Chapter 15 Lecture- Chemical Equilibrium

Chapter 15 - Chemical Equilibrium

Sections 15.1 - 15.4 Equilibrium &
the Equilibrium Constant (K)

The Concept of Equilibrium

Chemical equilibrium occurs when a
reaction and its reverse reaction proceed
at the same rate.

• As a system
approaches
equilibrium, both the
forward and reverse
reactions are
occurring.

• As a system
approaches
equilibrium, both the
forward and reverse
reactions are
occurring.
• At equilibrium, the
forward and reverse
reactions are
proceeding at the
same rate.

A System at Equilibrium

Once
equilibrium is
achieved, the
amount of
each reactant
and product
remains
constant.

Depicting Equilibrium
In a system at equilibrium, both
the forward and reverse
reactions are being carried out;
as a result, we write its equation

N2O4 (g) 2 NO2 (g)

N2(g) + 3H2(g) ⇌ 2NH3(g)

1M N2 + 3M H2 2M NH3

The Equilibrium Constant
• Forward reaction:
N2O4 (g) → 2 NO2 (g)
✤Rate law: Rate = kf [N2O4]

• Forward reaction:
N2O4 (g) → 2 NO2 (g)
✤Rate law: Rate = kf [N2O4]

• Reverse reaction:
2 NO2 (g) → N2O4 (g)
✤Rate law: Rate = kr [NO2]2

• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

• Rewriting this, it becomes
kf [NO2]2
=
kr [N2O4]


The ratio of the rate constants is a
constant at that temperature, and the
expression becomes
kf [NO2]2
Keq = =
kr [N2O4]

For the gas-phase reaction A ⇌ B
the forward reaction rate is 3.0 × 10−4 s−1 and
the reverse reaction rate is 1.5 × 10−2 s−1.
What is the value of the equilibrium
constant, Keq?

1. 0.02
2. 50
3. 0.0004
4. 2500

Correct Answer:

1. 0.02
2. 50
3. 0.0004
4. 2500

N2O4 (g) → 2 NO2 (g) Eq 15.1

kf [NO2]2 Eq 15.5
Keq = k = [N O ]
r 2 4
1. moles of reactant always equals moles of
product.
2. mass of reactant always equals mass of product.
3. the rates of opposing reactions are equal.
4. cannot be determined without temperature
information.

N2O4 (g) → 2 NO2 (g) Eq 15.1

kf [NO2]2
Keq = k = [N O ] Eq 15.5

r 2 4

b) the constant will be______ than 1

N2O4 (g) → 2 NO2 (g) Eq 15.1

kf [NO2]2
Keq = k = [N O ] Eq 15.1

r 2 4

b) the constant will be greater than 1

the concentrations of reactants
and products no longer
change.

• To generalize this expression, consider
the reaction
aA + bB cC + dD

the reaction
aA + bB cC + dD

• The equilibrium expression for this
reaction would be
[C]c[D]d
Kc =
[A]a[B]b

the reaction
aA + bB cC + dD

• The equilibrium expression for this
reaction would be
[C]c[D]d
Kc =
[A]a[B]b
a.k.a. the LAW of
MASS ACTION

products
reactants
=K

SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions

Write the equilibrium expression for Kc for the following
reactions:

Kc is independent of the starting
concentrations of reactants
and products.

Because pressure is proportional to
concentration for gases in a closed
system, the equilibrium expression
can also be written in terms of partial
pressures
(PC)c (PD)d
Kp =
(PA)a (PB)b

Relationship between Kc and Kp
• From the ideal gas law we know that

PV = nRT
• Rearranging it, we get

n
P= RT
V

Relationship between Kc and Kp
Plugging this into the expression for
Kp for each substance, the
relationship between Kc and Kp
becomes
Kp = Kc (RT)Δn
Where Δn = (moles of gaseous product) −
(moles of gaseous reactant)

Kc represents the equil.
constant when equilibrium
concentrations are
expressed in molarity and Kp
represents the equilibrium
constant when equilibrium
concentrations are
expressed in atm.

