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.
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.
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.
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
• 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
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
• 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)
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)
• So the products are two gases. These two gases react together to give back NH4Cl.
Let us do another experiment to confirm that 'NH3 and HCl reacts together to give back NH4Cl'.
We have completed the present discussion reversible and irreversible reactions. In the next section, we will see Chemical equilibrium.
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.
• 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:
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
• 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)
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)
• 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.
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.
Examine the chemical equations given below and write the forward and backward reactions in each:
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.
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)
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)
We have completed the present discussion reversible and irreversible reactions. In the next section, we will see Chemical equilibrium.
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