2.3.14 – 2.3.15 dynamic equilibrium le chatelier’s principle effect of pressure, concentration...

2.3.14 – 2.3.15

•Dynamic equilibrium

•Le Chatelier’s principle

•Effect of pressure, concentration and temperature on systems at equilibrium

•Optimum conditions – the balance between rate, yield and cost, e.g. in the Haber process

https://www.youtube.com/watch?v=g5wNg_dKsYY

Students should be able to:

Explain that dynamic equilibrium occurs when the rates of the forward and reverse reaction are equal.

State le Chatelier’s principle.

Apply le Chatelier’s principle to deduce the qualitative effect of a change in concentration or pressure on a homogeneous system in equilibrium.

Apply le Chatelier’s principle to deduce the qualitative effect of a change in temperature on a homogeneous system in equilibrium.

Explain, from given data, the importance in the chemical industry of a compromise between chemical equilibrium and reaction rate.

Reactions and arrows

The reaction of sodium and chlorine is irreversible.

2Na(s) + Cl2(g) 2NaCl(s)

Irreversible reactions are represented by a single arrow: .

The reaction of nitrogen and hydrogen to form ammonia is reversible.

Ammonia can also decompose to form nitrogen and hydrogen. Reversible reactions that can reach equilibria are represented by two half arrows: .

N2(g) + 3H2(g) 2NH3(g)

What is equilibrium?

If a reversible reaction is carried out in a closed container so that the reactants and products cannot escape, a state of dynamic equilibrium can be established.

This state is dynamic because both the forward and reverse reactions are ongoing.

It is an equilibrium because:

the net concentrations of the components of the reaction mixture remain constant.

the rates of the forward and reverse reactions are the same

A + B C + D

Dynamic equilibrium

This graph illustrates the dynamic nature of equilibrium. It shows that both the forward and back reactions are taking place: they both have non-zero rate. When their rates are equal, equilibrium is reached.

time

rate

forward reaction: 2SO2(g) + O2(g) 2SO3(g)

reverse reaction: 2SO3(g) 2SO2(g) + O2(g)

rates are equal at equilibrium

Composition of the reaction mixture

At equilibrium, the proportions of reactants and products present may not be a 50:50 mix.

The proportion of reactants and products depend on the particular reaction, as well as factors such as temperature, concentration and pressure.

However, for a given set of conditions, a particular reaction will always have the same proportions of reactants and products at equilibrium.

Iodine dissolved in potassium iodide solution a. 3 cm3 in two test tubes . .. b. add dilute alkali - NaOH to one. . drop wise . . slowly . . c. add dilute acid – HCl/H2SO4 to the other . . drop wise . . slowly . .d. Now add the opposite drop wise . . . . Note any observations

Chromate / Dichromatea.put 3 cm3 of chromate in one test tube and 3 cm3 of dichromate in anotherb.add dilute alkali - NaOH to one. . drop wise . . slowly . . c. add dilute acid – HCl/H2SO4 to the other . . drop wise . . slowly . .d. Now add the opposite drop wise . . . . Note any observations

Cobalt chlorideDissolve some cobalt chloride in a little water . . Now . .a.add conc acid - HCl . . drop wise . . slowly . .b.add water again . . drop wise . . slowly . .c.cool the mixture in ice . . . d.warm the mixture again . . Saturated solution of Potassium Nitrate

Look at the beaker of this saturated solutionDraw a diagram . . . showing the ions that explains what is happening .

Cobalt chloride – Acid/water

Cobalt chloride – temperature

Chromate / dichromate

2CrO4 2– (aq) + 2H+(aq) Cr2O7

2– (aq) + H2O(l)chromate(VI) ion dichromate(VI) ion

yellow orange

Start Acid added

Alkali added

+ HCl + NaOH

Temperature and equilibrium

K+ NO3-

(s) ↔ K+ (aq)

+ NO3- (aq)

Dynamic Equilibrium

A reaction in a closed system at constant temperature is in a state of dynamic equilibrium when:

rate of forward reaction = rate of reverse reaction

A reaction in a closed system at constant temperature is in a state of dynamic equilibrium when:

rate of forward reaction = rate of reverse reaction

Dynamic: means the reaction is occurring in both directions.

