Equilibrium is a state of balance.
Chemical equilibrium is state of balance between the reactants and products.
As reactants form products, some products form back the reactants.
Reactions in which the reactants form products to completion are said to be reversible i.e.
A + B -> C + D
Reactions in which the reactants form products and the products can reform the reactants are said to be reversible.
A + B C + D
Reversible reactions may be:
(a)Reversible physical changes
(b)Reversible chemical changes
(c)Dynamic equilibrium
(a)Reversible physical changes
Reversible physical change is one which involves:
(i) change of state/phase from solid, liquid, gas or aqueous solutions. States of matter are interconvertible and a reaction involving a change from one state/phase can be reversed back to the original.
(ii) colour changes. Some substances/compounds change their colours without change in chemical substance.
Examples of reversible physical changes
(i) colour change on heating and cooling:
I. Zinc(II)Oxide changes from white when cool/cold to yellow when hot/heated and back.
ZnO(s) ZnO(s)
(white when cold) (yellow when hot)
II. Lead(II)Oxide changes from yellow when cold/cool to brown when hot/heated and back.
PbO(s) PbO(s)
(brown when hot) (yellow when cold)
(ii)Sublimation
I. Iodine sublimes from a grey crystalline solid on heating to purple vapour. Purple vapour undergoes deposition back to the grey crystalline solid.
I2(s) I2(g)
(grey crystalline solid (purple vapour
undergo sublimation) undergo deposition)
II. Carbon (IV)oxide gas undergoes deposition from a colourless gas to a white solid at very high pressures in a cylinder. It sublimes back to the colourless gas if pressure is reduced
CO2(s) CO2(g)
(white powdery solid (colourless/odourless gas
undergo sublimation) undergo deposition)
(iii)Melting/ freezing and boiling/condensation
Ice on heating undergo melting to form a liquid/water. Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid.
| H2O(s) |
| H2O(l) |
| H2O(s) |
Melting boiling
Freezing condensing
(iv)Dissolving/ crystallization/distillation
Solid crystals of soluble substances (solutes) dissolve in water /solvents to form a uniform mixture of the solute and solvent/solution. On crystallization /distillation /evaporation the solvent evaporate leaving a solute back. e.g.
NaCl(s) + aq NaCl(aq)
(b)Reversible chemical changes
These are reactions that involve a chemical change of the reactants which can be reversed back by recombining the new substance formed/products.
Examples of Reversible chemical changes
(i)Heating Hydrated salts/adding water to anhydrous salts.
When hydrated salts are heated they lose some/all their water of crystallization and become anhydrous.Heating an unknown substance /compound that forms a colourless liquid droplets on the cooler parts of a dry test/boiling tube is in fact a confirmation inference that the substance/compound being heated is hydrated.
When anhydrous salts are added (back) some water they form hydrated compound/salts.
Heating Copper(II)sulphate(VI)pentahydrate and cobalt(II)chloride hexahydrate
(i)Heat about 5.0g of Copper(II)sulphate(VI) pentahydrate in a clean dry test tube until there is no further colour change on a small Bunsen flame. Observe any changes on the side of the test/boiling tube. Allow the boiling tube to cool.Add about 10 drops of distilled water. Observe any changes.
(ii)Dip a filter paper in a solution of cobalt(II)chloride hexahydrate. Pass one end the filter paper to a small Bunsen flame repeatedly. Observe any changes on the filter paper. Dip the paper in a beaker containing distilled water. Observe any changes.
Sample observations
| Hydrated compound | Observation before heating | Observation after heating | Observation on adding water |
| Copper(II)sulphate (VI) pentahydrate | Blue crystalline solid | (i)colour changes from blue to white. (ii)colourless liquid forms on the cooler parts of boiling / test tube | (i)colour changes from white to blue (ii)boiling tube becomes warm /hot. |
| Cobalt(II)chloride hexahydrate | Pink crystalline solid/solution | (i)colour changes from pink to blue. (ii) colourless liquid forms on the cooler parts of boiling / test tube (if crystal are used) | (i)colour changes from blue to pink (ii)boiling tube becomes warm/hot. |
When blue Copper(II)sulphate (VI) pentahydrate is heated, it loses the five molecules of water of crystallization to form white anhydrous Copper(II)sulphate (VI).Water of crystallization form and condenses as colourless droplets on the cooler parts of a dry boiling/test tube.
