1 / 81

Section 14 Chemical Equilibrium

Chemical Equilibrium

- Previously we have assumed that chemical

reactions results in complete conversion of

reactants to products - A B C D
- No A or B remaining or possibly an excess of A

or B but not both and eventually reaction stops

Many chemical reactions do not completely convert

reactants to products. Stop somewhere between no

rxn and complete rxn. A B C D some

left some formed reversible (both

directions)

The concentration of all species level off to

some individual constant value that persists

indefinitely as long as no stress is added to

system. The state that is reached when the

concentrations of R and P remain a constant in

time is called state of chemical equilibrium and

the mixture is considered an equilibrium

mixture. Individual molecules are continually

reacting even though the overall composition of

the reaction does not change. There is a

exchange going on A B forms C D but then C

D breaks down into A B at the same rate

therefore, concentration of each component

remains a constant and the rate of exchange is

constant at equilibrium. Reaction has not

stopped just the rates of the forward and

reverse reactions are equal.

Chemical Equilibrium

- Therefore, many reactions do not go to completion

but rather form a mixture of products and

unreacted reactants, in a dynamic equilibrium.

- A dynamic equilibrium consists of a forward

reaction, in which substances react to give

products, and a reverse reaction, in which

products react to give the original reactants.

- Chemical equilibrium is the state reached by a

reaction mixture when the rates of the forward

and reverse reactions have become equal.

Graphically we can represent this A B C

D The concentrations and reaction rate (less

collisions, less component) of A and B decreases

over time as the concentrations and reaction rate

of C and D increases (more collisions, more

component) over time until the rates are equal

and the concentrations of each components reaches

a constant. This occurs at what we call

equilibrium -- Rf Rr. If the rates are equal,

then there must be a relationship to show

this.

Figure Catalytic methanation reaction

approaches equilibrium. CO 3 H2 CH4

H2O

If we assume these reactions are elementary rxns

(based on collisions), we can write the rate laws

directly from the reaction A B

C D Rf kf AB

For the reverse reaction we have, C D

A B Rr kr CD We know at

equil that Rf Rr therefore, we can set these

two expressions as equal kf AB kr

CD Rearrange to put constants on one side we

get

Constant divided by constant just call a new

constant K. This ratio is given a special name

and symbol called equilibrium constant K relating

to the equilibrium condition at a certain

temperature (temp dependent) for a particular

reaction relating conc of each component. This

is basically a comparison between forward and

reverse reaction rates. At equilibrium, the

ratio of conc of species must satisfy K.

The Equilibrium Constant

- Every reversible system has its own position of

equilibrium- K- under any given set of

conditions.

- The ratio of products produced to unreacted

reactants for any given reversible reaction

remains constant under constant conditions of

pressure and temperature. If the system is

disturbed, the system will shift and all the

concentrations of the components will change

until equilibrium is re-established which occurs

when the ratio of the new concentrations equals

what -- "K". Different constant conc but ratio

same as before. - The numerical value of this ratio is called the

equilibrium constant for the given reaction, K.

The Equilibrium Constant

- The equilibrium-constant expression for a

reaction is obtained by multiplying the equil

concentrations ( or partial pressures) of

products, dividing by the equil concentrations

(or partial pressures) of reactants, and raising

each concentration to a power equal to its

coefficient in the balanced chemical equation.

- The molar concentration of a substance is denoted

by writing its formula in square brackets for aq

solutions. For gases can put Pa - atm. As long

as use M or atm, K is unitless. - Temp dependent any changes, ratio must equal K

when equil established

The Equilibrium Constant

- The equilibrium constant, K, is the value

obtained for the equilibrium-constant expression

when equilibrium concentrations (not just any

conc but equil conc) are substituted.

- A large K indicates large concentrations of

products at equilibrium. - A small K indicates large concentrations of

unreacted reactants at equilibrium.

The Equilibrium Constant

- Do same set up for all equil equations (future

chapters), just different subscript describing

the reaction in question i.e. acid hydrolysis -

acid water - Ka, Kb, Kc, Kp, Ksp - Kc is based on conc (M) and Kp is based on

pressures (atm). There is a difference between Kc

and Kp and to jump between these two, there is an

equation to use. Let's show how to jump

between.