Microscopic processes

When a system is in a state of dynamic equilibrium, there is a continual interchange between reactants and products at the molecular level.

These microscopic processes are characterised by constant change.

Macroscopic properties

Macroscopic properties are those we can see (colour) and measure (mass). These remain constant when a system has reached dynamic equilibrium.

Examples of situations . ..

An open system: If wet swimming trunks are hung outside, higher energy water molecules will be removed in the breeze. This physical process, which continue until the trunks are dry, can be represented by the equation:

H2O(l) H2O(g)

A closed system: If the swimming trunks are placed in a sealed plastic bag, they do not dry out. Initially, high energy water molecules evaporate. They remain in the atmosphere around the trunks until the low energy vapour molecules condense.

The rate of evaporation exceeds the rate of condensation until the vapour becomes saturated. At this point a balance is reached where:

rate of evaporation = rate of condensation

The system is in a state of dynamic equilibrium, shown by the symbol :

H2O(l) H2O(g)

A similar situation arises when the stopper is left off a bottle of a volatile liquid such as ethanol. This creates an open system from which the ethanol evaporates.

C2H5OH(l) C2H5OH(g)

When the stopper is replaced, a closed system is formed. Initially the rate of evaporation exceeds the rate of condensation:

C2H5OH(l) C2H5OH(g)

Eventually a state of dynamic equilibrium is reached between the liquid and its saturated vapour.

C2H5OH(l) C2H5OH(g)

Evaporating liquids . . . . ethanol

Several days later…


Pb2+

Pb2+

Pb2+

*I-

*I-

*I-

*I-

*I-

*I-

I-

Pb2+

* ** * *

Pb2+

Pb2+

I-

I-

I-

I-

I-

*

Pb2+

** *

Pb2+

Pb2+

I-

I-

*I-

*I-I-

*I-

A system in a state of dynamic equilibrium can be approached from either direction.

A system in a state of dynamic equilibrium can be approached from either direction.

Experiment 1

Non-radioactive PbI2(s) is dissolved to form a saturated solution.

Some radioactive Pb*I2(s) is added.

Experiment 2

Radioactive Pb*I2(s) is dissolved to form a saturated solution.

Some non-radioactive PbI2(s) is added.

After several days, dynamic equilibrium is reached between the solid and aqueous phases, and radioactive *I-(aq) ions are detected in both the solid and solution in both experiments.

• How do you account for this?• What physical factor must be constant during of the experiment

An example of this type of reaction is:

acid + alcohol ester + water

( A + B C + D )

a specific example of which is:

CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)

After several days equilibrium was reached. Analysis revealed that in both experiments, reactants and products were present in identical (but not equi-) molar ratios.

Count the particles in the flasks above.

In experiment 1, what was the original molar ratio of A : B? .........................

In experiment 2, what was the original molar ratio of C : D? .........................

At equilibrium, what was the molar ratio of A : B : C : D? ...........................

At equilibrium, are reactants or products in excess? ......................................



B A

A

A

B A

B

B B

A A

B

C D

A

C

C D

D

B D

A C

B

D

C C

C D C

D

D

D C C

D

Experiment 1 Equimolar quantities of an acid

(A) and an alcohol (B) were mixed together in a stoppered

bottle with a catalyst.

Experiment 2 Equimolar quantities of an ester (C) and water (D) were mixed together in a stoppered bottle

with a catalyst.

Copy this !

An example of this type of reaction is:

acid + alcohol ester + water

( A + B C + D )

a specific example of which is:

CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)

After several days equilibrium was reached. Analysis revealed that in both experiments, reactants and products were present in identical (but not equi-) molar ratios.

Count the particles in the flasks above.

In experiment 1, what was the original molar ratio of A : B? .........................

In experiment 2, what was the original molar ratio of C : D? .........................

At equilibrium, what was the molar ratio of A : B : C : D? ...........................

At equilibrium, are reactants or products in excess? ......................................



B A

A

A

B A

B

B B

A A

B

C D

A

C

C D

D

B D

A C

B

D

C C

C D C

D

D

D C C

D

Dynamic equilibrium does not necessarily mean the reactants and products are present in equimolar quantities.

