Question

How can I balance this equation? {\text{_}}{{\text{N}}_{\text{2}}}{\text{ + _}}{{\text{H}}_{\text{2}}} \to {\text{N}}{{\text{H}}_{\text{3}}}

Answer 1

For the balancing of an equation we have to maintain equal molecularity of each atom in the left hand side or reactant side as well as in the right hand side or product side.

Complete Solution :
In the given equation {\text{_}}{{\text{N}}_{\text{2}}}{\text{ + _}}{{\text{H}}_{\text{2}}} \to {\text{N}}{{\text{H}}_{\text{3}}}
-Nitrogen (${{\text{N}}_{\text{2}}}$) and hydrogen (${{\text{H}}_{\text{2}}}$) gas reacts together for the formation of ammonia (${\text{N}}{{\text{H}}_3}$) gas.
-On the reactant side two nitrogen atoms are present but on the product side only one nitrogen atom is present. So for balancing the amount of nitrogen on both the sides, we have to multiply nitrogen by 2 on the right hand side of the given chemical reaction.
-Similarly, the amount of hydrogen is also not equal on both sides of the chemical reaction as two hydrogen atoms are present on the reactant side and three are present on the product side. So for balancing the amount of hydrogen on both the sides, we have to multiply hydrogen by 3 on the left hand side and by 2 on the right hand side of the given chemical reaction.
-So, balanced form of above given chemical reaction is as follow:
${{\text{N}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}} \to 2{\text{N}}{{\text{H}}_{\text{3}}}$

Note: Some of you may get confused in between the order of reaction and molecularity, but these two things are totally different. As order of reaction is an experimental quantity and molecularity is a theoretical quantity. But these two things are the same for elementary reactions or single step reactions.