Equilibrium slides

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Equilibrium
E ilib i

AP Ch i t R id Learning Series
Chemistry Rapid L
i
S i
Wayne Huang, PhD
Kelly Deters, PhD
Russell Dahl, PhD
Elizabeth James, PhD
Debbie Bilyen, M.A.

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© Rapid Learning Inc. All rights reserved.

1


Learning Objectives
By completing this tutorial you will learn about…
Dynamic equilibrium
Equilibrium constants
E ilib i
t t
Reaction Quotients
Solubility equilibrium
How to solve equilibrium
problems
Le Chatelier’s Principle
p

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Concept Map
Previous content
Chemistry

New content

Equilibrium
Studies

When Forward &
reverse are equal

Reaction
Rates

Dissolution
Reaction

Equation with
Ratio of
products :
reactants

Equilibrium
Constant
Expression

Solubility
Product

Can reach

Matter
Undergo

When it’s
disturbed,
follow

Chemical
Reactions

With values
plugged in

Equilibrium
Constant

Le Chatelier’s
Principle

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2


Equilibrium

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Definition - Equilibrium
Reversible Reaction – A
c e ca eact o that can
chemical reaction t at ca
proceed in both directions
(represented by a “ ”).
Equilibrium – When the rate
of the forward reaction
equals the rate of the reverse
l th
t f th
reaction.

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3


Establishing Equilibrium
It takes time to establish equilibrium.
At first, there are only reactants
present. Only the forward reaction is
p
possible.

Reactants

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Products

But once there are products as well, they can begin to reform
p
,
y
g
reactants.
The reverse reaction becomes possible. Forward rate slows and
reverse rate increases until they are the same.
Once the rate of the forward and reverse process are equal, it is
at equilibrium.
When equilibrium is established, the number of products and
reactants doesn’t change…but the reaction keeps going.

Definition - Dynamic Equilibrium
Dynamic Equilibrium – The
reaction continues to
proceed in both directions,
but at the same rate.
The number of products and
reactants no longer change,
it may look as thought the
reaction has stopped…
But the reaction continues!
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4


Equilibrium
Constants

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Definition - Equilibrium Constant
Equilibrium Constant Expression – Equation
showing the ratio of the concentrations of
products to reactants at equilibrium.
Concentration is symbolized with brackets “[A]”.

Equilibrium Constant (K) – The number
calculated from the equilibrium constant
expression.
“K” is different for every reaction at every
temperature!
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5


Writing Equilibrium Constant Expressions
To write an equilibrium constant expression:

1

Write the concentration of products on the top—
take each one to a power of the coefficient in the
balanced equation.
b l
d
ti

2

Write the concentration of reactants on the
bottom—also take each to the power of the
balanced equation coefficient.

Example: Write the equilibrium constant expression for the following:
2 H2 (g) + O2 (g) 2 H2O (g)

K=

[H2O]
[H2] [O2]

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Definition: Homo- and Heterogeneous
Equilibrium
Homogeneous Equilibrium
– All of the species are the
same state of matter.
2 H2 (g) + O2 (g)

2 H2O (g)

Heterogeneous Equilibrium
– There are at least 2 states
of matter.
2 H2 (g) + O2 (g)

2 H2O (l)

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6


Concentrations of Solids and Liquids
Pure solids and pure liquids have constant “concentrations”.
If concentration (Molarity) = mole
liters
And Density = grams
liters
And Molar Mass = grams
mole
Then f
Th for a pure solid or liquid, Molarity = grams / liters
lid
li id M l it
grams / mole
Or, Molarity =

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Density .
Molar Mass

Both Density and Molar Mass are constants—they don’t change.
Therefore, “concentration” of a pure solid or liquid is a constant.

“K” Expressions with Solids or Liquids
How does this affect the writing of Equilibrium Constant
Expressions?
If the “concentration” of a pure solid or liquid is constant,
then it will not change during equilibrium and it is not written
in the “K” expression.
2 H2 (g) + O2 (g)

2 H2O (g)

K=

[ H 2 O ]2
[ H 2]2 [O2 ]

2 H2 (g) + O2 (g)

2 H2O (l)

K=

1
[ H 2]2 [O2 ]

H2O is not included in this “K” expression because it’s a liquid.