HW 14

The Equilibrium Constant, Kp

- In discussing gas-phase equilibria, it is often

more convenient to express concentrations in

terms of partial pressures rather than molarities.

The Equilibrium Constant, Kp

- The equilibrium-constant expression in terms of

partial pressures becomes (same way but partial

pressures instead of M)

The Equilibrium Constant, Kp

- In general if you need to jump between K's, the

numerical value of Kp differs from that of Kc.

A Problem to Consider

- Consider the reaction

- Kc for the reaction is 280 at 1000. K . Calculate

Kp for the reaction at this temperature.

A Problem to Consider

- Consider the reaction at 1000. K and Kc 280

- We know that

A Problem to Consider

- Consider the reaction

- Since

The Equilibrium Constant

- The law of mass action states that the value of

the equilibrium constant expression K is constant

for a particular reaction at a given temperature,

whenever equilibrium concentrations are

substituted. - When at equil, conc ratio equals K and concs are

constant but if disturbed, values change until

equil reached, different constant conc but same

ratio of K).

Obtaining Equilibrium Constants for Reactions

- Equilibrium concentrations for a reaction must be

obtained experimentally and then substituted into

the equilibrium-constant expression in order to

calculate Kc.

Obtaining Equilibrium Constants for Reactions

- Consider the reaction below

- Suppose we started with initial concentrations of

CO and H2 of 0.100 M and 0.300 M, respectively. - Obviously shift to right, decrease CO and H2 and

increase CH4 and water until equil established.

Obtaining Equilibrium Constants for Reactions

- When the system finally settled into equilibrium

we determined the equilibrium concentrations to

be as follows.

Obtaining Equilibrium Constants for Reactions

- If we substitute the equilibrium concentrations,

we obtain

Obtaining Equilibrium Constants for Reactions

- Regardless of the initial concentrations (whether

they are reactants or products or mixture), the

law of mass action dictates that the reaction

will always settle into an equilibrium where the

equilibrium-constant expression equals Kc.

Obtaining Equilibrium Constants for Reactions

- As an example, lets repeat the previous

experiment, only this time starting with initial

concentrations of products (note if only

products to start shift left until equil

established)

- CH4initial 0.2000 M and H2Oinitial

0.2000 M - Obviously shift to left decrease CH4 and H2O and

increase CO and H2 until equil established.

Obtaining Equilibrium Constants for Reactions

- We find that these initial concentrations result

in the following equilibrium concentrations.

Obtaining Equilibrium Constants for Reactions

- Substituting these values into the

equilibrium-constant expression, we obtain the

same result.

- Whether we start with reactants or products at

any initial conc, the system establishes the same

ratio.

A Problem to Consider

- Applying Stoichiometry to an Equilibrium Mixture

(basic setup for future problems).

- What is the composition of the equilibrium

mixture if it contains 0.080 mol NH3 at

equilibrium?

- Using the information given, set up the following

table. (ratio works for atm, M, mols, etc.)

Initial, no

Change, Dn

Equil, neq

- This procedure for many types of problems,

however in this problem given equil quantity of

NH3 therefore, figure out rest. - The equilibrium amount of NH3 was given as 0.080

mol. Therefore, 2x 0.080 mol NH3 (x 0.040

mol).

A Problem to Consider

- Using the information given, set up the following

table.

Starting 1.000 3.000 0

Change

Equilibrium

Equilibrium amount of N2 Equilibrium amount of

H2 Equilibrium amount of NH3

HW 15

Equilibrium Constant for the Sum of Reactions

- Similar to the method of combining reactions that

we saw using Hesss law in Chapter 6, we can

combine equilibrium reactions whose K values are

known to obtain K for the overall reaction.

- With Hesss law, when we reversed reactions

(change sign) or multiplied them prior to adding

them together (mult by factor). We had to

manipulate the DHs values to reflect what we had

done. - The rules are a bit different for manipulating K.

Equilibrium Constant for the Sum of Reactions

- If you reverse a reaction, invert the value of K.

