Reversible

1.1 Reversible Reactions

A reversible reaction is a chemical change in which the products can be converted back to the original reactants under suitable

conditions. A reversible reaction is shown by the sign , a half-arrow to the right (forward reaction, L to R), and, a half-arrow to

the left (backward reaction, R to L).

Reactions which are not reversible (irreversible) have the usual complete arrow only pointing to the right.

Example 1.1.1 On heating strongly, the white solid, ammonium chloride, thermally decomposes into a mixture of two colourless gases,

ammonia and hydrogen chloride. On cooling the reaction is reversed and solid ammonium chloride reforms. This is also an example

of sublimation and involves both a physical of state as well as a chemical change.

When a substance sublimes it changes directly from a solid into a gas without melting and on cooling reforms the solid without

condensing to form a liquid.

o Ammonium chloride + heat ammonia + hydrogen chloride

o NH4Cl(s) NH3(g) + HCl(g)

o Note: Reversing the reaction conditions reverses the direction of chemical change, typical of a reversible reaction.

o Thermal decomposition means using 'heat' to 'break down' a molecule into smaller ones. The decomposition of NH4Cl is

endothermic, ΔH +ve (heat absorbed, taken in from the surroundings) and the formation of NH4Cl is exothermic, ΔH -ve (heat

released, given out to the surroundings).

o This means if the direction of chemical change is reversed, the energy change sign must also be reversed but its numerical

value stays the same.

Example 1.1.2 On heating the blue solid, hydrated copper(II) sulphate, steam is given off and the white solid of anhydrous copper(II)

sulphate is formed. When the white solid is cooled and water added, blue hydrated copper(II) sulphate is reformed.

o blue hydrated copper(II) sulphate + heat white anhydrous copper(II) sulphate + water

o CuSO4.5H2O(s) CuSO4(s) + 5H2O(g)

o Note: The crystal structure is broken down on heating and the water of crystallisation is given off. Thermal decomposition is

endothermic (ΔH +ve) as heat is absorbed to drive off the water. The reverse reaction is exothermic (ΔH +ve) i.e. on adding water

to cold white anhydrous copper(II) sulphate the mixture heats up as the blue crystals reform.

o These are typical examples you encounter at an earlier study level, but it begs the question, "is it possible to have a situation,

under suitable conditions, in which the reaction does not completely go in one direction or the other and both reactants

and products co-exist?", and the answer is yes! and the situation is called a dynamic chemical equilibrium. The word dynamic

is used because the 'forward' (L to R) and 'backward' (R to L) reactions do not cease but match each other in rate so the

concentrations of reactants and products are constant when the equilibrium is established.

1.2 Reversible reactions and the concept of a dynamic chemical equilibrium

Although most reactions you have encountered at an earlier academic level did go to 100% completion, it is a fact that many reactions

do NOT go to completion i.e. 100% yield from the forward reaction.

If ammonium chloride were heated in a closed system, over a certain temperature range, some of the NH4Cl will be sublimed into the

gases NH3 and HCl and some of the solid salt remains.

o A closed system means nothing can enter or leave the system.

http://www.docbrown.info/page07/equilibria1.htm

o When a reversible reaction occurs in a closed system, depending on conditions, a chemical equilibrium is formed, in which

the original reactantsand products formed coexist. In other words the reaction (i.e. from left to right as the equation is

written) never goes to completion.

o Eventually the 'system settles down' and the net concentrations of the reactants and products remain constant i.e. a state of

concentration balance exists.

o BUT the reactions don't stop! Reactants are continually forming products, and the products are continually re-forming the original

reactants, hence the termdynamic equilibrium.

o In terms of kinetics ('rates of reaction'), it means that the

rate of formation of product = rate of re-formation of reactants,

or the rate of the forward reaction = rate of the backward reaction

Example 1.2.1 The formation/decomposition of hydrogen iodide.

o hydrogen + iodine hydrogen iodide (2 mol gas ==> 2 mol gas)

o H2(g) + I2(g) 2HI(g) (all gases above 200oC)

o L to R forward reaction: If you start with pure hydrogen and pure iodine, so much of them combines to form hydrogen iodide.

o R to L backward reaction: If you start with pure hydrogen iodide, some, but not all of it, will decompose into hydrogen and

iodine.

o Starting with the same total number of moles of either H2 + I2 or HI, the final equilibrium concentrations will be the same at

the same temperature, volume and pressure. This is illustrated in the diagram below showing the fate of 2 mol of reacting gases.

o

o Graph lines (1) and (2) show what happens if you start with 2.0 mol of pure hydrogen iodide which decomposes 50%, for the

sake of argument and mathematical simplicity, into hydrogen and iodine.