Only gases and solutions are included in “K” expressions!
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7


Calculating “K” Example
Example: Solve for equilibrium constant for
Fe2O3 (s) + 3 H2 (g) 2 Fe (s) + 3 H2O (g) if the following
are concentrations at equilibrium: [H2] = 0.45 M and
[H2O] = 0.18 M
[H2]eq = 0.45 M

K=

[ H 2O ]3
[ H 2 ]3

K=

[H2O]eq = 0.18 M

[0.18]3
[0.45]3

K=?

Note that Fe2O3 and Fe were
not included in the K
expression as they are solids!

K = 0.064

Most instructors and textbooks do not require units for “K” as each
one would be different.
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Meaning of Equilibrium Constant
What general meaning can you get from the magnitude
of the equilibrium constant?
If K is very large…

[Products]
[Reactants]

There is a much larger ratio of products to reactants at
equilibrium.
The reaction is said to “lie to the right” (products are on
the right).
If K is very small
small…

[Products]

[Reactants]
There is a much smaller ratio of products to reactants at
equilibrium.
The reaction is said to “lie to the left”.
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8


Using “K” to Find Equilibrium Concentration
Example: Find the equilibrium concentration for NO if the
equilibrium constant for N2 (g) + O2 (g) 2 NO (g) is
1.24×10-4, and the other equilibrium concentrations are
[N2] = 0.166 M and [O2] = 0.145 M
[N2]eq = 0.166 M

K=

[O2]eq = 0.145 M
K = 1.24×10-4
[NO]eq = ? M

1.24 ×10 −4 =

[ NO ]2
[ N 2 ][O2 ]

[ NO ]2
(0.166 M )(0.145M )

(1.24 ×10 )(0.166M )(0.145M ) = [ NO]
−4

[NO]eq = 0.00173 M

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Reaction
Quotient

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9


What is the Reaction Quotient?
Reaction Quotient is “Q”
K

Q

Equilibrium Constant

Reaction Quotient

Expression is ratio of
products to reactants with
balanced equation
coefficients as powers

Expression is ratio of
products to reactants with
balanced equation
coefficients as powers

Only includes gases and
solutions

Only includes gases and
solutions

To solve for K, plug in
concentrations at
equilibrium

To solve for Q, plug in
concentrations at any time

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The Difference between K and Q
What exactly is the difference?
2 H2 (g) + O2 (g)

2 H2O (g)

K=

[ H 2 O ]2
[ H 2]2 [O2 ]

2 H2 (g) + O2 (g)

2 H2O (g)

Q=

[ H 2 O ]2
[ H 2]2 [O2 ]

The expressions for K and Q are the same.
To solve for “K”, plug in concentrations at equilibrium only.
To solve for “Q”, plug in concentrations at any time.

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10


Using Reaction Quotient
Reaction Quotient is used to determine if a system
is at equilibrium…and if it’s not, which way does it
need to go to get there.
[products now]

= Q

K =

[reactants at equilibrium]

[reactants now]
Q=K

[products at equilibrium]

[now] = [equilibrium]
[Products now] too
large

Q>K

[Reactants now] too
small
[Products now] too
small

Q<K

[Reactants now] too
large

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System is at equilibrium
System will make more
S t
ill
k
reactants to reach
equilibrium

System will make more
products to reach
equilibrium

Reaction Quotient Example
Example: For N2 (g) + O2 (g) 2 NO (g), if [N2] = 0.81 M, [O2] = 0.75 M
and [NO] = 0.030 M, is the reaction at equilibrium if
K = 0.0025? If not, which way will it go to reach equilibrium?

