(c) doc b

REVERSIBLE REACTIONS (c) doc b

CHEMICAL EQUILIBRIUM

Doc Brown's Chemistry KS4 science GCSE/IGCSE/O level Revision Notes

PART B Reversible reactions and CHEMICAL EQUILIBRIUM

What is a chemical equilibrium? Why can all components of a reaction co-exist? Section B moves on from the idea of a reversible reaction to look at what we mean by a chemical equilibrium. The rules governing the position of a chemical equilibrium are described and explained with examples. The effects of changing temperature, pressure or concentration on the position of a chemical equilibrium are discussed. These revision notes on chemical equilibrium, should prove useful for the new AQA GCSE chemistry, Edexcel GCSE chemistry & OCR GCSE chemistry (Gateway & 21st Century) GCSE (91), (9-5) & (5-1) science courses.

Index: A Reversible Reactions  *  B Reversible reactions & equilibrium (this page)  *  C Haber Synthesis of ammonia

D(a) The manufacture and uses of ammonia-nitric acid-fertilisers  *  D(b) Fertilisers-environmental problems

E The nitrogen cycle 

Advanced A Level Notes - Chemical Equilibrium * Advanced A Level Notes - Nitrogen & Ammonia


PART B 2. Reversible reactions and Equilibrium

  • REMINDERS about reversible reactions before learning about CHEMICAL EQUILIBRIUM
    • A reversible reaction is a chemical change in which the products can be converted back to the original reactants under suitable conditions.
    • This means the reaction can go in either direction i.e.
      • A + B ==> C + D or C + D ==> A + B
    • In a reversible reaction, changing the reaction conditions e.g. concentration, pressure or temperature will change the net direction the reaction goes i.e. more to the right (forward) or more to left (backward).
      • This idea becomes important in understanding chemical equilibrium, which is all about the state of balance between relative amounts of reactants and products in a reversible reaction situation.
    • A reversible reaction is shown by the sign (c) doc b, but on this page you must now think of it as an equilibrium sign too.
    • Conventions used in talking about reversible reactions and a chemical equilibrium
      • It is really important you understand that the terms right & left AND forward & backward are used in the context of how the equation of a reversible reaction is presented,
      • and a half-arrow to the right means the direction of the forward reaction,
      • and a half-arrow to the left means the direction of the reverse or backward reaction.
      • If you say the equilibrium is over to the right, you mean there are more products and than reactants.
      • If you say the equilibrium is over to the left, you mean there are more of the original reactants and than products formed.
  • So, WHAT IS A CHEMICAL EQUILIBRIUM?
    • When a reversible reaction occurs in apparatus which stops the escape of any reactants or products, an equilibrium (a chemical balance) is reached when the forward and reverse reactions occur at exactly the same rate and all of the concentrations of reactants and products stay the same.
      • This experimental situation is called a closed system, it may be a sealed container of gases or a non-volatile liquid mixture or a solution.
  • When a reversible reaction occurs in a closed system an equilibrium is formed, in which the original reactants and products formed coexist in the same reaction mixture AND the concentrations of all components in the mixture remain constant.
    • The 'closed system' might be a beaker of a solution containing a reaction mixture in the school laboratory or gaseous reactants in an enclosed reactor chamber in the chemical industry.
    • A closed system quite simply means nothing can escape from the reaction mixture.
  • In an equilibrium there is a state of balance between the concentrations of the reactants and products and once a state of chemical equilibrium is reached there is no further change in concentrations BUT the reactions don't stop!
    • Because neither the forward reaction or backward reaction stops, and the concentrations do not change, the situations is referred to as a dynamic equilibrium.
  • At equilibrium 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.
    • In other words the rate of the forward reaction (L to R) is equal to the rate of the backward reaction (R to L),
    • AND, these forward and backward reactions go on all the time, but sort of cancel each other out because the two rates of reaction are equal.
  • However the final relative equilibrium amounts/concentrations of the reactants and products depends on the reaction conditions e.g. the temperature and pressure.
  • A good example of a chemical equilibrium is the reversible reaction formation of an ester in organic chemistry
  • Ethyl ethanoate, an ester,  is formed by the reaction of ethanoic acid with ethanol e.g.
  • ethanoic acid + ethanol (c) doc b ethyl ethanoate + water
  • (c) doc b + (c) doc b (c) doc b (c) doc b + H2O
  • Its an equilibrium, and starting with the pure acid plus pure alcohol you get about 2/3rds conversion to the ester, and the reaction is catalysed by a few drops of concentrated sulphuric acid.
    • Conversely, if you start with equal amounts ester and water, the reaction will go 'backwards' and about 1/3rd of the ester and water will change back to the alcohol and acid.
    • The point here is that you end up with a chemical equilibrium involving all four reactants & products.
  • As the equation is written, left to right is called the forward reaction, and right to left the backward reaction.
    • The terms forward (L to R) and backward (R to L) must be used in the context of the direction the reversible reaction equation is written i.e. for the above chemical equilibrium

