1 Chapter 31 Chemical Equilibrium Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved

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Chapter 31

Chemical Equilibrium

Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved.

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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)

Chemical EquilibriumPreviously 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.

exchange, constant conc., Ratef = Rater “equilibrium”

A + B C + D

A + B C + D

4

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.

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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.

3H2 + CO CH4 + H2O

3H2 + CO CH4 + H2O rate decreases over time

CH4 + H2O 3H2 + CO rate increases over time

Rf=Rr

Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005.

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For the reverse reaction we have,

C + D A + B Rr = kr [C][D]

We know at equil that Rf = Rr ;therefore, we can set these two expressions as equal

kf [A][B] = kr [C][D]

Rearrange to put constants on one side we get

]][[

]][[

BA

DC

k

kK

r

f

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 [A][B]

8

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.

]][[

]][[

BA

DC

k

kK

r

f

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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 "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.

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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.

dDcC bBaA

ba

dc

BA

DCK

][][

][][ b

Ba

A

dD

cC

p PP

PPK

)()(

)()(

c

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; liquids and solids = 1; setup same for all K’s (Ka, Kb, etc.)

Temp dependent; any changes, ratio will still equal K when equil established

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31.1 The Equilibrium Constant, K

• 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).

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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.

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• Consider the reaction below

O(g)H (g)CH (g)H 3 )g(CO 242

Suppose we started with initial concentrations of CO and H2 of 0.100 M and 0.300 M, respectively.

0.100 M 0.300 M 0 0

Obviously shift to right, decrease CO and H2 and increase CH4 and water until equil established.

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– When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows.

Reactants Products[CO]eq = 0.0613 M[H2] eq = 0.1839 M

[CH4] eq = 0.0387 M[H2O] eq = 0.0387 M


O(g)H (g)CH (g)H 3 )g(CO 242

15

– If we substitute the equilibrium concentrations, we obtain:

93.3)1839.0)(0613.0(

)0387.0)(0387.0(3cK


O(g)H (g)CH (g)H 3 )g(CO 242

eqeq

eqeqc

HCO

OHCHK

32

24

][][

][][

16

– 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.


O(g)H (g)CH (g)H 3 )g(CO 242

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As an example, let’s repeat the previous experiment, only this time starting with initial concentrations of products (note: if only products to start shift left until equil established):

[CH4]initial = 0.2000 M and [H2O]initial = 0.2000 M

Obviously shift to left decrease CH4 and H2O and increase CO and H2 until equil established.


O(g)H (g)CH (g)H 3 )g(CO 242 0 0 0.2000 M 0.2000 M

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– We find that these initial concentrations result in the following equilibrium concentrations.

Reactants Products

[CO] eq = 0.0990 M

[H2] eq = 0.2970 M

[CH4] eq = 0.1010 M

[H2O] eq = 0.1010 M


O(g)H (g)CH (g)H 3 )g(CO 242

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– Substituting these values into the equilibrium-constant expression, we obtain the same result.

93.3)2970.0)(0990.0(

)1010.0)(1010.0(3cK

– Whether we start with reactants or products at any initial conc, the system establishes the same ratio at same temperature.


O(g)H (g)CH (g)H 3 )g(CO 242

eqeq

eqeqc

HCO

OHCHK

32

24

][][

][][

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• 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, K>1, indicates large concentrations of products at equilibrium.

– A small K, K<1, indicates large concentrations of unreacted reactants at equilibrium.

HW 14code: horse

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– 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.

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31.1.1 The Equilibrium Constant in Terms of Pressure, Kp

• In discussing gas-phase equilibria, it is often more convenient to express concentrations in terms of partial pressures rather than molarities.

– It can be seen from the ideal gas equation (PV =nRT) that the partial pressure of a gas is proportional to its molarity

– P proportional to M - n/V ; therefore handled the same way.

MRTRTVn

P )(

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The Equilibrium Constant, Kp

– Consider the reaction below.

O(g)H (g)CH (g)H 3 )g(CO 242 – The equilibrium-constant expression in terms of partial

pressures becomes (same way but partial pressures instead of M):

3)(

2

24

HCO

OHCHpK

PP

PP 3

2

24

][H ][

O][ ][

CO

HCHKc

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The Equilibrium Constant, Kp

• In general if you need to jump between K's, the numerical value of Kp differs from that of Kc.

where ng is the sum of the moles of gaseous

products in a reaction minus the sum of the moles of gaseous reactants.

ngcp RTKK )(

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A Problem to Consider

• Consider the reaction

– Kc for the reaction is 280 at 1000. K . Calculate Kp for the reaction at this temperature.

