Equilibrium slides

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E ilib iEquilibrium

AP Ch i t R id L i S i

Rapid Learning Centerwww.RapidLearningCenter.com/© Rapid Learning Inc. All rights reserved.

AP Chemistry Rapid Learning Series

Wayne Huang, PhDKelly Deters, PhDRussell Dahl, PhD

Elizabeth James, PhDDebbie Bilyen, M.A.



Learning Objectives

Dynamic equilibrium

E ilib i t t

By completing this tutorial you will learn about…

Equilibrium constants

Reaction Quotients

Solubility equilibrium

How to solve equilibrium problems

Le Chatelier’s Principle

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p

Concept MapChemistry

Studies

Previous content

New content Equilibrium

Equation withWhen Forward &

Matter

ReactionsChemicalReactions

Undergo

Can reach

ReactionRates

EquilibriumConstant

Expression

Equation with Ratio of products : reactants

When Forward & reverse are equal

When it’s disturbed, follow

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EquilibriumEquilibriumConstant

DissolutionDissolutionReaction

Le Chatelier’sLe Chatelier’sPrinciple

With values plugged in

Solubility Product

follow



Equilibrium

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Definition - Equilibrium

Reversible Reaction – A chemical reaction that can c e ca eact o t at caproceed in both directions (represented by a “ ”).

Equilibrium – When the rate of the forward reaction

l th t f th

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equals the rate of the reverse reaction.



Establishing EquilibriumIt takes time to establish equilibrium.

At first, there are only reactants present. Only the forward reaction is possible.

Reactants Products

p

But once there are products as well, they can begin to reform

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p , y greactants.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

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it may look as thought the reaction has stopped…

But the reaction continues!



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.

Equilibrium Constant (K) – The number

Concentration is symbolized with brackets “[A]”.

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calculated from the equilibrium constant expression.

“K” is different for every reaction at every temperature!



Writing Equilibrium Constant Expressions

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

1

To write an equilibrium constant expression:

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

2

Example: Write the equilibrium constant expression for the following:2 H2 (g) + O2 (g) 2 H2O (g)22

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[H2O]

[H2] [O2]K =

Definition: Homo- and Heterogeneous Equilibrium

Homogeneous EquilibriumAll of the species are the– All of the species are the

same state of matter.

Heterogeneous EquilibriumThere are at least 2 states

2 H2 (g) + O2 (g) 2 H2O (g)

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– There are at least 2 states of matter.

2 H2 (g) + O2 (g) 2 H2O (l)



Concentrations of Solids and Liquids Pure solids and pure liquids have constant “concentrations”.

If concentration (Molarity) = molelitersliters

And Density = gramsliters

And Molar Mass = gramsmole

Th f lid li id M l it grams / liters

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Then for a pure solid or liquid, Molarity = grams / litersgrams / mole

Or, Molarity = 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)][][

][

22

2

22

OHOHK =

1

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2 H2 (g) + O2 (g) 2 H2O (l) ][][1

22

2 OHK =

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

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



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

32

32

][][

HOHK =

[H2]eq = 0.45 M[H2O]eq = 0.18 MK = ?

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

3

3

]450[]18.0[

=K K = 0.064

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3]45.0[

Most instructors and textbooks do not require units for “K” as each one would be different.

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 [Products]

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If K is very small… [Products]

[Reactants]

There is a much smaller ratio of products to reactants at equilibrium.

The reaction is said to “lie to the left”.



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

]][[][

22

2

ONNOK =

[N2]eq = 0.166 M[O2]eq = 0.145 MK = 1.24×10-4

[NO]eq = ? M

)145.0)(166.0(][1024.1

24

MMNO

=× −

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[NO]eq = 0.00173 M

( )( )( ) ][145.0166.01024.1 4 NOMM =× −

Reaction Quotient

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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

Only includes gases and

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

Only includes gases and

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Only includes gases and solutions

To solve for K, plug in concentrations at equilibrium

Only includes gases and solutions

To solve for Q, plug in concentrations at any time

The Difference between K and QWhat exactly is the difference?

2 H2 (g) + O2 (g) 2 H2O (g)][][

][2

22

OHOHK =

][][ 22 OH

2 H2 (g) + O2 (g) 2 H2O (g)][][

][

22

2

22

OHOHQ =

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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.



Using Reaction QuotientReaction 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]

[reactants now]= Q

[products at equilibrium]

[reactants at equilibrium]=K

Q = K [now] = [equilibrium] System is at equilibrium

[Products now] too S t ill k

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Q > Klarge[Reactants now] too small

System will make more reactants to reach equilibrium

Q < K

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

System will make more products to reach equilibrium

Reaction Quotient ExampleExample: 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?

