Chapter 11.2 - Equilibrium in Reversible Reactions

In the previous section, we saw the factors which influence the rate of a chemical reaction. We also saw some basic details about reversible and irreversible reactions. In this section we will learn about the Equilibrium in Reversible Reactions.

Let us do an experiment. 
1. Take the solutions of potassium nitrate (KNO₃), potassium thiocyanate (KCNS) and ferric nitrate (Fe(NO₃)₃).
2. We are going to do some chemical reactions with these solutions. So we will want to note down the initial colors. For that, observe the color of each solution and note them down:
• KCNS solution - initial color: colourless
• Fe(NO₃)₃ solution - initial color: Light yellow
• KNO₃ solution - initial color: colourless
3. Take a little of dilute ferric nitrate solution in a test tube and add a few drops of potassium thiocyanate to it. 
• The solution becomes red in color. 
• The chemical equation of the reaction is:
Fe(NO₃)₃ (aq) + 3KCNS (aq) ⟶ Fe(CNS)₃ (aq) + 3KNO₃ (aq) 
• The formation of ferric thiocyanate (Fe(CNS)₃) by the combination of Fe(NO₃)₃ and KCNS is the cause for the red color.
4. Keep the solution without disturbing. Examine after some time. 
• We can see that the red color neither increases or decreases. 
5. Dilute the solution and keep it without disturbing. Examine after some time. 
• This time also, we can see that, the red color neither increases or decreases. 
6. Transfer the dilute solution to three test tubes in equal amounts. We are going to do some tests to the tree test tubes separately 
7. Test tube 1: Add Fe(NO₃)₃ to test tube 1. 
• We can see that the red color becomes deep. 
What may be the reason? 
Ans: We have seen earlier that, formation of Fe(CNS)₃ is responsible for the red color. Now, we have added more Fe(NO₃)₃ to the solution. 
• This newly added Fe(NO₃)₃ combines with KCNS to produce more Fe(CNS)₃. So naturally, the red color deepens. 
8. Test tube 2: Add some KCNS to the second test tube.
• We can see that the red color becomes deep.
What may be the reason?
Ans: We have seen earlier that, formation of Fe(CNS)₃ is responsible for the red color. Now, we have added more KCNS to the solution.
• This newly added KCNS combines with Fe(NO₃)₃ to produce more Fe(CNS)₃. So naturally, the red color deepens.
9. Test tube 3: Add a drop of concentrated KNO₃ to the third test tube.
• We can see that the intensity of the red color decreases significantly.
What may be the reason?
Ans: The newly added KNO₃ reacts with the Fe(CNS)₃ to give back the original reactants.
• So some of the Fe(CNS)₃ is used up to give back the original reactants.
• As a consequence, only a less Fe(CNS)₃ is available now.
• This causes the decrease in the intensity of red color.

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Consider the third test tube in the above experiment:

• We added KNO₃. It is one of the products. It reacted with the other product Fe(CNS)₃ to give back the reactants. 

■ So this is a reversible reaction.

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• Once we conclude that it is a reversible reaction, we can give a detailed explanation as given below:

1. We now know that the reaction that we saw is reversible. So we can represent it using '⇌' symbol. This is shown below:
Fe(NO₃)₃ (aq) + 3KCNS (aq) ⇌ Fe(CNS)₃ (aq) + 3KNO₃ (aq) 

2. Let us start a stop watch at the instant when KCNS is added to Fe(NO₃)₃.

• As the time passes by, more and more quantities of the products [Fe(CNS)₃ and KNO₃] will be produced

• In the first few seconds, the concentrations of the products will be very low. At the same time, the concentrations of the reactants will be high.

• As time passes by, the newly formed products will begin to react together to give back the original reactants. This is the backward reaction

    ♦ But as the concentrations of the products is low at the early stages, the rate of this back ward reaction will also be low in the early stages.

    ♦ As time passes by, the concentrations of the products increases, and the rate of backward reaction also increases.
• This 'increasing rate' is represented by the rising red curve in fig.11.7 below. We can see that, the red curve rises with the increase in time. If we mark a point 'C' on that curve, we can read off two values: 

    ♦ From the X axis, we can read off a 'particular time' corresponding to point 'C'

    ♦ From the Y axis, we can read off the 'rate of the backward reaction' at that 'particular time'

3. Now consider the forward reaction. As time passes by, the concentrations of the reactants decreases. This is because, more and more of the reactants are being converted into products. 

