Equilibrium AP Chem Mr. Nelson. The Concept of Equilibrium The term reversible reaction is used to...
-
Upload
gwenda-stevens -
Category
Documents
-
view
217 -
download
0
Transcript of Equilibrium AP Chem Mr. Nelson. The Concept of Equilibrium The term reversible reaction is used to...
EquilibriumAP Chem
Mr. Nelson
The Concept of EquilibriumThe term reversible reaction is used to describe reactions that can go in either the forward or the reverse direction
Equilibrium is used to specify that a reversible reaction has reached an equal rate for both directions
The Concept of EquilibriumChemical equilibrium occurs when opposing reactions are proceeding at equal rates◦Concentrations of the reactants/products remains constant
The Concept of Equilibrium
Its important to note that forward and reverse reactions still occur, although outwardly no apparent change occurs
)(2)(3)( 322 gNHgHgN
The Concept of EquilibriumNote that the equilibrium concentrations are independent of initial concentrations
A mathematical relationship exists between concentration of the reactants and products once equilibrium has been reached
The Concept of EquilibriumFish Tank Example
Connection between Equilibrium & KineticsFor generalized reaction:
A⇋BForward reaction:
AB◦Forward Rate Law: Ratef = kf[A]
Reverse reaction:
BA◦Reverse Rate Law: Rater = kr[B]
Connection between Equilibrium & KineticsBy definition, at equilibrium
forward and reverse rates must be equal, so:
Ratef = Rater
kf[A] = kr[B]Also, concentrations remain
constant,so: eqK constant A
B
r
f
k
k
The Equilibrium ConstantThe equilibrium constant (Keq)
can be determined at constant temperature for any reaction at equilibrium:
aA+ bB cC + dDKeq =
[C]c [D]d
[A]a [B]b
Keq is unitless
The Equilibrium ConstantNote that only equilibrium
concentrations can be placed into an equilibrium constant expression!
Also, no solids or liquids are included in an equilibrium constant expression
The reason for this is their concentrations remain fairly constant in equilibrium
Note that they are still required for the reaction to proceed however
The Equilibrium Constant
The Many Types of
Equilibrium Constants, Keq
Kc
Kp
KspKa/Kb
Kw
Concentration vs. Pressure EquilibriaEquilibrium expressions can be
written in terms of concentration (Kc) measured in molarity or pressure (Kp) measured in atm◦Example:
)(2)( 242 gNOgON
][
][
42
22
ON
NOKc
42
2
2
ON
NO
p P
PK or
Concentration vs. Pressure EquilibriaIn general, Kc is not equal to Kp
The following equation converts between the two:
n = moles of gas products – moles of gas reactants◦Note: When no change in moles,
Kp=Kc
ncp RTKK
Homogeneous vs. Heterogeneous Equilibria
The term homogeneous equilibrium is used to describe reactions where all the species are in the same phase◦ Example:
The term heterogeneous equilibrium is used to describe reactions where species are in different phases◦ Example:
H2CO3 (aq) + H2O (l) HCO3- (aq) + H3O+
(aq)
)(2)( 242 gNOgON
Multiple EquilibriaWhen the products of one
reaction are the reactants of another
Example: A + B C + D
C + D E + F
K1
A + B E + F
Kc
K2
Multiple EquilibriaUsing their equilibrium
expressions, we can derive a mathematical relationship:
]][[
]][[
]][[
]][[
]][[
]][[
DC
FEx
BA
DC
BA
FEKc
21 KKKc
Manipulating Equilibria Expressions
If an equilibrium is written in the reverse, the following relationship exists:
]][[
][
23
5
ClPCl
PClKc
][
]][[
5
23
PCl
ClPClKc
cc KK /1
PCl3 + Cl2 PCl5
PCl5 PCl3 + Cl2
Manipulating Equilibria Expressions
If an equilibrium is multiplied by a factor, the following relationship exists:
]][[
][
23
5
ClPCl
PClKc PCl3 + Cl2 PCl5
2PCl3 + 2Cl2 2PCl5 22
23
25
][][
][
ClPCl
PClKc
2cc KK
The Magnititude of Equilibrium Constants
What does Keq tell us?