In this case the equilibrium position favours the right hand side of the equation.

This is an example of homogeneous equilibrium as reactants and products are all in the same (in this case, liquid) phase.

If more than one phase (solid, liquid or gas) is present, the equilibrium mixture is described as heterogeneous.

Le Châtelier’s Principle

Le Châtelier’s Principle states that if a system in a state of dynamic equilibrium is disturbed, the equilibrium position shifts in the direction which tends to reduce the disturbance and restore the equilibrium.

Le Châtelier’s Principle states that if a system in a state of dynamic equilibrium is disturbed, the equilibrium position shifts in the direction which tends to reduce the disturbance and restore the equilibrium.

Systems in dynamic equilibrium are very sensitive to changes in concentration, pressure and temperature.

Le Châtelier’s Principle enables us to predict the effect of changes on the equilibrium mixture.

However, it does not explain why that change will occur, or what the extent of the change will be.

Temperature

A reaction which is exothermic in the forward direction is endothermic in the reverse direction.

If a chemical reaction is heated, the equilibrium position shifts in the direction that absorbs heat energy, that is in the endothermic direction.

If a chemical reaction is cooled, the equilibrium position shifts in the direction that releases heat energy, that is in the exothermic direction.

exothermic

endothermic A + B C + D

Pressure

2 A2(g) + 3 B2 (g) 2A2B3(g)

(LHS) (RHS)

If a gaseous reaction with different numbers of moles on the left hand side (LHS) and right hand side (RHS) is put under pressure, the equilibrium position shifts to the side with the fewer number of moles.

N2(g) + 2CO2(g) 2NO(g) + 2CO(g)

So increasing the pressure on this reaction will . . . . .

Decreasing the pressure on this reaction will . . . .

Increase pressure

decrease pressure

Concentration

If a reactant is added the equilibrium position will shift to the RHS, that is in the direction that will convert reactants to products.

If a product is removed the equilibrium position will shift to the RHS, that is in the direction that will replace the lost product.

concentration C or D

increases

A + B C + D

concentration A or B

increases

concentration C or D

decreases

concentration A or B

decreases

shift

shift

Pressure

2 A2(g) + 3 B2 (g) 2A2B3(g)

(LHS) (RHS)

If a gaseous reaction with different numbers of moles on the left hand side (LHS) and right hand side (RHS) is put under pressure, the equilibrium position shifts to the side with the fewer number of moles.

N2(g) + 2CO2(g) 2NO(g) + 2CO(g)

So increasing the pressure on this reaction will . . . . .

Decreasing the pressure on this reaction will . . . .

Increase pressure

decrease pressure

.

Catalysts

Catalysts do not alter the equilibrium position, only the rate at which that state is achieved.

The forward rate is increased

The reverse rate is increased

The same yield is obtained but at a faster rate.

A + B C + D

Reverse Rate increase

Forward rate increase

Catalysts and equilibriumA catalyst is a substance that speeds up the rate of reaction by providing an alternative reaction pathway of lower energy.

The use of catalysts is particularly important in industry.

equilibrium is reached faster.

there is no change to the position of the equilibrium

When added to a reversiblereaction, a catalyst increasesthe rate of both the forward and reverse reactions equally. This has two results:

time

ener

gy

with catalyst

without catalyst

Discussion topics

1 Discuss what you would observe when water is added to the following reaction that is in a state of dynamic equilibrium. It is initially blue in colour.

There are three possibilities: The solution turns pink The solution becomes deeper blue The solution remains unchanged

2 What is the effect of adding water to a solution of ethanoic acid, which is dissociated into ions and attains dynamic equilibrium:

CH3COOH(l) + H2O(l) CH3COO-(aq) + H3O+(aq)

Will the equilibrium position: Shift to the left (decrease degree of dissociation)? Shift to the right (increase degree of dissociation)? Remain unchanged?

Co(H2O)62+(aq) + 4Cl-(aq) CoCl4

2-(aq) + 6H2O(l)

pink blue

Discussion topics

1 Discuss what you would observe when water is added to the following reaction that is in a state of dynamic equilibrium. It is initially blue in colour.