(reciprocal rule, 1/K)

- If you multiply/divide each of the coefficients

in an equation by the same factor (2, 1/2, ),

raise Kc to the same power (2, 1/2, ). (known

as coefficient rule, Kn) - When you finally combine (that is, add) the

individual equations together, take the product

of the equilibrium constants to obtain the

overall K.(rule of multiple equilibria, K1 x K2 x

K3 KT)

Equilibrium Constant for the Sum of Reactions

- For example, nitrogen and oxygen can combine to

form either NO(g) or N2O (g) according to the

following equilibria.

Kc 6.4 x 10-16

(1)

Kc 2.4 x 10-18

(2)

overall

Kc ?

(1)

(2)

overall

HW 16

Heterogeneous Equilibria

- A heterogeneous equilibrium is an equilibrium

that involves reactants and products in more than

one phase. Up to now all our reactions have been

homogeneous - all gases or aqueous solutions.

- The equilibrium of a heterogeneous system is

unaffected by the amounts of pure solids or

liquids present, as long as some of each is

present. - The concentrations of pure solids and liquids are

always considered to be 1 activity and

therefore, do not appear in the equilibrium

expression. Solids and pure liquids have no

effect on conc or pressure.

Heterogeneous Equilibria

- Consider the reaction below.

HW 17

Predicting the Direction of Reaction

- How could we predict the direction in which a

reaction at non-equilibrium conditions will shift

to reestablish equilibrium? Remember did example

with only reactants, obviously had to go right

and if only products obviously must go left but

what if have some of R and P?

- To answer this question, substitute the current

concentrations into the reaction quotient

expression and compare it to Kc. - The reaction quotient, Qc, is an expression that

has the same form as the equilibrium-constant

expression but whose concentrations are not

necessarily at equilibrium.

Predicting the Direction of Reaction

- For the general reaction

Predicting the Direction of Reaction

- For the general reaction

- If Qc Kc, then the reaction is at equilibrium.
- If Qc gt Kc, the reaction will shift left toward

reactants until equil reached. - If Qc lt Kc, the reaction will shift right toward

products until equil reached.

A Problem to Consider

- Consider the following equilibrium.

- A 50.0 L vessel contains 1.00 mol N2, 3.00 mol

H2, and 0.500 mol NH3. Is the sytem at equil? If

not, in which direction (toward reactants or

toward products) will the system shift to

reestablish equilibrium at 400 oC? - Kc for the reaction at 400 oC is 0.500.

A Problem to Consider

- First, calculate concentrations from moles of

substances.

A Problem to Consider

- First, calculate concentrations from moles of

substances.

A Problem to Consider

- First, calculate concentrations from moles of

substances.

HW 18

Calculating Equilibrium Concentrations

- Once you have determined the equilibrium

constant, K, for a reaction, you can use it to

calculate the concentrations of substances in the

equilibrium mixture.

Calculating Equilibrium Concentrations

- For example, consider the following equilibrium

mixture.

- Suppose a gaseous mixture contained 0.30 mol CO,

0.10 mol H2, 0.020 mol H2O, and an unknown amount

of CH4 per liter at equilibrium. - What is the concentration of CH4 at equilibrium

in this mixture? The equilibrium constant Kc

equals 3.92. - Note the amounts given are equil amounts

therefore can plug into K eq.

Calculating Equilibrium Concentrations

- First, calculate concentrations from moles of

substances.

Calculating Equilibrium Concentrations

- First, calculate concentrations from moles of

substances.

Calculating Equilibrium Concentrations

- First, calculate concentrations from moles of

substances.

Calculating Equilibrium Concentrations

- Suppose we begin a reaction with known amounts of

starting materials and want to calculate the

quantities at equilibrium.

Calculating Equilibrium Concentrations

- Consider the following equilibrium.

- Suppose you start with 1.000 mol each of carbon

monoxide and water in a 50.0 L container.

Calculate the molarity of each substance in the

equilibrium mixture at 1000 oC. - Kc for the reaction is 0.58 at 1000 oC.
- Which way will reaction shift?

Calculating Equilibrium Concentrations

- First, calculate the initial molarities of CO and

H2O.

Calculating Equilibrium Concentrations

- First, calculate the initial molarities of CO and

H2O.

- The starting concentrations of the products are

0. - We must now set up a table of concentrations

(starting, change, and equilibrium expressions in

x).