    • forward: CH3COOH + CH3CH2OH ==> CH3COOCH2CH3 + H2O (esterification)

    • backward: CH3COOCH2CH3 + H2O ==> CH3COOH + CH3CH2OH (hydrolysis)

  • In this example all four components of the equilibrium co-exist with unchanging concentrations unless some change is imposed on the system
    • e.g. change in temperature or addition of any of the four components of the reaction.
    • If any such change is imposed on the system, then the reaction will go more to the right or more to the left to re-establish the equilibrium.
    • All four components can co-exist because the energy changes involved are not sufficient to promote the reaction 100% one way or the other.
    • The relative amounts of the components in an equilibrium depends on the conditions e.g. the particular temperature, pressure or concentrations.
  • WHAT CHANGES THE POSITION OF AN EQUILIBRIUM?
    • By 'position of an equilibrium' we mean what are the relative amounts of reactants and products.
    • AND, importantly, what changes the position of an equilibrium?
    • In other words, what factors affect the position of an equilibrium?
    • In a reversible reaction, changing the reaction conditions e.g. concentration, pressure or temperature will change the net direction the reaction goes i.e. more to the right (forward) or more to left (backward) and this must inevitably change the position of the equilibrium.
    • If you enforce a change on a chemical system at equilibrium, then the system will respond to alter the equilibrium position, BUT the system responds in a way to minimise the enforced change.
    • The system continues to change until equilibrium is re-established.
    • The most important factors to consider that strongly influence the position of an equilibrium are temperature, pressure (if gases) and concentration (if solution).

     

What are the RULES GOVERNING THE POSITION OF A CHEMICAL EQUILIBRIUM?

For industrial processes, it is important to maximise the concentration of the desired products and minimise the 'leftover' reactants. A set of rules can be used to predict the best reaction conditions to give the highest possible yield of product.

The three rules outlined below are known as Le Chatelier's Principle.

This essentially states that if a change is imposed on a system, the system will change to minimise the enforced change to re-establish equilibrium e.g. if you increase pressure, the system will try to reduce the number of gas molecules by moving in the direction of the side of the equation with the least number of gaseous molecules.


Le Chatelier's Principle: Rule 1 The effect of temperature change on the position of an equilibrium

Reminder: If a forward reaction is exothermic, the reverse backward reaction is endothermic and vice versa.

If the temperature of a chemical system at equilibrium is increased then the relative amount of products at equilibrium increases in the direction of the endothermic reaction, but the relative amount of products at equilibrium decreases for an exothermic reaction.

If the temperature of a chemical system at equilibrium is decreased: then the relative amount of products at equilibrium decreases in the direction of the endothermic reaction, but the relative amount of products at equilibrium increases for an exothermic reaction because it move in that direction on cooling.

Rule 1a: If the forward reaction forming the products is endothermic, raising the temperature favours its formation increasing the yield of product (lowering the temperature decreases the yield).

So increasing temperature favours the endothermic direction reaction.

The system attempts to absorb the heat and minimise the increase in temperature.

Rule 1b: If the forward reaction forming the products is exothermic, decreasing the temperature favours its formation (increasing temperature decreases the yield).

So decreasing temperature favours the exothermic direction reaction

The system attempts to release heat to minimise the temperature decrease.

Rule 1 examples

The equilibrium between hydrogen gas, gaseous iodine and gaseous hydrogen iodide.

H2(g) + I2(g) (c) doc b 2HI(g) (plus 10 kJ of heat energy, exothermic L to R)

Increasing temperature favours the endothermic direction, backward reaction, some hydrogen iodide will decompose.

Decreasing temperature favours the exothermic reaction, so more hydrogen and iodine react to form hydrogen iodide.


Le Chatelier's Principle: Rule 2 The effect of changing pressure on the position of an equilibrium

You can increase/decrease the pressure by decreasing/increasing the volume of the gases OR increasing/decreasing the concentration of gases in the same volume.

For reactions involving gases at equilibrium, an increase in pressure causes the equilibrium position to shift towards the side with the smallest number of gaseous molecules as indicated by the balanced symbol equation for that reaction.

A decrease in pressure of chemical reaction system involving gases, causes the equilibrium position to shift towards the side with the larger number of gaseous molecules as indicated by the balanced symbol equation for that reaction.

A correctly balanced equation is important, because ALL gaseous molecules shown in the equation must be taken into account.

Rule 2a: Increasing the pressure favours the side of the equilibrium with the least number of gaseous molecules as shown by the balanced symbol equation.