Graph line (1) shows the gradual 50% reduction of HI from 2 mol to 1 mol.

Graph line (2) shows the gradual formation, from 0 mol of each, of 0.5 mol H2 and 0.5 mol I2.

o Graph lines (3) and (4) show what happens if you start with 1.0 mol of hydrogen plus 1.0 mol of iodine and no hydrogen iodide.

Graph line (3) shows the 50% reduction of 1.0 mol of H2 or I2 to 0.5 mol of each.

Graph line (4) shows the formation of 1.0 mol of HI from the net reaction of 0.5 mol H2 and 0.5 mol I2.

o Note:

1. The final equilibrium composition is the same in each case no matter which direction you started from for the same total

moles of gas.

2. Where the graph lines first become horizontal, meaning no further net change in concentration, the equilibrium point

was first reached i.e. here, after about 32 minutes.

3. See also K c expression and application of Le Chatelier's Principle

Example 1.2.2 The formation and hydrolysis of the ester ethyl ethanoate.


o ethanoic acid + ethanol ethyl ethanoate + water (hydrolysis <==> esterification)

o CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)

o L to R forward reaction: If equimolar amounts of pure ethanoic acid and pure ethanol are refluxed with a few drops of conc.

H2SO4(l) catalyst, about 2/3rds of the initial reactants are converted into the ester.

o The forward or backward reactions are slow at room temperature without heating and employing a catalyst, but equilibrium would

eventually be reached at room temperature even in the absence of a catalyst.

o R to L backward reaction: If equimolar amounts of pure ethyl ethanoate and pure water are mixed, eventually about 1/3rd of the

ester reverts back to ethanoic acid and ethanol.

o Note: If the ester is refluxed with lots of acidified water, the ester is 100% hydrolysed back to the original acid and alcohol.

o See also Equilibria Part 2 K c expressions

Example 1.2.3 The formation/decomposition of ammonia.

o nitrogen + ammonia ammonia (Haber synthesis)

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

o L to R forward reaction: This is the direction of interest to the chemical industry as ammonia is an important chemical for fertiliser

and nitric acid manufacture.

o For reasons explained later (The Haber Synthesis in Equilibria Part 3) a mere 8% yield (L to R) is commercially acceptable,

even if it doesn't sound much.

o See also Equilibria Part 2 K c expression , or Kp expression and Le Chatelier's Principle (this page)

Some important outcomes from experimentally studying dynamic equilibrium reactions

1. It doesn't matter whether you start with the 'reactants' or the 'products', either way, if the conditions are suitable, both are

present when a state of equilibrium exists.

2. Eventually the net concentrations of the reactants and products remain the same BUT the forward and backward

reactions don't stop.

3. When a dynamic equilibrium is achieved there is a state of balance between the constant concentrations of the reactants and

products because the rate at which the reactants change into products is exactly equal to the rate at which the products

change back to the original reactants.

4. However, the actual relative amounts of the original reactants left, and products formed, at equilibrium, depend on the

particular reaction and reaction conditions e.g. the initial concentrations, temperature and pressure (if gaseous reactants or

products are involved) and the value of the equilibrium constant (see on this page Le Chatelier's Principle below and Equilibria

Part 2).

5. A catalyst does not affect the position of the equilibrium, i.e. the final constant concentrations are the same with or without a

catalyst, you simply get to the equilibrium point faster with a catalyst!

6. In some cases you can adjust reaction conditions sufficiently to make the reaction go virtually 100% in one direction e.g.

example 1.2.2, the hydrolysis of an ester.

7. At a given constant temperature, all the final equilibrium concentrations are mathematically governed by the equilibrium

expression and the equilibrium constant and these are dealt with in detail in Equilibria Part 2 .

1.3 Le Chatelier's Principle

This is initially described without reference to equilibrium constants and equilibrium expressions which are fully explained and

described inEquilibria Part 2.


Le Chateliers's Principle states that if an 'instantaneous' change is imposed on an equilibrium, the position of the equilibrium

will further change to minimise the 'enforced' change.

In other words, if a change is 'instantaneously' imposed, the equilibrium attempts to restore the original situation, but it cannot do this

completely BUT the change 'trend' can be predicted. Also, when considering the equilibrium rules outlined below, any change affects

BOTH the rates of the forward and backward reactions.

o Rule 1 - Temperature and energy changes (ΔH)

1a. Raising the temperature favours the endothermic direction (ΔH +ve). The system absorbs the heat energy from

the surroundings to try to minimise the temperature increase.

1b. Decreasing the temperature favours the exothermic direction (ΔH -ve). The system releases heat energy to the

surroundings to try to minimise the temperature decrease.

o Rule 2 - Gas pressure changes at constant temperature (ΔV)

2a. Increasing the pressure favours the side of the equilibrium with the least number of gaseous molecules as

indicated by the balanced symbol equation. The system attempts to reduce the number of gas molecules present to

reduce the pressure increase.

2b. Decreasing the pressure favours the side of the equilibrium with the most number of gaseous molecules as

indicated by the balanced symbol equation. The system attempts to increase the number of gas molecules to minimise

the pressure decrease.

NOTE:

States symbols (g/l/s/aq) are particularly important when considering equilibrium equations, if no (g) the

pressure rule doesn't apply since solids and liquids are virtually incompressible.

Rule 2 ONLY applies to a reaction with one or more gaseous reactants or products because pressure has

no real effect on the 'concentration' on the virtually incompressible liquids or solids.

If there is NO net change in the number of gas molecules, gas pressure has NO effect on the position of

the equilibrium, though pressure increase effectively increases gas concentration so both the forward and

backward reactions will be speeded up.

Over and above rule 2, all the individual partial pressures of the gases, must comply with the mathematics of

the Kp equilibrium expression described and explained in Equilibria Part 2.1b (also explains what partial

pressures are).

The rest of the rules 1, 3 and 4 apply to ANY reaction, whatever the physical states of the reactants and

products.

o Rule 3 - Concentration changes at constant temperature

3a. If the concentration of a reactant (on the left) is increased, then some of it must change to the products (on the right)

to maintain a balanced equilibrium position.

3b. If the concentration of a reactant (on the left) is decreased, then some of the products (on the right) must change

back to reactants to maintain a balanced equilibrium position.

This means if you change ANY concentration, all the other concentrations must change too (see Example

1.4.2 below).

Also, any net concentration changes must comply with the Kc equilibrium expression (fully explained in Equilibria

Part 2.1a)

o Rule 4 - Using a catalyst

A catalyst does NOT affect the position of an equilibrium.

BUT it does is enable you get to the point of equilibrium faster!

A catalyst speeds up both the forward and reverse reactions by providing a mechanistic pathway with a lower

activation energy, but there is no way it can influence the final 'balanced' concentration ratios.

The importance of a catalyst lies with economics of chemical production e.g. bringing about reactions with high

activation energies at lower temperatures and so reducing energy requirements and time, and both reductions save

money!

1.4 Applying Le Chatelier's Principle and the equilibrium rules

These are initially described without reference to equilibrium constants and equilibrium expressions which are fully explained

and described inEquilibria Part 2 (where some examples are repeated with their Kc or Kp equilibrium expressions)

Example 1.4.1

o The thermal decomposition of calcium carbonate (limestone) to make calcium oxide (quicklime):

o CaCO3(s) CaO(s) + CO2(g) (ΔH = +178 kJ mol-1)

Note: By convention, the ΔH value quoted corresponds to the forwards reaction (L to R), reversing the sign gives the

ΔH for the backward reaction (R to L).

For more details of the industrial process see Equilibria Part 3.1

o Rule 1 - temperature and energy change (ΔH)

Increasing temperature favours the endothermic direction (RHS) so more quicklime is formed.

o Rule 2 - gas pressure (ΔV)

Decreasing the partial pressure of carbon dioxide increases the yield of quicklime (0 ==> 1 mole gas).


o Rule 3 - concentration

Not applicable to the reactants because you can't decrease or increase the concentration of a solid but you can reduce

the concentration of carbon dioxide by venting the gases to increase the yield of quicklime.

o Rule 4 - catalyst: Not applicable.