There are three possibilities: The solution turns pink The solution becomes deeper blue The solution remains unchanged

2 What is the effect of adding water to a solution of ethanoic acid, which is dissociated into ions and attains dynamic equilibrium:

CH3COOH(l) + H2O(l) CH3COO-(aq) + H3O+(aq)

Will the equilibrium position: Shift to the left (decrease degree of dissociation)? Shift to the right (increase degree of dissociation)? Remain unchanged?

3 What is the effect of adding water to the following reaction which has reached dynamic equilibrium:

H2S(aq) + Zn2+(aq) ZnS(s) + 2H+(aq)

Will the equilibrium position: Shift to the left? Shift to the right? Remain unchanged?

4 In a 10cm3 sample of a solution of sodium dichromate, Na2Cr2O7, of concentration 0.5 mol dm-3, the following dynamic equilibrium is established:

If an additional 10 cm3 of the original sodium dichromate solution is added, what would you observe? There are three options: The solution becomes yellow The solution becomes deeper orange The solution remains unchanged

CrO42-(aq) + 2H+(aq) Cr2O7

2-(aq) + H2O(l)

yellow orange

Equilibrium reactions

ingaseous systems

At a high temperature, iodine molecules partially dissociate

into atoms:

I2(g) 2I(g)

H = +151 kJ mol-1 (endothermic)

At a high temperature, hydrogen iodide is in

dynamic equilibrium with its constituent elements:

2HI(g) H2(g) + I2(g)

H = -52 kJ mol-1 (exothermic)

Temperature: increase

decrease

Pressure:

increase

decrease

Addition of I2

Removal decrease of I2

Addition of catalyst

Equilibrium reactions

inaqueous

solutions

Ethanoic acid has a low degree of dissociation. It ionises according to

the equation:

CH3COOH(l) + H2O(l)

↔

CH3COO-(aq) + H3O+(aq)

H = +78 kJ mol-1 (endothermic)

In aqueous solution, hydrogen sulphide and zinc

ions react to form a precipitate of zinc sulphide:

H2S(aq) + Zn2+(aq)

↔

ZnS(s) + 2H+(aq)

H = -9 kJ mol-1 (exothermic)

Temperature: increase

decrease

Addition of:H3O+(aq)/ H+(aq)

OH-(aq)

catalyst

Making ethanol from ethene

Traditionally, ethanol is made by fermentation of biomass using yeast. However, this process is not suitable for meeting industrial ethanol needs in countries like the UK.

The forward reaction is exothermic (∆H = -45 kJ mol-1) and is catalyzed by phosphoric acid.

Industrially, ethanol is manufactured by hydration of ethene, which is a product of cracking crude oil.

C2H4(g) + H2O(g) C2H5OH(g)

Hydration of ethene: conditions

Hydration of ethene: true or false?

Formation of methanol

Methanol is an important chemical used in the manufacture of other chemicals. Methanol is made industrially by the reaction of carbon monoxide and hydrogen.

A mixture of copper, zinc oxide and aluminium oxide is used as a catalyst.

The mixture of carbon monoxide and hydrogen is called synthesis gas. It is made from methane and water by a process called steam reforming.

CO(g) + 2H2(g) CH3OH(g)

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

Formation of methanol: conditions

Carbon neutral activities in industry

Ethanol and methanol can be used as liquid fuels; for example, for specially-adapted motor cars.

Methanol produced from carbon monoxide and hydrogen can be a carbon neutral fuel.

A carbon neutral activity is one that has no net annual carbon (greenhouse gas) emissions to the

atmosphere.

Greenhouse gas emissions are a major cause of global warming. Aiming for carbon neutral status is one way in which industries can try to reduce their impact on the environment.

Formation of methanol: true or false?

What’s the keyword?

Multiple-choice quiz

Industrial processes

Le Châtelier’s Principle will now be applied to important industrial processes to determine the most favourable conditions under which the best yield can be obtained. High yield often has to be balanced by economic factors.

The Haber Process: production of ammonia

For this process, hydrogen is produced by reacting methane with steam at high temperature and under high pressure over a nickel catalyst:

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

Nitrogen is obtained from the air.

Temperature

The reaction is exothermic so a high temperature will favour an equilibrium position to the left hand side, that is in the endothermic direction, which will absorb heat energy.