• So the rate of the forward reaction decreases. This 'decreasing rate' is represented by the declining green curve in the fig.11.7.
• We can see that, the green curve declines with the increase in time. If we mark a point 'B' on that curve, we can read off two values: 

    ♦ From the X axis, we can read off a 'particular time' corresponding to point 'B'

    ♦ From the Y axis, we can read off the 'rate of the forward reaction' at that 'particular time' 

4. So we have a rising curve and a declining curve. At some point in time, they will obviously meet. This meeting point is marked as 'A' in the fig.11.7

5. It is interesting to note that: 

• after the 'time corresponding to A', the rate of forward reaction does not decrease

• after the 'time corresponding to A', the rate of backward reaction does not increase

6. What may be the reason? Let us analyse:
• Let R₁ and R₂ be the reactants. Also, let P₁ and P₂ be the products
    ♦ As time passes by, the concentrations of R₁ and R₂ decreases
    ♦ As time passes by, the concentrations of P₁ and P₂ increases
• At the meeting point A, let the concentrations be [R₁]_A, [R₂]_A, [P₁]_A and [P₂]_A.
• At the meting point A, rate of forward reaction = rate of backward reaction
    ♦ Rate of forward reaction is the 'quantity of R₁ and R₂ consumed in a particular interval of time'
    ♦ Rate of backward reaction is the 'quantity of P₁ and P₂ consumed in that particular interval of time'
• At the meeting point, the rates are equal. So:
Quantity of R₁ and R₂ consumed = Quantity of P₁ and P₂ consumed
• So, after the point A, there is no change in the quantities of reactants and products. That is:
After point A, [R₁]_A, [R₂]_A, [P₁]_A and [P₂]_A will not change
    ♦ If there is no change in the quantities of reactants, the rate of forward reaction will not change
    ♦ If there is no change in the quantities of products, the rate of backward reaction will not change
• That means:
    ♦ After point A, the rate of forward reaction will not decline
    ♦ After point A, the rate of backward reaction will not rise up
• So, after point A, the rate will be represented by the horizontal yellow line. We know that, all 'y values' on a horizontal graph will be the same.
At point A, the reaction is said to have attained Chemical Equilibrium 

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Chemical Equilibrium is a stage at which the rate of forward reaction becomes equal to the rate of backward reaction in a chemical reaction

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• Even after the reaction has reached the equilibrium stage, both the forward and backward reactions will continue. But at the same rate.
    ♦ This continuing reactions are not visible to us because there is no color change after the equilibrium stage.
    ♦ There is no color change because, after equilibrium stage, the concentrations of reactants and products does not change.
• At equilibrium, both the reactants and products are present in the system
• As both the reactions are continuing even at equilibrium, it is called a dynamic equilibrium

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Now we will consider the three test tubes

1. The resulting solution was transferred to three test tubes at the time when the reaction was at the equilibrium state. This is clear from the fact that, we had kept it with out disturbing for some time and no color change was observed. (see (5) at the beginning of this section)
• Further more, the transfer was made in equal amounts. (see (6) at the beginning of this section). So the concentrations in the three test tubes are the same.
2. To the test tube 1, more Fe(NO₃)₃ was added. This is shown in fig.11.8(b) below. 
• This extra Fe(NO₃)₃ reacts with KCNS which is already present, to form more Fe(CNS)₃. So the red colour increases. 
• Thus there is an increase in rate of the forward reaction. 
• But soon the rates of both forward and backward reaction will become equal, and the system will attain a new equilibrium state

3. To the test tube 2, more KCNS was added. This is shown in fig.11.8(c) above. 

• This extra KCNS reacts with Fe(NO₃)₃ which is already present, to form more Fe(CNS)₃. So the red colour increases
• Thus there is an increase in rate of the forward reaction. 
• But soon the rates of both forward and backward reaction will become equal, and the system will attain a new equilibrium state  
4. To the test tube 3, more KNO₃ was added. This is shown in fig.11.8(d) above. 
• This extra KCNS reacts with Fe(CNS)₃ which is already present, to form more reactants. That is., some of the Fe(CNS)₃ is used up. So the red colour decreases
• Thus there is an increase in rate of the backward reaction. 
• But soon the rates of both forward and backward reaction will become equal, and the system will attain a new equilibrium state.

