Chemical Reaction Or Chemical Change

2.4   Chemical Reaction Or Chemical Change

Any change, which alters the composition of a substances, is a chemical change In this type of change one or more new substances are formed from the original substances, for example, when iron (Fe) rusts, it reacts with oxygen (O) of air in presence of moisture to form red brown iron oxide (rust). Similarly, when coal burns, it forms smoke, gaseous products and ashes. The burning of coal is a chemical reaction (change) in which it combines with oxygen in air to form entirely new substances.    

2.4.1 Types of Chemical Reactions:     

Chemical reactions can be divided commonly into five different types.

1.       Decomposition reactions.

2.       Addition reaction (combination reaction).

3.       Single displacement reaction.

4.       Double displacement reaction.

5.       Combustion reaction.

 

1.  Decomposition Reactions:

A reaction in which a chemical substance breaks down to form two or more simpler substances is called a decomposition reaction. These reactions require some energy for decomposition.

For example: Calcium carbonate decomposes into calcium oxide and carbon dioxide in presence of heat.

CaCO3(s)    Heat     CaO(s) + CO2 (g)

Similarly, potassium chlorate (KClO3) on heating produces two simpler substances, potassium chloride and oxygen gas.

2KClO3(s)    Heat     2KCl(s) +3O2 (g)

2.  Addition or Combination Reaction:

A reaction in which two or more substances combine to form a single substance is called an addition or combination reaction. These reactions are reverse of decomposition reactions.

For example: Calcium oxide (CaO) reacts with carbon dioxide (CO2) to form calcium carbonate (CaCO3).

CaO(s) + CO2 (g)    Heat     CaCO3 (s)

Another example is, when sodium reacts with chlorine gas it gives a new substance called common salt (NaCl).

2Na(s) + Cl2 (g)    Heat     2NaCl(s)

1.      Single Replacement (Displacement) Reaction:

A reaction in which one atom or group of atoms of a compound is replaced by another atom or group of atoms is defined as displacement reaction. Some metals react with acids, bases or even water to displace hydrogen (H2) gas.

For example: Zinc replaces hydrogen in hydrogen chloride (HCl).to give zinc chloride.

Zn + 2HCl    Heat     ZnCl2 + H2 (g)

Similarly sodium (Na) metal reacts with water to give sodium hydroxide and hydrogen gas.

2Na + 2(H -OH)    Heat     2NaOH+ H2 (g)

When chlorine reacts with a solution of potassium bromide, Chlorine replaces bromine to form KCl and Br2 vapors.

Cl2 + 2KBr(aq)    Heat     2KCl(aq) + Br2(g).

2.      Double Displacement Reaction:

It is a reaction in which two compounds exchange their partners, so that two new compounds are formed. In double displacement reaction usually there is an exchange of ionic radicals.

For example: When sodium chloride (NaCl) is reacted with silver nitrate (AgNO3) solution, they exchange their partners to form two different compounds silver chloride (AgCl) and sodium nitrate (NaN03).

NaCl(aq)+ AgNO3 (aq)   NaNO3 (aq) + AgCl(s) (insoluble white precipitate)

Consider, another example, when, calcium chloride (CaCl2) is reacted with sodium carbonate (Na2CO3)they exchange their partners to form two new compounds, sodium chloride and calcium carbonate (CaC03)(s).

CaCl2 (aq) + Na2CO3 (aq)     2NaCl(aq) + CaCO3 (s) (insoluble white precipitate)

Remember that neutralization and hydrolysis reactions are also double displacement reactions. These reactions will be discussed in the 9th chapter on Acids, Bases and Salts.

3.      Combustion Reaction:

A reaction in which substances react with either free oxygen or oxygen of the air, with the rapid release of heat and flame, is called combustion reaction.

For example, when methane (CH2), gas burns in air, it forms carbon dioxide gas (CO2), water (H2O) and

CH4 (g) + 2O2 (g)   CO2 (g) + 2H2O(g) + AH (Heat)

Similarly, when carbon (C) burns in air, it produces carbon dioxide

C(s) + O2 (g)   CO2+ AH (Heat)

 

2.4.2 Chemical Equation:

Chemical equation is short hand method of describing (expressing) the chemical reaction, in terms of symbols and formulae of the substances involved in a chemical reaction.