◦If Keq is very large, products will be favored This implies the numerator is large in the Keq
expression compared to denominator
◦If Keq is very small, reactants will be favored This implies the denominator is large in the
Keq expression compared to numerator
◦ If Keq is close to 1, then roughly equal amounts of reactants and products are present at equilibrium
The Reaction Quotient (Q)
Recall: In electrochemistry, we utilized “Q” which is similar in all respects to “Keq” except in definition◦Q is utilized when given initial concentrations, not concentrations at equilibrium
Predicting the Direction of Reversible ReactionsIf given an initial set of conditions for a reversible reaction, utilizing Q and Keq, a prediction of the direction of the reaction (towards reactants/towards products) can be made
Comparing Kc and Qc
Qc<Kc
• Ratio of prds to rcts is too small• To reach equilibrium, rcts must go
towards prds• Shifts to the right
Qc=Kc•Initial concentrations are at equilibrium concentrations•System is at equilibrium
Qc>Kc
• Ratio of prds to rcts is too large• To reach equilibrium, prds must go
towards rcts• Shifts to the left
Comparison of Kc and Qc
Reactants → Products
Equilibrium: no net change
Reactants ← Products
Qc
Kc KcKc
Qc
Qc
Calculating Equilibrium ConcentrationsTabulate all known initial and
equilibrium concentrations◦When both are known, calculate the
changeUsing coefficients, calculate
changes in concentration for all participants in a rxn
Calculate equilibrium concentrations
The ICE MethodThis is the process used for
determining equilibrium concentrations
Initial molarityChange in molarityEquilibrium molarity
Example #1: Calculating Equilibrium Concentrations
A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00 L stainless-steel flask at 430 °C. The equilibrium constant Kc for the reaction:
H2 (g) + I2 (g) 2 HI (g)
is 54.3 at this temp. Calculate the concentrations of H2, I2, and HI at equilibrium.
Initial (M)0.500 0.500 0
Change (M) -x -x +2x
Equilibrium (M) 0.500 – x 0.500 – x 2x
H2 (g) + I2 (g) 2 HI (g)
]][[
][3.54
22
2
IH
HIKc
Substituting, we get:
Taking the square root of both sides:
)500.0)(500.0(
)2(3.54
2
xx
x
)500.0(
237.7
x
x
x = 0.393 M
So, the concentrations at equilibrium would be the following:◦[H2] = (0.500 – 0.393) M = 0.107 M
◦[I2] = (0.500 – 0.393) M = 0.107 M
◦[HI] = 2 x 0.393 M = 0.786 M
Example #2: Calculating Equilibrium ConcentrationsWhen 3.0 mol of I2 and 4.0 mol of
Br2 are placed in a 2.0 L reactor at 150oC, the following reaction occurs until equilibrium is reached:
I2(g) + Br2(g) 2IBr(g)
Chemical analysis then shows that the reactor contains 3.2 mol of IBr. What is the value of the equilibrium constant Kc for the reaction?
Initial (M)1.5 2 0
Change (M) -x -x +2x
Equilibrium (M) 1.5 – x 2 – x 1.6
I2 (g) + Br2 (g) 2 IBr (g)
]][[
][
22
2
BrI
IBrKc
When initial and final concentrations are known, calculate the change:◦0 + 2x = 1.6◦x = 0.8
Then determine all equilibrium concentrations:◦[I2] = 1.5 – 0.8 = 0.7 M
◦[Br2] = 2 – 0.8 = 1.2 M
◦[IBr] = 1.6 M
Plug equilibrium concentrations into equation and solve:
]][[
][
22
2
BrI
IBrKc
05.3]2.1][7.0[
]6.1[ 2
cK
Calculating Equilibrium Concentrations when Keq is small
When solving equilibrium concentrations and Keq is small, assume the effect of the change (“x”) is negligible (for rcts only)
Example: Calculating Equilibrium Concentrations when Keq is small
At 100 °C the equilibrium constant (Kc) for the reaction:
COCl2 (g) 2 CO (g) + Cl2 (g)
is 2.19 x 10-10. If the initial concentration of [COCl2] = 1.5 M, calculate the concentrations of all species at equilibrium.
