Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9

Download Presentation

Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9

Loading in 2 Seconds...

- 182 Views
- Uploaded on
- Presentation posted in: General

Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Predicting if a Reaction is in Equilibrium

Trial Keq

Lesson 9

The Keqis a constant- a number that does not change

Changing the volume, pressure, or any concentration, does notchange the Keq.

Onlytemperature changes the Keq

Increasing a [Reactant] shifts right and maintains the Keq

Increasing the temperature of an endothermic equilibrium shifts right andincreases the Keq

Ktrial

How can you tell if a system is in equilibrium or not?

Calculatea trial Keq.

Use initial concentrations in the equilibrium expression- evaluate.

How can you tell if a system is in equilibrium or not?

Calculatea trial Keq. Put initial concentrations into the equilibrium expression and evaluate.

If Kt < Keq

If Kt = Keq

If Kt > Keq

Shifts right

equilibrium

Shifts left

Kt

Kt

Kt

Keq

5

20

35

1.10.0 moles of NH3, 15.0 moles of N2, and 10.0 moles of H2 are put in a 5.0 L container. Is the system in equilibrium and how will it shift if it is not?

2NH3(g)⇄N2(g) + 3H2(g) Keq = 10

2.0 M3.0 M2.0 M

[N2][H2]3

Kt=

=(3)(2)3

= 6

(2)2

[NH3]2

Not in equilibrium

Kt < Keq

Shifts right!

2.4.56 x 10-5 moles of NH3, 5.62 x 10-4 moles of N2, and 2.66 x 10-2 moles of H2 are put in a 500.0 mL container. Is the system in equilibrium and how will it shift if it is not?

2NH3(g)⇄N2(g) + 3H2(g) Keq = 10

9.12 x 10-5 M1.124 x 10-5 M5.32 x 10-2 M

[N2][H2]3

Kt=

=(1.124 x 10-3 )(5.32 x 10-2 )3

= 20.3

(9.12 x 10-5)2

[NH3]2

Not in equilibrium

Kt > Keq

Shifts left!

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.

CO(g) + H2O(g) ⇄ CO2(g) + H2(g) Keq = 1.0

2.00 M 2.00 M 3.00 M 3.00 M

Is the system at equilibrium?

+x +x -x -x

(3)(3)

Kt =

= 2.25

(2)(2)

2.00 + x 2.00 + x 3.00 - x 3.00 - x

Not in equilibrium

Shifts left!

[CO2][O2]

Keq =

[CO][H2O]

Calculate all equilibrium concentrations.

(3 - x)2

= 1.0

(2 + x)2

3 - x

= 1.0

2 + x

3 - x = 2 + x

1 =2x

x = 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

products

Keq=

reactants

Keq= 10

Little Keq

products

Keq=

reactants

Keq=0.1

Note that the keq cannot be a negative number!

Keq=1

products

Keq=

reactants