The starting substances are called reactants and are always written on the left hand side, where as the substances which are produced (formed) are known as products and are. Always written on the right hand side of an equation. The reactants and products are separated from one another by using the single arrow ( ) or double headed arrow (<==>), depending upon the kind (type) of reaction.

For illustration, when two molecules of hydrogen (H2) combine (react) with one molecule of oxygen (O2) to give two molecules of water (H2O), instead of writing the full names of substances, chemist represents this chemical reaction in the form of following equation.

e.g.    2H2 (g) O2 (g)   2H2O(l)

Here, hydrogen and oxygen are called "Reactants" (substances, which are present before the chemical reaction) and water is the product (substances that result from the chemical reaction).

The numbers, in front of the formulas in a chemical equation are called co-efficient (they show the number of molecules that react with each other) Where no co-efficient appears, only one number is considered.

                                    One coefficient

 2H2 + O  2H2O

Coefficients                                                Coefficients

The expressions (g), (1) and (s) placed some times as subscript after the formulas of the reactants and products, indicating the state, gaseous, liquid and solid of reactants and products.

e.g.    C(s)+O2 (g)   CO2 (g)

Another expression, frequently used is (aq) for aqueous, showing that the substance is in the form of water solution.

e.g.    NaCl(s) H2O  NaCl (aq)

Sometimes, reaction conditions are written over the arrow, for example, when reactants are heated, a capital Greek letter delta (Δ) may be placed over the arrow.

e.g.    CaCO3(s)     Δ    CaO(s) + CO2 (g)

Similarly, if catalyst is used, this catalyst is shown over the arrow.                 

e.g.    2SO2(g) + O2(g)  ___Pt___   2SO3(g)

If the reactants are heated in the presence of catalyst, both the symbols are placed on the arrow.                      

eg.    2KC103(s)   Δ MnO2   2KCl(s) + 3O2(g)

2.4.3 Writing of Chemical Equation:

Before we write a chemical equation, we must know the composition of all reactants and products. Otherwise, we cannot introduce the proper formulae into equation. Knowing the formulae of all reactants and products, we balance the equation by knowing the fact that in a chemical reaction atoms are neither created nor destroyed (to conform with the law of conservation of mass). Thus the number of atoms of each element must be same on both sides of the arrow in a chemical equation.

Consider, when zinc (Zn) reacts with sulphuric acid (H2SO4) to form zinc sulphate ZnS04 and hydrogen (H2) gas.

The chemical equation for this reaction can be represented as:

Zn(s) + H2SO4(aq)   ZnSO4(aq) + H2(g)

Where the arrow (-à) is read as "gives" "produces", "yields" or "forms". The (+) sign on the left side of equation appears for "reacts with" and on the right side of an equation is read as "and". The reaction is assumed to proceed from left to right, as the arrow indicates.

2.4.2 The meaning of Chemical Equation:

The chemical equation gives the following important information about the chemical reactions.

i)             The nature of reactants and products.

ii)            The relative number of each i.e. reactants and products.

2.4.5   Balancing of Chemical Equation:

All chemical equations must be balanced in order to comply with the law of conservation of mass. In balancing the chemical equation, to make the number of atoms of each element the same on both sides of the equation,

we can change the co-efficient (the number in front the formulas) but not subscript (the number within formulas). Remember that changing a subscript in a formula e.g. from H2O to H2O2 change the identity of chemical compound. The substance hydrogen peroxide (H2O2) is quite different from water (H2O). In contrast, changing a co-efficient, only changes the amount but not the identity of the substance. 2H2O means two water molecules and 3H2O means three water molecules and so on. Reduce the co-efficient to their smallest whole number values, if necessary, dividing them by common divisor. Thus the formulas of the reactants and the products in an equation remain same i.e. fixed and cannot be altered; so the only way of balancing an equation by taking appropriate numbers of molecules of the reactants and the products concerned.

Most chemical equations can be balanced by inspection method, that is, by trial and error method, with the experience; you should be able to balance any equation quickly. In general we can balance the equation by the following steps.

1. Write the correct formulae of all reactants on the left side and the formulae of products on the right side of an equation.

2. Balance the number of atoms on each side.

3. If the number of atoms may appear more on one side than the other, balance the equation by inspection method for this, multiply the formula by co-efficient so as to make the number of atoms, same on both sides of an equation.

4. The covalent molecules of hydrogen, oxygen, nitrogen and chlorine exist as di-atomic molecules e.g. H2, O2, N2 and Cl2, rather than isolated atoms, hence we must write them as such in chemical equation.

5. Finally, check the balanced equation, to be sure that the number and kind of atoms are the same on both sides of equation.

Let us consider a specific example. In the laboratory, oxygen (O2) gas is conveniently prepared by heating potassium chlorate (KClO3). The products are potassium chloride (KCl) and oxygen (O2) gas. To balance the equation.

Write correct formulae of reactants on left side and formulae of the products on right side of an equation.

 

KClO3(s)   KCl(s) + O2(g)

Balance the number of atoms on each side

Reactants                    Products

K (1)                                        K(1)

C (1)                                         C (1)

O (3)                                         O (2)

We see that (K) and (CI) elements have the same number of atoms on both sides of the equation, but there are three oxygen atoms on the left and two oxygen atoms on the right side of the equation- We can balance the oxygen atoms by placing a co-efficient 2 in front of KClO3 and 3 in front of O .

2 KClO3                 KCl + 3O2

Reactants              Products

K (2)                                 K (1)

CI (2)                            CI (1)

O (6)                              O (6)

We balance the (K) and (CI) atom by placing 2 in front of KCl.

 

2KClO3                  2KCl + 3O2

Reactants              Products

K (2)                                         K (2)

CI (2)                                        Cl (2)

O (6)                                         O (6)

This equation is now balanced, because the numbers of atoms of each element are same on both sides of the equation.

Note that this equation could also be balanced with co-efficient, those are multiple of 2 of each, for example:

4 KClO3 (s)    4KCl(s) + 6 O2 (g)

However, it is common practice to reduce the co-efficient to their smallest whole number value, here, this equation is divided by 2, so as to get smaller whole number ratio.

2 KClO3 (s)    2KCl(s) + 3 O2 (g)

Consider, another example, when hydrogen burns in air (which contains oxygen) to form water, we can write the equation.

H2 (g) + O2 (g)     H2O(1)

Balance the number of atoms on each side.

Reactants              Products

H(2)                                          H(2)

O (2)                             O(1)

We see that hydrogen element has the same number of atoms on both sides of the equation, but there are two oxygen atoms on left side and one oxygen atom on the right side of the equation. We can balance the oxygen atoms by placing co-efficient 2, in front of water (H2O).

 

H2 + O2                             2H2O

Reactant                            Products

H(4)                                           H(4)

O(2)                                          O(2)

Now we see that the number of oxygen atoms is same on both sides, but the number of hydrogen atoms can be balanced by placing 2 in front of (H2) on left side of the equation, we get:

 

2 H2 (g) + O                      2H2O(l)

Reactants                          Products

H(4)                                          H(4)

O(2)                                          O(2)

 

The equation is now balanced, because the numbers of atoms of each element are same on both sides of the equation.

2 H2 (g) + O2 (g)    2H2O(l)

For further illustration, let us consider the combustion of natural gas methane (CH4) in oxygen or air, which yield carbon dioxide (CO2) and water (H2O). We write from this information.

CH4 (g) + O2 (g)    CO2 (g) + H2O(g).

 Balance the no. of atoms on each side

Reactants                          Products

C (1)                                         C (l)

H (4)                                         H (2)

O (2)                                         O (3)

We see that the number of C-atoms is same on both sides but the number of hydrogen atoms on left side is (4), where as on right side is (2), we can balance it by placing a co-efficient 2 in front of water on right side, thus we get.

