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Manipulating Keq
Lecture 16
There are three ways in which Keq
can be manipulated.
Equilibrium II
1) Coefficient Rule
Manipulating Keq
The Reaction Quotient (Q)
Le Chatelier’s Principle
2) Reciprocal Rule
3) Multiple Equilibria Rule
Coefficient Rule
Reciprocal Rule
When coefficients are changed by a factor of
n, Keq is raised to the power of n.
When a reaction is reversed the new Keq value
is the inverse of the old Keq value.
(i)
(i)
2A(g) + B(aq)
6D(aq)
2A(g) + B(aq)
6
K ci =
½(i) = (ii)
[D]
[A]2 [B]
A(g) + ½ B(aq)
K ci =
Kci =
3D(aq)
⎡ [D] ⎤
=⎢ 2 ⎥
[A][B] ⎣ [A] [B] ⎦
3
6
[D]
(i)-1 = (iii)
1
2
Multiple Equilibria Rule
When two or more reactions are combined,
the new Keq is the product of the Keq
values from the individual reactions.
2A(g) + B(aq)
3E(aq) + B(aq)
3E(aq) + 2B(aq)
[D]6
[A]2 [B]
x
Kci
x
[A]2 [F]
Kciii =
[D]6
[A]2 [B]
2A(g) + B(aq)
[A]2 [B]
[D]6
Kciii = 1 / Kci
Kcii = (Kci)½
(i)
(iv)
(v)
6D(aq)
1
2
6D(aq)
6D(aq)
2A(g) + F(aq)
6D(aq) + F(aq)
[D]6 [F]
[E]3 [B]
=
[E]3 [B]2
Kciv
=
Kcv
Ex) The Reaction Quotient (Q)
How can we tell if a system is at equilibrium?
Ex) Kc = 0.042 at 227oC for the following
reaction:
SO2(g) + Cl2(g)
SO2Cl2(g)
A chemist tests this system and finds that:
[SO2] = 4.62 x 10-4 M, [Cl2] = 3.85 x 10-3 M,
[SO2Cl2] = 2.78 x 10-5 M
a) Is the system at equilibrium?
b) If not, in which direction will it proceed?
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Ex) The Reaction Quotient (Q) (cont)
(Q) is used to determine if the reaction is at
equilibrium.
Q=
[SO2Cl2]
[SO2] [Cl2]
(2.78 x 10-5)
=
SO2(g) + Cl2(g)
The system proceeds in this direction until
Q = Kc
(4.62 x 10-4) (3.85 x 10-3)
Q = 15.6
Q > Kc ; the system still has too many products.
The reaction will proceed to the left until the
system reaches equilibrium.
The Reaction Quotient (Q)
Q > Keq (The reaction will proceed to the left.)
» It will continue to increase the concentration of
reactants until equilibrium is established.
Q < Keq (The reaction will proceed to the right.)
» It will continue to increase the concentration of
products until equilibrium is established.
Q = Keq (The system is at equilibrium.)
Stress from Increasing Pressure
[SO2] and [Cl2] increase, and [SO2Cl2] decreases
until the system reaches equilibrium.
Le Chatelier’s Principle
When a system at equilibrium is subjected
to a stress, the equilibrium will shift in
order to reduce that stress.
The only three stresses are, changes in:
1) Pressure
2) Concentrations
3) Temperature
Stress from Increasing Pressure
N2(g) + 3H2(g)
If the pressure on a system at equilibrium is
increased the equilibrium will shift toward
the side with fewer moles of gas to reduce
that stress.
This changes the equilibrium concentrations, but it
does not change the Equilibrium Constant (Keq).
SO2Cl2(g)
2 NH3(g)
Equilibrium shifts this way
14 gaseous molecules
10 gaseous molecules
2 NH3, 5 N2, 7 H2
6 NH3, 3 N2, 1 H2
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Stress from Decreasing Pressure
If the pressure on a system at equilibrium is
decreased the equilibrium will shift toward
the side with more moles of gas to reduce
that stress.
This changes the equilibrium concentrations, but it
does not change the Equilibrium Constant (Keq).
Stress from Increasing Concentrations
Stress from Decreasing Pressure
N2(g) + 3H2(g)
2 NH3(g)
Equilibrium shifts this way
10 gaseous molecules
14 gaseous molecules
Adding more NH3 to the system
N2(g) + 3H2(g)
2 NH3(g)
Equilibrium shifts this way
If the concentration of one of the species in an
equilibrium system is increased, the
equilibrium will shift in the direction that will
reduce the concentration of that species.
