Question

Equivalent weight of ${{N}_{2}}$ in the change ${{N}_{2}}\text{ }\to N{{H}_{3}}~~$is:
A $\dfrac{28}{6}$
B 28
C $\dfrac{28}{2}$
D $\dfrac{28}{3}$

Answer 1

The mass of one equivalent, that is, the mass of a given material that will mix with or displace a specified quantity of another substance, is known as equivalent weight. The mass of an element that combines with or displaces 1.008 gram of hydrogen, 8.0 gram of oxygen, or 35.5 gram of chlorine is its equivalent weight.

Complete answer:
The mass that provides or interacts with one mole of hydrogen cations (${{H}^{+}}$) is the equivalent weight of an acid or base in acid–base reactions. In a redox reaction, one mole of electrons (${{e}^{-}}$) is supplied or interacts with the equivalent weight of each reactant. Unlike atomic weight, which is dimensionless, equivalent weight contains dimensions and units of mass. Molar masses are now used to calculate equivalent weights, which were formerly established through experiment. The equivalent weight of a compound may also be determined by dividing the molecular mass by the amount of positive or negative electrical charges produced by the molecule's dissolution.
${{\text{N}}_{2}}\to \text{N}{{\text{H}}_{3}}$
Oxidation no. of $\mathbf{N}$ in $\mathbf{N}_{2}=0$
Oxidation no. of $\mathrm{N}$ in $\mathrm{NH}_{3}=-3$
Valence factor = no. of atoms $\times$ change in oxidation no. $=2 \times|0-(-3)|=6$
The Molecular weight of $\mathrm{N}_{2}=\mathbf{2 8} \mathrm{g}$
As we know that, equivalent weight. $=\dfrac{\text{ molecular weight }}{\text{ valence factor }}$
$\therefore$ Equivalent weight of $\mathrm{N}_{2}$ for the given change $=\dfrac{28}{6}$
Hence for the given change, the equivalent weight of $\mathrm{N}_{2}$ will be $\dfrac{28}{6}$.

Hence option a is correct.

Note:
Equivalent weights, on the other hand, have their own set of issues. For starters, the hydrogen-based scale proved impractical since most elements do not form simple compounds when they react directly with hydrogen. One grams of hydrogen, on the other hand, interacts with 8 gram of oxygen to produce water or with 35.5 gram of chlorine to produce hydrogen chloride, therefore 8 gram of oxygen and 35.5 gram of chlorine may be used to calculate comparable weights. Different acids and bases can be used to expand this system further.