This is a chemical change that produces a new substance. On adding drops of water to an anhydrous white copper(II)sulphate(VI) the hydrated compound is formed back. The change from hydrated to anhydrous and back is therefore reversible chemical change.Both anhydrous white copper(II)sulphate(VI) and blue cobalt(II)chloride hexahydrate are therefore used to test for the presence of water when they turn to blue and pink respectively.
CuSO4(s) + 5H2 O(l) CuSO4.5H2 O(s/aq)
(white/anhydrous) (blue/hydrated)
CoCl2(s) + 6H2 O(l) CoCl2.6H2 O(s/aq)
(blue/anhydrous) (pink/hydrated)
(ii)Chemical sublimation
Some compounds sublime from solid to gas by dissociating into new different compounds. e.g.
Heating ammonium chloride
(i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made.
When a glass rod containing hydrogen chloride gas is placed near ammonia gas, they react to form ammonium chloride solid that appear as white fumes.
This experiment is used interchangeably to test for the presence of hydrogen chloride gas (and hence Cl– ions) and ammonia gas (and hence NH4+ ions)
(ii)Put 2.0 g of ammonium chloride in a long dry boiling tube. Place wet / moist /damp blue and red litmus papers separately on the sides of the mouth of the boiling tube. Heat the boiling tube gently then strongly. Explain the observations made.
When ammonium chloride is heated it dissociates into ammonia and hydrogen chloride gases. Since ammonia is less dense, it diffuses faster to turn both litmus papers blue before hydrogen chloride turn red because it is denser. The heating and cooling of ammonium chloride is therefore a reversible chemical change.
NH3(g) + HCl(g) NH4Cl(s)
(Turns moist (Turns moist (forms white fumes)
litmus paper blue) litmus paper red)
(c)Dynamic equilibria
For reversible reactions in a closed system:
(i) at the beginning;
-the reactants are decreasing in concentration with time
-the products are increasing in concentration with time
(ii) after some time a point is reached when as the reactants are forming products the products are forming reactants. This is called equilibrium.
| Reactants concentration decreases to form products |
Sketch showing the changes in concentration of reactants and products in a closed system
For a system in equilibrium:
(i) a reaction from left to right (reactants to products) is called forward reaction.
(ii) a reaction from right to left (products to reactants) is called backward reaction.
(iii)a reaction in which the rate of forward reaction is equal to the rate of backward reaction is called a dynamic equilibrium.
A dynamic equilibriumis therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered.
The influence of different factors on a dynamic equilibrium was first investigated from 1850-1936 by the French Chemist Louis Henry Le Chatellier. His findings were called Le Chatelliers Principle which states that:
“if a stress/change is applied to a system in dynamic equilibrium, the system readjust/shift/move/behave so as to remove/ reduce/ counteract/ oppose the stress/change”
Le Chatelliers Principle is applied in determining the effect/influence of several factors on systems in dynamic equilibrium. The following are the main factors that influence /alter/ affect systems in dynamic equilibrium:
(a)Concentration
(b)Pressure
(c)Temperature
(d)Catalyst
(a)Influence of concentration on dynamic equilibrium
An increase/decrease in concentration of reactants/products at equilibrium is a stress. From Le Chatelliers principle the system redjust so as to remove/add the excessreduced concentration.
Examples of influence of concentration on dynamic equilibrium
(i)Chromate(VI)/CrO42- ions in solution are yellow. Dichromate(VI)/Cr2O72- ions in solution are orange. The two solutions exist in equilibrium as in the equation:
2H+ (aq) + 2CrO42- (aq) Cr2O72- (aq) + H2O(l)
(Yellow) (Orange)
I. If an acid is/H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already H+ ions. The equilibrium shift forward to the right to remove/reduce the excess H+ ions added. Solution mixture becomes MoreCr2O72- ions formed in the solution mixture make it to be more orange in colour.
II. If a base/OH– (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH– (aq) to form water.
H+ (aq) +OH– (aq) -> H2O(l)
The equilibrium shift backward to the left to add/replace the H+ ions that have reacted with the OH– (aq) ions . More of the CrO42- ions formed in the solution mixture makes it to be more yellow in colour.