[N2] = 0.81 M
[O2] = 0.75 M

Q=

[ NO ]2
[ N 2 ][O2 ]

Q=

(0.030M ) 2
(0.81M )(0.75M )

[NO] = 0.030 M
K = 0.0025
At equilibrium = ?

Q = 0.0015

Q<K
Reaction is not at equilibrium
More products will need to be made (and also thereby
reducing reactants) to have Q = K
Reaction will go to the right (products) to reach equilibrium
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11


Solubility
Equilibrium
Constants

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Definition - Solubility Product
Solubility Product (Ksp) – Equilibrium
Constant for a dissolution equation.
Equation showing a solid dissolving and producing ions

NaCl (s)

Na+ (aq) + Cl- (aq)

Ksp describes the equilibrium between the solid
forming dissociated ions and the dissociated ions
joining back together to reform the solid.
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12


Breaking Compounds into Electrolytes
How do you break up a compound when forming
electrolytes?
1

Do not break up polyatomic ions

2

Use subscripts that are not a part of a polyatomic
ion as coefficients
e.g. CaCl2 doesn’t have “Cl2” ions, it has 2 “Cl” ions

Example:

Break up the following strong electrolytes:

Na3PO4
(NH4)2CO3

3 Na+ + PO432 NH4+ + CO32-

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Writing Solubility Product Expressions
Writing solubility products expressions are just like
writing equilibrium constant expressions

1

Break the solid into it’s electrolytes.

2

Write the concentration of products on the top—
take each one to a power of the coefficient in the
balanced equation.

3

Write the concentration of reactants on the
bottom—Except for dissolution equations, it’s
always a solid so it’s “1”.
solid…so it s 1

Example: Write the equilibrium constant expression for the following:
CaCl2 (s) Ca2+ (aq) + 2 Cl- (aq)

K sp =

[Ca 2+ ][Cl − ]2
1

or

K sp = [Ca 2+ ][Cl − ]2

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13


Solubility Product Example
Example: Find the solubility product of Cd(OH)2 if at equilibrium
the solution is 1.7×10-5 M Cd2+ and 3.4×10-5 M OHCd(OH)2 (s)

Cd2+ (aq) + 2 OH- (aq)

[Cd2+]eq = 1.7×10-5 M
[OH-]eq = 3.4×10-5 M
Ksp = ?

K sp = [Cd 2+ ][OH − ]2
K sp = (1.7 ×10 −5 M )(3.4 ×10 −5 M ) 2
Ksp = 2.0×10-14

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Definition - Saturated Solution

Saturated Solution – Th
S t
t d S l ti
The
solution has reached the
equilibrium between the
solid and the dissociated
particles.

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14


Using Reaction Quotient with Solubility
Reaction Quotient can be used to determine if a
solution is saturated or not.
[dissolved ions now] = Qsp

Ksp = [dissolved ions at saturation]

Qsp = Ksp

[now] = [equilibrium]

Equilibrium (saturated
solution)

Q>K

Too many ions
dissolved

Solution will be saturated.
Extra ions will precipitate
back out (go back to the
solid form)

Q<K

Too few ions dissolved

System is not saturated (it
could hold more)

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Using Reaction Quotient with Ksp
Example: Is a solution of Mg(OH)2 saturated if 1.8×10-4 M Mg2+ and
3.0×10-4 M OH- is present? Ksp is 8.9×10-12
Mg(OH)2 (s)
[Mg2+] = 1.8×10-4 M
[OH-] = 3.0×10-4 M
Ksp = 8.9×10-12
At equilibrium = ?

Mg2+ (aq) + 2 OH- (aq)

Qsp = [ Mg 2+ ][OH − ]2
Qsp = (1.8 ×10 −4 )(3.0 ×10 −4 M ) 2
Qsp = 1.62×10-11

Qsp > Ksp
Solution is saturated
There are extra ions that will reform a solid
So there is a saturated solution with a precipitate
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15


Solving Equilibrium
Problems

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ICE Charts in Equilibrium Problems
Equilibrium problems are often difficult for people
to solve. Using “ICE” charts is an excellent
technique.

