Consider the given reaction:

${\mathrm{N}}_2+3{\mathrm{H}}_2⟶2{\mathrm{NH}}_3$

If the molecular weight of NH₃ and N₂ are x₁ and x₂, respectively. Their equivalent weights are y₁ and y₂, respectively. Then, determine the value of (y₁-y₂).

Reaction that shows the formation of ammonia when nitrogen gas reacts with hydrogen gas is:

${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$

Equivalent weight of an element is determined when the molecular weight of the element divided by its n-factor.

$\mathrm{Equivalent} \mathrm{weight}=\frac{\mathrm{Molecular} \mathrm{weight}}{\mathrm{n}-\mathrm{factor}}$

In the given reaction, oxidation state of nitrogen in N₂ is zero and in NH₃, it is -3. As there are two nitrogen atoms in N₂, thereby the n-factor of N₂ is six.

Equivalent weight of N₂ (y₂) is determined when the molecular weight of N₂ (x₂) is divided by 6.

${\mathrm{y}}_2=\frac{{\mathrm{x}}_2}{6}...\left(1\right)$

There are two moles of ammonia and the n-factor of 2 moles of ammonia is also 6 because the oxidation state of N in the given reaction changes from zero to -3. Therefore, the n-factor for one mole of ammonia is 3.

Equivalent weight of NH₃ (y₁) is determined when the molecular weight of NH₃ (x₂) is divided by 3.

${\mathrm{y}}_1=\frac{{\mathrm{x}}_1}{3}...\left(2\right)$

Value of y₁-y₂ is determined using equation (1) and equation (2).

$\begin{array}{ccc}{\mathrm{y}}_1-{\mathrm{y}}_2& =& \frac{{\mathrm{x}}_1}{3}-\frac{{\mathrm{x}}_2}{6}\\ & =& \frac{2{\mathrm{x}}_1-{\mathrm{x}}_2}{6}\end{array}$