2OH– (aq) + 2Cr2O72- (aq) CrO42- (aq) + H2O(l)
(Orange) (Yellow)
I. If an acid/ H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side on the OH– (aq). H+ ions react with OH– (aq) to form water.
H+ (aq) +OH– (aq) -> H2O(l)
The equilibrium shift backward to the left to add/replace the 2OH– (aq) that have reacted with the H+ (aq) ions . More Cr2O72- (aq)ions formed in the solution mixture makes it to be more Orange in colour.
II. If a base /OH– (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already OH– (aq) ions. The equilibrium shift forward to the right to remove/reduce the excess OH– (aq) ions added. Moreof the Cr2O72- ions are formed in the solution mixture making it to be more orange in colour.
(i)Practical determination of the influence of alkali/acid on Cr2O72- / CrO42- equilibrium mixture
Measure about 2 cm3 of Potassium dichromate (VI) solution into a test tube.
Note that the solution mixture is orange.
Add three drops of 2M sulphuric(VI) acid. Shake the mixture carefully.
Note that the solution mixture is remains orange.
Add about six drops of 2M sodium hydroxide solution. Shake carefully.
Note that the solution mixture is turns yellow.
Explanation
The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium. Dichromate(VI) ions are stable in acidic solutions while chromate(VI)ions are stable in basic solutions. An equilibrium exist thus:
| Cr2O72- |
| CrO42- |
OH-
H+
When an acid is added, the equilibrium shift forward to the right and the mixture become more orange as more Cr2O72- ions exist.
When a base is added, the equilibrium shift backward to the left and the mixture become more yellow as more CrO42- ions exist.
(ii)Practical determination of the influence of alkali/acid on bromine water in an equilibrium mixture
Measure 2cm3 of bromine water into a boiling tube. Note its colour.
Bromine water is yellow
Add three drops of 2M sulphuric(VI)acid. Note any colour change
Colour becomes more yellow
Add seven drops of 2M sodium hydroxide solution. Note any colour change.
Solution mixture becomes colourless/Bromine water is decolourized.
Explanation
When added distilled water,an equilibrium exist between bromine liquid (Br2(aq)) and the bromide ion(Br–), hydrobromite ion(OBr–) and hydrogen ion(H+) as in the equation:
H2O(l) + Br2(aq) OBr– (aq) + H+ (aq) + Br– (aq)
If an acid (H+)ions is added to the equilibrium mixture, it increases the concentration of the ions on the product side which shift backwards to the left to remove the excess H+ ions on the product side making the colour of the solution mixture more yellow.
If a base/alkali OH– is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water.
H+ (aq)+ OH–(aq) -> H2O(l)
This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized)
(iii)Practical determination of the influence of alkali/acid on common acid-base indicators.
Place 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes.
To each test tube add two drops of water. Record your observations in Table 1 below.
To the same test tubes, add three drops of 2M sulphuric(VI)acid. Record your observations in Table 1 below.
To the same test tubes, add seven drops of 2M sodium hydroxide solution. Record your observations in Table 1 below.
| To the same test tubes, repeat adding four drops of 2M sulphuric(VI)acid. Table 1Indicator | Colour of indicator in | ||
| Water | Acid(2M sulphuric (VI) acid) | Base(2M sodium hydroxide) | |
| Phenolphthalein | Colourless | Colourless | Pink |
| Methyl orange | Yellow | Red | Orange |
| Litmus solution | Colourless | Red | Blue |
Explanation
An indicator is a substance which shows whether another substance is an acid , base or neutral.
Most indicators can be regarded as very weak acids that are partially dissociated into ions.An equilibrium exist between the undissociated molecules and the dissociated anions. Both the molecules and anions are coloured. i.e.
HIn(aq) H+ (aq) + In– (aq)
(undissociated indicator (dissociated indicator
molecule(coloured)) molecule(coloured))
When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears.
When a base/alkali OH– is added to the indicator, the OH– reacts with H+ ions from the dissociated indicator to form water.
H+ (aq) + OH–(aq) -> H2O(l)
(from indicator) (from alkali/base)
The equilibrium shift forward to the right to replace the H+ ion and therefore the colour of dissociated (In–) molecule shows/appears.
The following examples illustrate the above.
(i)Phenolphthalein indicator exist as:
HPh H+ (aq) + Ph–(aq)
(colourless molecule) (Pink anion)
On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore colourless.
