Trung Tâm Luyện Thi Đại Học

Recall the law of conservation of mass that you studied in Class IX; mass can neither be created nor destroyed in a chemical reaction. That is, the total mass of the elements present in the products of a chemical reaction
has to be equal to the total mass of the elements present in the reactants.
In other words, the number of atoms of each element remains the same, before and after a chemical reaction. Hence, we need to balance a skeletal chemical equation. Is the chemical Eq. (1.2) balanced? Let us learn about balancing a chemical equation step by step.
The word-equation for Activity 1.3 may be represented as –

 \( Zinc+Sulphuric\,\,acid\to Zinc\,\,sulphate+Hydrogen \)

The above word-equation may be represented by the following chemical equation –

 \( Zn+{{H}_{2}}S{{O}_{4}}\to ZnS{{O}_{4}}+{{H}_{2}}\,\,\,\,\,\,\,(1.3) \)

Let us examine the number of atoms of different elements on both sides of the arrow.

As the number of atoms of each element is the same on both sides of the arrow, Eq. (1.3) is a balanced chemical equation.

Let us try to balance the following chemical equation –

 \( Fe+{{H}_{2}}O\to F{{e}_{3}}{{O}_{4}}+{{H}_{2}}\,\,\,\,\,\,\,(1.4) \)

Step I: To balance a chemical equation, first draw boxes around each formula. Do not change anything inside the boxes while balancing the equation.

 \( Fe+{{H}_{2}}O\to F{{e}_{3}}{{O}_{4}}+{{H}_{2}}\,\,\,\,\,\,\,(1.5) \)

Step II: List the number of atoms of different elements present in the unbalanced equation (1.5).

Step III: It is often convenient to start balancing with the compound that contains the maximum number of atoms. It may be a reactant or a product. In that compound, select the element which has the maximum number of atoms. Using these criteria, we select Fe₃O₄ and the element oxygen in it. There are four oxygen atoms on the RHS and only one on the LHS.
To balance the oxygen atoms –

To equalise the number of atoms, it must be remembered that we cannot alter the formulae of the compounds or elements involved in the reactions. For example, to balance oxygen atoms we can put coefficient ‘4’ as 4 H₂O and not H₂O₄ or (H₂O)₄. Now the partly balanced equation becomes –

 \( Fe+4{{H}_{2}}O\to F{{e}_{3}}{{O}_{4}}+{{H}_{2}}\,\,\,\,\,\,\,\,(1.6)\,\,(partly\,\,balanced\,\,equation) \)

Step IV: Fe and H atoms are still not balanced. Pick any of these elements to proceed further. Let us balance hydrogen atoms in the partly balanced equation.
To equalise the number of H atoms, make the number of molecules of hydrogen as four on the RHS.

The equation would be –

 \( Fe+4{{H}_{2}}O\to F{{e}_{3}}{{O}_{4}}+4{{H}_{2}}\,\,\,\,\,\,\,\,\,(1.7)\,\,(partly\,\,banlanced\,\,equation) \)

Step V: Examine the above equation and pick up the third element which is not balanced. You find that only one element is left to be balanced, that is, iron.

To equalise Fe, we take three atoms of Fe on the LHS.

 \( 3Fe+4{{H}_{2}}O\to F{{e}_{3}}{{O}_{4}}+4{{H}_{2}}\,\,\,\,\,\,\,\,(1.8) \)

Step VI: Finally, to check the correctness of the balanced equation, we count atoms of each element on both sides of the equation.

 \( 3Fe+4{{H}_{2}}O\to F{{e}_{3}}{{O}_{4}}+4{{H}_{2}}\,\,\,\,\,\,\,\,\,(1.9)\,\,(balanced\,\,equation) \)

The numbers of atoms of elements on both sides of Eq. (1.9) are equal. This equation is now balanced. This method of balancing chemical equations is called hit-and-trial method as we make trials to balance the equation by using the smallest whole number coefficient.

Step VII: Writing Symbols of Physical States Carefully examine the above balanced Eq. (1.9). Does this equation tell us anything about the physical state of each reactant and product? No information has been given in this equation about their physical states.
To make a chemical equation more informative, the physical states of the reactants and products are mentioned along with their chemical formulae. The gaseous, liquid, aqueous and solid states of reactants
and products are represented by the notations (g), (l), (aq) and (s), respectively. The word aqueous (aq) is written if the reactant or product is present as a solution in water.
The balanced Eq. (1.9) becomes

 \( 3Fe(s)+4{{H}_{2}}O(g)\to F{{e}_{3}}{{O}_{4}}(s)+4{{H}_{2}}(g)\,\,\,\,\,(1.10) \)

Note that the symbol (g) is used with H₂O to indicate that in this reaction water is used in the form of steam.
Usually physical states are not included in a chemical equation unless it is necessary to specify them.
Sometimes the reaction conditions, such as temperature, pressure, catalyst, etc., for the reaction are indicated above and or below the arrow in the equation. For example

 \( CO(g)+2{{H}_{2}}(g)\xrightarrow{340\,atm}C{{H}_{3}}OH(\ell )\,\,\,\,\,\,\,(1.11) \)

 \( \begin{align}  & 6C{{O}_{2}}(aq)+12{{H}_{2}}O(\ell )\xrightarrow[Chlorophyll]{Sunlight}{{C}_{6}}{{H}_{12}}{{O}_{6}}(aq)+6{{O}_{2}}(aq)+6{{H}_{2}}O(\ell )\,\,\,\,\,\,\,(1.12) \\  &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\text{glucose}) \\ \end{align} \)

Using these steps, can you balance Eq. (1.2) given in the text earlier?