(g)SO 2 )g(O)g(SO2 322

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– We know that

From the equation we see that ng = 2 – 3 = -1. We can simply substitute the given reaction temperature and the value of R (0.0821 L.atm/mol.K) to obtain Kp.

• Consider the reaction at 1000. K and Kc = 280

(g)SO 2 )g(O)g(SO2 322

ngcp RTKK )(

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– Since

3.4 K) 1000 0821.0( 280 -1

KmolatmL

pK

• Consider the reaction

(g)SO 2 )g(O)g(SO2 322

ngcp RTKK )(

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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?

– Suppose we place 1.000 mol N2 and 3.000 mol H2 in a reaction vessel at 450 oC and 10.0 atmospheres of pressure. The reaction is

(g)2NH )g(H3)g(N 322

322

23

c ]H][N[

]NH[K

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– This procedure for many types of problems, however in this problem given equil quantity of NH3 ;therefore, figure out rest.

• Using the information given, set up the following table. (ratio works for atm, M, mols, etc.)

(g)2NH )g(H3 )g(N 322 Initial, no

Change, n

Equil, neq

1.000 3.000 0

- - +

– The equilibrium amount of NH3 was given as 0.080 mol. Therefore, 2x = 0.080 mol NH3 (x = 0.040 mol).

x 3x 2x

1.000 - x 3.000 - 3x 2x = 0.080 mol

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A Problem to Consider• Using the information given, set up the

following table.

Equilibrium amount of N2 = 1.000 - 0.040 = 0.960 mol N2

Equilibrium amount of H2 = 3.000 - (3 x 0.040) = 2.880 mol H2

Equilibrium amount of NH3 = 2x = 0.080 mol NH3

(g)2NH )g(H3 )g(N 322 Initial 1.000 3.000 0

Change -x -3x +2xEquilibrium1.000 - x 3.000 - 3x 2x = 0.080 mol

HW 15

x = 0.040 mol

code: rhino

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31.2 Changing the Chemical Equation

• Similar to the method of combining reactions that we saw using Hess’s law in thermochemistry, we can combine equilibrium reactions whose K values are known to obtain K for the overall reaction.

– With Hess’s law, when we reversed reactions (change sign) or multiplied them prior to adding them together (mult by factor). We had to manipulate the H’s values to reflect what we had done.

– The rules are a bit different for manipulating K.

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2. 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)

3. 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)

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

(reciprocal rule, 1/K)

Equilibrium Constant for the Sum of Reactions

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• For example, nitrogen and oxygen can combine to form either NO(g) or N2O (g) according to the following equilibria.

NO(g) )g(O2

1)g(N

2

122

O(g)N )g(O)g(N 2221

2

Kc = 6.4 x 10-16

Kc = 2.4 x 10-18

(1)

(2)

Kc = ?

– Using these two equations, we can obtain K for the formation of NO(g) from N2O(g):

NO(g) 2 )g(O)g(ON 221

2 overall

Equilibrium Constant for the Sum of Reactions

reverse and multiply by factor of 1

same and multiply by factor of 2

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NO(g) 2 )g(O)g(ON 221

2

NO(g) 2 )g(O)g(N 22 Kc = (6.4 x 10-16)2 =

4.1 x 10-31

(1)

)g(O (g)N O(g)N 221

22 Kc = (2)

overall

18-10 4.2

1

131731 107.1)102.4()101.4()( overallKc

HW 16

NO(g) )g(O2

1)g(N

2

122

O(g)N )g(O)g(N 2221

2

Kc = 6.4 x 10-16

Kc = 2.4 x 10-18

= 4.2 x 1017

code: ally

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31.1.2 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.

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Heterogeneous Equilibria

• Consider the reaction below.

(g)H CO(g) )()( 22 lOHsC

2

2 ]][[

HCOp

c

PPK

HCOK

HW 17

code: mike

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31.3 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.

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Predicting the Direction of Reaction

• For the general reaction

dDcC bBaA the Qc expresssion would be (i=initial):

ba

dc

c ]B[]A[

]D[]C[Q

ii

ii

ba

dc

c BA

DCK

eqeq

eqeq

][][

][][

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Predicting the Direction of Reaction

• For the general reaction

dDcC bBaA – If Qc = Kc, then the reaction is at equilibrium.

