Question

How do you balance${\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$?

Answer 1

Step 1: Note down the reaction.

Write the unbalanced reaction as it is.

$\underset{\mathrm{Ammonia}}{{\mathrm{NH}}_3\left(\mathrm{g}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Nitrogen}\mathrm{dioxide}}{{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Step 2: Find the oxidation state of each atom.

Write the oxidation numbers of each atom.

On the reactant side, the oxidation state of nitrogen in ammonia is $-\frac{1}{3}$ and in the product nitrogen oxide, the oxidation state changes to $+4$ as it donates four electrons to two oxygen atoms.

On the reactant side, the oxidation state of oxygen in gaseous form is $0$ and in the product nitrogen oxide and water the oxidation state changes to $-2$ as it accepts electrons from nitrogen and hydrogen atoms.

The oxidation state of hydrogen in both ammonia and water is $+1$, as nitrogen and oxygen are more electronegative than hydrogen, it donates electrons.

$$\begin{gathered} \mathrm{On}\mathrm{RHS},\mathrm{N}=-\frac{1}{3},\mathrm{H}=+1,\mathrm{O}=0 \\ \mathrm{On}\mathrm{LHS},\mathrm{N}=+4,\mathrm{H}=+1,\mathrm{O}=-2 \end{gathered}$$

Step 3: Separate the oxidation and reduction reactions.

Write the oxidation and reduction couplets.

Reduction: On the reactant side the oxygen atom is in the gaseous state, hence its oxidation state is zero but when there is the formation of water in the product, the oxidation state of oxygen changes to $-2$. Therefore, it is a reduction-half reaction.

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

Oxidation: On the reactant side the nitrogen is present in Ammonia with $-3$an oxidation state, but when there is the formation of nitrogen oxide in the product, the oxidation state of nitrogen changes to $+4$. Therefore, it is an oxidation-half reaction.

${\mathrm{NH}}_3\to {\mathrm{NO}}_2$

Step 4: Balancing Oxygen and Hydrogen.

Balance all other elements except oxygen and hydrogen, but here all other atoms in the reaction are balanced. So, there is no need to balance them.

Balance Oxygen and Hydrogen.

Balance Oxygen with Water molecules.

$$\begin{gathered} {\mathrm{NH}}_3+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{NO}}_2 \\ {\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Balance Hydrogen with proton ions.

$$\begin{gathered} {\mathrm{NH}}_3+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{NO}}_2+7{\mathrm{H}}^{+} \\ {\mathrm{O}}_2+4{\mathrm{H}}^{+}\to 2{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Charge balancing using electrons.

$$\begin{gathered} {\mathrm{NH}}_3+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{NO}}_2+7{\mathrm{H}}^{+}+7{\mathrm{e}}^{-}............\mathrm{equation}1 \\ {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}.......................\mathrm{equation}2 \end{gathered}$$

Step 5: Writing resulting reaction

Multiply with the suitable numbers to balance the coefficients.

Multiply equation $1$ by $4$ and equation $2$ by $7$.

After the multiplication and addition of both equations, we get the final balanced equation.

$\underset{\mathrm{Ammonia}}{4{\mathrm{NH}}_3\left(\mathrm{g}\right)}+\underset{\mathrm{Oxygen}}{7{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Nitrogen}\mathrm{dioxide}}{4{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Therefore, the balanced chemical reaction is

$\underset{\mathrm{Ammonia}}{4{\mathrm{NH}}_3\left(\mathrm{g}\right)}+\underset{\mathrm{Oxygen}}{7{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Nitrogen}\mathrm{dioxide}}{4{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$