Question

Write a balanced equation for the reaction of ammonia and oxygen in the presence of a catalyst:
(a)- \[N{{H}_{3}}+5{{O}_{2}}\xrightarrow[Pt]{800K}NO+{{H}_{2}}O+\Delta \]
(b)- \[4N{{H}_{3}}+5{{O}_{2}}\xrightarrow[Pt]{800K}4NO+6{{H}_{2}}O+\Delta \]
(c)- \[4N{{H}_{3}}+3{{O}_{2}}\xrightarrow[Pt]{800K}4NO+{{H}_{2}}O+\Delta \]
(d)- \[N{{H}_{3}}+3{{O}_{2}}\xrightarrow[Pt]{800K}NO+{{H}_{2}}O+\Delta \]

Answer 1

Hint: First we have to find out the oxidation number of all atoms other than oxygen and hydrogen. Since only nitrogen is present other than hydrogen and oxygen, hence oxygen is used as another atom. Identify the oxidizing and reducing agents. Balancing of oxygen and hydrogen is done by hit and trial method.

Complete step by step answer:
When the ammonia reacts with oxygen it forms nitric oxide and water. Since this is an exothermic reaction, there is an evolution of heat.
The equation is balanced by the Oxidation number method.
Step by step we can balance the equation. In the first step write the skeletal equation and write the oxidation number of each atom.
$\begin{align}
  & -3\text{ +1 0 +2 -2 +1 -2} \\
 & \text{N }{{\text{H}}_{3}}\text{ + }{{\text{O}}_{2}}\text{ }\to \text{ N O + }{{\text{H}}_{2}}\text{ O + }\Delta \\
\end{align}$
Next, find out the elements which change oxidation number.
Since only nitrogen is present other than O and H. O is used as another atom for finding the oxidizing and reducing agent.

For the easy calculation we use the oxygen of water, not of nitric oxide.
The oxidation number of N increases from -3 to +2. And the oxidation number of O decreases from 0 to -2.
Next, find out the total change in oxidation number.
Since there is only one nitrogen in both L.H.S and R.H.S, the total increase in oxidation number is 5. The number of O atoms on L.H.S is 2 and on R.H.S is 1, so the total decrease of O.N will be 2 x 2 = 4.
Next, balance the oxidation number.
Since the total increase in the oxidation number is 5 and the total decrease is 4, therefore multiply $N{{H}_{3}}\text{ and NO}$ by 4 and multiply ${{O}_{2}}\text{ and }{{\text{H}}_{2}}O$ with 5.
$\text{4N}{{\text{H}}_{3}}\text{+ 5}{{\text{O}}_{2}}\text{ }\to 4\text{NO + 5}{{\text{H}}_{2}}\text{O }$
Next balance all atoms other than H and O.
No need to balance the equation because there are 4 N atoms on both sides.
Now balance the oxygen and hydrogen by hit and trial method.
There are 10 oxygen atoms on the L.H.S and 9 oxygen atoms on the R.H.S, therefore, add one${{H}_{2}}O$ molecule to the R.H.S.
$\text{4N}{{\text{H}}_{3}}\text{+ 5}{{\text{O}}_{2}}\text{ }\to 4\text{NO + 6}{{\text{H}}_{2}}\text{O }$
The hydrogen atoms get balanced automatically.
Hence, the balanced equation is: $\text{4N}{{\text{H}}_{3}}\text{+ 5}{{\text{O}}_{2}}\text{ }\to 4\text{NO + 6}{{\text{H}}_{2}}\text{O }$

Hence, the correct option is (b).

Note: The reaction takes place in platinum as a catalyst and at temperature 800 K. This reaction is used in the preparation of nitric acid which is called Ostwald's process.