– If Qc > Kc, the reaction will shift left toward reactants until equil reached.

– If Qc < Kc, the reaction will shift right toward products until equil reached.

ba

dc

c ]B[]A[

]D[]C[Q

ii

ii

ba

dc

c ]B[]A[

]D[]C[Q

ii

ii

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– 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.

(g)2NH )g(H3 )g(N 322

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First, calculate concentrations from moles of substances.

(g)2NH )g(H3 )g(N 322

322

23

][][

][

ii

ic

HN

NHQ

Next, plug into Q expression

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(g)2NH )g(H3 )g(N 322

– Substituting these concentrations into the reaction quotient gives:

1.23)0600.0)(0200.0(

)0100.0(Q 3

2

c

Note: you cannot just look at number of mols or molarity to decide direction of reaction because must account for the size of K.

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0.0100 M0.0600 M0.0200 M


(g)2NH )g(H3 )g(N 322

– Because Qc = 23.1 is greater than Kc = 0.500, the reaction will go to the left (toward reactants - consume products, produce reactants) as it approaches equilibrium.

HW 18

code: josh

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31.4 Calculating K and Equilibrium Quantities

• 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.

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Calculating Equilibrium Concentrations

– 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.

O(g)H (g)CH (g)H 3 )g(CO 242

– For example, consider the following equilibrium mixture.

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– First, calculate concentrations from moles of substances then plug into K expression.

O(g)H (g)CH (g)H 3 )( 242 gCO

eqeq

eqeqc

HCO

OHCHK

32

24

][][

][][

47


– Substituting the known concentrations and the value of Kc gives:

34

)10.0)(30.0(

)020.0]([92.3

CH

??0.30 M 0.10 M 0.020 M

O(g)H (g)CH (g)H 3 )g(CO 242

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– You can now solve for [CH4].

MCH 059.0)020.0(

)10.0)(30.0)(92.3(][

3

4

– The concentration of CH4 in the equil mixture is

0.059 mol/L.

??0.30 M 0.10 M 0.020 M

O(g)H (g)CH (g)H 3 )g(CO 242

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31.4.1 Calculating Equilibrium Quantities from Initial Values (Perfect Square)

• Suppose we begin a reaction with known amounts of starting materials and want to calculate the quantities at equilibrium.

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• 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.

(g)H(g)CO )g(OH)g(CO 222 0 0

- +

Which way will reaction shift?

51


– First, calculate the initial molarities of CO and H2O.

(g)H(g)CO )g(OH)g(CO 222

0 0

52


• The starting concentrations of the products are 0.• We must now set up a table of concentrations (initial, change,

and equilibrium expressions in x, ICE table).

0.0200 M 0.0200 M 0 M 0 M


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

Initial 0.0200 0.0200 0 0

ChangeEquil

– The equilibrium-constant expression is:

]OH][CO[]H][CO[

K2

22c


- - + +x x x x 0.0200-x 0.0200-x x x

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– Solving for x.

Initial 0.0200 0.0200 0 0Change -x -x +x +x

Equilibrium0.0200-x 0.0200-x x x

– Substituting the values for equilibrium concentrations, we get:

)x0200.0)(x0200.0()x)(x(

58.0


]OH][CO[]H][CO[

K2

22c

55


– Solving for x.



2

2

)x0200.0(

x58.0


56


– Solving for x.



– Taking the square root of both sides because perfect square, we get:

x

xx

xx

x

x

76.10152.0

76.00152.0

)0200.0(76.0

)0200.0(76.0


2

2

)0200.0(58.0

x

x

Mx 0086.076.1

0152.0

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– Solving for equilibrium concentrations.



– 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

[CO]eq = 0.0200 – x = 0.0200 – 0.0086 = 0.0114 M[H2O]eq = 0.0200 – x = 0.0200 – 0.0086 = 0.0114 M[CO2]eq = x = 0.0086 M[H2]eq = x = 0.0086 M

code: katie

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• The preceding example illustrates the three steps in solving for equilibrium concentrations.

1. Set up a table of concentrations (initial, change, and equilibrium expressions in x – ICE table).

2. Substitute the expressions in x for the equilibrium concentrations into the equilibrium-constant equation.

3. Solve the equilibrium-constant equation for the values of the equilibrium concentrations.

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Another example: If the initial pressure of C is 1.0 atm, what would be the partial pressures of each species at equil.