So increasing pressure favours the reaction direction to reduce the number of gaseous molecules.

The system is changing to minimise the impact of the increase in pressure by removing some gas molecules.

Rule 2b: Decreasing the pressure favours the side of the equilibrium with the most number of gaseous molecules as shown by the balanced symbol equation.

So decreasing pressure favours the reaction direction to produce the most gaseous molecules.

The system is changing to minimise the impact of the decrease in pressure by increasing the number of gas molecules.

Rule 2 examples

(i) The formation of ammonia from nitrogen and hydrogen

N2(g) + 3H2(g) (c) doc b 2NH3(g)

4 gas molecules ==> 2 gas molecules, so to re-establish a dynamic equilibrium ...

Increase in pressures favours the forward reaction to reduce the number of gas molecules, so more ammonia formed.

Decrease in pressure encourages the formation of more gas molecules, so some of the ammonia decomposes into nitrogen and hydrogen.

(ii) The thermal decomposition of dinitrogen tetroxide into nitrogen dioxide

N2O4(g) (c) doc b 2NO2(g)

1 gas molecule ==> 2 gas molecules, so to re-establish a dynamic equilibrium ...

Increase in pressure favours backward direction to reduce the number of gaseous molecules and give more dinitrogen tetroxide.

Decrease in pressure encourages more gas molecules to form, so the forward reaction gives more nitrogen dioxide.

(iii) The combining of nitrogen and oxygen to form nitrogen monoxide

N2(g) + O2(g) (c) doc b 2NO(g)

2 gas molecules ==> 2 gas molecules

Change in pressure has no effect on equilibrium position.

Note: Nitrogen monoxide is also known as 'nitric oxide' or 'nitrogen(II) oxide'

PLEASE NOTE

Rules 1 on temperature and rule 3 on concentration (below), apply to any reaction, BUT rule 2 on pressure above, ONLY applies to a reaction with one or gaseous reactants or gaseous products.

Increase in pressure does not influence the concentration of substances in a solution or solid mixture because they are too dense to be significantly compressed i.e. no effective change in concentration.

The situation is quite different in gases where is a lot of space between the molecules to compress them closer together.

If a reaction involves gases BUT there are equal numbers of gaseous molecules on each side of the equation, increasing or decreasing pressure has no effect on the position of the equilibrium.

e.g the equilibrium position of the reaction to form hydrogen iodide from hydrogen and iodine

H2(g) + I2(g) (c) doc b 2HI(g)

is unaffected by change in pressure, the are two molecules (or moles) of gas on each side of the equation.


Le Chatelier's Principle: Rule 3 The effect of changing concentration on the position of an equilibrium

If the concentration of any of the reactants or products is changed, the system cannot any longer be at equilibrium.

The concentrations of all the substances will change until equilibrium is reached again.

If the concentration of a reactant is increased, more products will be formed until equilibrium is reached again.

If the concentration of a product is decreased, more reactants will react until equilibrium is reached again.

Rule 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.

Rule 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.

Rule 3 examples

  • e.g. nitrogen + hydrogen (c) doc b ammonia
    • or N2(g) + 3H2(g) (c) doc b 2NH3(g)
    • If the nitrogen or hydrogen concentration was increased, some of this extra gas would change to ammonia.
    • If the nitrogen or hydrogen concentration was decreased, some of ammonia would change back to nitrogen and hydrogen.
    • At advanced A level things can get more complicated e.g. can you figure out why in terms of concentration to maintain the equilibrium balance? (and if a gcse student, don't worry if you can't) ...
    • So in terms of enforced change ==> system response:
    • Increasing nitrogen concentration ==> decreases hydrogen concentration and increases ammonia concentration
    • Increasing hydrogen concentration ==> decreases nitrogen concentration and increases ammonia concentration
    • Increasing ammonia concentration ==> increases both nitrogen and hydrogen concentrations
    • Decreasing ammonia concentration ==> decreases both nitrogen and hydrogen concentration
    • Decreasing nitrogen concentration ==> increases hydrogen concentration and decreases ammonia concentration
    • Decreasing hydrogen concentration ==> increases nitrogen concentration and decreases ammonia concentration

Le Chatelier's Principle: Rule 4 The effect of using a catalyst on the position of an equilibrium

A catalyst does NOT affect the position of an equilibrium.

You just get to the equilibrium position here faster!

A catalyst usually speeds up both the forward and reverse reaction but there is no way it can influence the final 'balanced' concentrations.

However, the importance of a catalyst lies with economics e.g.

(i) bringing about reactions with high activation energies at lower temperatures and so saving the cost on energy,

(ii) and saving time is saving money, i.e. a catalyst increases the efficiency of the chemical process e.g. the Haber synthesis of ammonia.