COCl2 (g) 2 CO (g) + Cl2 (g)Initial (M)
1.5 0 0
Change (M) -x +2x +x
Equilibrium (M) 1.5 – x +2x +x
][
][][
2
22
COCl
ClCOKc
)5.1(
)()2( 2
x
xxKc
)5.1(
)()2( 2 xx
5.1
41019.2
310 x
x
4
4
4
10285.3 310 xx
3 33 11102125.8 xx
Le Chatelier’s Principle
If a stress is applied to a system in equilibrium, the equilibrium shifts to relieve that stress
Forms of “stress”:◦Concentration◦Pressure◦Temperature
Change in Reactant or Product ConcentrationAdding a substance shifts a system at equilibrium away from the added substance◦System must consume added item
Removing a substance from a system at equilibrium shifts the reaction toward the removed substance◦System must replace removed item
Changes in Pressure
Increase in pressure forces an equilibrium to shift in the direction that reduces the number of moles of gas in the system◦Example:◦An increase in pressure causes a shift to the left
N2O4 (g) 2 NO2 (g)
Changes in PressureRecall: Changes in volume can also
affect changes in pressure◦Increase volume = decrease pressure◦Decrease volume = increase pressure
Also the addition of an inert gas (one not in the reaction) can cause an increase in pressure as well
Changes in TemperatureEnthalpy of the reaction plays a major roll in how heat effects the equilibrium◦Recall: Endothermic (+∆H) means heat is a reactant, while exothermic (- ∆H) means it is a product
Treat changes in heat in the same manner that you treat changes in concentration
Effects of Catalysts on EquilibriumRecall: Catalysts lower energy
barrier between reactants and products◦Ea for both forward and reverse are
lowered to the same extentCatalysts increase the rate at
which equilibrium is reachedHowever, it does NOT change the
composition of the equilibrium mixture.
Solubility Equilibria
Saturated solutions of salts are another type of chemical equilibria
Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and anions in solution
The Solubility Product, Ksp
The equilibrium constant, the Ksp, is no more than the product of the ions in solution (Recall: solids do not appear)
For a saturated solution of AgCl, the equation would be:
The solubility product expression would be:
)()()( aqClaqAgsAgCl
]][[ ClAgK sp
Sequence of Steps for Solubility Equilibria
Solubility
of compoun
d
Molar solubility of
compound
Concentrations of cations and anion
s
Ksp of
compoundKsp
of compound
Concentrations of cations and anion
s
Molar solubility of
compound
Solubility
of compoun
d
SolubilitySolubility can be expressed two ways:◦Molar solubility is the moles of solute per 1L of saturated solution (mol/L)
◦Solubility is the number of grams of solute per 1 L of saturated solution (g/L)
Ksp expressions require molar solubility
Determining Ksp by Experimental MeasurementsUse ICE table, but with “s” for change (instead of “x”)
“s” is not only the change, but the molar solubility
Determining Ksp: Example #1
The solubility of CaSO4 is found to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate.
Solubility of CaSO4 in g/L
Molar solubility of CaSO
4
[Ca2
+] and [SO4
2-]
Ksp of
CaSO4
Determining Ksp: Example #1
)()()( 24
24 aqSOaqCasCaSO
Initial (M)0 0
Change (M) -s +s +s
Equilibrium (M) s s
224
2 ]][[ sSOCaK sp
Determining Ksp: Example #1
As CaSO4 dissolves, we get 1 mole Ca2+, and 1 mole SO4
2-, so at equilibrium:
sMxCaSOg
CaSOmol
solutionL
CaSOg 3
4
44 109.42.136
1
1
67.0
MxSOCa 324
2 109.4][][ 533 104.2)109.4)(109.4( xxxK sp
Determining Ksp: Example #2
Lead (II) chloride dissolves to a slight extent in water according to the equation:
Calculate the Ksp if the solubility of lead (II) chloride has been found to be 1.62 x 10-2 M
)(2)()( 22 aqClaqPbsPbCl
Determining Ksp: Example #2Molar
solubility of PbCl2
[Pb2+
] and [Cl-]
Ksp of PbCl2
)(2)()( 22 aqClaqPbsPbCl
Initial (M)0 0
Change (M) -s +s +2s
Equilibrium (M) s 2s
Determining Ksp: Example #2
Mxs 5108.1 35 )108.1(4 xK sp
141033.2 xK sp
3222 4)2)((]][[ sssClPbK sp
Calculating Solubility from Ksp: Example #1
The Ksp for CaCO3 is 3.8 x 10-9 at 25 °C. Calculate the solubility (in g/L) of calcium carbonate in pure water.