I nitial
C hange
ilib i
E quilibrium
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16


How to Use an ICE Chart
The ICE chart is an efficient way to organize the
information in an equilibrium problem.
“ICE”
Place it down the side

Reactants & Products
Place each reactant and then each
product in a column
H2

I2

HI

1.0×10-3 M

Initial

2.0×10-3 M

0M

Change
1.87×10-3 M

Equilibrium

Use Equilibrium Values
Find a way to determine them and use them
in the K expression

Given information
Fill in any information given

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Determining the “Change”
The balanced equation gives the stoichiometric
change for the reaction.
2 H2 (g) + O2 (g)

2 H2O (g)

For every 2 H2’s that react, 1 O2 will react and 2 H2O will form.
H2

O2

H2O

Initial
-2x

Change

-x

+2x

Equilibrium

Mg(OH)2 (s)

Mg2+ (aq) + 2 OH- (aq)

For every 1 Mg(OH)2 that reacts, 1 Mg2+ and 2 OH- will form.
Mg(OH)2

Mg2+

OH-

Initial
Change

-x

+x

+2x

Equilibrium
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17


Finding Equilibrium Constant - 1
Example: H2 (g) + I2 (g) 2 HI (g) A mixture of 1.0 × 10-3 M H2 and
2.0 × 10-3 M I2 is placed in a container. At equilibrium, the
mixture shows the concentration of HI is 1.87 × 10-3 M.
Find the equilibrium constant.
H2
Initial
Change
Equilibrium

1.0×10-3

I2
M

-x
6.0×10-5

HI

2.0×10-3 M

0M

-x
M

+2x

1.1×10-3

M

1.87×10-3 M

If you know an initial, change and equilibrium for one species, you
can solve for the “x”.
l f
h “ ”

x=

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1.87 × 10 −3 M − 0 M
2

x = 9.4×10-4 M

Once the “x” is known, the equilibrium concentrations can be
determined.

Finding Equilibrium Constant - 2
Example: H2 (g) + I2 (g) 2 HI (g) A mixture of 1.0 × 10-3 M H2 and
2.0 × 10-3 M I2 is placed in a container. At equilibrium, the
mixture shows the concentration of HI is 1.87 × 10-3 M.
Find the equilibrium constant.
H2
Initial
Change
Equilibrium

I2

HI

1.0×10-3 M

2.0×10-3 M

0M

-x
6.0×10-5 M

-x

+2x

1.1×10-3 M

1.87×10-3 M

Once all equilibrium concentrations are known, you can write the
equilibrium constant expression and solve for K.
ilib i
i
d l f K

K=

[ HI ]2
[ H 2 ][ I 2 ]

K = 53

(1.87 ×10 −3 M ) 2
K=
(6.0 ×10 −5 M )(1.1× 10 −3 M )
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18


Definition: Quadratic Formula
Quadratic formula –
formula used to find
“x”
“ ” in an equation that
eq ation
contains “x” and “x2”.

− b ± b 2 − 4ac
x=
2a
where ax 2 + bx + c = 0
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Quadratic Formula Example
When working equilibrium problems, you may end up
with equations that contain an “x” and an “x2
Example:

Solve for x (“x” represents “concentration”)

1x 2 + 5 x − 8 = 0
x=

− b ± b 2 − 4ac
2a

From above: a = 1, b = 5, c = -8

x=

− 5 ± 52 − (4 ×1× −8)
2×
2 ×1

Using the + give x = 1.27
Using the – gives x = -6.27

x=

− 5 ± 57
2

A negative answer isn’t possible for concentration
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19


Finding Equilibrium Concentrations #1
Example: H2 (g) + I2 (g) 2 HI (g) A flask is filled with 1.0 M H2
and 2.0 M I2. The value of K at this temperature is 50.5.
What are the concentrations of H2, I2 and HI in the flask
at equilibrium?
H2