Initial, Po 0 0 1.0Change, P

Equil, Peq

HW 20

A + B 2CKp = 9.0

+ + -x x 2x

x x 1.0-2x

code: chris

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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.

Initial, Po 0.10 1.00 1.00Change, P

Equil, Peq

HW 21

2C A + B Kp = 0.016

Q > K therefore, shift left

+ - -2x x x 0.10 + 2x 1.00 – x 1.00 - x

code: three

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31.4.2 Calculating Equilibrium Quantities from Initial Values (Quadratic Formula)

• 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.

62



• 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.

HI(g)2 )g(I)g(H 22

63


– The concentrations of substances are as follows.

HI(g)2 )g(I)g(H 22 Initial, Po 1.00 2.00 0

Change, PEquil, Peq

- - +x x 2x

1.00 - x 2.00 – x 2x

64


– The concentrations of substances are as follows.

Po 1.00 2.00 0P -x -x +2xPeq 1.00-x 2.00-x 2x

– Substituting our equilibrium concentration expressions gives:

7.49)x00.2)(x00.1(

)x2(K

2

p

HI(g)2 )g(I)g(H 22

65


– Solving for x.

– Because the right side of this equation is not a perfect square (sq over sq), you must solve the quadratic equation.

HI(g)2 )g(I)g(H 22 Initial 1.00 2.00 0

Change -x -x +2x

Equilibrium 1.00-x 2.00-x 2x

a

acbbx

cbxax

2

4

0

2

2

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49.7(1.00-x)(2.00-x) = (2x)2

49.7(2.00 - 3.00x + x2) = 4x2

99.4 - 149.1x + 49.7x2 = 4x2

45.7x2 - 149.1x + 99.4 = 0

7.49)x00.2)(x00.1(

)x2(K

2

p

a b c

a

acbbx

cbxax

2

4

0

2

2

)7.45(2

)4.99)(7.45(4)1.149()1.149( 2 x

4.91

32.18170)81.22230()1.149( x

4.91

49.4060)1.149( x

4.91

72.631.149 x

934.04.91

38.85x

33.24.91

82.212x

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– However, x = 2.33 gives a negative value to 1.00 - x (the equilibrium concentration of H2), which is not possible.

possible 0.934Only x If you do the shift correctly, you may 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 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 and selecting proper shift makes life easier (use Q when necessary).

68


• Solving for equilibrium pressures.

– If you substitute 0.934 for x in the last line of the table you obtain the following equilibrium concentrations.

HI(g)2 )g(I)g(H 22 Initial 1.00 2.00 0

Change -x -x +2x

Equilibrium 1.00-x 2.00-x 2x

HW 22code: two

69

31.5 Effect of Changing the Reaction Conditions Upon Equilibrium

• Obtaining the maximum amount of product from a reaction depends on the proper set of reaction conditions which gets us to:– Le Chatelier’s 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 until equilibrium is established again.

70

31.5.1 Adding or Removing a Species

• 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.

71

31.5.2 Compression or Expansion• 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 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.

72

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.

73

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.

74

SO2 (g) + 1/2 O2 (g) SO3 (g)

Increase P?

Shift to right toward smaller mols gas

N2 (g) + 3 H2 (g) 2NH3 (g)

Decrease P?

N2 (g) + O2 (g) 2NO (g)

Decrease P?

C (s) + H2O(g) CO (g) + H2 (g)

Increase P?

Shift to left toward larger mols gas

No shift

Shift to left toward smaller mols gas; note that C is a solid.

75

31.5.3 Temperature

• 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.

76

Effect of Temperature Change

• Let’s 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.

A + B C + D + heat exo

A + B + heat C + D endo

77


• 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)

negative) is H( heat""products reactants

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• 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)

positive) is H( products reactantsheat""

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• In summary:

– For an exothermic reaction (H is negative) the amounts of reactants are increased (shift left) at equilibrium by an increase in temperature (Kc is smaller at higher temperatures).

– For an endothermic reaction (H positive) the amounts of products are increased (shift right) at equilibrium by an increase in temperature (Kc is larger at higher temperatures).

Note: opposite occurs when lower temperature.

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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.

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

Add gas products, shift left until equil established.

Increase P, decrease V, shift towards smaller mols of gas; therefore to the left until equil.

Endo reaction, therefore similar to adding reactants and rxn shifts to the right

Producing more products (shift right) and less reactants, therefore larger K

No effect, if reaction wasn’t at equil just getting there faster.

code: reaction

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1 Chapter 31 Chemical Equilibrium Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved