Question

Write the steps to prepare $\mathrm{Hydrogen}$ by Bosch's process.

Answer 1

Hydrogen

• $\mathrm{Hydrogen}$$\left(\mathrm{H}\right)$ is the most abundant element in the stars, sun, and universe.
• About 93% of atoms in the universe are $\mathrm{Hydrogen}\mathrm{atoms}$.
• $\mathrm{Hydrogen}$ is the first element the periodic table.

Preparation of Hydrogen gas by Bosch's process.

Step 1

• Production of water gas: A mixture of $\mathrm{Carbon}\mathrm{Monoxide}\left(\mathrm{CO}\right)\mathrm{and}\mathrm{Hydrogen}\left({\mathrm{H}}_2\right)$ being called water gas. They are produced by passing steam over a heated coke at $1000°\mathrm{C}$ in a converter. This is an endothermic reaction.

$\underset{\mathrm{Coke}}{\mathrm{C}\left(\mathrm{s}\right)}+\underset{\mathrm{Steam}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}\stackrel{1000°\mathrm{C}}{\to }\underset{\mathrm{Carbon}\mathrm{monoxide}}{\mathrm{CO}\left(\mathrm{g}\right)}+\underset{\mathrm{Hydrogen}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}$

Step 2

• Separation of Hydrogen: Water gas formed from step 1 is mixed with steam and they are passed through a heated $\mathrm{Iron}\left(\mathrm{III}\right)\mathrm{oxide}\left({\mathrm{Fe}}_2{\mathrm{O}}_3\right)$ and $\mathrm{Chromium}\left(\mathrm{III}\right)\mathrm{oxide}\left({\mathrm{Cr}}_2{\mathrm{O}}_3\right)$ at a temperature of $450°\mathrm{C}-500°\mathrm{C}$ to get a mixture of $\mathrm{Carbon}\mathrm{dioxide}\left({\mathrm{CO}}_2\right)\mathrm{and}\mathrm{Hydrogen}\left({\mathrm{H}}_2\right)$. The reaction is exothermic in nature.

$\underset{\mathrm{Carbon}\mathrm{monoxide}}{\mathrm{CO}\left(\mathrm{g}\right)}+\underset{\mathrm{Hydrogen}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\underset{{\mathrm{Fe}}_2{\mathrm{O}}_3}{\overset{450-500°}{\to }}\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Hydrogen}}{2{\mathrm{H}}_2\left(\mathrm{g}\right)}+\mathrm{Heat}$

Step 3

• Removal of Carbon monoxide and Carbon dioxide from the mixture

(1) Removal of $\mathrm{Carbon}\mathrm{dioxde}\left({\mathrm{CO}}_2\right)$ : The mixture of ${\mathrm{CO}}_2+{\mathrm{H}}_2$ is passed through cold water or Caustic potash ($\mathrm{KOH}$) under pressure, it dissolves the

$\mathrm{Carbon}\mathrm{dioxde}\left({\mathrm{CO}}_2\right)$ leaving behind $\mathrm{Hydrogen}\left({\mathrm{H}}_2\right)$.

$\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Cold}\mathrm{water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\stackrel{}{\to }\underset{\mathrm{Carbonic}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{CO}}_3}$(l)

$\underset{\mathrm{Caustic}\mathrm{potash}}{2\mathrm{KOH}\left(\mathrm{aq}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxde}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Potassium}\mathrm{carbonate}}{{\mathrm{K}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

(2) Removal of $\mathrm{Carbon}\mathrm{monoxide}$($\mathrm{CO}$): The solution is passed through the solution of $\mathrm{Ammonical}\mathrm{copper}\left(\mathrm{I}\right)\mathrm{Chloride}$, which dissolves the unreacted

$\mathrm{Carbon}\mathrm{monoxide}\left(\mathrm{CO}\right)$ and leaving behind $\mathrm{Hydrogen}\left({\mathrm{H}}_2\right)$.

$\underset{\mathrm{Ammonium}\left(\mathrm{I}\right)\mathrm{chloride}}{\mathrm{CuCl}\left(\mathrm{s}\right)}+\underset{\mathrm{Carbon}\mathrm{monoxide}}{\mathrm{CO}\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\to \underset{\mathrm{Addition}\mathrm{compound}}{\mathrm{CuCl}.\mathrm{CO}.2{\mathrm{H}}_2\mathrm{O}}$