I2

HI

Initial

1.0 M

2.0 M

0M

Change

-x

-x

Equilibrium

1.0 M - x

+2x

2.0 M - x

0 + 2x

No equilibrium concentrations are known, so the “x” cannot be
determined yet. Therefore, only expressions may be written for the
d
i d
Th f
l
i
b
i
f
h
equilibrium concentrations.
4x2
[ HI ]2
( 2 x) 2
50.5 =
50.5 =
K=
[ H 2 ][ I 2 ]
(1.0 − x)(2.0 − x)
2.0 − 3.0 + x 2

50.5 × (2.0 − 3.0 + x 2 ) = 4 x 2

101 − 151.5 + 50.5 x 2 = 4 x 2

0 = −46.5 x + 151.5 − 101
2

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at equilibrium?
H2

I2

HI

Initial

1.0 M

2.0 M

0M

Change

-x

-x

+2x

1.0 M - x

2.0 M - x

0 + 2x

Equilibrium

Use quadratic formula to solve:
q

x=

40/66

− 151.5 ± 151.12 − (4 × −46.5 × −101)
2 × −46.5

x = 0.94 M or 2.31 M

0 = −46.5 x 2 + 151.5 − 101


20


at equilibrium?
H2

I2

HI

Initial

1.0 M

2.0 M

0M

Change

-x

-x

+2x

1.0 M - x

2.0 M - x

Equilibrium

0 + 2x

Using 2.31 M for “x” would result in negative values for equilibrium
concentration, which isn’t possible.
isn t
Therefore, 0.94 M is the correct value for “x”
Use it to find equilibrium concentrations

x = 0.94 M or 2.31 M

[ H 2 ]eq = 1.0 M − 0.94 M = 0.06 M
[ I 2 ]eq = 2.0 M − 0.94 M = 1.06 M
41/66

[ HI ]eq = 2 × 0.94 M = 1.88M

Making Approximations for Tiny K’s
Approximations can be made when the equilibrium
constant is very small.

Very tiny
equilibrium
constant

Only a very small
amount of
reactants
react and
produce
products

The concentration
of the reactant
before is
approximately
the same as the
concentration
after

For K’s that are ×10-5 and smaller you can approximate any
number being added or subtracted by an “x” as the number
itself.
e.g.
0.25 M + x ≅ 0.25 M
because “x” is so tiny.
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21


Keq = 1.6 × 10-5. If the
Example: 2NOCl (g) 2 NO (g) + Cl2 (g)
initial concentration of NOCl is 0.50 M, what are the
concentrations of all three at equilibrium?

NOCl

Change

Cl2

0.50 M

Initial

NO
0M

0M

-2x

Equilibrium

+2x

+x

Approximations: 0.50 M

0 M + 2x

0M+x

0 M + 2x

0.50 M - 2x

0M+x

K i very small—you may approximate that 0.50 M – 2 ≅ 0 50 M
is
ll
i t th t 0 50
2x 0.50

K=

[ NO ]2 [Cl2 ]
[ NOCl ]2

1.6 × 10 −5 =

3

[ 2 x ]2 [ x ]
[0.50 M ]2

1.6 ×10 −5 =

1.6 ×10 −5 × 0.25
=x
4

4 x3
0.25

x = 0.01 M

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Definition - Solubility

Solubility – Amount of a solid
y
that will dissolve in a solution.
The maximum initial amount
of solid that would go to 0
moles per liter.
liter

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22


Finding Solubility
Example: Find the solubility of CaF2 if the Ksp is 4.0×10-11
CaF2 (s)

Ca2+ (aq) + 2 F- (aq)
CaF2

Initial

Ca2+

F-

x

0M

0M

Change

-x

+x

Equilibrium

0M

0M+x

+2x
0 M + 2x

You’re trying to find out how much solid can go to “0”, so set it’s
equilibrium value to 0.