When a base/alkali OH– is added to the indicator, the OH– reacts with H+ ions from the dissociated indicator to form water.
H+ (aq) + OH–(aq) -> H2O(l)
(from indicator) (from alkali/base)
The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The pink colour of dissociated (Ph–) molecule shows/appears.
(ii)Methyl Orange indicator exists as:
HMe H+ (aq) + Me–(aq)
(Red molecule) (Yellow/Orange anion)
On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore red.
When a base/alkali OH– is added to the indicator, the OH– reacts with H+ ions from the dissociated indicator to form water.
H+ (aq) + OH–(aq) -> H2O(l)
(from indicator) (from alkali/base)
The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The Orange colour of dissociated (Me–) molecule shows/appears.
(b)Influence of Pressure on dynamic equilibrium
Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules. Decrease in pressure shift the equilibrium towards the side with more volume/molecules. More yield of products is obtained if high pressures produce less molecules / volume of products are formed.
If the products and reactants have equal volume/molecules then pressure has no effect on the position of equilibrium
The following examples show the influence of pressure on dynamic equilibrium:
(i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture
Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas.
Chemical equation : 2NO2(g) ===== N2 O4 (g)
Gay Lussacs law 2Volume 1Volume
Avogadros law 2molecule 1molecule
2 volumes/molecules of Nitrogen(IV)oxide form 1 volumes/molecules of dinitrogen tetraoxide
Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules.The equilibrium mixture become more yellow.
Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The equilibrium mixture become more brown.
(ii)Iodine vapour-Hydrogen gas/Hydrogen Iodide mixture.
Pure hydrogen gas reacts with Iodine vapour to form Hydrogen Iodide gas.
Chemical equation : I2(g) + H2(g) ===== 2HI (g)
Gay Lussacs law 1Volume 1Volume 2Volume
Avogadros law 1molecule 1molecule 2molecule
(1+1) 2 volumes/molecules of Iodine and Hydrogen gasform 2 volumes/molecules of Hydrogen Iodide gas.
Change in pressure thus has no effect on position of equilibrium.
(iii)Haber process.
Increase in pressure of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process.
The yield of ammonia is thus favoured by high pressures
Chemical equation : N2(g) + 3H2 (g) -> 2NH3 (g)
Gay Lussacs law 1Volume 3Volume 2Volume
Avogadros law 1molecule 3molecule 2molecule
(1 + 3) 4 volumes/molecules of Nitrogen and Hydrogen react to form 2 volumes/molecules of ammonia.
Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules.
The yield of ammonia increase.
Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules.
The yield of ammonia decrease.
(iv)Contact process.
Increase in pressure of the Sulphur(IV)oxide/Oxygen mixture favours the formation of more molecules of Sulphur(VI)oxide gas in Contact process. The yield of Sulphur(VI)oxide gas is thus favoured by high pressures.
Chemical equation : 2SO2(g) + O2 (g) -> 2SO3 (g)
Gay Lussacs law 2Volume 1Volume 2Volume
Avogadros law 2molecule 1molecule 2molecule
(2 + 1) 3 volumes/molecules of Sulphur(IV)oxide/Oxygen mixture react to form 2 volumes/molecules of Sulphur(VI)oxide gas.
Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of Sulphur(VI)oxide gas increase.
Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of Sulphur(VI)oxide gas decrease.
(v)Ostwalds process.
Increase in pressureof the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures.
Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O(g)
Gay Lussacs law 4Volume 5Volume 4Volume 6Volume
Avogadros law 4molecule 5molecule 4molecule 6Molecule
(4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 10 volumes/molecules of Nitrogen(II)oxide gas and water vapour.
Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour decrease.
Decrease in pressure shift the equilibrium forward to the right where there is more volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour increase.
Note
If the water vapour is condensed on cooling, then:
Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O(l)
Gay Lussacs law 4Volume 5Volume 4Volume 0Volume
Avogadros law 4molecule 5molecule 4molecule 0Molecule
(4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 4 volumes/molecules of Nitrogen(II)oxide gas and no vapour.
Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules. The yield of Nitrogen(II)oxide gas increase.
Decrease in pressure shift the equilibrium backward to the left where there is more volume/molecules. The yield of Nitrogen(II)oxide gas decrease.
(c)Influence of Temperature on dynamic equilibrium
A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΔH).
An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΔH).
