Chapter 11.3 - Factors which affect Equilibrium of a Chemical Reaction

In the previous section, we saw the equilibrium state of a chemical reaction. In this section we will learn about the factors which affect the Equilibrium.

Influence of concentration on Chemical equilibrium

■ Consider a system in equilibrium. If some quantities of a reactant is added to the system, what will happen? 
• Ans: The concentration of that reactant must decrease back to the previous level.  For that, the rate of forward reaction will increase, so as to convert the extra reactant to product.
Consider the example: N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
■ If we add some nitrogen to the system, what will happen?
Ans: The rate of forward reaction will increase so as to convert more nitrogen into ammonia. As a result, production of ammonia will increase.
■ If we add some ammonia to the system, what will happen?
Ans: The rate of backward reaction will increase so as to convert more ammonia back into nitrogen and hydrogen.
■ If we remove ammonia continuously from the system, what will happen?
Ans: The rate of forward reaction will increase so as to produce more ammonia

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• Effects of change in concentrations in a system producing ammonia can be written in a tabular form as follows:

Action	Change of Concentration	Change in rate
More hydrogen is added	Increases the concentration of the reactant	Rate of forward reaction increases
More ammonia is added	Increases the concentration of the product	Rate of backward reaction increases
Ammonia is removed	Decreases the concentration of the product	Rate of forward reaction increases
More nitrogen is added	Increases the concentration of the reactant	Rate of forward reaction increases

• The above discussion gives us an idea about the influence of concentrations on chemical equilibrium.

Influence of pressure on Chemical equilibrium

• Influence of pressure is felt in reactions involving gases. It can be analyzed using an example:
• Consider the reaction between nitrogen and hydrogen which gives ammonia as the product. The balanced chemical equation is: 
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
• One molecule of nitrogen reacts with 3 molecules of hydrogen to produce two molecules of ammonia. 
• So, if we take N_A (Avogadro number) of molecules of nitrogen, we must take 3N_A molecules of hydrogen. The result will be 2N_A molecules of ammonia.
• It is clear that, there are greater number of molecules on the reactant side than in the product side.
• In gaseous systems, larger number of molecules will produce greater pressure.

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Consider a state of equilibrium in the manufacture of ammonia.
■ If at this equilibrium we increase the pressure, what will happen?
Ans: The system will try to reduce that pressure. For that, the number of molecules must decrease.
• For a decrease in the number of molecules, more quantities of the product should be formed. Because, in this reaction, number of molecules on the product side is lower.
• So rate of the forward reaction will increase
■ If at this equilibrium we decrease the pressure, what will happen?
Ans: The system will try to increase the pressure. For that, the number of molecules must increase.
• For an increase in the number of molecules, more quantities of the reactants should be formed. Because, in this reaction, number of molecules on the reactants side is higher.
• So rate of the backward reaction will increase
■ In the manufacture of ammonia, why is a pressure of 200 - 900 atm used?
Ans: In the manufacture of ammonia, high pressure is used so that, the rate of the forward reaction will increase and more ammonia will be formed.

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Is an increase in pressure always helpful? Let us see another example:
Consider the balanced equation:
H₂ (g) + I₂ (g) ⇌ 2HI (g)
• One molecule of hydrogen reacts with 1 molecule of iodine to produce two molecules of hydrogen iodide. 
• So, if we take N_A (Avogadro number) of molecules of hydrogen, we must take N_A molecules of iodine. The result will be 2N_A molecules of hydrogen iodide.
• It is clear that, there are equal number of molecules on the reactant side and in the product side.
■ If there is no change in the number of reactant or product molecules as a result of forward and backward reactions, in gaseous reactions, pressure will not have any effect on the equilibrium state.