]][[][

22

2

ONNOQ =

[N2] = 0.81 M[O2] = 0.75 M[NO] = 0.030 MK = 0.0025At equilibrium = ? Q = 0.0015)75.0)(81.0(

)030.0( 2

MMMQ =

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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



SolubilitySolubility 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)

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



Breaking Compounds into Electrolytes

How do you break up a compound when forming electrolytes?

Do not break up polyatomic ions1

Example: Break up the following strong electrolytes:

Use subscripts that are not a part of a polyatomic ion as coefficients2

e.g. CaCl2 doesn’t have “Cl2” ions, it has 2 “Cl” ions

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Na3PO4

(NH4)2CO3

3 Na+ + PO43-

2 NH4+ + CO3

2-

Writing Solubility Product Expressions

Writing solubility products expressions are just like writing equilibrium constant expressions

Break the solid into it’s electrolytes.1

Write the concentration of products on the top—take each one to a power of the coefficient in the balanced equation.Write the concentration of reactants on the bottom—Except for dissolution equations, it’s always a solid so it’s “1”

2

3

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always a solid…so it s 1 .

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

2 2

1[ ][ ]

spCa ClK

+ −

=2 2[ ][ ]spK Ca Cl+ −=or



Solubility Product ExampleExample: 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 OH-

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

2 2[ ][ ]spK Cd OH+ −=[Cd2+]eq = 1.7×10-5 M[OH-]eq = 3.4×10-5 MKsp = ?

Ksp = 2.0×10-14

255 )104.3)(107.1( MMKsp−− ××=

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Definition - Saturated Solution

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

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Using Reaction Quotient with Solubility

Reaction Quotient can be used to determine if a solution is saturated or not.

[dissolved ions now] = Qsp [dissolved ions at saturation]=Ksp

Qsp = Ksp [now] = [equilibrium] Equilibrium (saturated solution)

Q > K Too many ions Solution will be saturated. Extra ions will precipitate

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Q > K dissolved back out (go back to the solid form)

Q < K Too few ions dissolved System is not saturated (it could hold more)

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+ (aq) + 2 OH- (aq)

2 2[ ][ ]spQ Mg OH −+=[Mg2+] = 1.8×10-4 M[OH-] = 3.0×10-4 MKsp = 8.9×10-12

At equilibrium = ?

Qsp = 1.62×10-11

244 )100.3)(108.1( MQsp−− ××=

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Qsp > Ksp

Solution is saturated

There are extra ions that will reform a solid

So there is a saturated solution with a precipitate



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 nitialChangeE ilib i

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Equilibrium



How to Use an ICE ChartThe ICE chart is an efficient way to organize the information in an equilibrium problem.

Reactants & Products“ICE”

H2 I2 HI

Initial 1.0×10-3 M 2.0×10-3 M 0 M

Change

Equilibrium 1.87×10-3 M

Reactants & ProductsPlace each reactant and then each product in a column

Place it down the side

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Equilibrium 1.87×10 M

Given informationFill in any information given

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

Determining the “Change”

The balanced equation gives the stoichiometric change for the reaction.

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

H2 O2 H2OInitialChangeEquilibrium

-2x -x +2x

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

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Mg(OH)2 (s) Mg (aq) + 2 OH (aq)

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

Mg(OH)2 Mg2+ OH-

InitialChangeEquilibrium

-x +x +2x



Finding Equilibrium Constant - 1Example: 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 I2 HI

Initial 1.0×10-3 M 2.0×10-3 M 0 M

Change


-x -x +2x

If you know an initial, change and equilibrium for one species, you l f h “ ”

6.0×10-5 M 1.1×10-3 M

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can solve for the “x”.

201087.1 3 MMx −×

=−

x = 9.4×10-4 M

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

Finding Equilibrium Constant - 2Example: 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.

Once all equilibrium concentrations are known, you can write the ilib i i d l f K

H2 I2 HI

Initial 1.0×10-3 M 2.0×10-3 M 0 M

Change


-x -x +2x

6.0×10-5 M 1.1×10-3 M

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equilibrium constant expression and solve for K.

