" APPLICATION OF POAC ( PRINCIPAL OF ATOMIC CONSERVATION) AND LAW OF CHEMICAL EQUIVALENCE IN NUMERICALS OF PHYSICAL CHEMISTRY OF JEE / ISEET.

1.    Mole concept

2.    POAC

3.    Equivalent method ( law of chemical equivalence)

POAC & equivalent method does not depend on stoichiometric coefficient, but one necessary condition is that we must have given reactant from which products are to be produced. i.e. balancing of reaction is not necessary for applying POAC and eq. method. POAC is used for simpler reaction.

For saving time Eq. Method should apply for solving complex/ redox reaction rather than POAC method.

POAC method:- principle of atomic conservation

Through the chemical reaction, total no. of atoms of reactant must be equal to products. If atoms are conserved, moles of those atoms shall also be conserved. Examples:-

NaNO₃     →   NaNO₂ + O₂

Reactant & product contains only 1 Na atom. So we can write as

1 * mole of NaNO₃ = 1 * mole of NaNO₂

1*[ wt. Of NaNO₃/ m.wt of NaNO₃] = 1*[ wt. Of NaNO₂/m.wt. of NaNO₂]

Similarly we can apply POAC on N & O atoms

1* mole of NaNO₃    = 1* mole of NaNO₂   (for N atoms)

3* mole of NaNO₃    = 2* mole of NaNO₂ + 2 * mole of O₂   (for O atoms)

Since one mole of NaNO₃, NaNO₂ & O₂ contain 3, 2, and 2 moles of O atoms respectively.

KClO₃ → KCl + O₂

For   K atom

1* mole of KClO₃ = 1* mole of KCl

3 * mole of KClO₃ = 2 * mole of O₂

Q.1 calculates the no. of O & P atoms in 1420 gm of P₄O₁₀. M.wt of P₄O₁₀  = 568

Mole of P₄O₁₀ = 1420/ 568 = 5 moles

1* mole of P₄O₁₀ = 4* mole of P

5 moles of   P₄O₁₀ = 20 moles of P

20 moles of P   =   20 * 6.022* 10 ²³ atoms of P = 12.044 * 10 ²⁴ atoms of P

Similarly O atom

1* mole of P₄O₁₀ = 10* mole of O

5 mole of P₄O₁₀ = 50 mole of O

50 mole of O = 50* 6.022 * 10²³ atoms of O

= 30.11 * 10²⁴ atoms of O

Equivalent method:-

Sum of equivalents( Eqs.) or milliequivalents ( meqs.) of all oxidising agents is equal to sum of Eqs. Or  meqs. Of all reducing agents.

i.e. sum of equivalents of oxidising agents = sum of equivalents of reducing agents.

Example{ 1} Suppose we have a mixture of  FeO+ Fe₃O₄ ,in this mixture we add acidic  KMnO₄.

Here Fe is in oxidation state of +2 in FeO and +8/3 in Fe₃O₄ i.e. less than the maximum Oxidation state of Fe (less than +3). Hence in the presence of oxidising agent, it has a tendency to oxidise itself. In KMnO_4,   Mn has +7 oxidation state {max. Oxd. State} so it can not be further oxidise. It can be only reduce to itself in the presence of any reducing agent so 

Meq. of FeO + meq. of Fe₃O₄ = meq. Of KMnO₄

Or n₁ * mole of Fe₃O₄ + n₂ * mole of FeO = n₃ * mole of KMnO₄

Or           N₁V₁       +     N₂V₁      =      N₃V₂

Example 2. If a mixture contains FeO+ Fe₃O₄ + Fe₂O₃ (which is not reacted to each other) and we add KMnO₄  in it initially, which oxidise Fe of mixture to completely to its higher oxidation state ( +3) then we write

Meq. Of FeO+ meq. of Fe₃O₄ = meq. Of KMnO₄

Because Fe is in max. Oxidation state(+3)  in  Fe₂O₃. So it can’t be further oxidise. so Fe₂O₃ does not react with  KMnO₄ .

If after this oxidation process we add excess of KI (reducing agent), it will reduce Fe₂O₃ present in the initial solution. In this case Fe₂O₃  is react as oxidising agent because of max. Oxd. State (+3) and  KI as reducing agent