K sp = [Ca +2 ][ F −1 ]2

4.0 ×10 −11 = [ x][2 x]2

4.0 ×10 −11 = 4 x 3
3

4.0 ×10 −11
=x
4

x = 2.2×10-4 M
The “x” is the maximum solid
that would dissolve—that is
the “solubility”

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Le Chatelier’s
Principle

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23


Definition—Le Chatelier’s Principle
Le Chatelier’s Principle – If a system
at equilibrium is disturbed, it will
shift to re-establish equilibrium.
hift t
t bli h
ilib i
A system will try to undo whatever you’ve
done.

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Increasing Concentrations
How does adding a reactant or product affect a
system at equilibrium?
Reaction shifts
to right
Adding a
reactant

Q becomes too
small

(get rid of extra
reactants and
make more
products)

Reaction shifts
to left
Adding a product

Q becomes too
large

(get rid of extra
products and
make more
reactants)

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24


Decreasing Concentrations
How does removing a reactant or product affect a
system at equilibrium?

Removing a
reactant

Q becomes too
large

Removing a
product

Q becomes too
small

Reaction shifts
to left
(make more
reactants)

Reaction shifts
to right
(make more
products)

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Changes in Pressure
How does changing the pressure affect a system at
equilibrium?

Decrease volume

Increase volume

Pressure
increases

Reaction shifts
to the side
with least
moles of gas
to decrease
pressure

Pressure
decreases

Reactions shifts
to the side
with the most
moles of gas
to increase
pressure

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25


Definition - Endo & Exothermic Reactions
Endothermic Reaction – The reaction
gy
products have
takes in energy…the p
more energy than the reactants.
Energy is a reactant in the reaction.

Exothermic Reaction – The reaction
gives off energy…the products have
less energy than the reactants.
Energy is a product in the reaction.
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Temperature and Endothermic
For endothermic, think of temperature (or energy)
as a reactant.
Reaction shifts
to right

Increase
temperature of
endothermic
reaction

Increasing a
reactant

Decrease
temperature of
endothermic
reaction

Remove a
reactant

(get rid of extra
reactants and
make more
products)

Reaction shifts
to left
(make more
reactants)

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26


Temperature and Exothermic
For exothermic, think of temperature (or energy) as
a product.
Increase
temperature of
exothermic
reaction

Decrease
temperature of
exothermic
reaction

Reaction shifts
to left
Increasing a
product

Remove a
product

(get rid of extra
products and
make more
reactants)

Reaction shifts
to right
(make more
products)

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Changes that Have No Effect
Some changes have no effect because they do not
affect the value of “Q”.
Adding a pure solid or liquid reactant or
g p
q
product.
Increasing pressure by adding an inert gas.
Changing the volume of a reaction with an
equal number of moles of gas on each side of
the reaction.
Adding a catalyst
A catalyst will speed up how fast
equilibrium is established—but not the
number of reactants and products once it’s
at equilibrium.
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27


Le Chatelier’s Examples
Example: Which way will the reaction shift for each of the
following changes:
NH4Cl (s)
Removing NH4Cl

NH3 (g) + HCl (g)
No change (it s a solid)
(it’s

Adding HCl

(Adding a product)

Adding Ne (g)

No change (it’s an inert gas)

Decreasing volume

(Goes to side with least gas moles)

Example: Which way will the reaction shift for each of the
following changes:
2 SO2 (g) + O2 (g) 2 SO3 (g) an exothermic reaction
Increasing volume

(Goes to side with most gas moles)

Raising temperature

(Energy is a product)

Adding O2

(Adding a reactant)

Removing SO2

(Removing a reactant)

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Le Chatelier in Industry
When companies need to make large amounts of
product, a reaction with a very small K is a
problem.

Small K

Small ratio of
products to
reactants

Lots of reactants
left over
(wasting
money) and
few products
made (not
making
y)
money)

They can push the reaction towards the products.
e.g. Remove the products as they’re made, adjust pressure or
temperature as needed to push it to the right.
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28


Haber Process
The reaction to produce ammonia, NH3, is very
important to manufacturing.
N2 + 3 H2

2 NH3

(an exothermic reaction)

In order for the reaction to occur at a reasonable rate, the
temperature must be very high.
But when the temperature is high, the equilibrium constant is very low.
A compromise is made and a catalyst is added to increase the rate
at the lower temperature.