SAMPLE EXERCISE 15.2 Converting Between Kc and Kp
In the synthesis of ammonia from nitrogen and
hydrogen, Kc = 9.60 at 300°C.
Calculate Kp for this reaction at this temperature.

SAMPLE EXERCISE 15.2 Converting Between Kc and Kp
In the synthesis of ammonia from nitrogen and
hydrogen, Kc = 9.60 at 300°C.
Calculate Kp for this reaction at this temperature.

Solve:
Δn = 2 – 4 = –2 (Remember that Δ functions are always products minus reactants)
T = 273 + 300 = 573 K
R = 0.0821 L-atm/mol-K
Kc = 9.60

Equilibrium Can Be Reached from
Either Direction
N2O4 (g) → 2 NO2 (g)

As you can see, the ratio of [NO2]2 to
[N2O4] remains constant at this
temperature no matter what the initial
concentrations of NO2 and N2O4 are.

Either Direction
This is the data from
the last two trials
from the table on the
previous slide.

N2O4 (g) → 2 NO2 (g)

Either Direction

It does not matter whether we start
with N2 and H2 or whether we start
with NH3. We will have the same
proportions of all three substances
at equilibrium.

What Does the Value of K Mean?
• If K >> 1, the reaction
is product-favored;
product
predominates at
equilibrium.

What Does the Value of K Mean?
• If K >> 1, the reaction
is product-favored;
product
predominates at
equilibrium.

• If K << 1, the reaction
is reactant-favored;
reactant predominates
at equilibrium.

SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Equilibrium Constant

The reaction of N2 with O2 to form NO might
be considered a means of “fixing” nitrogen:
The value for the equilibrium constant for this
reaction at 25°C is Kc = 1 × 10–30. Describe the
feasibility of fixing nitrogen by forming NO at
25°C.

SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Equilibrium Constant

The reaction of N2 with O2 to form NO might
be considered a means of “fixing” nitrogen:
The value for the equilibrium constant for this
reaction at 25°C is Kc = 1 × 10–30. Describe the
feasibility of fixing nitrogen by forming NO at
25°C.
Solve: Because Kc is so small, very little
NO will form at 25°C. The equilibrium lies
to the left, favoring the reactants.
Consequently, this reaction is an extremely
poor choice for nitrogen fixation, at least at
25°C.

Manipulating Equilibrium Constants
The equilibrium constant of a
reaction in the reverse reaction is
the reciprocal of the equilibrium
constant of the forward reaction.
[NO2]2
N2O4 (g) 2 NO2(g) Kc = = 0.212 at 100°C
[N2O4]

The equilibrium constant of a
reaction in the reverse reaction is
the reciprocal of the equilibrium
constant of the forward reaction.
[NO2]2
N2O4 (g) 2 NO2(g) Kc = = 0.212 at 100°C
[N2O4]
[N2O4] 1
2 NO2 (g) N2O4(g) Kc = =
[NO2]2 0.212
= 4.72 at 100°C

The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to a power that is equal to that
number.
[NO2]2
N2O4(g) 2 NO2(g) Kc = = 0.212 at 100°C
[N2O4]

The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to a power that is equal to that
number.
[NO2]2
N2O4(g) 2 NO2(g) Kc = = 0.212 at 100°C
[N2O4]

[NO2]4
2 N2O4(g) 4 NO2(g) Kc = = (0.212)2 at 100°C
[N2O4]2


The equilibrium constant for a net
reaction made up of two or more steps
is the product of the equilibrium
constants for the individual steps.

the new equilibrium constant is
equal to (Kp)3.