Ksp of CaCO3
Concentrations of [Ca2+]
and [CO32-]
Molar solubility of
CaCO3
Solubility of CaCO3
Calculating Solubility from Ksp: Example #1
223
2 ]][[ sCOCaK sp
)()()( 23
23 aqCOaqCasCaCO
Initial (M)0 0
Change (M) -s +s +s
Equilibrium (M) s s
Calculating Solubility from Ksp: Example #1
3
335
1
100
1
1016.6
CaCOmol
CaCOg
solutionL
CaCOmolx
29108.3 sx
Mxs 51016.6
Lgx /1016.6 3
Predicting Precipitation Reactions using QUsing the reaction quotient, Q, we can
◦Decide whether a ppt will form◦Calculate ion concentration required
to begin the ppt. of an insoluble salt
Q < Ksp
• System has too much reactant
• No ppt
Q = Ksp
• System at equilibrium
• No ppt
Q > Ksp
• System has too much product
• Forms ppt!
Predicting Precipitation ReactionsSteps
1. Determine insoluble product using solubility rules
2. Calculate cation and anion concentration of insoluble product (if mixed)
3. Calculate Q4. Compare to Ksp
Example: Predicting Precipitation
Exactly 200 mL of 0.0040 M BaCl2 are mixed with exactly 600 mL of 0.0080 M K2SO4. Will a precipitate form? (Ksp of insoluble salt is 1.1 x 10-10)
The Common Ion EffectLe Chatelier’s Principle can describe the effect of an added common ion◦For example, if NaCl were added to a saturated solution of PbCl2, what effect would occur?
◦Increase product concentration shifts equilibrium towards reactants, forming more ppt.
Solubility and the Common Ion EffectThe presence
of a common ion will cause the equilibrium to shift so that even less of the substance (with smaller Ksp value) will dissolve
The Common Ion EffectBe aware: pH can affect solubility
◦The addition of an acid or base could have a side reaction, decreasing concentration of an ion
◦If an acid were added to the following, what effect would be observed?
H+ combines with OH-, so [OH-] is decreased, forcing equilibrium to the right
)(2)()()( 22 aqOHaqMgsOHMg
Example: Common Ion Effect
Calculate the solubility of silver chloride (in g/L) in a 6.5 x 10-3 M silver nitrate solution. (Ksp = 1.6 x 10-10)
◦Since silver nitrate is a soluble, strong electrolyte: )()()( 33
2 aqNOaqAgsAgNO OH
6.5 x 10-3 M
6.5 x 10-3 M
Example: Common Ion Effect
)()()( aqClaqAgsAgCl Initial (M)
6.5 x 10-3 0.00
Change (M) -s +s +s
Equilibrium (M)
(6.5 x 10-3 + s)
s
sxxK sp )105.6(106.1 310
Because Ksp is so small, s must be very small in comparison to 6.5 x 10-3, so we approximate that 6.5 x 10-3 + s 6.5 x 10-3
Mxs 8105.2
Selective PrecipitationIn certain problems, you will have
to determine which of two possible precipitates will form◦The salt with the lower Ksp will
precipitate firstTo determine the amount of an ion
needed to precipitate, recall that Ksp concentrations are saturated, so anything greater than this will form a ppt.
Example: Selective Precipitation
A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp = 1.4 x 10-8) or CuI (Ksp = 5.3 x 10-12) precipitate first? Specify the concentration of I- necessary to begin precipitation of each salt