Example 1.4.2

o The synthesis of ammonia: nitrogen + hydrogen ammonia

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


The forward, and desirable reaction, to form ammonia, is exothermic, so lowering the temperature favours its formation.


Increase in pressure favours ammonia formation since 4 mol of gaseous reactants ==> 2 mol gaseous products.

o Rule 3 - concentration: In terms of enforced change => system response

If the nitrogen or hydrogen concentration was increased, some of this 'extra' gases would change to ammonia.

If the nitrogen or hydrogen concentration was decreased, some of ammonia would change to nitrogen and hydrogen.

A 1:3 ratio N2:H2 mixture is used in industry, but for academic reasoning practice, specifically ...

Increasing nitrogen ==> decreases hydrogen and increases ammonia.

Increasing hydrogen ==> decreases nitrogen and increases ammonia.

Decreasing ammonia ==> decreases nitrogen and hydrogen.

Decreasing nitrogen ==> increases hydrogen and decreases ammonia.

Decreasing hydrogen ==> increases nitrogen and decreases ammonia.

o Rule 4 - catalyst: An iron oxide catalyst is used, time = money for industrial chemical production!

o More details of the Haber process are given in Equilibria Part 3 .

Example 1.4.3

o The formation of hydrogen iodide from hydrogen and iodine:

o H2(g) + I2(g) 2HI(g) (ΔH = -10 kJ mol-1, iodine gaseous above 200oC)


Increasing temperature favours the LHS, i.e. increases the endothermic decomposition of hydrogen iodide.

o Rule 2 - gas pressure (ΔV):

No effect on position of equilibrium, 2 mol gas ==> 2 mol gas, no net change in gas moles.


e.g. if more iodine was added to a constant volume container, the hydrogen concentration or partial pressure would

decrease as some reacts with added iodine to give more hydrogen iodide as the system tries to minimise the iodine

increase. Please note that there would still be an overall increase in iodine at the new equilibrium point.


o See also section 1.2.1, 2.1a.1 on equilibrium-rates connection and calculation 2.2a.1

Example 1.4.4

o Esterification: e.g. ethanoic acid + ethanol ethyl ethanoate + water

o CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) (ΔH = -2 kJ mol-1)


Decrease in temperature favours the ester formation (ethyl ethanoate, RHS) but in practice a small decrease in yield at

higher temperatures is acceptable because heating under reflux is needed to efficiently prepare the ester.

o Rule 2 - gas pressure (ΔV): Not applicable, no gases involved.


Sometimes it is desirable to add a large excess of the alcohol to ensure most of the acid is converted into ester. The

alcohol may be cheaper than the acid and the ester and unreacted alcohol separated by distillation and the latter recycled.

o Rule 4 - catalyst

The forward esterification reaction is catalysed by acids e.g. a few drops of conc. sulphuric acid.


The reverse reaction i.e. hydrolysis of the ester back to the acid and alcohol is catalysed by dilute acids.

Example 1.4.5

o The equilibria between oxygen O2, nitrogen (II) oxide NO, nitrogen(IV) oxide NO2 and its dimer N2O4.

o NO2 can be made from the irreversible thermal decomposition of lead(II) nitrate in a pyrex boiling tube connected to a 100

cm3 gas syringe in a fume cupboard.

lead(II) nitrate ==> lead(II) oxide + nitrogen(IV) oxide + oxygen

2Pb(NO3)2(s) ==> 2PbO(s) + 4NO2(g) + O2(g)

o (a) 2NO2(g, brown) 2NO(g, colourless) + O2(g, colourless) (ΔH = +113 kJ mol-1)

(a) The temperature effect can be observed by strongly heating the gases in the pyrex tube above 400oC.

o (b) 2NO2(g, brown) N2O4(g, colourless) (ΔH = -58 kJ mol-1)

(b) The temperature effect can be observed by cooling and warming below 100oC.

(b) The pressure effect can be observed by sealing the cool gases in the gas syringe and compressing and

decompressing it.


(a) Increases in temperature favours the endothermic decomposition of NO2 to NO and O2, so at high temperatures the

brown colour fades.