Influence of temperature on equilibrium

We have seen endothermic reactions and exothermic reactions in our earlier classes.
• Endothermic reactions are those in which heat is absorbed during the reaction
• Exothermic reactions are those in which heat is liberated during the reaction
Consider the equation of a reversible reaction:

1. In the forward reaction, N₂O₄ changes to NO₂. This forward reaction is endothermic. That is., if this forward reaction is to take place, we have to supply heat
2. In the backward reaction, NO₂ changes to N₂O₄. This backward reaction is exothermic. That is., when this backward reaction takes place, we will get heat.
3. Consider a system in which the above reaction is at equilibrium.
• If this system at equilibrium is placed in ice, the temperature will be lowered.
• The system will then try to increase the temperature by producing more heat.
• To produce heat, exothermic reaction must take place.
• So rate of the backward reaction increases. Thus So more NO₂ will be converted to N₂O₄.
4. Consider a system in which the above reaction is at equilibrium.
• If this system at equilibrium is placed in hot water, the temperature will be increased.
• The system will then try to decrease the temperature.
• To reduce temperature, heat must be absorbed. For heat to get absorbed, endothermic reaction must take place.
• So rate of the forward reaction increases. Thus So more N₂O₄ will be converted to NO₂.
■ Note that nitric acid (HNO₃) is a very corrosive acid and Nitrogen dioxide (NO₂) is a poisonous gas. All safety precautions should be taken while doing experiments.

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Let us now see the effect of temperature in the manufacture of ammonia.
1. We know that ammonia is manufactured by Haber process. It is a reversible reaction. It can be represented by the following balanced equation:
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
2. Here the forward reaction is exothermic in nature. That is., the forward reaction releases heat. So we can rewrite the equation as:
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g) + Heat
3. If we increase the temperature, the system will try to reduce it.
• For that, less heat must be produced. So the rate of backward reaction will increase. We will be getting more nitrogen and hydrogen instead of ammonia. So it is not advisable to increase the temperature.
4. If we cannot increase the temperature, can we decrease it?
• Reducing the temperature to low values has adverse effects. When the temperature is low, the rates of both forward and backward reactions become very low. (This is because, the molecules must have a minimum amount of energy for the collisions to take place)
• As a result, the system will take more time to reach equilibrium. Hence in the manufacture of ammonia, 450^o c is taken as optimum temperature.

Influence of catalyst on equilibrium

• We know that Positive catalysts are those substances which increase the rate of a reaction.
• In reversible reactions, there is forward as well as backward reactions taking place.
• A positive catalyst will increase the rate of both forward and backward reactions. So equilibrium is reached in a lesser time.
• Once equilibrium is reached, the rates of forward or backward reactions can be controlled by altering the other factors like temperature, pressure etc.,
• So, in many reversible reactions, positive catalyst is used at the beginning of the reaction.

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The French scientist Henry-Louis Le Chatelier put forward a principle regarding the effects of various factors on chemical equilibrium.
Le Chatelier's Principle:
When the concentration, pressure or temperature of a system at equilibrium is changed, the system will readjust itself so as to nullify the effect of that change. As a result of the readjustment, the system will attain a new state of equilibrium

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Now we will see a solved example:

Solved example 11.1

The manufacture of sulphuric acid by contact process has different stages. The chemical equation of one such stage is:

2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g) + heat

■ Identify how the factors given below influence this reaction

• Increase in the amount of oxygen

• Increase in pressure

• Maintains the temperature at optimum level

• Catalyst (V₂O₅) is added

• SO₃ is removed

Solution:

1. Increase in the amount of oxygen:

• Oxygen is one of the reactants.
• The system will try to nullify the effect of increase in concentration of a reactant by increasing the rate of the forward reaction.
• So more products will be formed
2. Increase in pressure:
• From the balanced equation, it is clear that, there are greater number of molecules on the reactant side than in the product side.
• In gaseous systems, larger number of molecules will produce greater pressure.
• The system will increase the rate of the forward reaction. So the number of molecules will become less and pressure will become low.
• Thus the effect of increase in pressure will be nullified
3. Maintains temperature at optimum level
• From the given balanced equation it is clear that, the forward reaction is exothermic
• If we increase the temperature, the system will try to nullify it by increasing the rate of the backward reaction. In that case we will get lesser quantities of the product.
• On the other hand, a decrease in temperature is also not advisable. Because, the molecules need a minimum amount of energy for the collisions to take place. 
• So, if we maintain an optimum temperature, we will get more products
4. Catalyst V₂O₅ is added
• The catalyst increases the rate of both forward and backward reactions. So it helps to achieve equilibrium in a lesser duration of time.
5. SO₃ is removed
• When SO₃ is removed, concentration of product decreases. To nullify the effect of this 'decrease in concentration', more SO₃ has to be produced. For that, the rate of forward reaction increases

Solved example 11.2
(a) In which of the following reactions does the change in pressure not influence equilibrium? What is the reason?
(i) H₂ (g) + I₂ (g) ⇌ 2HI (g)
(ii) N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
(b) What is the use of applying high pressure duriing the formation of ammonia from nitrogen and hydrogen?
Solution:
(a) In reaction (i), increase or decrease in pressure will not influence equilibrium. Because, the number of molecules in the reactants side and products side are equal
(b) If we increase the pressure, the system will try to reduce that pressure. For that, the number of molecules must decrease.
• For a decrease in the number of molecules, more quantities of the product should be formed. Because, in this reaction, number of molecules on the product side is lower.
• So rate of the forward reaction will increase and more ammonia will be formed.

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We have completed the present discussion on Chemical equilibrium. In the next chapter, we will see Reactivity Series and Electrochemistry. 

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Chapter 11.1 - Reversible and Irreversible reactions

In the previous section, we saw how concentration, nature of reactants and pressure influence the rate of a chemical reaction. In this section we will see a few more factors. Later in this section we will see some basic details about reversible and irreversible reactions.

Influence of surface area of solids on the speed of reaction. 

Consider the chemical reactions in which solid substances are used. If those solid pieces are broken down into small pieces, or used in powder form, the surface area will increase. This can be seen from fig.11.4 below. 

Fig.11.4(a) shows a rectangular block. It's base is a square of side 2 cm. It has a height of 5 cm. Let us calculate it's surface area:
• Base and top:
2×2×2 = 8 cm²
Lateral faces: 4×2×5 = 40 cm²
Total surface area = 8 + 40 = 48 cm²
• Fig(b) shows the block split into two equal pieces. Let us calculate the total surface area of the two pieces:
Bottom piece:
Base and top:
2×2×2 = 8 cm²
Lateral faces: 4×2×2.5 = 20 cm²
Total surface area = 8 + 20 = 28 cm²
• Since both pieces are identical, the total surface area of the two pieces = 2×28 = 56 cm²
■ So, when the rectangular block in fig(a) is split, the surface area increases from 48 cm² to 56 cm²

■ If the solids are made into powder form, a large increase in surface area can be obtained. When there is a greater surface area, greater number of molecules become exposed to the other reactant. So the number of collisions will increase. This will increase the speed of the reaction. Let us do an experiment:
1. Take dilute HCl of the same concentration in two beakers. See fig.11.5 below:

2. Take two marble pieces of the same size. Break one into small pieces. 
3. Put the large piece in beaker 1 and the broken pieces in beaker 2. 
Note down the observations:
• We can see that reaction in the beaker 2 is faster than that in the beaker 1. 
• More CO₂ gas is formed in the beaker 1. We can write down the reason:
■ In beaker 1, there is a greater chance for the acid molecules to get in contact with the marble molecules. So the number of collisions will increase.