At low temperatures a better yield is obtained as the equilibrium position shifts to the right hand side but the rate is too slow to be economically viable.

A temperature of 450ºC is used which is a compromise between yield and rate.

Comment on the bond energies of N?N and H-H:

............................................................................................................................

How will the strength of these bonds influence the activation energy?

............................................................................................................................

What influence does the activation energy have on the reaction rate?

............................................................................................................................

Hydrogen and nitrogen react in an exothermic reaction to form ammonia:

N2(g) + 3H2(g) 2NH3(g) H = -92 kJ mol-1

Temperature

The reaction is exothermic so a high temperature will favour an equilibrium position to the left hand side, that is in the endothermic direction, which will absorb heat energy.

At low temperatures a better yield is obtained as the equilibrium position shifts to the right hand side but the rate is too slow to be economically viable.

A temperature of 450ºC is used which is a compromise between yield and rate.

Comment on the bond energies of N≡N , H-H and H-N:

.............................................................................................................................

How will the strength of these bonds influence the activation energy?

.............................................................................................................................

What influence does the activation energy have on the reaction rate?

.............................................................................................................................

N≡N : 934 kj moleH-H : 435 kj moleN-H : 391 kj mole

Pressure

N2(g) + 3H2(g) 2NH3(g)

Total number of moles: ……… ……… reactants products

Le Châtelier’s Principle indicates that high pressure will shift the equilibrium position to the right hand side, that is the side with the fewer number of moles. A pressure of 200 to 250 atm is used.

Catalyst

The mixture of nitrogen and hydrogen (in a 1 : 3 molar ratio) is passed over finely divided iron, a transition metal. The efficiency of this catalyst is enhanced by traces of aluminium oxide or potassium hydroxide.

Why is the iron finely divided? .........................................................................

.............................................................................................................................

Recovery of ammonia

The gaseous mixture is then cooled until the ammonia liquefies. It is separated from the unreacted nitrogen and hydrogen.

Identify the intermolecular forces between nitrogen, hydrogen and ammonia molecules and give a reason why ammonia liquefies at a temperature at which nitrogen and hydrogen remain gaseous.

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

Recycling nitrogen and hydrogen

After the ammonia has been recovered, the unreacted nitrogen and hydrogen gases are recycled back into the main reaction chamber.

In fact, equilibrium is never reached in the Haber Process as reactants (nitrogen, hydrogen) are continually being added and the product (ammonia) is removed.

Thermodynamic stability

Draw an energy level diagram to represent the Haber Process reaction.

Label the axis, reactants, products, activation energy and -H.

Comment on the thermodynamic stability of the product with respect to the reactants and hence the feasibility of the reaction.

.............................................................................................................................

Reactions are feasible when they are exothermic and also when they become more disordered (the entropy increases), that is, when the number of particles increases.

If the enthalpy change of the reaction is favourable, give one reason why such drastic conditions need to be employed to achieve a reasonable yield of ammonia.

.............................................................................................................................

Economics

The unused nitrogen and hydrogen are recycled.

The heat energy from the exothermic Haber reaction and also from the initial reactions to produce the feedstock (nitrogen and hydrogen) is used to heat the temperature to 450ºC.

At higher pressures the cost of the plant and the complexity of the engineering become very expensive.

At higher temperatures the catalyst would corrode more quickly so would need to be replaced more often.

So why do this reaction . . ?Give three uses of ammonia . . . . . . .

The Contact Process: production of sulphuric acid

At a high temperature, molten sulphur reacts with dry air to form sulphur dioxide:

S(l) + O2(g) SO2(g) H = -297 kJ mol-1

The sulphur dioxide is oxidised by oxygen in the air to sulphur trioxide:

2SO2(g) + O2(g) 2SO3(g) H = -196 kJ mol-1

It is this second reaction for which the optimum conditions need to be determined in order to maximise the yield. As the system is in a state of dynamic equilibrium, Le Châtelier’s Principle is applied.