(b) Decrease in temperature favours the exothermic formation of the dimer N2O4 from NO2, so the brown colour fades on

cooling the gas mixture.


(a) Increase in pressure favours the LHS, more NO2, because 2 mol gas <== 3 mol gas, so theoretically the mixture would

get darker.

(b) Increase in pressure favours N2O4 formation from NO2, 2 mol gas ==> 1 mol gas, so the mixture would get lighter in

colour.

(b) This can be demonstrated by compressing/decompressing the gas mixture in the syringe to see the brown

colour intensity increase/decrease.

In fact you can even see the dynamic equilibrium 'kinetics' in operation here. There is a time lag of about 1-2

seconds before the new equilibrium position is established as the 'imposed' colour intensity change becomes

constant.

o Rule 3 - concentration: e.g.

(a) Theoretically an increase in O2 would lead to decrease in NO and increase in NO2, so the mixture would get darker.

(b) Increase in NO2 would increase N2O4, but overall the colour would still be darker because not all of the 'extra' NO2 can

be converted to maintain the equilibrium.


Example 1.4.6

o Pink hexa-aqua cobalt(II) ions form a complex with chloride ions, the blue tetrachlorocobaltate(II) ion.

o [Co(H2O)6]2+(aq, pink) + 4Cl-

(aq) [CoCl4]2-(aq, blue) + 6H2O(l) (ΔH +ve, don't know value)


Increase in temperature favours the endothermic blue complex formation, cooling favours the exothermically form pink

ion.

If prepared at higher temperature, with just enough chloride added to turn the solution blue, on cooling it becomes pink.

o Rule 2 - gas pressure (ΔV): Not applicable, no gases involved.


Increase in chloride concentration decreases the pink ion concentration and increases the blue ion concentration i.e. shifts

the equilibrium position from left to right.

Diluting with water shifts the equilibrium to the left so solution is less blue and more pink.

o Rule 4 - catalyst: Not applicable, despite involving transition metal complexes!

o For more details of other equilibrium see Inorganic Chemistry Part Transition Metals (In preparation)

Example 1.4.7

o The formation of nitrogen(II) oxide.

o N2(g) + O2(g) 2NO(g) (ΔH = +181 kJ mol-1)


Increase in temperature favours the endothermic formation of NO.

This reaction does not happen at room temperature but is formed at the high temperatures in car engines.

Unfortunately when released through the car exhaust, it cools to normal temperatures when NO irreversibly reacts with

oxygen in air to form nitrogen(IV) oxide, NO2, which is acidic, a lung irritant and a reactive free radical molecule involved in

the chemistry of photochemical smog not good!


Since 2 mol gaseous reactants gives 2 mol gaseous products, pressure does not affect the position of the equilibrium.


The concentration of nitrogen is high from air, but although the concentration of oxygen is low in the exhaust gases, there

is sufficient present in the combustion process to ensure a small % of NO is formed.


Example 1.4.8

o The oxidation of sulphur dioxide to sulphur trioxide

o e.g. in the Contact Process for manufacturing sulphuric acid, see Equilibria Part 3.3 .

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


The exothermic formation of sulphur trioxide is favoured by low temperature.


Higher pressure favours a higher yield of sulphur trioxide as 3 gas moles ==> 2 gas moles, though 1-2 only atm is used in

practice because the equilibrium is already so far to the right (about 99%).



Air is used as the source of oxygen and despite its dilution with nitrogen the concentration, the oxygen concentration is

high enough to move the equilibrium very much to the RHS.

o Rule 4 - catalyst

A vanadium(V) oxide, V2O5, catalyst ensures the high yield of 99% SO3 is attained fast, but no more!

Example 1.4.9

o One way to produce hydrogen for the Haber synthesis of ammonia is to react methane gas with steam.

o CH4(g) + H2O(g) 3H2(g) + CO(g) (ΔH = +206 kJ mol-1)


Increase in temperature favours the endothermic formation of hydrogen (and carbon monoxide).


For the desired forward reaction, 2 mol of reactant gases ==> 4 mol of product gases, so the increase in product volume

is favoured by lower pressure.


Theoretically increase in methane and steam concentrations will increase the hydrogen concentration, but this essentially

means increasing pressure favouring the LHS, so you might not gain as much hydrogen as you like to!

o Rule 4 - catalyst

Reversible

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