Chapter 16: Chemical Equilibrium- General Concepts WHAT IS EQUILIBRIUM?
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Transcript of Chapter 16: Chemical Equilibrium- General Concepts WHAT IS EQUILIBRIUM?
![Page 1: Chapter 16: Chemical Equilibrium- General Concepts WHAT IS EQUILIBRIUM?](https://reader030.fdocuments.in/reader030/viewer/2022020106/56649cff5503460f949cff2c/html5/thumbnails/1.jpg)
Chapter 16: Chemical Equilibrium- General Concepts
• WHAT IS EQUILIBRIUM?
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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.
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• The equilibrium mixture is independent of whether we start on the “reactant side” or the “product side”
The equilibrium between N2O4 and NO2.
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The same equilibrium composition is reached from either the forward or reverse direction, provided the overall system composition is the same.
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• 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
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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.
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• 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
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• 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
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• 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
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• 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
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• 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
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– 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
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– 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
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• 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 (
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• 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
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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
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• 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
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• 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
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• 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
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– 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.
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• 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
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• 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
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– 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
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– 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
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• 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
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• 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
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• 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
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– 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
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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
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• 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
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• 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
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• 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
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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
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• Substituting:[N2]=0.033-x=0.033 M
[O2]=0.00810-x=0.030810 M
[NO]=2x=1.60x10-17 M