Partial method how to solve?

In Partial equation method, equation is first divided into partial equation, which are simply probable steps that might occur in the chemical reaction. These probable steps are then balanced by hit and trail method and finally added. Following steps are applied in balancing chemical equation by partial equation method.

- At first, different probable steps are written for the given chemical equation. These probable steps are simply the partial chemical equation which we can write for easily balancing the reactants and product. Note that the probable steps may or may not occur in real.
- After breaking down the probable steps for the chemical equation, the partial equations are individually balanced by using Hit and Trial Method.
- Such balanced partial equations are multiplied by suitable integer. This is done if required, so that the elements which are not formed in the product side of the overall chemical equation is canceled out.
- Finally, the balanced partial equations are added to get the final overall balanced equation.

Lets’ illustrate some examples:

PbS + O_{3} → PbSO_{4} + O_{2}

In the given reaction, the atom of Pb and S is balanced in both reactant and product side. But there is 3 atom of oxygen in reactant side and 6 oxygen atom in product side. So let’s balance this chemical equation by partial equation method.

- First let’s think the possible step for occurring of the given chemical reaction. There is lead sulphide reacting with ozone, it means ozone must be acting as oxidizing agent which liberates nascent oxygen. So the first step can be generation of nascent oxygen by decomposition of ozone molecule.

**Step-1**

O_{3} →O_{2} + [O]

This liberated nascent oxygen can react with PbS to give lead sulphate and oxygen as;

**Step-2**

PbS + [O] → PbSO_{4}

This partial step is not balanced reaction. There is equal number of Pb and S atom, but no. of oxygen is one in reactant side but four in product side. So we can multiply nascent oxygen in reactant side by four.

PbS + 4[O] → PbSO_{4}

If we make four nascent oxygen in second probale step, then we must also need to make it equal number of nascent oxygen liberated in step one. So we need to multiply step one with integer four

{ O_{3} →O_{2} + [O] } × 4

Now lets add both partial equation:

{ O_{3} →O_{2} + [O] } × 4

PbS + 4[O] → PbSO_{4}

PbS + 4O_{3} PbSO_{4} + 4O_{2}

This is the balanced chemical equation.