Endothermic reaction are thus favoured by high temperature/heating
Exothermic reaction are favoured by low temperature/cooling.
If a reaction/equilibrium mixture is neither exothermic or endothermic, then a change in temperature/cooling/heating has no effect on the equilibrium position.
(i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture
Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas.
Chemical equation : 2NO2(g) ===== N2 O4 (g)
On heating /increasing temperature, the mixture becomes more brown. On cooling the mixture become more yellow.
This show that
(i)the forward reaction to the right is exothermic(-ΔH).
On heating an exothermic process the equilibrium shifts to the side that generate /liberate less heat.
(ii)the backward reaction to the right is endothermic(+ΔH).
On cooling an endothermic process the equilibrium shifts to the side that do not generate /liberate heat.
(c)Influence of Catalyst on dynamic equilibrium
A catalyst has no effect on the position of equilibrium. It only speeds up the rate of attainment. e.g.
Esterification of alkanols and alkanoic acids naturally take place in fruits.In the laboratory concentrated sulphuric(VI)acid catalyse the reaction.The equilibrium mixture forms the ester faster but the yield does not increase.
CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l)
(d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes
Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium.
For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount.
The conditions required to obtain the highest yield of products within the shortest time at minimum cost are called optimum conditions
Optimum condition thus require understanding the effect of various factors on:
(i)rate of reaction(Chemical kinetics)
(ii)dynamic equilibrium(Chemical cybernetics)
1.Optimum condition in Haber process
Chemical equation
N2 (g) + 3H2 (g) ===Fe/Pt=== 2NH3 (g) ΔH = -92kJ
Equilibrium/Reaction rate considerations
(i)Removing ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia. More/higher yield of ammonia is attained.
(ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of ammonia is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 500atmospheres is normally used.
(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -92kJ) . Ammonia formed decomposes back to Nitrogen and Hydrogen to remove excess heat therefore a less yield of ammonia is attained. Very low temperature decrease the collision frequency of Nitrogen and Hydrogen and thus the rate of reaction too slow and uneconomical.
An optimum temperature of about 450oC is normally used.
(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation.
2.Optimum condition in Contact process
Chemical equation
2SO2 (g) + O2 (g) ===V2O5/Pt=== 2SO3 (g) ΔH = -197kJ
Equilibrium/Reaction rate considerations
(i)Removing sulphur(VI)oxide gas once formed shift the equilibrium forward to the right to replace the sulphur(VI)oxide. More/higher yield of sulphur(VI) oxide is attained.
(ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of sulphur(VI)oxide is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 1-2 atmospheres is normally used to attain about 96% yield of SO3.
(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -197kJ) . Sulphur(VI)oxide formed decomposes back to Sulphur(IV)oxide and Oxygen to remove excess heat therefore a less yield of Sulphur(VI)oxide is attained. Very low temperature decrease the collision frequency of Sulphur(IV)oxide and Oxygen and thus the rate of reaction too slow and uneconomical.
An optimum temperature of about 450oC is normally used.
(iv)Vanadium(V)Oxide and platinum can be used as catalyst. Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation.
3.Optimum condition in Ostwalds process
Chemical equation
4NH3 (g) + 5O2 (g) ===Pt/Rh=== 4NO (g) + 6H2O (g) ΔH = -950kJ
Equilibrium/Reaction rate considerations
(i)Removing Nitrogen(II)oxide gas once formed shift the equilibrium forward to the right to replace the Nitrogen(II)oxide. More/higher yield of Nitrogen(II) oxide is attained.
(ii)Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules . Less/lower yield of Nitrogen(II)oxide is attained. Very low pressures increases the distance between reacting NH3and O2 molecules.
An optimum pressure of about 9 atmospheres is normally used.
(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -950kJ) . Nitrogen(II)oxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of Nitrogen(II)oxide is attained. Very low temperature decrease the collision frequency of Ammonia and Oxygen and thus the rate of reaction too slow and uneconomical.
An optimum temperature of about 900oC is normally used.
(iv)Platinum can be used as catalyst. Platinum is very expensive.It is:
-promoted with Rhodium to increase the surface area/area of contact.
-added/coated on the surface of asbestos to form platinized –asbestos to reduce the amount/quantity used.
The catalyst does not increase the yield of Nitrogen (II)Oxide but it speed up its rate of formation.
