Question

What is the $n$ factor or valency factor of ozone during the change $2{O_3} \to 3{O_2}$
A. $2$
B. $4$
C. $6$
D. $8$

Answer 1

Hint: $n$ factor or valency factor can be found by calculating how many equivalents of the product is formed per mole of the reactant. Thus, first we must find the number of equivalents of oxygen gas formed, and then divide it with the number of moles of the reactant to get the $n$ factor.

Formulas used: $eq.wt = \dfrac{M}{v}$
Where $eq.wt$ is the equivalent weight, $M$ is the atomic weight and $v$ is the valency.
$N = \dfrac{W}{{eq.wt}}$
Where $N$ is the number of equivalents and $W$ is the given mass.

Complete step by step answer:
We are given that two moles of ozone react to give three moles of oxygen. To find the valency factor, we must first compute the number of equivalents of oxygen formed. For this, we need to first determine the equivalent weight of oxygen using the formula:
$eq.wt = \dfrac{M}{v}$
Where $eq.wt$ is the equivalent weight, $M$ is the atomic weight and $v$ is the valency.
As we know, for oxygen, $M = 16g/mol$ and its valency is $2$. Substituting these values, we get:
$eq.wt = \dfrac{{16}}{2} = 8g$
Now we need to find the number of equivalents, which is given by the formula:
$N = \dfrac{W}{{eq.wt}}$
Where $N$ is the number of equivalents and $W$ is the given mass.
The molecular weight of oxygen gas (${O_2}$) is $(16 \times 2) = 32g/mol$. Since there are three moles in the equation, given mass of oxygen, $W = 3 \times 32 = 96g$ and we have found out that $eq.wt = 8g$. Substituting these, we get:
$N = \dfrac{{96}}{8} = 12$
Therefore, according to the reaction, two moles of ${O_3}$ is equivalent to $12$ ${O_2}$.We need to find the number of equivalents per mole of ${O_3}$ to get the $n$ factor. Thus, we have:

Moles of ${O_3}$	Equivalents of ${O_2}$
$2$	$12$
$1$	$n$

Using the cross-multiplication rule, we have:
$2 \times n = 1 \times 12$
$ \Rightarrow n = \dfrac{{12}}{2} = 6$
Thus, the $n$ factor of ozone in this reaction is $6$.

So, the correct answer is Option C .

Note: $n$ factor of a compound is accurately defined as the number of moles of electrons gained or lost per mole of that compound. Thus, it is similar to the concept of valency of elements, and is based on the principle of equivalence. Note that $n$ factor plays a major role in the balancing of redox reactions, as it also signifies the change in oxidation number.