### Equilibrium

```Introduction To Dynamic Chemical
Equilibria (Ch 19)
Suggested HW (Ch 19): 4, 10, 14, 18, 26, 36, 46, 52
Recap: Gibbs Free Energy and the Criteria for Spontaneity
• In the last lecture, we were introduced to Gibbs free energy (ΔG)
• The sign of ΔG tells us if a reaction will occur spontaneously, and in
which direction.
• A negative value of ΔG means that a reaction is favored and
spontaneous in the forward direction.
∆G = ∆H − T∆S
Recap: Rate Laws
• Recall that for any reaction, the rate of the reaction is dependent on the
reactant concentration.
• For a reaction A  B, the rate law is written as:
= k[A]m
• As expressed by the equation above, as the concentration of A decreases,
the rate (speed) of the reaction slows down.
• The reaction order, m, expresses exactly how the rate depends on [A]
Dynamic Equilibria
• To date, we have discussed reactions occurring in only one direction.
However, in many cases, the forward and back reactions both occur at
significant rates, and are important. For example:
kF
N2O4 g
kB
2NO2 (g)
As the forward reaction proceeds, [N2O4] decreases, and [NO2]
increases.
The rate of the forward reaction slows down, while the rate of the
back reaction increases.
Once the rates are equal, the system is in equilibrium.
Dynamic Equilibria
kF
N2O4 g
kB
2NO2 (g)
• Equilibrium reaction are elementary, meaning that the reaction order
follows the stoichiometry
rate F = rate B
[  ] =   [ ]
• Chemical equilibrium is dynamic because the reactions never actually stop
– At equilibrium, NO2 is formed and consumed at the exact same rate, so
there is no net change in the concentration of the products or reactants
Dynamic Equilibria
• Both NO2 and N2O4 are in the system at equilibrium.
• In this instance, the vast majority of gas in the system is the product, NO2
(forward direction is the preferred direction). There is always some reactant
remaining in any chemical reaction, no matter how small the amount. All
reactions approach equilibrium.
Why Do Reactions Want to Approach Equilibrium?
• Equilibrium is a counter-intuitive concept. Why wouldn’t a reaction just ‘go all
the way’? If every reaction has a preferred direction, what is the driving force of
the back reaction?
• The answer has to do with ENTROPY and GIBBS FREE ENERGY
• If the standard free energy of the products is less than the reactants, ΔG will be
negative and the reaction will proceed to the right.
• However, a point in every reaction is reached where it becomes entropically
unfavorable (ΔS starts to become negative) to form more product, which causes
G to become more positive.
• Since the reaction can’t stop, the back reaction proceeds to prevent any further
formation of product and to keep G at a minimum (maximize ΔG)
100%
Reactants
0%
equilibrium
S
G
ΔG if reaction
proceeds 100%
ΔG at
equilibrium
0%
Forward reaction
Back reaction
Products
100%
A mix of reactants and products is
entropically favored. No reaction
goes 100% to completion.
Equilibrium Constant, Kc
• For any equilibrium reaction:
+

+
• The rates of the forward and back reaction are equal when equilibrium is
established. The reactions in an equilibrium process are elementary
reactions. The rates can be described as:
=  [] []
=  [] []
[] [] =  [] []
• The equilibrium constant, K, is equal to the ratio of the rate constants of the
forward and reverse reactions. It can be expressed in terms of either
concentrations or pressures (subscripts are used to distinguish).
[] []
=
[] []
Kc
Equilibrium Constant, Kc
• The value of Kc is constant at a given temperature.
• Reactants and products that are pure solids or pure liquids DO NOT
appear in the equilibrium constant expression.
Examples
• Write equilibrium constant expressions of the following reactions
N2 g + 3H2 g
2NH3 (g)
1
3
N g + H2 g
2 2
2
NH3 (g)
PCl3 L + Cl2 g
PCl5 (s)
CH4 (g) + 2O2 (g)
CO2 g + 2H2 O(L)
[NH3 ]2
Kc =
N2 [H2 ]3
Kc =
[NH3 ]
[N2 ]1/2 [H2 ]3/2
Kc =
1
[Cl2 ]
[CO2 ]
Kc =
[CH4 ] [O2 ]2
I.C.E. Tables: Calculating Equilibrium Concentrations
• Suppose you are carrying out a reaction, and you wish to determine either
the value of the equilibrium constant, or the equilibrium concentrations of
your products and reactants (if K is given);
• To do this, you would plug the starting concentrations into an I.C.E (initial,
change, equilibrium) table and tabulate the change in each concentration
as the reaction proceeds
I.C.E. Table Calculations: Finding K
+
()
• A closed system initially contains 1.0 x 10-3 M H2 and 2 x 10-3 M I2 at 448oC.