The reaction yields 20%...the leftover reactants are recycled and
put back into the reaction again.
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Equilibrium &
The AP Exam

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29


Equilibrium in the Exam
Common equilibrium problems:
The first free response question is always
equilibrium:
ilib i
Either K, Ksp, Ka, or Kb (Ka and Kb are in the next
tutorial)
Determine the solubility or solubility product
Determine equilibrium concentrations given K
Determine K given equilibrium concentrations
Use Q to determine which way a reaction will proceed
Use Le Chatelier’s Principle to determine which way a
reaction will proceed
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Multiple Choice Questions
Occasionally, equilibrium is found in the multiple choice
section.
Example:

Which is the correct equilibrium constant expression for
HOBr (aq) H+ (aq) + OBr- (aq)?
A.
B.
C.
D.
E.

[HOBr] / [H+][OBr-]
[H+][OBr-] / [HOBr]
[H+] / [HOBr]
[OBr-] / [HOBr]
None of the above
Answer: B

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30


Free Response Questions
The equilibrium free response question is often about
solubility.
At 25°C the value of Ksp for PbCl2 is 1 6 × 10-5 and the value
25°C,
1.6
of Ksp for AgCl is 1.8 × 10-10.
A. If 30.0 mL of 0.060 M NaCl is added to 30 mL of 0.030 M
Pb(NO3)2, will a precipitate form. Assume that the
volumes are additive.
B. Calculate the equilibrium value of [Pb+2] in a saturated
solution of PbCl2.
C. If NaCl is added slowly to a beaker containing equal
y
g q
concentrations of Pb+2 and Ag+, which will precipitate
first?

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Answering Free Response Questions
At 25°C, the value of Ksp for PbCl2 is 1.6 × 10-5 and the value
A. If 30.0 mL of 0.060 M NaCl is added to 30 mL of 0.030 M
Pb(NO3)2, will a precipitate form. Assume that the
(
p
p
volumes are additive.
New volume = 60 mL
New [NaCl] = 30 mL × 0.060 M / 60 mL = 0.030 M
New [Pb(NO3)2] = 30 mL × 0.030 M / 60 mL = 0.015 M
PbCl2

Pb+2 + 2 Cl-1

Q = [Pb+2][Cl-1]2
Q = (0.015 M)(0.030 M)2 = 1.35 × 10-5
Q < K, it will not precipitate out
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Answering Free Response Questions
At 25°C, the value of Ksp for PbCl2 is 1.6 × 10-5 and the value
B. Calculate the equilibrium value of [Pb+2] in a saturated
solution of PbCl2.
PbCl2

Pb+2 + 2 Cl-1

Ksp = [Pb+2][Cl-1]2
1.6 × 10-5 = (x)(2x)2
x = 0.016 M
[Pb+2] = 0.016 M

C. If NaCl is added slowly to a beaker containing equal
concentrations of Pb+2 and Ag+, which will precipitate
first?
AgCl is less soluble (Ksp is smaller), therefore it will
precipitate first.
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Learning Summary
The ICE chart is a
good technique
to use to solve
equilibrium
problems.

Dynamic
equilibrium is
established when
the rates of the
forward and
reverse reactions
are equal.

The
Th equilibrium constant
ilib i
t t
give the ratio of product:
reactants with the
stoichiometric ratios as
the powers.

Le Chatelier’s
Principle governs
how a reaction at
h
ti
t
equilibrium will
change when
disturbed.

The solubility product
is the equilibrium
constant for a
dissolution reaction.

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32


Congratulations
You have successfully completed
the rapid tutorial

Equilibrium

Chemistry :: Biology :: Physics :: Math

What’s N t
Wh t’ Next …

Step 1: Concepts – Core Tutorial (Just Completed)
Step 2: Practice – Interactive Problem Drill
Step 3: Recap – Super Review Cheat Sheet

Go for it!

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