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NEET · Chemistry · STD 11 - 9. Hydricarbon

Methane reacts with steam at 1273 K in the presence of nickel catalyst to form :

  1. A CO and \(H _2 O\)
  2. B \(CO _2\) and \(H _2\)
  3. C CO and \(H _2\)
  4. D \(CO _2\) and \(H _2 O\)
Verified Solution

Answer & Solution

Correct Answer

(C) CO and \(H _2\)

Step-by-step Solution

Detailed explanation

(C) CO and \(H _2\)
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:
\(CH_4(g)+H_2 O(g) \xrightarrow[Ni]{1273 K} CO(g)+3 H_2(g)\)
The products formed are carbon monoxide \(( CO )\) and hydrogen \(\left( H _2\right)\).