For the following hypothetical reaction:
Y(g) + Z(g) 2 A(g)
Keq = 4.0 × 10−2. Given that, what is the
value of Keq for the reaction:
4 A(g) 2 Y(g) + 2 Z(g)

1. 25 3. 0.016
2. 5.0 4. 6.3 × 102

Correct Answer:
1. 25
2. 5.0
3. 0.016
4. 6.3 × 102

The second equation is twice the first and reversed.
Thus, Keq of the second equation is related to the
first as shown below:

K2 = 630

SAMPLE EXERCISE 15.5 Combining Equilibrium Expressions
Given the following information,

determine the value of Kc for the reaction



Solve: Multiplying the first equation by 2 and raising K to the power 2, we get




Reversing the second equation and taking the reciprocal of K gives




Reversing the second equation and taking the reciprocal of K gives

With two equations that sum to give the net equation, we multiply the
individual Kc values to get the desired equilibrium constant.

The Concentrations of Solids and
Liquids Are Essentially Constant

Both can be obtained by dividing
the density of the substance by its
molar mass—and both of these are
constants at constant temperature.

The Concentrations of Solids and
Liquids Are Essentially Constant

Therefore, the concentrations of solids
and liquids do not appear in the
equilibrium expression

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

Kc = [Pb2+] [Cl−]2

CaCO3 (s) CO2 (g) + CaO(s)
As long as some CaCO3 or CaO remain in the
system, the amount of CO2 above the solid
will remain the same.

PRACTICE EXERCISE
Write the following equilibrium-constant expressions:

SAMPLE EXERCISE 15.7 Analyzing a Heterogeneous Equilibrium

Each of the following mixtures was placed in a closed container
and allowed to stand. Which is capable of attaining the equilibrium

(a) pure CaCO3,
(b) CaO and a CO2 pressure greater than the value of Kp,
(c) some CaCO3 and a CO2 pressure greater than the value of Kp,
(d) CaCO3 and CaO?

SAMPLE EXERCISE 15.7 Analyzing a Heterogeneous Equilibrium

Each of the following mixtures was placed in a closed container
and allowed to stand. Which is capable of attaining the equilibrium

(a) pure CaCO3,
(b) CaO and a CO2 pressure greater than the value of Kp,
(c) some CaCO3 and a CO2 pressure greater than the value of Kp,
(d) CaCO3 and CaO?
Solve: Equilibrium can be reached in all cases except (c) as long as sufficient
quantities of solids are present.
(a) CaCO3 simply decomposes, forming CaO(s) and CO2(g) until the equilibrium
pressure of CO2 is attained. There must be enough CaCO3, however, to allow the
CO2 pressure to reach equilibrium.
(b) CO2 continues to combine with CaO until the partial pressure of the CO2
decreases to the equilibrium value.
(c) There is no CaO present, so equilibrium can’t be attained because there is no way
the CO2 pressure can decrease to its equilibrium value (which would require some of
the CO2 to react with CaO).
(d) The situation is essentially the same as in (a): CaCO3 decomposes until


Sections 15.5 & 15.6
Equilibrium calculations

Bellwork
Write the equilibrium expression for the
following equilibrated systems.

2 NO(g) + O2(g) ⇌ 2NO2(g)

6 CO2(g) + 6 H2O(l) ⇌ C6H12O6(s) + 6 O2(g)

2 Y(aq) + 3 Z(aq) A(aq)
Calculate Keq given the following
equilibrium concentrations: [A] = 0.50 M,
[Y] = 0.10 M, [Z] = 1.0 M
[C ]c[D ]d
1. 2.5 K eq =
2. 5.0
[A ]a[ ]b
3. 0.40 B
4. 0.025
5. 50.

Correct Answer:

[0.50]
K eq = 2 3
1. 2.5 [0.10] [ ]
1.0
2. 5.0
3. 0.40
4. 0.025
5. 50.

Equilibrium Calculations
A closed system initially containing
1.000 x 10−3 M H2 and 2.000 x 10−3 M I2 at
448°C is allowed to reach equilibrium.
Analysis of the equilibrium mixture shows
that the concentration of HI is 1.87 x 10−3
M. Calculate Kc at 448°C for the reaction

H2 (g) + I2 (g) 2 HI (g)

What Do We Know?

