Chapter 16: Chemical Equilibrium- General Concepts

• WHAT IS EQUILIBRIUM?

The decomposition of N2O4(g) into NO2(g). The concentrations of N2O4 and NO2 change relatively quickly at first, but eventually stop changing with time when equilibrium is reached.

• The equilibrium mixture is independent of whether we start on the “reactant side” or the “product side”

The equilibrium between N2O4 and NO2.

The same equilibrium composition is reached from either the forward or reverse direction, provided the overall system composition is the same.

• There is a simple relationship among the concentrations of the reactants and products for any chemical system at equilibrium

• Consider the equilibrium:

)(HI2)(I)(H 22 ggg

Four experiments to study the equilibrium among H2, I2, and HI gases. Different amounts of the reactants and products are placed in a 10.0 L reaction vessel at 440oC where the gases establish equilibrium. When equilibrium is reached, different amounts of reactants and products remain.

• This average value is called the reaction quotient, Q

49.5 Average

49.5 0.311 0.0442 0.0442 IV

49.4 0.100 0.0135 0.0150 III

49.8 0.280 0.0450 0.0350 II

49.4 0.156 0.0222 0.0222 I

]][I[H

[HI] ][HI ][I ][H Expt.

22

2

22

• The reaction can be evaluated at any concentrations

• At equilibrium (and 440oC) for this reaction the reaction quotient has the value 49.5 (a unitless number)

• This relationship is called the equilibrium law for the system

]I][H[

[HI]

22

2

Q

• The value 49.5 is called the equilibrium constant, Kc, and characterizes the system

• For chemical equilibrium to exist, the reaction quotient Q must be equal to the equilibrium constant Kc

• Consider the general chemical equation

C)440(at 5.49]I][H[

[HI] o

22

2

cK

gGfFeEdD

• The exponents in the mass action expression are the same as the stoichiometric coefficients

• At equilibrium

• The form is always “products over reactants” raised to the appropriate powers

c][][

][][K

ED

GFed

gf

• Various operations can be performed on equilibrium expressions– Changing the direction of equilibrium – when

the direction of an equilibrium is reversed, the new equilibrium constant is the reciprocal of the original

c

c

KPCl

ClPClKClPClPCl

ClPCl

PClKPClClPCl

c

1

][

]][[

]][[

][

5

23'235

23

5523

– Multiplying the coefficients by a factor – when the coefficients in an equation are multiplied by a factor, the equilibrium constant is raised to a power equal to that factor

22

5

22

23"

235

23

5523

][

][][ 222

]][[

][

c

c

KPCl

ClPClKClPClPCl

ClPCl

PClKPClClPCl

c

– Adding chemical equilibria – when chemical equilibria are added, their equilibrium constants are multiplied

2142

22

42

3222

32

22

42

2222

22

2

22

1222

][][

][ 442

:gives Adding

][][

][ 432

][][

][ 22

ccc

c

c

KKON

NOKNOON

OON

NOKNOOON

ON

ONKONON

• The gas law can be used to write the equilibrium constant in terms of partial pressures

• Equilibrium constants written in terms of partial pressures are given the symbol Kp

RTRTV

nP

nRTPV

ion)concentratmolar (

• The size of the equilibrium constant gives a measure of how the reaction proceeds

• General statements can be made about the equilibrium constant (either Kc or KP)

and ]][[

][

)(2)(3)(

3

2

322

23

322

22

3

HN

NHPc PP

PK

HN

NHK

gNHgHgN

The magnitude of K and the position of equilibrium. A large amount of product and very little reactant at equilibrium gives K>>1 (large K). When , approximately equal amounts of reactant and product are present at equilibrium. When K<<1, mostly reactant and very little product are present at equilibrium.

1K

• The two different forms of the equilibrium constants can be related

reactants) gaseous of (moles

products) gaseous of moles(n

where)(

thatso ion)concentratmolar (

g

-

RTKK

RTRTV

nP

nRTPV

gncP

• In a homogeneous reactions, all the reactants and products are in the same phase

• Heterogeneous reactions involve more than one phase

• For example the thermal decomposition of sodium bicarbonate (baking soda)

• Heterogeneous reactions can come to equilibrium just like homogeneous systems

)()()()(2 22323 gCOgOHsCONasNaHCO heat

• If NaHCO3 is placed in a sealed container, homogeneous equilibrium is established

• The equilibrium law involving pure liquids and pure solids can be simplified

KNaHCO

OHCOCONa

gOHgCOsCONasNaHCO

23

2232

22323

][

]][][[

:isconstant mequilibriu The

)()()()(2

– For a pure liquid or solid, the ratio of amount of substance to volume of substance is constant

The concentration of a substance in a solid is constant. Doubling the number of moles doubles the volume, but the ratio of moles to volume remains the same.

