a more appropriate definition of k (for the same chemical reaction discussed previously) is with...
TRANSCRIPT
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A more appropriate definition of K (for the same chemical reaction discussed previously) is
With this definition, each concentration term is divided by the units of molarity, M, so K is then dimensionless.
ba
dc
[B]/M[A]/M
[D]/M[C]/MK
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A more appropriate definition of K (for the same chemical reaction discussed previously) is
With this definition, each concentration term is divided by the units of molarity, M, so K is then dimensionless. Because it is tedious to write the above equation, most people simply write
to save writing time.
ba
dc
[B]/M[A]/M
[D]/M[C]/MK
ba
dc
[B][A]
[D][C]K
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Relationship between Q and K
N2O4(g) 2 NO2(g)
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Different ways of expressing the equilibrium constant
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Different ways of expressing the equilibrium constant
Homogeneous equilibrium: A state of equilibrium between reactants and products in the same phase.
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Different ways of expressing the equilibrium constant
Homogeneous equilibrium: A state of equilibrium between reactants and products in the same phase.
Example: CH3CO2H(aq) CH3CO2-(aq) + H+
(aq)
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Different ways of expressing the equilibrium constant
Homogeneous equilibrium: A state of equilibrium between reactants and products in the same phase.
Example: CH3CO2H(aq) CH3CO2-(aq) + H+
(aq)
H]CO[CH][H]-CO[CHK
23
23c
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Example: For the gas phase reaction N2O4(g) 2 NO2(g)
]O[N][NOK42
22c
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Example: For the gas phase reaction N2O4(g) 2 NO2(g)
When working with gaseous reactions, it is frequently more convenient to use different units.
]O[N][NOK42
22c
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For the preceding reaction
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2
ON
2NO
p p
pK
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For the preceding reaction
The subscript p on K indicates partial pressures are used in the mass action expression.
42
2
ON
2NO
p p
pK
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For the preceding reaction
The subscript p on K indicates partial pressures are used in the mass action expression.
is the equilibrium partial pressure of N2O4.
is the equilibrium partial pressure of NO2.
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2
ON
2NO
p p
pK
42ONp
2NOp
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For the preceding reaction
The subscript p on K indicates partial pressures are used in the mass action expression.
is the equilibrium partial pressure of N2O4.
is the equilibrium partial pressure of NO2.
In general, Kp and Kc (for the same reaction) are not equal.
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2
ON
2NO
p p
pK
42ONp
2NOp
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Exercise: For the reaction a A(g) b B(g) find a connection between Kp and Kc. Assume the gases can be treated as ideal gases.
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Exercise: For the reaction a A(g) b B(g) find a connection between Kp and Kc. Assume the gases can be treated as ideal gases.
aA
bBp p
pK
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Exercise: For the reaction a A(g) b B(g) find a connection between Kp and Kc. Assume the gases can be treated as ideal gases.
Now from the ideal gas equation PV = n RT,
aA
bBp p
pK
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Exercise: For the reaction a A(g) b B(g) find a connection between Kp and Kc. Assume the gases can be treated as ideal gases.
Now from the ideal gas equation PV = n RT,
aA
bBp p
pK
[A]RTVRTnpA
AA [B]RT
VRTnpB
BB
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Exercise: For the reaction a A(g) b B(g) find a connection between Kp and Kc. Assume the gases can be treated as ideal gases.
Now from the ideal gas equation PV = n RT,
Plug these two results into the expression for Kp.
aA
bBp p
pK
[A]RTVRTnpA
AA [B]RT
VRTnpB
BB
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Hence:
aA
bBp p
pK
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Hence:
aA
bBp p
pK
a
b
p[A]RT
[B]RTK
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Hence:
aA
bBp p
pK
a
b
p[A]RT
[B]RTK
a
b
a
b
RT
RT[A][B]
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Hence:
aA
bBp p
pK
a
b
p[A]RT
[B]RTK
a
b
a
b
RT
RT[A][B]
a - ba
bRT
[A][B]
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Hence:
where
aA
bBp p
pK
a
b
p[A]RT
[B]RTK
a
b
a
b
RT
RT[A][B]
a - ba
bRT
[A][B]
abΔn na
bRT
[A][B]
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Hence:
where and therefore
aA
bBp p
pK
a
b
p[A]RT
[B]RTK
a
b
a
b
RT
RT[A][B]
a - ba
bRT
[A][B]
abΔn na
bRT
[A][B]
ncp RTKK
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Sample Problems
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Sample Problems Example: 2 NO(g) + O2(g) 2 NO2(g)
At a temperature of 230 oC the concentrations of the various species are [NO] = 0.0542 M, [O2] = 0.127 M, and [NO2] = 15.5 M at equilibrium. Calculate the equilibrium constant Kc at 230 oC.
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Sample Problems Example: 2 NO(g) + O2(g) 2 NO2(g)
At a temperature of 230 oC the concentrations of the various species are [NO] = 0.0542 M, [O2] = 0.127 M, and [NO2] = 15.5 M at equilibrium. Calculate the equilibrium constant Kc at 230 oC.