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change

At 1.87 x 10-3
equilibrium

[HI] Increases by 1.87 x 10-3 M

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At 1.87 x 10-3
equilibrium

Stoichiometry tells us [H2] and [I2]
decrease by half as much

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At 1.87 x 10-3
equilibrium

We can now calculate the equilibrium
concentrations of all three compounds…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3
equilibrium

…and, therefore, the equilibrium
constant

[HI]2
Kc =
[H2] [I2]

(1.87 x 10-3)2
=
(6.5 x 10-5)(1.065 x 10-3)

= 51

PRACTICE EXERCISE
Sulfur trioxide decomposes at high
temperature in a sealed container:

Initially, the vessel is charged at 1000 K with
SO3(g) at a partial pressure of 0.500 atm.
At equilibrium the SO3 partial pressure is
0.200 atm.
Calculate the value of Kp at 1000 K.

PRACTICE EXERCISE
Sulfur trioxide decomposes at high
temperature in a sealed container:

Initially, the vessel is charged at 1000 K with
SO3(g) at a partial pressure of 0.500 atm.
At equilibrium the SO3 partial pressure is
0.200 atm.
Calculate the value of Kp at 1000 K.

Answer: 0.338

The Reaction Quotient (Q)

• To calculate Q, one substitutes the
initial concentrations on reactants
and products into the equilibrium
expression.

The Reaction Quotient (Q)

• To calculate Q, one substitutes the
initial concentrations on reactants
and products into the equilibrium
expression.
• Q gives the same ratio the
equilibrium expression gives, but for
a system that is not at equilibrium.

If Q = K,
the system is at equilibrium.

If Q > K,
there is too much product and the
equilibrium shifts to the left.

If Q < K,
there is too much reactant, and the
equilibrium shifts to the right.

2 X(g) + Y(g) A(g)
Keq = 250. At a point during the reaction, the
concentrations are [A] = 1.0 M, [X] = 0.50 M,
and [Y] = 0.10 M. How will the reaction
proceed to achieve equilibrium?

1. From left to right (toward products)
2. From right to left (toward reactants)
3. Already at equilibrium

Correct Answer:
3. Already at equilibrium

[ ]
1.0
Qc =
2
[0.50] [0.10]
Qc = 40
Because Qc < Keq, the reaction will shift from
reactants toward products to achieve equilibrium.

SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations

For the Haber process,
at 500°C. In an equilibrium mixture of the three gases at 500°C, the partial
pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial
pressure of NH3 in this equilibrium mixture?

SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations

For the Haber process,
at 500°C. In an equilibrium mixture of the three gases at 500°C, the partial
pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial
pressure of NH3 in this equilibrium mixture?

Comment: We can always check our answer by using it to recalculate the value of
the equilibrium constant:

SAMPLE EXERCISE 15.12 Calculating Equilibrium Concentrationsfrom Initial Concentrations

A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at
448°C. The value of the equilibrium constant Kc for the reaction
at 448°C is 50.5. What are the equilibrium concentrations of H2 ,
I2 , and HI in moles per liter?

If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this
expression to obtain a quadratic equation in x:

Solving the quadratic equation (Appendix A.3) leads to two solutions for x:

If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this
expression to obtain a quadratic equation in x:

Solving the quadratic equation (Appendix A.3) leads to two solutions for x:

When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find negative
concentrations of H2 and I2. Because a negative concentration is not chemically meaningful, we reject
this solution. We then use x = 0.935 to find the equilibrium concentrations:

Check: We can check our solution by putting these numbers into the equilibrium-constant
expression:


Section 15.7 Le Chatelier’s Principle

BELLWORK
Equilibrium is established between gases in the
reaction 2A(g) + 3B(g) ⇌ 1C(g) + 2D(g)

a) Write an expression for the equilibrium constant
b) What happens to the forward reaction rate if
more A is added? All reactants are involved in
the forward reaction’s rate law.
c) Would the equilibrium concentrations be the
same as they were before more A was added?
Explain.
d) Would [D] increase, decrease, or stay the same?

a) [C][D]2
[A]2[B]3
b) Forward reaction rate increases if A is added
c) No, to keep K constant when more A is added,
new equilibrium concentrations will develop.
If all added A stays as A the ratio is “bottom heavy”.
If all added A reacts to form products the ratio is
“top heavy”
So SOME A will react to form products but not all.
All of the concentrations will change, the
amount of change will depend on K,
stoichiometry, and the amount of A added.
USE ICEBOX!
d) D will increase as more products are formed.

