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Ch. 15 Chemical Equilibrium Consider colorless frozen N2O4. At room temperature, it

decomposes to brown NO2:N2O4(g) p 2NO2(g).

At some time, the color stops changing and we have a

mixture of N2

O4

and NO2

. We now write it like this

Each is constantly being formed at the same rate that it is

being consumed. It is therefore a dynamic equilibrium.

Chemical equilibrium is the point at which theconcentrations of all species are constant.

N2O4(g) 2NO2(g)

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The Equilibrium Constant Consider the following reaction:

There are two reactions going on, forward and reverse. The rate of

each reaction can be expressed separately:

Ratef

= kf

[A]a[B]b

Rater= kr[C]c[D]d

At equilibrium, Rf= Rrorkf[A]a[B]b = kr[C]

c[D]d

We can rearrange this equation and combine the rate constants into

a single constant. We end up with this

where Keq is called the equilibrium constant.

aA + bB cC + dD

? A ? A

? A ? Aba

dc

eqKBA

DC!

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The Equilibrium Constant If the substances in the reaction are gases, then their

concentrations are proportional to their partial pressuresaccording to PV=nRT, so we can express Keq in terms of

pressures as well

(PC)c (PD)d(PA)

a (PB)b

(Reminder Molarity = n/V =P/RTand R= 0.0821 Latm/molK)

(Changes in pressure at equilibrium will be explored later in the notes.)

The actual value for Keq is determined experimentally at a specific

temperature for a specific reaction.

Keq =

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The Magnitude of Equilibrium Constants

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N2O4(g) 2NO2(g)

More Facts About Keq The equilibrium constant of a reaction in the reverse direction is

the inverse of the equilibrium constant of the reaction in theforward direction Keq(forward) = 1/Keq(reverse)

For example:

At 100 C, Keq(forward) = 6.49

At 100 C, Keq(reverse) = 1/6.49 = 0.154 Keq does notdepend on the reaction mechanism.

The value of Keq varies with temperature. (We will see this later.)

The stoichiometry of a reaction that has been multiplied by a

number changes the equilibrium constant. Keq gets raised to thepower equal to that number.

For example: 2 4 Keq = (6.49)2 = 42.12

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More Facts About Keq The equilibrium constant for a net reaction made up of two or more

steps is the productof the equilibrium constants for the individual

steps.

For example: A + B X + C Keq(1) = 2.0

X+ B

D Keq(2) = 5.0

A + 2B C + D K eq = K1 x K2 = 10.0

When this is written in terms of concentrations at equilibrium

Keq=10.0 = [C][D]/[A][B]

2

Keq usually is not written with units. Why? The Molarity &

Pressures are based on a standard: (1 M or 1 atm.)

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More Facts About Keq If concentrations are used in the equilibrium expression, Keq is

sometimes written as Kc.

When pressures are used in the expression, Keq it sometimes written

as Kp.

T

he equilibrium expression only contains the concentrations of gasesor aqueous substances and NEVER solids or pure liquids. Why?

- Consider the decomposition of calcium carbonate:

CaCO3(s) CaO(s) + CO2(g)

- The concentrations of solids and pure liquids are constant.The amount of CO2 formed will not depend greatly on the

amounts of CaO and CaCO3present.

Keq = [PCO2]

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Note: Although the concentrations of these species are not included in

the equilibrium expression, they doparticipate in the reaction and must

be present for an equilibrium to be established!

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Changes in Equilibrium Le Chateliers Principle: If a stress is applied to a system that is

already at equilibrium, the equilibrium will shift to reduce the effect

of the stress.

We will now look at changing various things on a system at

equilibrium and discuss how the equilibrium will shift to relieve the

stress.

(1) Changing Concentrations: A + B C + D

If more [A] is added, the rate of the forward reaction increases

to relieve the stress. As this occurs [C] and [D] increase, the rate

of the reverse reaction increases and quickly equilibrium is re-

established. At equilibrium Rf= Rr. The concentrations of A, B,

C, D have changed but:

[C][D] remains constant at a given temperature.

[A][B]

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Changes in EquilibriumChanging pressure continued

Example: 2H2(g) + O2(g) 2H2O(g) If pressure is increased by decreasing the volume of the gases in the

container at equilibrium, then the forward reaction is favored. Why?

- This reduces the # of gas particles from 3 to 2.

-Note: If the # of gas particles on both sides of the equation is thesame, then changing pressure has NO EFFECT on the equilibrium.

If an inert gas is added, it WILL NOT change the equilibrium.

(3) Temperature: All chemical reactions either give off heat

(exothermic) or take in heat (endothermic). An increase in

temperature favors the endothermic reaction; a decrease in

temperature favors the exothermic reaction.

