Transcript Document

Predicting if a Reaction is in Equilibrium
Trial Keq
Lesson 9
The Keq is a constant- a number that does not change.
Changing the volume, pressure, or any concentration, does not
change the Keq.
Only temperature changes the Keq
If the [Reactant] is increased the reaction will shift
The Keq is a constant- a number that does not change.
Changing the volume, pressure, or any concentration, does not
change the Keq.
Only temperature changes the Keq
If the [Reactant] is increased the reaction will shift right
The Keq will
The Keq is a constant- a number that does not change.
Changing the volume, pressure, or any concentration, does not
change the Keq.
Only temperature changes the Keq
If the [Reactant] is increased the reaction will shift right
The Keq will remain constant
If the temperature of an endothermic reaction is increased the
reaction will
The Keq is a constant- a number that does not change.
Changing the volume, pressure, or any concentration, does not
change the Keq.
Only temperature changes the Keq
If the [Reactant] is increased the reaction will shift right
The Keq will remain constant
If the temperature of an endothermic reaction is increased the
reaction will shift right
The Keq will
The Keq is a constant- a number that does not change.
Changing the volume, pressure, or any concentration, does not
change the Keq.
Only temperature changes the Keq
If the [Reactant] is increased the reaction will shift right
The Keq will remain constant
If the temperature of an endothermic reaction is increased the
reaction will shift right
The Keq will increase
1.
If 6.00 moles CO2, and 6.00 moles H2 are put in a 2.00 L
container at 670 oC, calculate all equilibrium concentrations.
CO(g) +
H2O(g) ⇄
CO2(g) +
H2(g)
Keq = 9.0
Initial concentrations means ICE!
Because we are starting with products it goes left
Add on left and subtract on right
I
0
0
3.00 M
3.00 M
C
+x
+x
-x
-x
E
x
x
3.00 - x
3.00 - x
[CO2][H2]
[CO][H2O]
=
Keq
=
9.0
Keq
=
(3 - x)2
x2
=
9
=
3
1
Square root both sides
3 - x
x
Cross multiply
3x
=
3 - x
4x
=
3
[CO] = [H2O]
=
x
=
[CO2] = [H2]
= 3.00 - 0.75 =
0.75 M
2.25 M
Ktrial
How can you tell if a system is in equilibrium or not?
Calculate a trial Keq. Put the concentrations into the equilibrium
expression and evaluate.
Ktrial
How can you tell if a system is in equilibrium or not?
Calculate a trial Keq. Put the concentrations into the equilibrium
expression and evaluate.
If Ktrial = Keq
Equilibrium
Kt
Keq
Ktrial
How can you tell if a system is in equilibrium or not?
Calculate a trial Keq. Put the concentrations into the equilibrium
expression and evaluate.
If Ktrial = Keq
Equilibrium
If Ktrial < Keq
Not at Equilibrium
Kt
Keq
Ktrial
How can you tell if a system is in equilibrium or not?
Calculate a trial Keq. Put the concentrations into the equilibrium
expression and evaluate.
If Ktrial = Keq
Equilibrium
If Ktrial < Keq
Not at Equilibrium
Kt
Keq
Shifts Right
Ktrial
How can you tell if a system is in equilibrium or not?
Calculate a trial Keq. Put the concentrations into the equilibrium
expression and evaluate.
If Ktrial = Keq
Equilibrium
If Ktrial < Keq
Not at Equilibrium
Shifts Right
If Ktrial > Keq
Keq
Kt
Ktrial
How can you tell if a system is in equilibrium or not?
Calculate a trial Keq. Put the concentrations into the equilibrium
expression and evaluate.
If Ktrial = Keq
Equilibrium
If Ktrial < Keq
Not at Equilibrium
Shifts Right
If Ktrial > Keq
Not at Equilibrium
Shifts Left
Keq
Kt
The following amounts of gases are placed into a 2.0 L container.
Determine if each system is at equilibrium or not. If not, determine
the direction that the equilibrium will shift in order to get to
equilibrium.
2NH3(g)
⇄
N2(g)
+
3H2(g)
Keq = 10
1.
2.0 moles NH3
Get concentrations.
2.0 moles N2
2.0 mole H2
1.0 M
Calculate Kt
1.0 M
1.0 M
Kt
=
Not in equilibrium
[N2][H2]3
[NH3]2
=
Kt < Keq
(1)(1)3
(1)2
= 1
Shifts right!
The following amounts of gases are placed into a 2.0 L container.
Determine if each system is at equilibrium or not. If not, determine
the direction that the equilibrium will shift in order to get to
equilibrium.
2NH3(g)
⇄
N2(g)
+
3H2(g)
Keq = 10
2.
2.0 moles NH3
Get concentrations.
4.0 moles N2
4.0 mole H2
1.0 M
Calculate Kt
2.0 M
2.0 M
Kt
=
Not in equilibrium
[N2][H2]3
[NH3]2
=
Kt > Keq
(2)(2)3
(1)2
= 16
Shifts left!
The following amounts of gases are placed into a 2.0 L container.
Determine if each system is at equilibrium or not. If not, determine
the direction that the equilibrium will shift in order to get to
equilibrium.
2NH3(g)
⇄
N2(g)
+
3H2(g)
Keq = 10
3.
2.0 moles NH3
Get concentrations.
2.5 moles N2
4.0 mole H2
1.0 M
Calculate Kt
1.25 M
2.0 M
Kt
=
In equilibrium
[N2][H2]3
[NH3]2
=
Kt = Keq
(1.25)(2)3
(1)2
= 10
4.
If 4.00 moles of CO, 4.00 moles H2O, 6.00 moles CO2, and
6.00 moles H2 are placed in a 2.00 L container at 670 oC,
Keq = 1.0
CO(g)
+
H2O(g) ⇄
CO2(g) +
H2(g)
Is the system at equilibrium?
If not, how will it shift in order to get there?
Calculate all equilibrium concentrations.
Get Molarities
2.00 M
2.00 M
3.00 M
(3)(3)
(2)(2)
=
3.00 M
Calculate a Kt
Kt
=
Not in equilibrium
Shifts left!
2.25
Do an ICE chart
I
C
E
CO(g)
+
2.00 M
+x
2.00 + x
Keq
H2O(g) ⇄
2.00 M
+x
2.00 + x
=
CO2(g) +
3.00 M
-x
3.00 - x
(3 - x)2
(2 + x)2
=
H2(g)
3.00 M
-x
3.00 - x
1.0
Square root
3
2
3
1
x
[CO2] =
[CO] =
[H2] =
[H2O] =
- x
=
+x
- x = 2 + x
=
2x
= 0.50 M
1.0
3.00 - 0.50 = 2.50 M
2.00 + 0.50 = 2.50 M
Size of the Keq
Big Keq
products
Keq
=
reactants
Keq
=
10
Little Keq-
products
Keq
=
reactants
Keq
=
0.1
Note that the keq cannot be a negative number!
Keq about 1
products
Keq
=
reactants
Keq
=
1
Which reaction favours the products the most?
Keq = 2.6 x 106
Keq = 3.5 x 102
Which reaction favours the reactants the most?
Keq = 2.6 x 10-6
Keq = 3.5 x 10-7
Which reaction favours the products the most?
Keq = 2.6 x 106
Keq = 3.5 x 102
Which reaction favours the reactants the most?
Keq = 2.6 x 10-6
Keq = 3.5 x 10-7