]][[][

22

2

IHHIK =

)101.1)(100.6()1087.1(

35

23

MMMK −−

−

×××

=

K = 53



Definition: Quadratic Formula

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

acbbx 42 −±−=

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a2

02 =++ cbxaxwhere

Quadratic Formula ExampleWhen working equilibrium problems, you may end up with equations that contain an “x” and an “x2

Example: Solve for x (“x” represents “concentration”)2

aacbbx

242 −±−

=

0851 2 =−+ xx

12)814(55 2

×−××−±−

=x

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

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12×

2575±−

=x

Using the + give x = 1.27

Using the – gives x = -6.27

A negative answer isn’t possible for concentration



Finding Equilibrium Concentrations #1Example: 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 0 M

Change

Equilibrium

-x -x +2x

1.0 M - x 2.0 M - x 0 + 2x

No equilibrium concentrations are known, so the “x” cannot be d i d Th f l i b i f h

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determined yet. Therefore, only expressions may be written for the equilibrium concentrations.

]][[][

22

2

IHHIK =

)0.2)(0.1()2(5.50

2

xxx

−−= 2

2

0.30.245.50

xx+−

=

22 4)0.30.2(5.50 xx =+−× 22 45.505.151101 xx =+−

1015.1515.460 2 −+−= x



H2 I2 HI


Change

Equilibrium

-x -x +2x

1.0 M - x 2.0 M - x 0 + 2x

Use quadratic formula to solve:

40/661015.1515.460 2 −+−= x

5.462)1015.464(1.1515.151 2

−×−×−×−±−

=x x = 0.94 M or 2.31 M

q





Using 2.31 M for “x” would result in negative values for equilibrium concentration, which isn’t possible.

H2 I2 HI


Change

Equilibrium

-x -x +2x

1.0 M - x 2.0 M - x 0 + 2x

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concentration, which isn t possible.

Therefore, 0.94 M is the correct value for “x”

Use it to find equilibrium concentrations

MMMeqH 06.094.00.1][ 2 =−=

MMMeqI 06.194.00.2][ 2 =−=

MMeqHI 88.194.02][ =×=

x = 0.94 M or 2.31 M

Making Approximations for Tiny K’s

Approximations can be made when the equilibrium constant is very small.

The concentration

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

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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.



Finding Equilibrium Concentrations #4Example: 2NOCl (g) 2 NO (g) + Cl2 (g) Keq = 1.6 × 10-5. If the

initial concentration of NOCl is 0.50 M, what are the concentrations of all three at equilibrium?

NOCl NO Cl2Initial 0.50 M 0 M 0 M

Change

Equilibrium

-2x +2x +x

0.50 M - 2x 0 M + 2x 0 M + x

Approximations: 0.50 M 0 M + 2x 0 M + x

K i ll i t th t 0 50 M 2 0 50 M

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22

2

][][][

NOClClNOK = 2

25

]50.0[][]2[106.1

Mxx

=× −

25.04106.1

35 x=× −

x=×× −

35

425.0106.1 x = 0.01 M

K is very small—you may approximate that 0.50 M – 2x ≅ 0.50 M

Definition - Solubility

Solubility – Amount of a solid ythat will dissolve in a solution.

The maximum initial amount of solid that would go to 0 moles per liter

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moles per liter.



Finding SolubilityExample: Find the solubility of CaF2 if the Ksp is 4.0×10-11

CaF2 Ca2+ F-

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

Initial x 0 M 0 M

Change

Equilibrium 0 M

-x +x +2x

0 M + x 0 M + 2x

212 ]][[ −+= FCaKsp

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

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]][[sp

211 ]2][[100.4 xx=× −

311 4100.4 x=× −

x=× −

311

4100.4

x = 2.2×10-4 M

The “x” is the maximum solid that would dissolve—that is the “solubility”

Le Chatelier’s Principle

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Definition—Le Chatelier’s Principle

Le Chatelier’s Principle – If a system at equilibrium is disturbed, it will hift t t bli h ilib ishift to re-establish equilibrium.

A system will try to undo whatever you’ve done.

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How does adding a reactant or product affect a system at equilibrium?

Increasing Concentrations

Reaction shifts

Adding a reactant

Q becomes too small

Reaction shifts to right

(get rid of extra reactants and make more products)

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Adding a product Q becomes too large

Reaction shifts to left

(get rid of extra products and make more reactants)



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

Decreasing Concentrations

Removing a reactant

Q becomes too large


(make more reactants)

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Removing a product

Q becomes too small


(make more products)

How does changing the pressure affect a system at equilibrium?

Changes in Pressure

Decrease volume Pressure increases

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

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Increase volume Pressure decreases

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



Definition - Endo & Exothermic Reactions

Endothermic Reaction – The reaction takes in energy…the products have gy pmore energy than the reactants.

Exothermic Reaction – The reaction

Energy is a reactant in the reaction.

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gives off energy…the products have less energy than the reactants.

Energy is a product in the reaction.

For endothermic, think of temperature (or energy) as a reactant.