Le Châtelier’s Principle

“If a system at equilibrium is disturbed
by a change in temperature, pressure,
or the concentration of one of the
components, the system will shift its
equilibrium position so as to
counteract the effect of the
disturbance.”

An equilibrium position is a set of
concentrations which fulfill an equilibrium
expression.

expression.

For a given reaction and temperature, there are
infinite equilibrium positions, yet the value of K
never changes.

expression.

never changes.

Some things can shift the equilibrium position, but
not the value of K.

expression.

never changes.

Some things can shift the equilibrium position, but
not the value of K.

Only a change in temperature will affect the value
of K.

Le Chatelier’s principle

If a change is applied to a system at
equilibrium, the position of the equilibrium
will shift in a direction which counteracts
the change.

The system will shift to undo the change

Changing concentration
A(g) + B(g) ⇌ C(g) + D(g)
Adding reactant causes a shift to the right as
the forward rate increases and reactants
form products.

A(g) + B(g) ⇌ C(g) + D(g)
form products.
Removing reactant causes a shift to the left
as the forward reaction rate slows, causing
a shift towards reactants.

A(g) + B(g) ⇌ C(g) + D(g)
form products.
Adding products causes a shift towards
reactants as the reverse reaction rate
increases.

A(g) + B(g) ⇌ C(g) + D(g)
form products.
Adding products causes a shift towards
reactants as the reverse reaction rate
increases.
Removing products causes a shift to the
right and more product formation.



A(g) + B(g) ⇌ C(g) + D(g)

Increase or decrease in concentration

Equilibrium position shift to right or left

a. the equilibrium shifts to the right,
using up some of the added O2.

b. the equilibrium shifts to the left,
forming more NO.

The following reaction is at equilibrium:
N2(g) + 3 H2(g) 2 NH3(g)
If we remove NH3(g), in what direction
will the reaction move to reestablish
equilibrium?
3. No change in equilibrium

Correct Answer:

If [NH3] is decreased, to
return to equilibrium,
more NH3 must be
produced from the
reactants. Thus, the
reaction shifts from the
reactants to the
products.


Changing volume
N2(g) + 3H2(g) ⇌ 2NH3(g)
4moles of gas 2moles of gas


Changing volume
N2(g) + 3H2(g) ⇌ 2NH3(g)

Increasing pressure (decreasing volume) will
shift equilibrium towards fewer moles of gas.
Ex. Shifts right to decrease pressure


Changing volume
N2(g) + 3H2(g) ⇌ 2NH3(g)

Increasing pressure (decreasing volume) will
shift equilibrium towards fewer moles of gas.
Ex. Shifts right to decrease pressure
Decreasing pressure shifts equilibrium towards
more moles of gas.
Ex. Shifts left to increase pressure


Changing volume
N2(g) + 3H2(g) ⇌ 2NH3(g)

Changing volume/pressure only affects
gaseous equilibria
Adding an inert gas does not change
the equilibrium position

the equilibrium shifts to the left.

N2(g) + 3 H2(g) 2 NH3(g)
If we increase the volume while holding
the temperature constant, in what
direction will the reaction move to
reestablish equilibrium?

Correct Answer:

Increasing the volume
causes a shift in the
equilibrium in the
direction that produces
more gas molecules, in
this case on the
reactants side.

N2(g) + O2(g) 2 NO(g)
If we increase the pressure, in what
direction will the reaction move to
reestablish equilibrium?