Calculating Equilibrium Concentrations

- The equilibrium-constant expression is

- Let x be the moles per liter of product formed.

o 0.0200 0.0200 0 0

D

eq

- Solving for x.

Starting 0.0200 0.0200 0 0

Change

Equilibrium

Calculating Equilibrium Concentrations

- Solving for x.

Starting 0.0200 0.0200 0 0

Change -x -x x x

Equilibrium 0.0200-x 0.0200-x x x

Calculating Equilibrium Concentrations

- Solving for x.

Starting 0.0200 0.0200 0 0

Change -x -x x x

Equilibrium 0.0200-x 0.0200-x x x

Calculating Equilibrium Concentrations

- Solving for equilibrium concentrations.

Starting 0.0200 0.0200 0 0

Change -x -x x x

Equilibrium 0.0200-x 0.0200-x x x

- If you substitute for x in the last line of the

table you obtain the following equilibrium

concentrations. If plug into Keq, should equal K

or close to it because of sign fig for a check

HW 19

Calculating Equilibrium Concentrations

- The preceding example illustrates the three steps

in solving for equilibrium concentrations.

- Set up a table of concentrations (starting,

change, and equilibrium expressions in x). - Substitute the expressions in x for the

equilibrium concentrations into the

equilibrium-constant equation. - Solve the equilibrium-constant equation for the

values of the equilibrium concentrations.

Another example If the initial pressure of C is

1.0 atm, what would be the partial pressures of

each species at equil.

A B

2C

Kp 9.0

Po 0 0 1.0

DP

Peq

HW 20

Another example If the initial pressure of C is

0.10 atm and A and B are 1.00 atm, what would be

the partial pressures of each species at equil.

Kp 0.016

2C A

B

Po 0.10 1.00 1.00

DP

Peq

HW 21

Calculating Equilibrium Concentrations

- In some cases it is necessary to solve a

quadratic equation to obtain equilibrium

concentrations - not a perfect square. - The next example illustrates how to solve such an

equation.

Calculating Equilibrium Concentrations

- Consider the following equilibrium.

- Suppose 1.00 atm H2 and 2.00 atm I2 are placed in

a 1.00-L vessel. What are the partial pressures

of all species when it comes to equilibrium at

458 oC? - Kp at this temperature is 49.7.

Calculating Equilibrium Concentrations

- The concentrations of substances are as follows.

Po 1.00 2.00 0

DP

Peq

Calculating Equilibrium Concentrations

- The concentrations of substances are as follows.

Po 1.00 2.00 0

DP -x -x 2x

Peq 1.00-x 2.00-x 2x

- Substituting our equilibrium concentration

expressions gives

Calculating Equilibrium Concentrations

- Solving for x.

Starting 1.00 2.00 0

Change -x -x 2x

Equilibrium 1.00-x 2.00-x 2x

- Because the right side of this equation is not a

perfect square (sq over sq), you must solve the

quadratic equation.

(No Transcript)

Calculating Equilibrium Concentrations

- However, x 2.33 gives a negative value to 1.00

- x (the equilibrium concentration of H2), which

is not possible.

Obviously if you get a negative and positive

number, take the positive number and if two

positive numbers, take the smaller number.

However, if you neglect the shift if all

components are present (you don't do Q - not only

initial reactants or products) and select it

incorrectly, you will end up with either a and

- or two -'s. In this case the negative

number will be correct or the smaller of the two

negative numbers will be correct. Bottom line

examine numbers carefully when selecting answer.

Calculating Equilibrium Concentrations

- Solving for equilibrium concentrations.

Starting 1.00 2.00 0

Change -x -x 2x

Equilibrium 1.00-x 2.00-x 2x

- If you substitute 0.934 for x in the last line of

the table you obtain the following equilibrium

concentrations.

HW 22

Le Chateliers Principle

- Obtaining the maximum amount of product from a

reaction depends on the proper set of reaction

conditions which gets us to

- Le Chateliers principle states that when a

system in a chemical equilibrium is disturbed by

a change of temperature, pressure, or

concentration, the equilibrium will shift in a

way that tends to counteract this change.