][O[NO]][NOK
22
22c
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Sample Problems Example: 2 NO(g) + O2(g) 2 NO2(g)
At a temperature of 230 oC the concentrations of the various species are [NO] = 0.0542 M, [O2] = 0.127 M, and [NO2] = 15.5 M at equilibrium. Calculate the equilibrium constant Kc at 230 oC.
= 6.44 x 105
][O[NO]][NOK
22
22c
(0.127)(0.0542)(15.5)
2
2
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Example: The equilibrium constant Kp for the reaction PCl5 (g) PCl3(g) + Cl2(g) is 1.05 at 250 oC.
If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2?
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Example: The equilibrium constant Kp for the reaction PCl5 (g) PCl3(g) + Cl2(g) is 1.05 at 250 oC.
If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2?
5
23
PCl
ClPClp p
ppK
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Example: The equilibrium constant Kp for the reaction PCl5 (g) PCl3(g) + Cl2(g) is 1.05 at 250 oC.
If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2?
This is shorthand for
5
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PCl
ClPClp p
ppK
atmp
atmpatmpK
5
23
PCl
ClPClp
/
//
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Example: The equilibrium constant Kp for the reaction PCl5 (g) PCl3(g) + Cl2(g) is 1.05 at 250 oC.
If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2?
This is shorthand for
so that
5
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PCl
ClPClp p
ppK
atmp
atmpatmpK
5
23
PCl
ClPClp
/
//
atmp
Katmpatmp
3
5
2PCl
pPClCl
/
//
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and so atm1.98p2Cl
atmatm0.463
(1.05)atmatm0.875atmp
2Cl /
//
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Heterogeneous Equilibria
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Heterogeneous Equilibria A reaction often involves reactants that are not
present in the same phase – this leads to a heterogeneous equilibrium.
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Heterogeneous Equilibria A reaction often involves reactants that are not
present in the same phase – this leads to a heterogeneous equilibrium.
Example: Heating CaCO3 in a closed vessel.
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Heterogeneous Equilibria A reaction often involves reactants that are not
present in the same phase – this leads to a heterogeneous equilibrium.
Example: Heating CaCO3 in a closed vessel.
CaCO3(s) CaO(s) + CO2(g)
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Heterogeneous Equilibria A reaction often involves reactants that are not
present in the same phase – this leads to a heterogeneous equilibrium.
Example: Heating CaCO3 in a closed vessel.
CaCO3(s) CaO(s) + CO2(g)
][CaCO][CO[CaO]K
3
2c
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The “concentration” of any pure solid is the ratio of the total number of moles present in the solid, divided by the volume of the solid.
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The “concentration” of any pure solid is the ratio of the total number of moles present in the solid, divided by the volume of the solid. If part of the solid is removed, the number of moles of solid will decrease – but so will it volume.
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The “concentration” of any pure solid is the ratio of the total number of moles present in the solid, divided by the volume of the solid. If part of the solid is removed, the number of moles of solid will decrease – but so will it volume. The ratio of moles to volume remains unchanged.
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Consider the ratio:
molesgramslitersgrams
mass molardensity
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Consider the ratio:
molesgramslitersgrams
mass molardensity
litersmoles
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Consider the ratio:
molesgramslitersgrams
mass molardensity
litersmoles
3
33 CaCOmass molar
CaCOdensity][CaCO
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Consider the ratio:
If the temperature is held fixed, then [CaCO3] is a constant. Similarly for [CaO], which is also a constant.
molesgramslitersgrams
mass molardensity
litersmoles
3
33 CaCOmass molar
CaCOdensity][CaCO
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So we can rewrite
][CaCO][CO[CaO]K
3
2c
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So we can rewrite
in the form
][CO[CaO]
][CaCOK 23c
][CaCO][CO[CaO]K
3
2c
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So we can rewrite
in the form
Now since [CaCO3] and [CaO] are both constant, the left-hand side of the preceding equation is constant.
][CO[CaO]
][CaCOK 23c
][CaCO][CO[CaO]K
3
2c
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So we can rewrite
in the form
Now since [CaCO3] and [CaO] are both constant, the left-hand side of the preceding equation is constant. Now set
][CO[CaO]
][CaCOK 23c
][CaCO][CO[CaO]K
3
2c
[CaO]
][CaCOKK 3cc
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Hence Kc = [CO2]
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Hence Kc = [CO2]
Notice that terms involving pure solids do not appear in the final equilibrium constant expression.
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Hence Kc = [CO2]
Notice that terms involving pure solids do not appear in the final equilibrium constant expression. This result generalizes to all chemical reactions.
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Hence Kc = [CO2]
Notice that terms involving pure solids do not appear in the final equilibrium constant expression. This result generalizes to all chemical reactions.
The corresponding expression for Kp for the decomposition of CaCO3 is:
Kp = 2COp
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The same concentration of CO2 exists in both containers (provided the temperature is the same), even though the amounts of CaO and CaCO3 are different.
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Summary Comment
Concentration factors for pure solids and pure liquids are ignored in the expression for the equilibrium constant for a chemical reaction.
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Multiple Equilibria
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Multiple Equilibria
Consider the two reactions:
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Multiple Equilibria
Consider the two reactions: A + B C + D C + D E + F
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Multiple Equilibria
Consider the two reactions: A + B C + D C + D E + F
For the first reaction
[A][B][C][D]Kc