Correct Answer:


The number of gas molecules on
both sides of the equation are
equal; thus, changing the
pressure will not change the
equilibrium position.

The equilibrium constant (K) does not
change with changing concentration
or pressure.

The equilibrium constant (K) does not
change with changing concentration
or pressure.

ONLY A
TEMPERATURE
CHANGE AFFECTS K

Changes that cause no shift

• A shift does not occur when a
catalyst is added because both
forward and reverse rates are
increased equally, but equilibrium
will be established faster.

Changes that cause no shift

• A shift does not occur when a
catalyst is added because both
forward and reverse rates are
increased equally, but equilibrium
will be established faster.

• Adding pure liquid, solid or an inert
gas to an equilibrium system will
not cause a shift.

A temperature change will change K

Rates for forward & reverse reactions are
affected differently by temperature
change because of activation energy
differences.

A temperature change will change K

Rates for forward & reverse reactions are
affected differently by temperature
change because of activation energy
differences.

Temperature change is the only thing that
will affect the value of K for an ideal
closed system.


Changing temperature
N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH=-92kJ
•Rewrite the equation with energy as a
reactant for endothermic processes or as a
product for exothermic processes.


N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH=-92kJ
•Rewrite the equation with energy as a
reactant for endothermic processes or as a
product for exothermic processes.
•Use Le Chatelier’s principle.
N2(g) + 3H2(g) ⇌ 2NH3(g) + 92kJ
Adding heat will shift the equilibrium to the left,
in the direction that consumes energy



N2O4(g) ⇌ 2NO2(g) ΔH = +57kJ

57kJ + N2O4(g) ⇌ 2NO2(g)
Heating the system shifts the
equilibrium right, consuming energy.



N2O4(g) ⇌ 2NO2(g) ΔH = +57kJ

57kJ + N2O4(g) ⇌ 2NO2(g)
Heating the system shifts the
equilibrium right, consuming energy.
Cooling the system shifts the
equilibrium towards the reactants

evaporation is an endothermic process
and increasing temperature shifts
equilibrium to the right.

N2(g) + O2(g) 2 NO(g)
ΔH° = +180.8 kJ. In what direction will the
reaction move to reestablish equilibrium if
the temperature is decreased?

Correct Answer:

As the temperature is lowered,
equilibrium will shift to the side of
the equation that produces heat.
Thus, for an endothermic
reaction, the equilibrium shifts
from right to left toward the
reactants.

Catalysts increase the rate of both the
forward and reverse reactions.

Equilibrium is achieved faster, but the
equilibrium composition remains
unaltered.

SAMPLE EXERCISE 15.13 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium

Consider the equilibrium
In which direction will the equilibrium shift when (a) N2O4 is
added, (b) NO2 is removed, (c) the total pressure is increased
by addition of N2(g), (d) the volume is increased, (e) the
temperature is decreased?

SAMPLE EXERCISE 15.13 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium

Consider the equilibrium
In which direction will the equilibrium shift when (a) N2O4 is
added, (b) NO2 is removed, (c) the total pressure is increased
by addition of N2(g), (d) the volume is increased, (e) the
temperature is decreased?
Solve: (a) The system will adjust to decrease the concentration of the added N2O4,
so the equilibrium shifts to the right, in the direction of products.
(b) The system will adjust to the removal of NO2 by shifting to the side that
produces more NO2; thus, the equilibrium shifts to the right.
(c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the
reaction. The partial pressures of NO2 and N2O4 are therefore unchanged, and there is
no shift in the position of the equilibrium.
(d) If the volume is increased, the system will shift in the direction that occupies a larger
volume (more gas molecules); thus, the equilibrium shifts to the right
(e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant
side of the equation. Decreasing the temperature will shift the equilibrium in the
direction that produces heat, so the equilibrium shifts to the left, toward the formation of
more N2O4. Note that only this last change also affects the value of the equilibrium
constant, K.

Chapter 15 Lecture- Chemical Equilibrium

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Chapter 15 Lecture- Chemical Equilibrium