Reaction Quotient-Q-Or

Trial K

The Keq is a constant- a number that does not change

Increasing the Temperature of an endothermic equilibrium shifts right and increases the Keq

Increasing a [Reactant] shifts right but maintains the Keq

Only temperature changes the Keq

Changing the volume, pressure, or any concentration, does not change the Keq.

K trial How can you tell if a system is in equilibrium or not? Calculate a trial Keq or Q- reaction quotient.

Use initial concentrations in the equilibrium expression to evaluate. 

Keq

Kt

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

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

If Kt = Keq

equilibrium

If Kt > Keq

Kt Kt

If Kt < Keq

Shifts right( to products) Shifts left (to reactants)

5 20 35

 

 

Shifts right!

2NH3(g) ⇄ N2(g) + 3H2(g) Keq = 10

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?

2.0 M 3.0 M 2.0 M

Kt = [N2][H2]3

[NH3]2

= (3)(2)3 = 6(2)2

Not in equilibriumKt < Keq

 

 

Shifts left!

2NH3(g) ⇄ N2(g) + 3H2(g) Keq = 10

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?

9.12 x 10-5 M 1.124 x 10-5 M 5.32 x 10-2 M

Kt = [N2][H2]3

[NH3]2

= (1.124 x 10-3 )(5.32 x 10-2 )3 = 20.3(9.12 x 10-5 )2

Not in equilibriumKt > Keq

3. 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

Is the system at equilibrium?

Calculate all equilibrium concentrations.

Shifts left!

Not in equilibrium

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

= 2.25+x +x -x -x

2.00 M 2.00 M 3.00 M 3.00 M

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

Keq =[CO2][H2]

[CO][H2O]

(3 - x)2 (2 + x)2

= 1.0

= 1.03 - x 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

Example FA 1.0 L reaction vessel contained 1.0 mol of SO2, 4.0 mol of NO2, 4.0 mol of SO3 and 4.0 mol of NO at equilibrium according to SO2(g) + NO2(g) SO3(g) + NO(g).

If 3.0 mol of SO2 is added to the mixture, what will be the new concentration of NO when equilibrium is re-attained?

SO2(g) + NO2(g) SO3(g) + NO(g).