Removing Products or Adding Reactants

- If a chemical system at equilibrium is disturbed

by adding a gaseous or aqueous species (not solid

or liquid R or P), the reaction will proceed in

such a direction as to consume part of the added

species. Conversely, if a gaseous or aqueous

species is removed (complex or escape gas), the

system shifts to restore part of that species.

This shift will occur until equilibrium is

re-established. - A B C D add - shift to opposite

side, remove - shift towards that

side.

Effects of Pressure Change

- A pressure change caused by changing the volume

of the reaction vessel can affect the yield of

products in a gaseous reaction only if the

reaction involves a change in the total moles of

gas present - Ex. N2O4 (g) 2NO2 (g)
- Suppose system is compressed by pushing down a

piston (decrease volume of space), which way

would the shift be that would benefit and use the

the available space wisely? - Shift to smaller number of gas molecules to pack

more efficiently and relieve the increase in

pressure due to piston coming down.

Effects of Pressure Change

- Basically the reactants require less volume (that

is, fewer moles of gaseous reactant) and by

decreasing the volume of the reaction vessel by

increasing the pressure, the rxn would shift the

equilibrium to the left (toward reactants) until

equil is established.

Effects of Pressure Change

- Literally squeezing the reaction (increase P)

will cause a shift in the equilibrium toward the

fewer moles of gas.

- Reducing the pressure in the reaction vessel by

increasing its volume would have the opposite

effect. - Decrease P, increase V, shift larger mols of gas

(L S not compressible) - Increase P, decrease V, shift to smaller mols of

gas - In the event that the number of moles of gaseous

product equals the number of moles of gaseous

reactant, vessel volume/pressure will have no

effect on the position of the equilibrium no

advantage to shift one way over the other.

SO2 (g) 1/2 O2 (g) lt--gt SO3 (g) N2 (g)

3 H2 (g) lt--gt 2NH3 (g) N2 (g) O2 (g) lt--gt

2NO (g) C (s) H2O(g) lt--gt CO (g) H2

(g)

Effect of Temperature Change

- Temperature has a significant effect on most

reactions.

- Reaction rates generally increase with an

increase in temperature. Consequently,

equilibrium is established sooner. - However, when you add or remove

reactants/products or change pressure, the result

is that the system establishes new conc of

species but ratio still equal to same K at that

temperature. For temp changes, the numerical

value of the equilibrium constant Kc varies with

temperature. K temp dependent.

Effect of Temperature Change

- Lets look at heat as if it were a product in

exothermic reactions and a reactant in

endothermic reactions.

- We see that increasing the temperature is related

to adding more product (in the case of exothermic

reactions) or adding more reactant (in the case

of endothermic reactions). - This ultimately has the same effect as if heat

were a physical entity.

Effect of Temperature Change

- For example, consider the following generic

exothermic reaction.

- Increasing temperature would be like adding more

product, causing the equilibrium to shift left. - Since heat does not appear in the

equilibrium-constant expression, this change

would result in a smaller numerical value for Kc

(numerator smaller and den larger)

Effect of Temperature Change

- For an endothermic reaction, the opposite is true.

- Increasing temperature would be analogous to

adding more reactant, causing the equilibrium to

shift right. - This change results in more product at

equilibrium, and a larger numerical value for Kc

(larger numerator, smaller den)

Effect of Temperature Change

- In summary

- For an endothermic reaction (DH positive) the

amounts of products are increased at equilibrium

by an increase in temperature (Kc is larger at

higher temperatures).

- For an exothermic reaction (DH is negative) the

amounts of reactants are increased at equilibrium

by an increase in temperature (Kc is smaller at

higher temperatures).

Effect of a Catalyst

- A catalyst is a substance that increases the rate

of a reaction but is not consumed by it.

- It is important to understand that a catalyst has

no effect on the equilibrium composition of a

reaction mixture. - A catalyst merely speeds up the attainment of

equilibrium but does not cause it shift one way

or the other just get to direction is was going

faster.

Consider the system I2 (g) 2I (g)

DH 151 kJ Suppose the system is at

equilibrium at 1000oC. In which direction will

rxn occur if a.) I atoms are added? b.) the

system is compressed? c.)the temp is increased?

d.)effect increase temp has on K? e.) add

catalyst?

HW 23