + dD (g)
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
14.2
Homogeneous Equilibrium
CH3COOH (aq) + H2O (l)
[CH3COO-][H3O+]
Kc‘ =
[CH3COOH][H2O]
CH3COO- (aq) + H3O+ (aq)
[H2O] = constant
[CH3COO-][H3O+]
= Kc‘ [H2O]
Kc =
[CH3COOH]
General practice not to include units for the
equilibrium constant.
14.2
The Equilibrium Expression
• Write the equilibrium expression for the
following reaction:
N2(g) + 3H2(g)
2NH3(g)
The equilibrium concentrations for the reaction between
carbon monoxide and molecular chlorine to form COCl2 (g)
at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] =
0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g)
COCl2 (g)
[COCl2]
0.14
=
= 220
Kc =
[CO][Cl2]
0.012 x 0.054
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1
R = 0.0821
T = 273 + 74 = 347 K
Kp = 220 x (0.0821 x 347)-1 = 7.7
14.2
The equilibrium constant Kp for the reaction
2NO2 (g)
2NO (g) + O2 (g)
is 158 at 1000K. What is the equilibrium pressure of O2 if
the PNO2 = 0.400 atm and PNO = 0.270 atm?
Kp =
2
PNO
PO2
2
PNO
2
PO2 = Kp
2
PNO
2
2
PNO
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm
14.2
Heterogenous equilibrium applies to reactions in which
reactants and products are in different phases.
CaCO3 (s)
[CaO][CO2]
Kc‘ =
[CaCO3]
[CaCO3]
Kc = [CO2] = Kc‘ x
[CaO]
CaO (s) + CO2 (g)
[CaCO3] = constant
[CaO] = constant
Kp = PCO2
The concentration of solids and pure liquids are not
included in the expression for the equilibrium constant.
14.2
CaCO3 (s)
CaO (s) + CO2 (g)
PCO 2 = Kp
PCO 2 does not depend on the amount of CaCO3 or CaO
14.2
Writing Equilibrium Constant Expressions
•
The concentrations of the reacting species in the
condensed phase are expressed in M. In the gaseous
phase, the concentrations can be expressed in M or in atm.
•
The concentrations of pure solids, pure liquids and solvents
do not appear in the equilibrium constant expressions.
•
The equilibrium constant is a dimensionless quantity.
•
In quoting a value for the equilibrium constant, you must
specify the balanced equation and the temperature.
•
If a reaction can be expressed as a sum of two or more
reactions, the equilibrium constant for the overall reaction is
given by the product of the equilibrium constants of the
individual reactions.
14.2
Calculating Equilibrium Concentrations
1. Express the equilibrium concentrations of all species in
terms of the initial concentrations and a single unknown x,
which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the
equilibrium concentrations. Knowing the value of the
equilibrium constant, solve for x.
3. Having solved for x, calculate the equilibrium
concentrations of all species.
14.4
At 12800C the equilibrium constant (Kc) for the reaction
Br2 (g)
2Br (g)
Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063
M and [Br] = 0.012 M, calculate the concentrations of these
species at equilibrium.
Let x be the change in concentration of Br2
Initial (M)
Change (M)
Equilibrium (M)
[Br]2
Kc =
[Br2]
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
Solve for x
14.4
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
-b ± b2 – 4ac
2
x=
ax + bx + c =0
2a
x = -0.0105 x = -0.00178
Initial (M)
Change (M)
Equilibrium (M)
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M
At equilibrium, [Br2] = 0.062 – x = 0.0648 M
14.4
Example Problem: Calculate Concentration
Note the moles into a 10.32 L vessel stuff ... calculate molarity.
Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M
2 HI
0.242 M
Initial:
Change: -2x
0.242-2x
Equil:
H2 + I2
0
+x
x
0
+x
x
[ H 2 ][ I 2 ]
Keq 
[ HI ]2
[ x][ x]
x2
3
Keq 


1
.
26
x
10
[0.242  2 x]2 [0.242  2 x]2
What we are asked for here is the equilibrium concentration of H2 ...
... otherwise known as x. So, we need to solve this beast for x.
Example Problem: Calculate Concentration
And yes, it’s a quadratic equation. Doing a bit of rearranging:
x2
3