Temperature and Endothermic

Reaction shifts

Increase temperature of endothermic reaction

Increasing a reactant


(get rid of extra reactants and make more products)

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Decrease temperature of endothermic reaction

Remove a reactant


(make more reactants)



For exothermic, think of temperature (or energy) as a product.

Temperature and Exothermic

Reaction shiftsIncrease

temperature of exothermic reaction

Increasing a product


(get rid of extra products and make more reactants)

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Decrease temperature of exothermic reaction

Remove a product


(make more products)

Some changes have no effect because they do not affect the value of “Q”.

Changes that Have No Effect

Adding a pure solid or liquid reactant or g p qproduct.

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.

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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.



Le Chatelier’s ExamplesExample: Which way will the reaction shift for each of the

following changes:NH4Cl (s) NH3 (g) + HCl (g)

Removing NH4Cl No change (it’s a solid)Removing NH4Cl

Adding HCl

Adding Ne (g)

Decreasing volume

Example: Which way will the reaction shift for each of the following changes:

2 SO (g) + O (g) 2 SO (g) an exothermic reaction

No change (it’s an inert gas)

No change (it s a solid)

(Adding a product)

(Goes to side with least gas moles)

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2 SO2 (g) + O2 (g) 2 SO3 (g) an exothermic reaction

Increasing volume

Raising temperature

Adding O2

Removing SO2

(Goes to side with most gas moles)

(Energy is a product)

(Adding a reactant)

(Removing a reactant)

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

Le Chatelier in Industry

Small K Small ratio of products to reactants

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

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y)

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.



The reaction to produce ammonia, NH3, is very important to manufacturing.

Haber Process

N + 3 H 2 NH (an exothermic reaction)

In order for the reaction to occur at a reasonable rate, the temperature must be very high.

N2 + 3 H2 2 NH3 (an exothermic reaction)

But when the temperature is high, the equilibrium constant is very low.

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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.

Equilibrium & The AP Exam

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Equilibrium in the Exam

The first free response question is always ilib i

Common equilibrium problems:

equilibrium:

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

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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

Multiple Choice QuestionsOccasionally, equilibrium is found in the multiple choice section.

Example: Which is the correct equilibrium constant expression for HOBr (aq) H+ (aq) + OBr- (aq)?

A. [HOBr] / [H+][OBr-]B. [H+][OBr-] / [HOBr]C. [H+] / [HOBr]D. [OBr-] / [HOBr]E. None of the above

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Answer: B



Free Response Questions

At 25°C the value of Ksp for PbCl is 1 6 10-5 and the value

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 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

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y g qconcentrations of Pb+2 and Ag+, which will precipitate first?

Answering Free Response QuestionsAt 25°C, the value of Ksp for PbCl2 is 1.6 × 10-5 and the value

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 ( 3)2, p pvolumes 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

PbCl Pb+2 + 2 Cl-1

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PbCl2 Pb + 2 Cl

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



Answering Free Response QuestionsAt 25°C, the value of Ksp for PbCl2 is 1.6 × 10-5 and the value

of Ksp for AgCl is 1.8 × 10-10.B. Calculate the equilibrium value of [Pb+2] in a saturated

solution of PbCl2.2

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

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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.

The ICE chart is a good technique

The ICE chart is a good technique

Dynamicequilibrium is

established when

Dynamicequilibrium is

established when

Le Chatelier’sPrinciple governs h ti t

Le Chatelier’sPrinciple governs h ti t

Learning Summary

The solubility productThe solubility product

to use to solve equilibrium problems.

to use to solve equilibrium problems.

the rates of the forward and

reverse reactions are equal.

the rates of the forward and

reverse reactions are equal.

how a reaction at equilibrium will change when

disturbed.

how a reaction at equilibrium will change when

disturbed.

Th ilib i t tTh ilib i t t

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The solubility product is the equilibrium

constant for a dissolution reaction.

The solubility product is the equilibrium

constant for a dissolution reaction.

The equilibrium constantgive the ratio of product:

reactants with the stoichiometric ratios as

the powers.

The equilibrium constantgive the ratio of product:

reactants with the stoichiometric ratios as

the powers.



Congratulations

You have successfully completed the rapid tutorial

Equilibrium

Rapid Learning Center

Rapid Learning Center

Wh t’ N t

Chemistry :: Biology :: Physics :: Math

What’s Next …

Step 1: Concepts – Core Tutorial (Just Completed)

Step 2: Practice – Interactive Problem Drill

Step 3: Recap – Super Review Cheat Sheet

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Equilibrium slides

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Transcript of Equilibrium slides