Question

What is the equivalent weight of ${\mathrm{H}}_2{\mathrm{O}}_2$ in decomposition reaction?

Answer 1

Decomposition reaction

• When one reactant breaks into two or more products, it is called a decomposition reaction.
• The decomposition of Hydrogen peroxide $\left({\mathrm{H}}_2{\mathrm{O}}_2\right)$ is as follows:

$2{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$

Calculate number of electron transfer per mole of ${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$

• The decomposition of Hydrogen peroxide involved $2\mathrm{mol}$ of ${\mathrm{H}}_2{\mathrm{O}}_2$.
• The reaction involves the transfer of $2\mathrm{mol}$ of electrons.
• The number of electron transfer per mole of reactant is as follows:

$n=\frac{n{e}^{-}}{nr}$

• Here, $n$ is the number of electron transfer per mole of reactant, $n{e}^{-}$ is the number of moles of electron transfer during reaction and $nr$ is the number of moles of reactant.
• The number of electron transfer per mole of ${\mathrm{H}}_2{\mathrm{O}}_2$ is as follows:

$\begin{array}{rcl}n& =& \frac{n{e}^{-}}{nr}\\ & =& \frac{2\bcancel{\mathrm{mol}}}{2\bcancel{\mathrm{mol}}}\\ & =& 1\end{array}$

Equivalent weight or mass

• The equivalent weight is calculated by dividing the molecular mass by the number of electrons transferred for each mole of reactant.
• The formula is as follows:

$W=\frac{n}{M}$

• Here, $W$ is the equivalent mass, and $M$ is the molecular mass.

Calculate equivalent weight of ${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$

• The molecular mass of ${\mathrm{H}}_2{\mathrm{O}}_2$, $M=34.01\mathrm{g}$.
• The equivalent weight of ${\mathrm{H}}_2{\mathrm{O}}_2$ is as follows:

$\begin{array}{rcl}W& =& \frac{n}{M}\\ & =& \frac{34.01\mathrm{g}}{1}\\ & =& 34.01\mathrm{g}\end{array}$

• The equivalent weight of ${\mathrm{H}}_2{\mathrm{O}}_2$ is $34.01\mathrm{g}$.