The equilibrium concentration of HI is 1.87 x 10-3 M. Calculate Kc.
Concentration (M)
H2
I2
2 HI
Initial
.001
.002
0
Change
-x
-x
+2x
Equilibrium
.001 - x
.002 - x
.00187
• To determine the equilibrium concentrations of H2 and I2, we must
determine the change in the initial concentration. We use the
stoichiometry to do this
• We see that the change in HI is twice the change in H2 and I2 (1:2 stochiometry).
I.C.E. Table Calculations
+
()
• x = .000935 M We can easily determine the equilibrium concentrations.
Concentration (M)
H2
I2
2 HI
Initial
.001
.002
0
Change
-(.000935)
-(.000935)
+2(.000935)
Equilibrium
.000065
.001065
.00187
• Now, we can calculate Kc
[]2
(.00187 )2
=
=
= 50.5
2 [2 ]
6.5  10−5  (1.065  10−3 )
• The magnitude of Kc (much greater than 1) suggests that there is
substantially more product than reactant at equilibrium. The
reaction is favored to the right
I.C.E Tables: Calculating Equilibrium Concentrations from K
2 4
22
= 0.20
• Suppose that 2.00 moles of N2O4 are injected into a 1.00 L reaction
vessel held at 100oC. Calculate the equilibrium concentrations.
• Set up the I.C.E. table. We don’t know the equilibrium concentrations,
so we leave them in terms of x.
Concentration (M)
N2O4
2NO2
Initial
0
Change
2.00
-x
+2x
Equilibrium
2.00-x
2x
2 4
22
Concentration (M)
N2O4
2NO2
Initial
2.00
0
Change
-x
+ 2x
Equilibrium
2.00 - x
2x
= 0.20
[2 ]2
=
[2 4 ]
• We plug in our equilibrium concentrations and the given value of Kc
(2)2
0.20 =
(2 − )
• Expand and cross multiply to get all terms on one side of the equation.
+ .   − .  =
• We are now left with a QUADRATIC EQUATION (ax2+bx+c = 0).
A Blast From the Past: Solving the Quadratic Equation
• To solve a quadratic equation, you must use the quadratic formula
− ± 2 − 4
=
2
• Since all quadratic formulas are of the form ax2 + bx +c = 0, we identify
these values and plug them into the formula. Quadratics yield two results.
a=4
b = 0.2
c = -0.4
=
−0.2 ± 2.53
=
8
−0.2 ±
(0.2)2 − 4(4)(−0.4) −0.2 ± 6.44
=
2(4)
8
= 0.29
X
= −0.34
Check your values of x to
determine the correct one.
Only one value will work.
I.C.E Tables: Calculating Equilibrium Concentrations from K
• Plug the calculated value of x back into the I.C.E. table to determine the
equilibrium concentrations (x = 0.29 M). DO NOT FORGET THIS STEP!
Concentration (M)
N2O4
2NO2
Initial
2.00
0
Change
- (.29)
+ 2 (.29)
Equilibrium
1.71
.58
Relating Concentrations to Partial Pressures
• Recall from the gas laws that concentration is related to pressure:

=

• Therefore, equilibrium constants can be expressed in terms of partial
pressures.
– Since solids and liquids do not contribute to the overall pressure, they
are not included in the expression (as we learned previously).
• When using pressures, we write the equilibrium expression as Kp. The
subscript p means that K is in terms of pressure.
Example
• A mixture of H2 and N2 is allowed to attain equilibrium at 472oC. The
equilibrium mixture was found to contain 7.38 atm H2, 2.46 atm N2, and
0.166 atm NH3. Find Kp?
2  + 32
23 ()
(3 )2
(.166 )2
=
=
= 2.79  10−5 −2
3
3
(2 ) (2 )
(7.38 ) (2.46 )
• The value of Kp is very small (less than 1), which tells us that the
reaction is favored to the left.
LeChatlier’s Principle
• LeChatlier’s Principle states that:
– If a chemical reaction at equilibrium is subjected to a change in
conditions that displaces it from equilibrium, the reaction adjusts
toward a new equilibrium state. The reaction will proceed in the
direction that offsets the change.
LeChatlier’s Principle
• For example, let’s say we have the following reaction at 25oC:
+ 2
2 ()
= 9
• Assume that our equilibrium concentration of CO is 6M, that of CO2 is 4M.
As we learned earlier, we can express Kc as:
[]2 62
=
=
=9
[2 ]
4
• Now, we add more CO2(g) to the system, until [CO2] = 8 M. We are no
longer in equilibrium.
– Thus, the state of the system is no longer described by the equilibrium
constant Kc, but by the reaction quotient, Q. The subscript ‘o’
indicates a non-equilibrium concentration
[]2
62
=
=
= 4.5
[2 ]
8
Reaction Quotients, Q
• The direction of spontaneity is always toward equilibrium.