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Based on the above experiment, we can write the characteristics of Chemical equilibrium:

• At equilibrium, both reactants and products coexist. That means, at equilibrium, both reactants and products will be present in the system. 
• At equilibrium, the rate of forward reaction is equal to the rate of backward reaction. 
• Since reactants and products are present, both forward and backward reactions are occurring even at equilibrium. 
• Since the reactions are continuing, it is called a dynamic equilibrium. 
• Once equilibrium is attained, there will be no change in the concentration of the reactants or products. 
• Chemical equilibrium is attained only in closed systems. That is., no substances should be added to or taken away from the system. Pressure and temperature of the system should not change.

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In the next section, we will see the factors which affect equilibrium of chemical reactions. 

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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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Chapter 11 - Rate of Chemical Reactions

In the previous section, we completed a discussion on Molarity. In this chapter we will see Rate of chemical reactions and Chemical equilibrium.

We see different chemical reactions in our daily life. Some examples are:
•Burning of firewood •Rusting of iron •Photosynthesis •Reaction between acid and alkali •Reaction between acid and metals •Reaction between cement and water and the setting of cement
■ Burning of firewood and rusting of iron are two different chemical reactions. Also, they occur at different speeds.  
• A small piece of iron reacts with the oxygen in the atmosphere for a long time to produce rust.
• But a small piece of firewood will burn out completely in a short time.
■ There are situations where we will want to increase or decrease the speed of chemical reactions.
• For example, if we can reduce the speed at which rusting of a piece of iron takes place, we can use that iron piece for a longer time.
• Similarly, if we can increase the speed at which the firewood burns, we will be able to cook food faster.   
■ Following are some methods which are commonly used to make the firewood burn faster:
• Provide more air
• Make the firewood into smaller pieces
• Remove moisture from the firewood by drying in sunlight before burning

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So factors like: provision of more air, making of smaller pieces, removal of moisture etc., helps to increase the rate (speed) of burning of firewood. In this way, there are many factors that influence a chemical reaction. Let us now learn about such factors:

Influence of Concentration on the rate of chemical reactions

Let us do a simple experiment:
1. Take magnesium ribbons of equal mass in two test tubes. 
2. Add concentrated HCl in one test tube (fig 11.1a) and dilute HCl to the other (fig 11.1b).

The volumes of HCl must be equal.
3. Let us note down the observations:
• Test tube 1: Many gas bubbles are formed quickly. The magnesium is consumed within a very short time.
• Test tube 2: Gas bubbles are formed slowly. It takes longer time for the magnesium to get consumed.
■ So we find that reaction is faster in one test tube and slower in the other. 
• We want to know the reason. For that, first we want to learn about effective collisions.

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Collision Theory
• According to this theory, reactant molecules have to collide against each other for chemical reaction to occur. 
• But all collisions between between reactant molecules need not result in chemical reaction.   
• If a collision results in a reaction, it is called an effective collision. 
■ But if a collision is to be effective, the colliding molecules should have an energy level above a specified level. It can be explained as follows:
• Consider two molecules. Let each of them possess an energy E. 
• Let Es be the specified energy level. 
• Let the two molecules collide. If E is less than Es, no reaction will take place between them. 
• If E is greater than Es, a reaction will take place. 
• So it is clear that, we must have a large number of effective collisions within a short span of time. Then only the reaction will take place at a greater speed. 
■ How can we increase the number of effective collisions?
If there are a large number of molecules, each of them with energy greater than Es, there will be an increased number of effective collisions.

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Now let us check our present experiment. The balanced chemical equation for the reaction is:
Mg + 2HCl ⟶ MgCl₂ + H₂
• One molecule of magnesium reacts with two molecules of HCl to produce one molecule of magnesium chloride and one molecule of hydrogen.
■ The amount of a substance contained in unit volume is called concentration of that substance.
• In the first test tube, the concentration of HCl was greater.  That means, the number of HCl molecules was greater.
• This resulted in the increase in number of effective collisions. So the speed of the chemical reaction increased.