Influence of temperature on the speed of reaction

When the temperature of a solution is increased, the kinetic energy of the molecules in that solution will increase. The speed with which the molecules move in the solution will also increase. This will increase the number of effective collisions. So the speed of the reaction will increase. Let us do an experiment:
1. Prepare a dilute solution of sodium thiosulphate in a beaker. 
2. Take equal volumes of this solution in two boiling tubes. 
3. Heat one boiling tube for some time. 
4. After heating, add equal volumes of  dilute hydrochloric acid to both the boiling tubes. Note down the observations:
• In the boiling tube containing heated solution, reaction takes place faster and a white precipitate is formed quickly. 
• In the test tube which is not heated, precipitate is formed very slowly. 
• Initially, the precipitate is white in colour. But it slowly changes to light yellow.
■ Temperature is an important factor which influences the rate of reaction. As temperature increases, the rate of reaction also increases.

Influence of Catalyst on speed of reaction

• Hydrogen peroxide is a compound that undergoes decomposition. The decomposition can be represented by the following equation:
2H₂O₂ (aq) ⟶ 2H₂O (l) + O₂ (g)

• But this decomposition is very slow. We will want it to take place a little faster. Let us do an experiment:
1. Take some hydrogen peroxide solution in a test tube. 
2. Show a glowing incense stick into the test tube. There is no change to the stick. 
3. Now add some manganese dioxide into the test tube. Again show the glowing incense stick. 
4. We can see that more gas bubbles are formed and the incense stick burns brightly.    
• We can infer that, when manganese dioxide is added, the speed of the decomposition increases and oxygen is formed faster. 
5. When the reaction is complete, filter the solution using a filter paper. The substance remaining in the filter paper is manganese dioxide itself. 
• When examined carefully, it becomes clear that, there is no change in the amount or property of manganese dioxide.
■ The presence of manganese dioxide has increased the speed of the reaction. Manganese dioxide acts as a catalyst in this reaction. 
■ Catalysts are substances which alter the rates of chemical reactions, without themselves undergoing any chemical change.      
• In this reaction, manganese dioxide acted as a catalyst by increasing the rate of the reaction. Such catalysts are called positive catalysts. 
• It is necessary to reduce the rate of decomposition of hydrogen peroxide while in storage. For this purpose, Phosphoric acid is added to hydrogen peroxide. 
• In this case, the phosphoric acid acts as a negative catalyst. Examples of some positive catalysts are given below:
    ♦ Vanadium pentoxide is used as a positive catalyst in the manufacture of sulphuric acid
    ♦ Iron is used as a positive catalyst in the manufacture of ammonia

Light and the rate of chemical reaction

There are many chemical reactions that take place by releasing light or absorbing light. Such reactions are known as photochemical reactions. Some examples of photochemical reactions are given below:
• Formation of HCl by the combination of H₂ and Cl₂
• Decomposition of silver bromide
• Photosynthesis
■ Light energy is a factor which influences the rate of chemical reactions

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So we have seen the different factors that influence the speed of chemical reactions. Let us make a list:
• Concentration of reactants
• Nature of the reactants
• Pressure (For reactions involving gases)
• Surface area (For reactions involving solid substances)
• Temperature
• Catalyst
• Light

Reversible and Irreversible reactions

Consider the reaction between sodium hydroxide and hydrochloric acid. The balanced chemical equation is:
• NaOH (aq) + HCl (aq) ⟶ NaCl (aq) + H₂O (l)
    ♦ The reactants are: sodium hydroxide (NaOH) and hydrochloric acid (HCl)
    ♦ The products are: sodium chloride (NaCl) and water (H₂O)
• In this reaction, NaOH and HCl react to give NaCl and H₂O.
    ♦ But the NaCl and H₂O does not react together to give back NaOH and HCl.
• So we can see that the reaction takes place in one direction only.
■ Reactions in which the reactants give products but the products do not give back the reactants under the same conditions are called irreversible reactions.
• Given below are some examples of irreversible reactions.
Zn (s) + 2HCl (aq) ⟶ ZnCl₂ (aq) + H₂ (g)
2Mg (s) + O₂ (g) ⟶ 2MgO (s)
CaCO₃ (s) + 2HCl (aq) ⟶ CaCl₂ (s) + CO₂ (g) + H₂O (l)