• The value of Q/Kc tells us the direction in which a system not at
equilibrium will proceed to reach equilibrium.
Back to the Example
• At a given temperature, the equilibrium constant DOES NOT
CHANGE. To reestablish equilibrium (Kc= 9), the reaction shifts right
• This will increase [CO] and decrease [CO2] until Kc equals 9 again.
When equilibrium is re-established, the new concentrations will be:
[]2 (6 + )2
=
=
=9
[2 ]
8−
x = .975M
= 7.95
2 = 7.025
LeChatlier’s Principle Applied To Volume and Pressure
• When you have gaseous reactants and products, changes to the volume
and pressure of the system induce shifts in the equilibrium.
• The general rule is, for any equilibrium involving gases, decreasing the
volume drives the equilibrium toward the side with the smaller number of
moles of gas. From Boyle’s Law, we know that pressure and volume are
inversely proportional, so increasing pressure has the same effect as
lowering volume.
kF
N2O4 g
kB
[2 ]2
=
[2 4 ]
2NO2 (g)
Equilibrium
Decreasing V
shifts
reaction left
Increasing V
shifts reaction
right
LeChatlier’s Principle Applied To Changing Temperature
• Remember, all reactions must be either endothermic or exothermic
• In an endothermic process, heat is absorbed into the system in order for
the reaction to proceed. Therefore, since heat is used in the reaction,
heat may be considered a reactant
• Let’s use our previous example.
2 4
22
= 0.20
∆ =
• We can rewrite this reaction as:
2 4  +
22
= 0.20
• Increasing the temperature of an endothermic process drives the
equilibrium to the right because you are adding a reactant.

LeChatlier’s Principle Applied To Changing Temperature
• The opposite must then be true for an exothermic process.
• Since heat leaves the system in an exothermic process, heat may be
treated as a product
• For the following exothermic reaction:
2  + 2
2
∆ = −
• We can rewrite the reaction as:
2  + 2
2  +
• Increasing the temperature of an exothermic process drives the
equilibrium to the left because you are adding a product.

Calculations Based On LeChatlier’s Principle
• Let’s take the reaction below. Assume that the reaction is at equilibrium.
The equilibrium concentrations are given.
2  + 2
0.75 M 0.60 M
+ 2 ()
0.10 M 0.25 M
• We can calculate Kc from the equilibrium values Kc = .056
• Now, additional CO2 is added to the system, raising [CO2] to 1.25 M. This
will cause the equilibrium to shift right. But how far? If we can ‘freeze
time’ right at the instant that the CO2 is added (before the shift), we
would have the following:
Concentration(M)
CO2
H2
CO
H2 O
Initial
1.25
0.60
0.10
0.25
Change
-x
-x
+x
+x
Equilibrium
1.25-x
0.60-x
0.10+x
0.25+x
Calculations Based On LeChatlier’s Principle
2  + 2
+ 2 ()
Concentration(M)
CO2
H2
CO
H2O
Initial
1.25
0.60
0.10
0.25
Change
-x
-x
+x
+x
Equilibrium
1.25-x
0.60-x
0.10+x
0.25+x
• Now we solve this as usual. Convert to quadratic form.
.10 +  (.25 + )
= .056
1.25 −  (.60 − )
0.944 2 + .4536 − .017 = 0
• Use quadratic formula. Solve for x
x = 0.0349M
• New equilibrium concentrations: [CO2]= 1.215 M, [H2]= 0.5651 M
[CO] = .1349 M, [H2O] = .2849 M
Group Example
+
2()
• The reaction above is enclosed in a vessel at 25oC. At equilibrium,
you have 0.171 atm of A and 4.89 atm of C. Then, the system is
disturbed by a 35% decrease in the pressure of A. What are the
new equilibrium pressures of A and C?
When performing an equilibrium calculation involving LeChatlier’s
principle, you must first answer the following
1. What is the equilibrium expression?
2. What is the value of Kp?
3. What are the initial pressures (at the instant before the reaction
proceeds)?
4. In which direction does the disturbance shift the equilibrium?
( )2
(4.89 )2
=
=
= 139.8

(.171 )
Pressure
A
2C
Initial
(.111)
4.89
Change
+x
-2x
Equilibrium
.111+x
4.89-2x
No longer at equilibrium. Must
shift LEFT to reestablish Kp
• When we plug our new equilibrium pressures into Kp, the value of Kp
MUST be 139.8
(4.89 − 2)2
= 139.8
(.111 + )
4 2 + 159.36 − 8.38 = 0
x = .0525 atm
PA = .1635 atm
PC = 4.785 atm
```