Question

Hydrogen peroxide (${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$) decomposes to form water and oxygen. How do you write the balanced equation for this reaction?

Answer 1

Hint For balancing any chemical equation, we have to maintain equal molecularity of each atom present on the reactant side as well as on the product side of the given chemical reaction.

Complete step by step solution: Given that, Hydrogen peroxide decomposes to produce water molecule and oxygen gas and chemical reaction for this is shown as follow:
${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} \to {{\text{H}}_{\text{2}}}{\text{O + }}{{\text{O}}_{\text{2}}}$
For balancing the above given equation we have to consider following points in mind:
-First we calculate the number of each atom on the left hand side as well as on the right hand side of the given chemical equation.
-So on the left hand side or reactant side two oxygen $\left( {\text{O}} \right)$ atoms and on the right hand side or product side of the chemical reaction three oxygen $\left( {\text{O}} \right)$ atoms are present, which is not correct.
-And on the left hand side as well as on the right hand side two hydrogen atoms (${\text{H}}$) are present which is correct.
So, we will balance the above given reaction by the following manner so that each atom will have the same molecularity on the reactant side as well as on the product of the reaction. And balanced equation is shown as:
${\text{2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} \to 2{{\text{H}}_{\text{2}}}{\text{O + }}{{\text{O}}_{\text{2}}}$
-In the above balanced equation four hydrogen atoms (${\text{H}}$) and four oxygen $\left( {\text{O}} \right)$ atoms are present on the left hand side as well as on the right hand side of the chemical reaction also.

Note: Here some of you may think order of reaction and molecularity of reaction are the same things but that will be true for the elementary (single step reaction) reactions only.