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Are all reactions irreversible? Let us do an experiment:
1. Heat a little ammonium chloride (NH₄Cl) taken in a boiling tube.
• We will get a characteristic odour. The gas formed is ammonia (NH₃)
2. Show a moist red litmus paper at the mouth of the boiling tube. 
• We can see that the red litmus becomes blue. (see details here)
    ♦ This color change shows the nature of ammonia. It is basic in nature. 
3. Keep the litmus at the mouth of the boiling tube for some more time and observe the change.
• The blue litmus becomes red again. 
    ♦ This color change shows that some acid has acted on the blue color and changed it into red. Where did this acid come from? 
• It is the presence of hydrogen chloride (HCl) gas which resulted in the moist litmus turning red again.
    ♦ When NH₄Cl is heated, the lighter NH₃ gas comes out first. This causes the initial color change
   ♦ The denser HCl gas comes out later. This causes the second color change
• The chemical equation for the reaction is:
NH₄Cl (s) ⟶ NH₃ (g) + HCl (g)

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• So the products are two gases. These two gases react together to give back NH₄Cl.

    ♦ This newly formed NH₄Cl is seen as a white powder sticking to the sides of the test tube. 

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Let us do another experiment to confirm that 'NH₃ and HCl reacts together to give back NH₄Cl'.

1. Take a glass tube. Place a piece of cotton dipped in HCl at one end of the glass tube. And another piece dipped in ammonia solution (NH₃) at the other end of the glass tube. The two cotton pieces should be well inside the glass tube. See fig.11.6 below:

2. Close both ends of the glass tube tightly using corks. Observe the changes inside the glass tube.
• Thick white fumes are formed.
    ♦ It is due to the combination of HCl vapour and NH₃ gas.
3. Heat the region of the glass tube where the white powder of ammonium chloride (NH₄Cl) is stuck.
• The white powder disappears.
    ♦ This is because, when ammonium chloride is heated, it gets converted to ammonia gas and hcl gas.
    ♦ These gases combine together to give back ammonium chloride.
• See the chemical equations for the two reactions:
NH₄Cl (s) ⟶ NH₃ (g) + HCl (g)
NH₃ (g) + HCl (g) ⟶ NH₄Cl (s)
• We can combine the two equations into one:
NH₄Cl (s) ⇌  NH₃ (g) + HCl (g)
    ♦ The '⇌' sign indicates that the reaction takes place in both directions.
■ Reactions taking place in both directions are called reversible reactions.
• In a reversible reaction, the reaction in which the reactants change to products is called the forward reaction.
• The reaction in which the products change back to the reactants is called the backward reaction.

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Examine the chemical equations given below and write the forward and backward reactions in each:

(i) N₂ (g) + 3H₂ (g)  ⇌  2NH₃ (g)

(ii) 2SO₂ (g) + O₂ (g)  ⇌  2SO₃ (g)

(iii) H₂ (g) + I₂ (g)  ⇌  2HI (g)
Solution:
(i) Forward reaction: N₂ (g) + 3H₂ (g)  ⟶  2NH₃ (g)
Backward reaction: 2NH₃ (g) ⟶ N₂ (g) + 3H₂ (g)
(ii) Forward reaction: 2SO₂ (g) + O₂ (g)  ⟶  2SO₃ (g)
Backward reaction: 2SO₃ (g) ⟶  2SO₂ (g) + O₂ (g)
(iii) Forward reaction: H₂ (g) + I₂ (g)  ⟶  2HI (g)
Backward reaction: 2HI (g)  ⟶  H₂ (g) + I₂ (g)

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We have completed the present discussion reversible and irreversible reactions. In the next section, we will see Chemical equilibrium. 

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