The Contact Process: production of sulphuric acid

At a high temperature, molten sulphur reacts with dry air to form sulphur dioxide:

S(l) + O2(g) SO2(g) H = -297 kJ mol-1

The sulphur dioxide is oxidised by oxygen in the air to sulphur trioxide:

2SO2(g) + O2(g) 2SO3(g) H = -196 kJ mol-1

Formation of oleum

Finally sulphur trioxide is absorbed by a fine spray of concentrated sulphuric acid to form oleum (H2S2O7).

The reaction between sulphuric acid and water is so exothermic that it would produce a fine mist of acid which would be difficult to collect.

SO3(g) + H2SO4(l) H2S2O7(l)

This oleum is diluted with water to produce concentrated sulphuric acid, which is approximately 98% H2SO4.

H2S2O7(l) + H2O(l) 2H2SO4(aq)

Give three uses of sulphuric acid:

.............................................................................................................................

Temperature

As the reaction is exothermic, what will happen to the equilibrium position at:

high temperatures ? . . . . . . . . . . .

low temperatures? . . . . . . . . . . . . The temperature used is 450ºC.

This is a compromise between . . . . . . and . . . . . . .

Pressure

2SO2(g) + O2(g) 2SO3(g)

Total number of moles: ……… ……… reactants products

Why does Le Châtelier’s Principle indicate that a high pressure will favour a high yield?

.............................................................................................................................

In fact a pressure of 2 - 5 atm is used as this gives a yield of 97% and is sufficient to force the gases through the plant.

A higher pressure is not justified on economic grounds.

Catalyst

The catalyst used is a transition metal oxide, vanadium(V) oxide, V2O5(s)

Identify the remaining oxidation numbers of sulphur and oxygen in the reaction:

Traces of vanadium in an oxidation state of +4 are found in the reaction mixture. This enables us to suggest a mechanism by which V2O5(s) acts as a catalyst.

The mechanism involves two steps, each of which is a redox reaction.

Fill in the missing oxidation states

2SO2(g) + O2(g) 2SO3(g)

S O O S O +4 -2 0 +6 -2

Catalyst

Fill in the missing ionic half equations in steps 1 and 2 below:

Step 1 Reduction: V(V) + e- V(IV)

Oxidation: ……………………………

Step 2 Oxidation: V(IV) V(V) + e-

Reduction: ……………………………

Comment on any economic factors that need to be taken into account

.............................................................................................................................

2SO2(g) + O2(g) 2SO3(g)

S O O S O +4 -2 0 +6 -2

S4+ → S6+ + 2e-2S4+ → 2S6+ + 4e-

O2 + 4e- → 2O2-

Formation of oleum

Finally sulphur trioxide is absorbed by a fine spray of concentrated sulphuric acid to form oleum (H2S2O7).

The reaction between sulphuric acid and water is so exothermic that it would produce a fine mist of acid which would be difficult to collect.

SO3(g) + H2SO4(l) H2S2O7(l)

This oleum is diluted with water to produce concentrated sulphuric acid, which is approximately 98% H2SO4.

H2S2O7(l) + H2O(l) 2H2SO4(aq)

Give three uses of sulphuric acid:

.............................................................................................................................

2.3.14 EXERCISE 1 – Equilibria 1. Consider the following exothermic reaction:

4HCl(g) + O2(g) ↔ 2Cl2(g) + 2H2O(g)

State, with a reason, what would happen to the amounts of chlorine and hydrogen chloride in the

system if the following changes were made after equilibrium had been established in a sealed

container:

a) water is removed from the system;

b) extra oxygen is added to the system;

c) the volume of the container was reduced;

d) the temperature of the container was increased;

e) a catalyst was added.

Answers to 2.3.14 Exercise 1

a) equilibrium moves to right to replace lost water

b) equilibrium moves to right to remove added oxygen

c) pressure is increased so equilibrium moves to right in

direction of fewer moles to reduce pressure

d) equilibrium moves to left in endothermic direction to

reduce temperature

e) equilibrium does not move as forward and reverse

reactions are getting faster by the same amount

1.

