Chemical Equilibrium

Advanced Higher Chemistry

Unit 2b

Revision

• Dynamic

• But to observe you would see no change

• Rate of forward reaction = rate of reverse reaction

• Concentrations of reactants and products are constant, but not (necessarily) equal

2Hl(g) H2(g) + I2(g)

To which side does the equilibrium lie?

How would the graph differ if you started with H2 and I2?

Calculating compositions

• To determine equilibrium composition of a mixture, you need equilibrium amount of only ONE of the species

• Example:

CO(g) + 3H2(g) CH4(g) + H2O(g)

You place 1.000 mol CO and 3.000 mol H2 in a reaction vessel at 1200 K and allow the reaction to come to equilibrium. The mixture is found to contain 0.387 mol H2O. What is the molar composition of the equilibrium mixture?

Calculation strategy

• You have starting amounts• These will change over time• They will remain constant at equilibrium

Amount (mol) CO 3H2 CH4 H2O

Starting

Change

Equilibrium

1.000

-x

1.000-x

3.000

-3x

3.000-3x

0

+x

x

0

+x

x = 0.387

Therefore: 0.613 mol CO, 1.839 mol H2, 0.387 mol CH4, 0.387 mol H2O

CO(g) + 3H2(g) CH4(g) + H2O(g)

Practice

1. You place 1.50 mol of dinitrogen trioxide into a flask, where it decomposes at 25ºC and 1.00 atm:

N2O3(g) NO2(g) + NO(g) What is the composition of the reaction mixture at

equilibrium if it contains 0.45 mol of nitrogen dioxide?

2. Nitric oxide, NO, reacts with bromine to give nitrosyl bromide, NOBr

2NO(g) + Br2(g) 2NOBr(g)

A sample of 0.0655 mol NO with 0.0328 mol Br2 gives an equilibrium mixture containing 0.0389 mol NOBr. What is the composition of the equilibrium mixture?

The Equilibrium Constant, Kc

• For the reaction:

aA + bB cC + dD

[C]c[D]d

[A]a[B]bKc =

• The value of the equilibrium constant is constant for a particular reaction at a given temperature, whatever the initial starting concentrations

Practice

Write the equilibrium constant expression for:

1. The synthesis of ammonia by the Haber process

2. The catalytic hydration of methane

Obtaining values for Kc

• Consider the reaction:CO(g) + 3H2(g) CH4(g) + H2O(g)

The equilibrium composition is:0.613 mol CO, 1.839 mol H2, 0.387 mol CH4, 0.387 mol H2O

The volume of the reaction vessel is 10.00 l.

So the equilibrium concentration of CO is 0.613/10 = 0.0613

What is the value of the equilibrium constant?

Practice

• Suppose for the same reaction the equilibrium compositions in a 10 l vessel are:

1.522 mol CO, 1.566 mol H2, 0.478 mol CH4, 0.478 mol H2O

What is the vale of Kc?

Homo- / Heterogeneous Equilibria

• Consider the reaction:

3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)

• Is this hetero- or homogeneous?

• For heterogeneous equilibria, pure solids and liquids are omitted from the Kc

• Write an expression for Kc for the above reaction

• This means that equilibrium is not affected by amounts of these substances

The Equilibrium Constant, Kp

• For the reaction: CO(g) + 3H2(g) CH4(g) + H2O(g)

• The equilibrium constant can be expressed in terms of partial pressures:

PCH4PH2O

PCOPH23

Kp =

In general, for a particular reaction the value of Kp will be different from Kc

Interpretation of K

• Does a large K mean that reactants or products are favoured?

• CH4(g) + 2O2(g) CO2(g) + 2H2O(g)• Kc = 1 x 10140 at 25ºC

• N2(g) + O2(g) 2NO(g)• Kc = 4.6 x 10-31 at 25ºC

• If K is around 1, neither reactants or products are favoured

Changing the reaction conditions

• Conditions can be changed to maximise product:– Concentration: adding reactants, removing

products– Pressure: change the volume of the vessel– Temperture

• NOT a catalyst

Concentration change

• Example:

CO(g) + 3H2(g) CH4(g) + H2O(g)

• If cooled, water can be condensed and removed

Moles

Stage of process CO H2 CH4 H2O

Original mixture 0.613 1.839 0.387 0.387

After removing water (before equilibrium readjusts)

0.613 1.839 0.387 0

Equilibrium re-established 0.491 1.473 0.509 0.122

Assuming a 1 l vessel, what is Kc for the original equilibrium and for the re-established equilibrium?

Pressure change• How can you change the pressure of a system?• What is meant by partial pressure?

Changing the volume of a reactant container changes the concentration of gaseous reactants and therefore their partial pressures

Equilibrium position will therefore moveThe value of Kc or Kp does NOT change

Changing pressure by adding more of an inert gas has no effect of the equilibrium position

- No effect on partial pressures

Pressure change

Temperature change

• Like concentration and pressure, temperature affects the position of equilibrium

• Unlike concentration and pressure, temperature, temperature does affect the size of the equilibrium constant

• Consider the following data:

Temperature (K) Kc

298

800

1000

1200

4.9 x 1027

1.38 x 105

2.54 x 102

3.92

Is this reaction exothermic or endothermic?

Summary

• K is temperature dependant, but is unaffected by changes in concentrations or partial pressures

ChangeEquilibrium

positionValue of K

Concentration

Pressure

Temperature

Catalyst

Changes

Changes

Changes

No change

No change

No change

Changes

No change