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 rightThe 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 rightThe 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 rightThe Keq will remain constant

If the temperature of an endothermic reaction is increased the reaction will shift rightThe 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 rightThe Keq will remain constant

If the temperature of an endothermic reaction is increased the reaction will shift rightThe 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 leftAdd 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

Keq = [CO2][H2] = 9.0

[CO][H2O]

Keq = (3 - x)2 = 9 x2

Square root both sides

3 - x = 3 x 1

Cross multiply

3x = 3 - x

4x = 3

[CO] = [H2O] = x = 0.75 M

[CO2] = [H2] = 3.00 - 0.75 = 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 

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   

KeqKt

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  

KeqKt

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 NH32.0 moles N2 2.0 mole H2

Get concentrations.

1.0 M 1.0 M 1.0 MCalculate Kt

Kt = [N2][H2]3 = (1)(1)3 = 1

[NH3]2 (1)2

Not in equilibrium Kt < Keq 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 NH34.0 moles N2 4.0 mole H2

Get concentrations.

1.0 M 2.0 M 2.0 MCalculate Kt

Kt = [N2][H2]3 = (2)(2)3 = 16

[NH3]2 (1)2

Not in equilibrium Kt > Keq 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 NH32.5 moles N2 4.0 mole H2

Get concentrations.

1.0 M 1.25 M 2.0 MCalculate Kt

Kt = [N2][H2]3 = (1.25)(2)3 = 10

[NH3]2 (1)2

In equilibrium Kt = Keq

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 = 1.0CO(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 3.00 M

Calculate a Kt

Kt = (3)(3) = 2.25(2)(2)

Not in equilibrium Shifts left!

Do an ICE chart

CO(g) + H2O(g) ⇄ CO2(g) + H2(g)

I 2.00 M 2.00 M 3.00 M 3.00 MC +x +x -x -xE 2.00 + x 2.00 + x 3.00 - x 3.00 - x

Keq = (3 - x)2 = 1.0(2 + x)2

Square root3 - x = 1.02 + x 3 - x = 2 + x1 = 2xx = 0.50 M

[CO2] = [H2] = 3.00 - 0.50 = 2.50 M

[CO] = [H2O] = 2.00 + 0.50 = 2.50 M

Size of the Keq

Big Keq

Keq =

Keq = 10

products

reactants

Little Keq-

Keq =

Keq = 0.1

Note that the keq cannot be a negative number!

products

reactants

Keq about 1

Keq =

Keq = 1

products

reactants

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