The temperature of a reaction will change the value of Keq, but for

now, lets just focus on how the equilibrium will shift.

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Changes in EquilibriumChanges in temperature continued

Example: H2 + I2 2HI H = - 25 kJ/mol (exothermic)Another way of looking at the reaction

H2 + I2 2HI + heat

Lowering the temperature favors HI formation. (You can

think of it as though we are removing the heat product fromthe equation.) Raising the temperature favors the reverse

reaction.

van Hoffs Law: In a system at equilibrium, an increase in heat

energy is displaced so that heat is absorbed.(4) Catalyst: A catalyst increases the rate of both the forward and the

reverse reaction by decreasing Ea. It has no effect on Keqbut does

cause equilibrium to be reached more quickly.

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Catalysts & Changes in Equilibrium

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Predicting the Direction of a Reaction When given the concentrations of reactants & products

and the value of Keq, you can easily predict the directionof a reaction.

- Simply plug the given concentrations into the

equilibrium equation and calculate the reaction

quotient, Q.

- IfQ > eq then the reverse reaction must occur to

reach equilibrium (i.e., products are consumed,

reactants are formed, the numerator in the equilibrium

constant expression decreases and Q decreases until it

equalsKeq).

- IfQ < Keq then the forward reaction must occur to

reach equilibrium. (S

ee Practice Problems for an example.)

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Predicting the Direction of a Reaction

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Calculating Keq Practice Problem: At a certain temperature, a 1.0 L flask initially

contains 0.298 moles of PCl3(g) and 0.0087 moles of PCl5(g). After the

system reaches equilibrium, 0.00200 moles of Cl2(g) was found in the

flask. Calculate the equilibrium concentrations of the gases in the

flask and also the value of Keq. PCl5 decomposes according to the

following reactionPCl5(g) PCl3(g) + Cl2(g)

Step 1-- Tabulate initial and equilibrium concentrations (or partialpressures) given.

This sort of table goes by the nickname ICEboxget it?

[PCl5] [PCl3] [Cl2]

Initial

Change

Equilibrium

0.0087 M 0.298 M 0

0.002 M

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Calculating Keq

[PCl5] [PCl3] [Cl2]

Initial

Change

Equilibrium

0.0087 M 0.298 M 0

0.002 M

Step 2- If an initial and equilibrium concentration is given for a

species, calculate the change in concentration.

Step 3- Use stoichiometry on the change in concentration line only to

calculate the changes in concentration of all species(Since

our reaction occurs in a 1:1:1 ratio, the changes for each

species is the same. Only the sign is different on the reactant.)

Step 4- Deduce the equilibrium concentrations of all species.

+ 0.002 M+ 0.002 M 0.002 M

0.0067 M 0.3 M

Step 5- Finally, plug the equilibrium concentrations into the

equilibrium equation and solve!

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Calculating Keq

[PCl5] [PCl3] [Cl2]

Initial

Change

Equilibrium

0.0087 M 0.298 M 0

0.002 M

Keq = [ PCl3][Cl2]/[PCl5]

Keq = [0.3][0.002]/[0.0067] = 0.08955 0.09

+ 0.002 M+ 0.002 M 0.002 M

0.0067 M 0.3 M

This system of tabulating data will allow you to solve for

equilibrium concentrations if you are given Keq

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Calculating KeqPractice Problem #2-- Given this equationH2(g)+ I2(g) 2 HI(g)

calculate all three equilibrium concentrations when:

[H2]o = [I2]o = 0.200 M and Kc = 64.0

[H2] [I2] [HI]

Initial

Change

Equilibrium

0.200 M 0.200 M 0

Heres where we use stoichiometry to calculate the changes in the concentrations

Now determine the equilibrium concentrations

Now plug them into the equilibrium expression and solve for x

x x + 2x

0.200 x 0.200 x + 2x

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Calculating KeqKc = [ HI]

2/[H2][I2]

64.0 = [2x]2/[0.200 x][0.200 x]

Solving this takes some good algebra skills & maybe even a quadratic equation will

have to be solvedYUCK!

Both sides are perfect squares, (done so on purpose), so we square root both sides

to get 8.00 = (2x) / (0.200 - x)

F

rom there, the solution should be easier, and so after some cross-multiplying anddividing, etc x = 0.160 M

This is notthe end of the solution since the question was asking for the equilibriumconcentrations, so

[H2] = 0.200 - 0.160 = 0.040 M

[I2] = 0.200 - 0.160 = 0.040 M[HI] = 2 (0.160) = 0.320 M

You can check for correctness by plugging back into the equilibrium expression:

Kc = (0.320)2 / (0.040)(0.040) = 64

Since Kc = 64.0 we know that the problem was correctly solved.

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The EndNow we get to work on more practice problems!!