Rule 4 examples

Iron catalyst in the synthesis of ammonia.

Vanadium pentoxide catalyst in the Contact Process for manufacturing sulfuric acid.

Both of these chemical processes are faster and made economically more efficient by use of a catalyst, but you don't get a greater % yield in the final reacted mixture because that's controlled by rules 1 to 3.

 

Advanced Level Notes on Chemical Equilibrium


Part B contd. Applying the rules 1 to 4 to some chemical processes

(a) The formation of calcium oxide (lime) and carbon dioxide from calcium carbonate (limestone)

CaCO3(s) (c) doc b CaO(s) + CO2(g)

The forward reaction is endothermic, 178kJ of heat energy is absorbed (taken in) for every mole of calcium oxide formed.

One mole of gas is formed in the process, so there is a net increase in the moles of gas in lime formation, since there are no gaseous reactants.

From rule 1: increasing the temperature will increase the yield of calcium oxide or lime, CaO which is endothermically formed.

From rule 2: decreasing the pressure will favour the formation of more gas molecules if possible, so more carbon dioxide formed, and hence more lime.

Lime is made commercially by heating limestone to a high temperature (e.g. 1000oC) in a limekiln that is well ventilated (this reduces the carbon dioxide pressure and so reduces the un-desired backward reaction).


(b) The formation of hydrogen chloride from hydrogen and chlorine

H2(g) + Cl2(g) (c) doc b 2HCl(g)

The forward reaction is very exothermic, 184kJ of heat energy is given out in forming hydrogen bromide according to the above equation (184/2 = 92kJ per mole of HCl formed).

There is no net change in the moles of gas (2 moles reactants (c) doc b 2 moles of product)

From rule 1: decreasing the temperature favours the exothermic formation of hydrogen chloride, so the equilibrium moves proportionately to the right-hand side (more HCl, less H2 or Cl2). If hydrogen chloride is heated to a very high temperature, endothermic direction, then more HCl decomposes into H2 or Cl2.

From rule 2: since there is no net change in the number of moles of gas on reaction, pressure has no effect on the yield of hydrogen chloride and the proportions of HCl, H2 and Cl2 stay the same.


(c) The formation of ammonia

See The Haber Synthesis of ammonia for a more detailed discussion,

but a summary of what the rules say is given below in terms of ammonia production!

nitrogen + hydrogen (c) doc b ammonia

N2(g) + 3H2(g) (c) doc b 2NH3(g)      (plus 92 kJ heat energy, exothermic)

Ammonia formation is favoured by ...

Rule 1. Lowering the temperature, because it is an exothermic reaction, but this may make it too slow, compromise required.

Rule 2. Increasing pressure, because there is a reduction in the molecules of gas, but the higher the pressure, the more costly the engineering.

Rule 4. An iron catalyst speeds up the reaction, but has no effect on the % ammonia in the reacted mixture exiting the reactor chamber in the chemical plant of a Haber synthesis process.

In the exam you may have to explain more about the rule and how they are used, which is what this page is all about, plus some more examples below


(d) The manufacture of sulfur trioxide

The process of making sulfur trioxide from sulfur dioxide is one stage in the manufacture of sulfuric acid by the Contact Process.

The equilibrium equation is given below.

2SO2(g) + O2(g) (c) doc b 2SO3(g)     (plus 95 kJ heat energy, exothermic)

3 gas molecules ==> 2 gas molecules.

Rule 1. The exothermic reaction is favoured by a lower temperature, but this may be too slow, so a compromise temperature of around 450oC is used, which gives a fast economic rate of sulphur trioxide production.

Rule 2. The reaction is favoured by high pressure (pressure equilibrium rule, 3 => 2 gas molecules, LHS ==> RHS), but only a small increase in pressure is used to give high yields of sulphur trioxide, because the formation of SO3 on the right hand side is so energetically favourable (approx. 99% yield, i.e. only about 1% SO2 unreacted).

Rule 4. The use of the V2O5 catalyst ensures a fast reaction without having to use too a higher temperature which would favour the left hand side and reduce the yield BUT it does not change the % of sulphur trioxide formed, you simply get there faster!


Index: A Reversible Reactions  *  B Reversible reactions and Chemical Equilibrium

 C The Haber Synthesis of ammonia  *  D(a) The Uses of ammonia-nitric acid-fertilisers 

D(b) Fertilisers-environmental problems  *  E The nitrogen cycle 

(c) doc b Foundation tier (easier) multiple choice QUIZ on ammonia, nitric acid and fertilisers etc.

(c) doc b Higher tier (harder) multiple choice QUIZ on ammonia, nitric acid and fertilisers etc.

Advanced A Level Notes on Equilibrium (use indexes)

Advanced A Level Chemistry Notes p-block nitrogen & ammonia


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