 

CH4  + O2                          CO2  + 2H2O

Reactants                          Products

C (1)                                         C (1)

H (4)                                         H (4)

O (2)                                         O (4)

 

Now, there is a difference in oxygen atoms, there are two oxygen atoms on left side but four oxygen atoms on right side, this can be balanced by placing co-efficient 2 in front of oxygen (O) on left side,

Thus we get

CH4 (g) + 2O (s)     CO2 (g) + 2H2O (g)

Reactants              Products

C (1)                                         C (l)

H (4)                             H(4)

O (4)                                         O(4)

The equation is now balanced, because the numbers of atoms of each element are same on both sides of equation

CH4 + 2O2    O2 + 2H2O

 

2.4.6 Concept of Mole-Ratio Based on Balanced Chemical Equation:

As we have seen, the co-efficient in a chemical reaction represents the number of moles (molecules) and not masses of molecules. However in chemical reaction, the amount of the reactants needed cannot be determined by counting molecules directly. Counting is always done by weighing. To find out the masses of reactants and products, in a balanced equation, the first step is to find out the mole-ratios in the balanced equation. Then convert the moles of reactants or products into mass.

The following steps will help in calculating the amount of reactant or products.

1.       Balance the equation for the given chemical reaction.

2.       Use the co-efficient in the balanced equation to get the mole ratio.

3.       Use the mole-ratio to calculate the number of moles of desired reactants or products.

4.       Convert the moles of reactants or products into mass, if required by the problem.

5.       The following example illustrates the use of the above four steps, in solving the mole-ratio problems.

 

Calculating the Amount of Reactants.

“Hydrogen (H2) reacts with oxygen (O2) to form wafer. The equation for this reaction is:

                                          2 H2 (g) + O2(g)      2H2O(l)

         i.            How many moles of oxygen are needed to react with 4.5 moles of hydrogen?

        ii.            How many grams of hydrogen will completely react with l00gof oxygen to form water?

Answer.

Step-(1) Balance the equation for chemical reaction.

2 H2 (g) + (O2) (g)             2H2O (l)

2(2)g + 2(16)g   2(2+16g)

4g+ 32g   36g

Step-(2) Set up the mole ratio, since 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

2 H2  =  O2 =  2H2O

2 moles     =   1 mole    =      2 moles

Step-(3) (i). Calculate the moles of oxygen as follows.

The balanced eq. shows that:

2 moles of hydrogen require 1 mole of oxygen

1 mole of hydrogen requires 1/2 mole of oxygen

 

4.5 moles of hydrogen require 1x4.5/2 =2.25 moles= 2.25 moles of oxygen

 

 ii) 32g of oxygen react with 4g of hydrogen

1g of oxygen reacts with 4g/32

100g of oxygen react with 4gx100/32 = 12.5g

Result:

i) No. of moles of oxygen = 2.25 moles.

ii) 100 g of oxygen require = 12.5g of hydrogen.

Calculating the Amount of Products:

Methane (CH4) is a common fuel, used for cooking. The chemical reaction is,

CH4 (g) + 2O2 (g)        CO (g) + 2H2O (g)

i) How many moles of CO2 will be produced by complete combustion of 10 moles of CH4?

ii) What mass of H2O vapors will be formed?

Answer.

Step (1) Balance the equation for chemical reaction

CH4 (g) + 2O2 (g)     CO2 (g) + 2H2O (g)

Step (2) Set up the mole ratio, the balance equation shows that 1 mole of methane (CH4) reacts with 2 moles of oxygen (O2) to produce 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O)

CH4    + 2O2         CO2   +    2H2O

1mole      2moles              lmole        2moles

Step (3)

i) Calculate the moles of (CO2) as follows The balanced equation shows that: 1 mole of CH4 on combustion produces 1 mole of CO2 10 moles of CH4 on combustion produce 1x10 = 10 moles of CO2

ii)     Mass of H2O

The balanced equation indicates that:

1 mole of CH4 on combustion produces 2 moles of H2O

10 moles of CH4 on combustion produce 2x10=20 moles of   H2O

Mass of H2O = no. of moles x m. mass

Mass of H2O    =??

No. of moles    = 20 moles

 M. mass of Hp = 2+16 = 18g

Mass of H2O = 20 x 18g = 360g

Result.  

No. of moles of CO2 = 10 moles

Mass of H2O formed = 360 grams

 
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