• The equilibrium law for a heterogeneous reaction is written without concentrations terms for pure solids or pure liquids.

• The equilibrium constants found in tables represent all the constants combined

][

][

]][[

32

23

22

CONa

NaHCOK

OHCOKc

• According to Le Châtelier’s principle: If an outside influence upsets an equilibrium, the

system undergoes a change in the direction that counteracts the disturbing influence and, if possible, returns the system to equilibrium

• We can consider some common “stresses” – Adding or removing a product or reactant

• The equilibrium shifts to remove reactants or products that have been added

• The equilibrium shifts to replace reactants or products that have been removed

– Changing the volume• Reducing the volume of a gaseous reaction causes

the reaction to decreases the number of molecules of gas, if it can

• Moderate pressure changes have a negligible effect on reactions involving only liquids or solids

– Changing the temperature• Increasing the temperature shifts a reaction in a

direction that produces an endothermic (heat-absorbing) change

• Decreasing the temperature shifts a reaction in a direction that produces an exothermic (heat-releasing) change

– Catalysts have no effect on the position of equilibrium

• Catalysts change how fast a system achieves equilibrium, not the relative distribution of reactants and products

– Adding an inert gas at constant volume• If the added gas cannot react with any reactants or

products it is inert towards the substances in the equilibrium

• No concentration changes occur, so Q still equals K and no shift in equilibrium occurs

• Equilibrium calculations can be divided into two main categories:

1) Calculating equilibrium constants from known equilibrium concentrations or partial pressures

2) Calculating one or more equilibrium concentrations or partial pressures using the known value of Kc or KP

• Consider the decomposition of N2O4

• Calculating the equilibrium constant this way is easy

32

42

22

c

2

42

o

42

242

1061.4(0.0292)

)0116.0(

][

][

and mol/L, 0116.0][

mol/L 0292.0][

mequilibriuat C,25at flask

liter 1 ain placed is mol 0.0350 If

)(2)(

ON

NOK

NO

ON

ON

gNOgON

• More commonly, you will have a set of initial conditions and an equilibrium constant

• If a KP describes the system, equilibrium will usually be described in terms of partial pressures

• If a Kc describes the system, equilibrium will usually be describe in terms of concentration (molarity, mol/L)

• The Initial, Change, Equilibrium or “ICE” table is a useful way to summarize the problem

– Example: Ethyl acetate, CH3CO2C2H5, is produced from acetic acid and ethanol by the reaction

At 25oC, Kc=4.10 for this reaction. Suppose 0.100 mol of ethyl acetate and 0.150 mol of water are placed in a 1.00 L reaction vessel. What are the concentrations of all species at equilibrium?

ANALYSIS: Use an ICE table and the equilibrium constant to find the concentrations.

)()()()( 252235223 lOHlHCCOCHlOHHClHCOCH

x

x0.250x-0.0150

x

x)-x)(0.150-(0.1004.10

x-0.150x -0.100 x x )( E

x- x - x x )( C

0.150 0.100 0 0 )( I

thenreacts,ion that concentrat

Let phase. liquid in the are species All

2

2

2c

252235222

K

M

M

M

OHHCCOCHOHHCHCOCH

x

• This can be solved by putting it in quadratic form:

:system For this2

4 ,0

form in theequation an For

22

a

acbbxcbxax

0401.0 xand 121.0x

and 0 0.0150- x 0.250 x3.10

or 4.10x

x0.250x-0.0150

-

2

2

2

c

K

• Negative concentrations are not allowed, so

• A similar procedure can be used to calculate partial pressures using KP

MxOH

MxHCCOCH

MxOHHC

MxHCOCH

x

011.0150.0][

060.0100.0][

0401.0][

0401.0][

mequilibriuat and 0401.0

2

5223

52

23

• Sometime simplifications can be madeExample: Nitrogen and oxygen react to form

nitrogen monoxide

with Kc=4.8x10-31. In air at 25oC and 1 atm, the N2 concentrations and O2 are initially 0.033 M and 0.00810 M. What are the equilibrium concentrations?

ANALYSIS: The equilibrium constant is very small, very little of the reactants will be converted into products

)(2)()( 22 gNOgOgN

18-

2

2

31c

22

108.01or x 00810)(0.033)(0.

4x

,x)-x)(0.00810-(0.033

(2x)

108.4

2xx -0.00810x -0.033 )E(

x x x )C(

0 0.00810 0.033 )( I

2

K

M

M

M

NOON

• Substituting:[N2]=0.033-x=0.033 M

[O2]=0.00810-x=0.030810 M

[NO]=2x=1.60x10-17 M