
The catalyst used in Bosch process of manufacture of ${{\text{H}}_{2}}$ is:
A. finely divided $\text{Ni}$
B. ${{\text{V}}_{2}}{{\text{O}}_{5}}$
C. $\text{Pb}$
D. $\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}+\text{C}{{\text{r}}_{2}}{{\text{O}}_{3}}$
Answer
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Hint: Catalysts are the substances that help in speeding up the rate of reaction by minimizing the activation energy required for the reaction. The Bosch process helps to manufacture hydrogen gas from water gas $\left( \text{CO}+{{\text{H}}_{2}} \right)$ and steam. Check out the options one-by-one to find the answer.
Complete step by step answer:
Let us discuss the options of the question to get the briefing of the catalyst used in Bosch process:
A. finely divided $\text{Ni}$: The catalysts like nickel $\left( \text{Ni} \right)$, platinum $\left( \text{Pt} \right)$ are used in hydrogenation of unsaturated hydrocarbons into saturated hydrocarbons or addition of hydrogen. The reaction is $\text{R}-\text{CH=C}{{\text{H}}_{2}}+{{\text{H}}_{2}}\xrightarrow{\text{Ni}}\text{R}-\text{C}{{\text{H}}_{2}}-\text{C}{{\text{H}}_{3}}$. The unsaturated compounds are alkenes and alkynes and saturated compounds are alkanes.
B. ${{\text{V}}_{2}}{{\text{O}}_{5}}$: The chemical name of ${{\text{V}}_{2}}{{\text{O}}_{5}}$ is vanadium pentoxide. It is used as a catalyst in many industrial processes. It mainly act as a catalyst in the manufacturing of sulphur trioxide $\left( \text{S}{{\text{O}}_{3}} \right)$ from sulphur dioxide $\left( \text{S}{{\text{O}}_{2}} \right)$ with reaction with oxygen gas $\left( {{\text{O}}_{2}} \right)$. The reaction is $\text{2S}{{\text{O}}_{2}}\left( \text{g} \right)+{{\text{O}}_{2}}\left( \text{g} \right)\xrightarrow{{{\text{V}}_{2}}{{\text{O}}_{5}}}2\text{S}{{\text{O}}_{3}}\left( \text{g} \right)$. This reaction comes under a process known as contact process (manufacture sulphuric acid).
C. $\text{Pb}$: Lead is not used as a catalyst in any reaction.
D. $\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}+\text{C}{{\text{r}}_{2}}{{\text{O}}_{3}}$: This catalyst is used to prepare hydrogen gas $\left( {{\text{H}}_{2}} \right)$ in Bosch process. The reactants of the reaction are water gas and steam $\left( {{\text{H}}_{2}}\text{O} \right)$. Water gas is the mixture of carbon monoxide $\left( \text{CO} \right)$ and hydrogen $\left( {{\text{H}}_{2}} \right)$ in 1:1 ratio. The reaction occurs at high temperature $\left( {{440}^{\text{o}}}\text{C}-{{550}^{\text{o}}}\text{C} \right)$ . The reaction is $\text{CO}+{{\text{H}}_{2}}+{{\text{H}}_{2}}\text{O}\xrightarrow{\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}-\text{C}{{\text{r}}_{2}}{{\text{O}}_{3}}}\text{C}{{\text{O}}_{2}}\uparrow +\text{2}{{\text{H}}_{2}}$. $\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}$ act as a catalyst and $\text{C}{{\text{r}}_{2}}{{\text{O}}_{3}}$ act as a promoter.
Note: It is a myth that catalyst poison destroys the reaction. These substances also prove to be important from the reaction and particular point of view. An example where poisons proved to be boon is:
Rosenmunds Reaction: When acyl halides $\left( \text{R}-\text{COCl} \right)$ reacts with hydrogen gas in the presence of palladium $\left( \text{Pd} \right)$ to form aldehydes $\left( \text{R}-\text{CHO} \right)$. The poison is used to stop the reaction at aldehydes stage otherwise, the reaction would have stopped at yielding alcohols. The poison used in this reaction is barium sulphate $\left( \text{BaS}{{\text{O}}_{4}} \right)$.
Complete step by step answer:
Let us discuss the options of the question to get the briefing of the catalyst used in Bosch process:
A. finely divided $\text{Ni}$: The catalysts like nickel $\left( \text{Ni} \right)$, platinum $\left( \text{Pt} \right)$ are used in hydrogenation of unsaturated hydrocarbons into saturated hydrocarbons or addition of hydrogen. The reaction is $\text{R}-\text{CH=C}{{\text{H}}_{2}}+{{\text{H}}_{2}}\xrightarrow{\text{Ni}}\text{R}-\text{C}{{\text{H}}_{2}}-\text{C}{{\text{H}}_{3}}$. The unsaturated compounds are alkenes and alkynes and saturated compounds are alkanes.
B. ${{\text{V}}_{2}}{{\text{O}}_{5}}$: The chemical name of ${{\text{V}}_{2}}{{\text{O}}_{5}}$ is vanadium pentoxide. It is used as a catalyst in many industrial processes. It mainly act as a catalyst in the manufacturing of sulphur trioxide $\left( \text{S}{{\text{O}}_{3}} \right)$ from sulphur dioxide $\left( \text{S}{{\text{O}}_{2}} \right)$ with reaction with oxygen gas $\left( {{\text{O}}_{2}} \right)$. The reaction is $\text{2S}{{\text{O}}_{2}}\left( \text{g} \right)+{{\text{O}}_{2}}\left( \text{g} \right)\xrightarrow{{{\text{V}}_{2}}{{\text{O}}_{5}}}2\text{S}{{\text{O}}_{3}}\left( \text{g} \right)$. This reaction comes under a process known as contact process (manufacture sulphuric acid).
C. $\text{Pb}$: Lead is not used as a catalyst in any reaction.
D. $\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}+\text{C}{{\text{r}}_{2}}{{\text{O}}_{3}}$: This catalyst is used to prepare hydrogen gas $\left( {{\text{H}}_{2}} \right)$ in Bosch process. The reactants of the reaction are water gas and steam $\left( {{\text{H}}_{2}}\text{O} \right)$. Water gas is the mixture of carbon monoxide $\left( \text{CO} \right)$ and hydrogen $\left( {{\text{H}}_{2}} \right)$ in 1:1 ratio. The reaction occurs at high temperature $\left( {{440}^{\text{o}}}\text{C}-{{550}^{\text{o}}}\text{C} \right)$ . The reaction is $\text{CO}+{{\text{H}}_{2}}+{{\text{H}}_{2}}\text{O}\xrightarrow{\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}-\text{C}{{\text{r}}_{2}}{{\text{O}}_{3}}}\text{C}{{\text{O}}_{2}}\uparrow +\text{2}{{\text{H}}_{2}}$. $\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}$ act as a catalyst and $\text{C}{{\text{r}}_{2}}{{\text{O}}_{3}}$ act as a promoter.
Note: It is a myth that catalyst poison destroys the reaction. These substances also prove to be important from the reaction and particular point of view. An example where poisons proved to be boon is:
Rosenmunds Reaction: When acyl halides $\left( \text{R}-\text{COCl} \right)$ reacts with hydrogen gas in the presence of palladium $\left( \text{Pd} \right)$ to form aldehydes $\left( \text{R}-\text{CHO} \right)$. The poison is used to stop the reaction at aldehydes stage otherwise, the reaction would have stopped at yielding alcohols. The poison used in this reaction is barium sulphate $\left( \text{BaS}{{\text{O}}_{4}} \right)$.
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