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Next we want a method to write the actual speed of a chemical reaction. We know that, the speed of a vehicle is written by considering the distance travelled in unit time. For example, if the vehicle travelled 1500 m in 20 seconds, it's speed is ¹⁵⁰⁰⁄₂₀ = 75 m/s

We follow a similar method for the 'speed of reactions' also. Consider the experiment that we did above. 
1. Equal masses of magnesium was taken in the test tubes. 
2. During the reaction, using a stop watch, find the time required for the complete consumption of magnesium in test tube 1. Then we will be able to write the following ratio:
^{Amount of magnesium consumed}⁄_{Time required for the consumption}
• Note that 'time' is in the denominator. So this ratio gives the amount of magnesium consumed in unit time. So this ratio is the 'speed (rate) of the reaction'.
■ In general, we can write:
Speed of reaction = ^{Amount of magnesium consumed}⁄_{Time required for the consumption}
• In test tube 1, the magnesium was consumed quickly. So the 'time' which comes in the denominator is less. Thus the ratio gives a large value. That means the speed is high.

Another method:
1. In the above experiment, hydrogen is one of the products. using special apparatus, we can collect all the hydrogen that is formed. 
2. We can also determine the actual quantity of hydrogen thus collected. Also, using a stop watch, we can determine the time required to collect that much hydrogen. 
3. So we can take the ratio:
^{Amount of hydrogen formed}⁄_{Time required for the formation}
• Note that 'time' is in the denominator. So this ratio gives the amount of hydrogen formed in unit time. So this ratio is the speed of the reaction.
■ In general, we can write:
Speed of reaction = ^{Amount of hydrogen formed}⁄_{Time required for the formation}
• In test tube 1, the hydrogen was formed quickly. So the time which comes in the denominator is less. Thus the ratio gives a large value. That means the speed is high.

Nature of reactants and the rate of chemical reaction

The ability of each substance to take part in chemical reaction is different. A substance A may readily react with another substance B. But a substance C may not show much readiness to react with substance C. This is shown in fig.11.2 below:

So the reaction between A and C will be faster than the reaction between B and C. This is because of the 'difference in nature' between A and B. Let us do an experiment:
1. Take equal volumes of dilute HCl in two test tubes.
2. Take magnesium metal and zinc metal of almost equal shape and mass. Put the magnesium and zinc into the two separate test tubes at the same time.
3. Let us note down the observations:
• The reaction between magnesium and HCl is faster.
We want to know the reason. Let us analyse:
• The balanced chemical equations are:
Test tube 1: Zn + 2HCl ⟶ ZnCl₂ + H₂
Test tube 2: Mg + 2HCl ⟶ MgCl₂ + H₂
• Dilute HCl was taken in both the test tubes. That means, the concentration of HCl was same in both cases.
• So it is clear that, the nature of metals influenced the rate of the chemical reactions. Zinc and magnesium has different natures. Magnesium reacts readily with HCl than zinc.

Influence of pressure on the speed of reactions

In those reactions in which gases are involved, pressure is an important factor which influences the speed. Let us see the reason:
1. When pressure is increased, the gas molecules come closer to each other. So the possibility of effective collision increases. Thus the speed of the reaction increases. Let us see an example:
2. Fig.11.3(a) shows nitrogen and hydrogen taken in a cylinder. The pressure can be varied using the piston. 

3. In fig.11.3(b), the pressure is increased to 2 atm. Because of that, the volume is reduced.
4. Now consider figs (c) and (d). In fig (c), a blue square is shown. It represents unit volume. It may be 1 mm³ or 1 cm³ or any other convenient unit volume.
• The number of molecules within the unit volume is low in fig (c).
• But in fig (d), it is high.
5. That means, when the pressure is increased, volume becomes less, and the number of molecules within unit volume is increased.
• When the number of molecules within unit volume increases, more effective collisions take place within a specified time. Thus the speed of reaction increases.

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In the next section, we will see a few more important factors. 

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