2. For each of the following reactions, state and explain whether a high or low temperature and a

high or low pressure should be used to maximize the yield of product:

a) 2SO2(g) + O2(g) ↔ 2SO3(g), ∆H = -ve

b) PCl5(g) ↔ PCl3(g) + Cl2(g), ∆H = +ve

c) H2(g) + I2(g) ↔ 2HI(g), ∆H = -ve

d) HCOOH(l) + CH3OH(l) ↔ HCOOCH3(l) + H2O(l), ∆H = 0

2.a)low temperature (moderate), as this will favour the exothermic direction which is the forward direction

high pressure, as this will favour the direction decreasing the gas moles which is the forward direction

b) high temperature, as this will favour the endothermic direction

which is the forward direction low pressure, as this will favour the direction increasing the gas moles which is the forward direction

c)low temperature (moderate), as this will favour the exothermic

direction which is the forward direction any pressure, as there is no change in the number of gas moles

d)any temperature, as there is no exothermic or endothermic direction

any pressure, as there is no change in the number of moles

3. The manufacture of ammonia by the Haber process is an important example of an industrial process

which involves an equilibrium reaction:

N2(g) + 3H2(g) ↔ 2NH3(g), ∆H = -ve

The reaction is carried out at 450 oC and 250 atm with an iron catalyst.

a) Give one reason why a higher temperature is not used.

b) Give one reason why a lower temperature is not used.

c) Give one reason why a higher pressure is not used.

d) Give two reasons why a lower pressure is not used.

e) Explain why a catalyst is used.

3.

a) yield is poor at high temperatures

b) reaction is slow at low temperatures

c) expensive equipment is needed for high pressures

d) yield is poor at low pressures reaction is slow at low

pressures

e) catalyst increases the rate of the reaction and reduces

costs be allowing a lower temperature to be used

Potential misconceptions

Students confuse rate and yield. Often they state that

‘the reaction is faster, so a bigger yield can be produced

in the same time’.

The effects of a catalyst are often confused. Students

think the catalyst is of benefit as it gives more product(s)

– the key point is that it gives the same amount of

product(s), only faster.

Nitric acid is made from the reaction of ammonia with oxygen.

The first stage produces nitrogen monoxide.(The same gas is formed as a pollutant in Petrol Engines).

A hot platinum catalyst is used (about 850 °C).The reaction is highly exothermic and the heat given out by

the reaction is sufficient to keep the reaction going.ammonia + oxygen nitrogen monoxide + water.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

The second stage involves the reaction ofnitrogen monoxide with more oxygen.

nitrogen monoxide + oxygen nitrogen dioxide.2NO(g) + O2(g) 2NO2(g)

Finally, nitrogen dioxide reacts with water and more oxygen.nitrogen dioxide + oxygen + water nitric acid.

4NO2(g) + O2(g) + 2H2O(l) 4HNO3(aq)

The overall reaction isammonia + oxygen nitric acid + water.

4NH3(g) + 8O2(g) 4HNO3(aq) + 4H2O(l)

This is the oxidation of ammonia to produce nitric acid.

Ammonia Fertilisers.Ammonia fertilisers are produced by the neutralisation of ammonia with a mineral acid.

Ammonia contains nitrogen and is used to make fertilisers.The main fertiliser is ammonium nitrate.This is produced by the reaction of ammonia with nitric acid

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nitric acid + ammonia ammonium nitrate.HNO3(aq) + NH3(aq) NH4NO3(aq)

The above reaction is an example of neutralisation, making a neutral salt.

The reaction of Ammonia with sulfuric acid would make ammonium sulfate (NH4)2SO4(aq)

Ammonium nitrate has a higher percentage of nitrogen than ammonium sulfate for the same mass of fertiliser.

The three main stages in the manufacture of nitric acid

ammonia oxidation 4NH3 + 5O2 4NO + 6H2O

H = - 900 kJmol-1

nitric oxide oxidation 2NO + O2 2NO2H = - 115 kJmol-1

2NO2 N2O4

H = - 58 kJmol-1

dinitrogen tetroxide absorption 3N2O4 + 2H2O 4HNO3 + 2NO H = - 103 kJmol-1

The three main stages in the manufacture of nitric acid

Oxidation of ammonia. Ammonia vapour is mixed with compressed air (10% and 90% respectively). The pressurised mixture is passed over a catalyst gauze of 90% platinum and 10% rhodium. About 96% conversion is achieved. #

2.3.14 – 2.3.15 dynamic equilibrium le chatelier’s principle effect of pressure, concentration...

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