1
.
26
x
10
[0.242  2 x]2
x 2  1.26 x103[0.242  2 x]2
 1.26 x103[0.0586  0.968 x  4 x 2 ]
 7.38 x105  1.22 x10 3 x  5.04 x103 x 2
0.995 x 2  1.22 x10 3 x  7.38 x10 5  0
 b  b 2  4ac
x
2a
x = 0.00802 or –0.00925
Since we are using this to
model a real, physical system,
we reject the negative root.
The [H2] at equil. is 0.00802 M.
Example Problem: Calculate Keq
This type of problem is typically tackled using the “three line” approach:
2 NO + O2
2 NO2
Initial:
Change:
Equilibrium:
Approximating
If Keq is really small the reaction will not proceed to
the right very far, meaning the equilibrium
concentrations will be nearly the same as the
initial concentrations of your reactants.
0.20 – x is just about 0.20 is x is really dinky.
If the difference between Keq and initial
concentrations is around 3 orders of magnitude or
more, go for it. Otherwise, you have to use the
quadratic.
Example
Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M
I2
2I
[ I ]2
0.20
0
Keq 
 2.94 x10 10
Initial
[I2 ]
-x
+2x
change
2
[
2
x
]
0.20-x
2x
equil:

 2.94 x10 10
[0.20  x]
More than 3
orders of mag.
between these
numbers. The
simplification will
work here.
With an equilibrium constant that small, whatever x is, it’s near
dink, and 0.20 minus dink is 0.20 (like a million dollars minus a
nickel is still a million dollars).
0.20 – x is the same as 0.20
[2 x]2
10
 2.94 x10
0.20
x = 3.83 x 10-6 M
Example
Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M
I2
2I
[ I ]2
These are too close to
Keq 
 0.209
0.20
0
Initial
each other ...
[I2 ]
-x
+2x
change
0.20-x will not be
2
[2 x]
trivially close to 0.20
0.20-x
2x
equil:

 0.209
[0.20  x]
here.
Looks like this one has to proceed through the quadratic ...
A+B
C+D
Kc‘
C+D
E+F
Kc‘‘
A+B
E+F
[C][D]
Kc‘ =
[A][B]
Kc
[E][F]
Kc‘‘ =
[C][D]
[E][F]
Kc =
[A][B]
Kc = Kc‘ x Kc‘‘
If a reaction can be expressed as the sum of
two or more reactions, the equilibrium
constant for the overall reaction is given by
the product of the equilibrium constants of
the individual reactions.
14.2
N2O4 (g)
K=
[NO2]2
[N2O4]
2NO2 (g)
= 4.63 x
10-3
2NO2 (g)
N2O4 (g)
[N2O4]
1
= 216
K‘ =
=
2
K
[NO2]
When the equation for a reversible reaction
is written in the opposite direction, the
equilibrium constant becomes the reciprocal
of the original equilibrium constant.
14.2
The reaction quotient (Qc) is calculated by substituting the
initial concentrations of the reactants and products into the
equilibrium constant (Kc) expression.
IF
•
Qc > Kc system proceeds from right to left to reach equilibrium
•
Qc = Kc the system is at equilibrium
•
Qc < Kc system proceeds from left to right to reach equilibrium
14.4
Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the
system adjusts in such a way that the stress is partially offset
as the system reaches a new equilibrium position.
• Changes in Concentration
N2 (g) + 3H2 (g)
2NH3 (g)
Equilibrium
shifts left to
offset stress
Add
NH3
14.5
Le Châtelier’s Principle
• Changes in Concentration continued
Remove
Add
Remove
Add
aA + bB
cC + dD
Change
Shifts the Equilibrium
Increase concentration of product(s)
Decrease concentration of product(s)
Increase concentration of reactant(s)
Decrease concentration of reactant(s)
left
right
right
left
14.5
Le Châtelier’s Principle
• Changes in Volume and Pressure
A (g) + B (g)
C (g)
Change
Shifts the Equilibrium
Increase pressure
Decrease pressure
Increase volume
Decrease volume
Side with fewest moles of gas
Side with most moles of gas
Side with most moles of gas
Side with fewest moles of gas
14.5
Le Châtelier’s Principle
• Changes in Temperature
Change
Increase temperature
Decrease temperature
Exothermic Rx
Endothermic Rx
K decreases
K increases
K increases
K decreases
colder
hotter
14.5
Le Châtelier’s Principle
• Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner
uncatalyzed
catalyzed
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
14.5
Example
Chemistry In Action
Life at High Altitudes and Hemoglobin Production
Hb (aq) + O2 (aq)
Kc =
HbO2 (aq)
[HbO2]
[Hb][O2]
Chemistry In Action: The Haber Process
N2 (g) + 3H2 (g)
2NH3 (g) DH0 = -92.6 kJ/mol
Le Châtelier’s Principle
Change
Shift Equilibrium
Change Equilibrium
Constant
Concentration
yes
no
Pressure
yes
no
Volume
yes
no
Temperature
yes
yes
Catalyst
no
no
14.5