Q74 / 180 · NEET 2026 (Cancelled) · Chemistry
Methane reacts with steam at $1273$ K in the presence of nickel catalyst to form :
- A$\text{CO}$ and $\text{H}_2\text{O}$
- B$\text{CO}_2$ and $\text{H}_2$
- Correct answerC$\text{CO}$ and $\text{H}_2$
- D$\text{CO}_2$ and $\text{H}_2\text{O}$
Solution
The reaction of methane with steam at $1273$ K in the presence of a nickel catalyst is known as steam reforming of methane. It is an industrial method for the preparation of hydrogen gas.
The balanced chemical equation is:
$\text{CH}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \xrightarrow[\text{Ni}]{1273 \text{ K}} \text{CO}(\text{g}) + 3\text{H}_2(\text{g})$
The products formed are carbon monoxide ($\text{CO}$) and hydrogen ($\text{H}_2$).
Answer: $\text{CO}$ and $\text{H}_2$
The balanced chemical equation is:
$\text{CH}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \xrightarrow[\text{Ni}]{1273 \text{ K}} \text{CO}(\text{g}) + 3\text{H}_2(\text{g})$
The products formed are carbon monoxide ($\text{CO}$) and hydrogen ($\text{H}_2$).
Answer: $\text{CO}$ and $\text{H}_2$
Hence, the correct answer is option: 'C'.