Q71 (2024)
Methane reacts with steam at 1273 K in the presence of nickel catalyst to form:
- (1) CO and $H_2$
- (2) CO and $H_2O$
- (3) $CO_2$ and $H_2O$
- (4) $CO_2$ and $H_2$
### Related Formula
$$CH_4(g) + H_2O(g) \xrightarrow[1273 \text{ K}]{Ni} CO(g) + 3H_2(g)$$
### Core Logic
Steam reforming of methane over nickel catalyst at 1273 K produces carbon monoxide and dihydrogen (water gas synthesis).
### Step 1: Conclusion
Products formed are CO and $H_2$.
### Pattern Recognition
Steam reforming of hydrocarbons always produces $CO + H_2$ mixture (syngas).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