This increases the equilibrium concentrations, but it
does not change the Equilibrium Constant (Keq).
Stress from Decreasing Concentrations
If the concentration of one of the species in an
equilibrium system is decreased, the
equilibrium will shift in the direction that will
increase the concentration of that species.
This decreases the equilibrium concentrations, but it
does not change the Equilibrium Constant (Keq).
at equilibrium
add 5 NH3(g)
back at equilibrium
no longer at equilibrium
1 NH3, 1 N2, 3 H2
6 NH3, 1 N2, 3 H2
4 NH3, 2 N2, 6 H2
Removing NH3 from the system
N2(g) + 3H2(g)
2 NH3(g)
Equilibrium shifts this way
at equilibrium
remove 7 NH3(g)
back at equilibrium
no longer at equilibrium
11 NH3, 4 N2, 12 H2
4 NH3, 4 N2, 12 H2
6 NH3, 3 N2, 9 H2
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Ex) Q and Le Chatelier
Stress from Changing Temperature
This is the only stress that changes the
equilibrium constant (Keq).
N2(g) + 3 H2(g)
2 NH3(g) + heat
Cooling (taking heat away) shifts the equilibrium
in the direction that produces heat.
N2(g) + 3 H2(g)
2 NH3(g) +heat
Adding heat (increasing the temperature) shifts the
equilibrium in the direction that absorbs heat.
Ex) Q and Le Chatelier (cont)
N2(g) + O2(g)
2 NO(g)
Ex) A chemist mixes 80% N2(g) and 20 % O2(g), by
moles, in a vacuum and find that the total
pressure is 1.0 atm at 25oC. (Kp = 1.0 x 10-30)
a) Find the initial partial pressure of each gas
b) Find the equilibrium partial pressures at 25oC.
c) If the temperature increases to 2127 oC
(Kp = 0.0025), find the new equilibrium
concentrations.
Ex) Q and Le Chatelier (cont)
N2(g) + O2(g)
a) Find initial partial pressures
2 NO(g)
b) Find the equilibrium partial pressures
PN = 1.0 atm x 0.80 = 0.80 atm
PN
PO
PNO
Initial
0.80
0.20
0
Change
-x
-x
+2x
0.20 - x
2x
2
2
2
PO = 1.0 atm x 0.20 = 0.20 atm
2
PNO = 0 atm
(no NO has been produced)
N2(g) + O2(g)
Kp =
(PNO)2
2 NO(g)
(2x)2
=
(PN ) (PO2)
2
(0.80 – x)(0.20 – x )
4x2
1.0 x 10-30 =
0.16
x = 2.0 x 10-16
2x = PNO = 4.0 x 10-16 atm
PN = 0.80 atm, PO = 0.20 atm
2
2
When Keq < 1 x 10-4
you can usually ignore
these (x)s to avoid
using the quadratic
equation.
Equilibrium
0.80 - x
Ex) Q and Le Chatelier (cont)
N2(g) + O2(g)
2NO(g)
c) Find the equilibrium partial pressures when
Kp = 0.0025
PN
PO
PNO
Initial
0.80
0.20
0
Change
-x
-x
+2x
0.20 - x
2x
2
Equilibrium
0.80 - x
2
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N2(g) + O2(g)
2 NO(g)
(PNO)2
Kp =
(2x)2
Ex) Q and Le Chatelier (cont)
Quadratic Formula
=
(PN ) (PO2)
2
(0.80 – x)(0.20 – x )
x=
−b ± b 2 − 4ac
2a
x=
−0.0025 ± 0.00252 − 4(3.9975)( −4.0 × 10−4 )
2(3.9975)
4x2
0.0025 =
0.16 – x + x2
4x2 = 0.0025x2 – 0.0025x +
0=
3.9975x2 +
0.0025x –
As Keq > 1x 10-4 in this
case, we cannot ignore
these (x)s. We must
use the quadratic
-4
4x10 equation.
x = + 9.7 x 10-3, or x = - 0.080
4x10-4
Ex) Q and Le Chatelier (cont)
c) Equilibrium partial pressures when Kp = 0.0025
PNO = 2x = 2(9.7 x 10-3) atm
PNO = 1.9 x 10-2 atm
PN = 0.80 – x = (0.80 – 0.0097) atm
2
PN = 0.79atm
2
PO = 0.20 – x = (0.20 – 0.0097) atm
2
PO = 0.19 atm
2
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Tutorials to assist you with this material are available online at www.apchemsolutions.com
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