Question

$4CH_4 + O_2 \xrightarrow{Cu \atop 573k/1100atm}$

Answer 1

Methanol ($CH_3OH$)

Answer 2

The given reaction is a controlled oxidation of methane.

Reaction Analysis:

The reaction involves methane ($CH_4$) and oxygen ($O_2$) under specific conditions:

• Catalyst: Copper (Cu)
• Temperature: 573 K
• Pressure: 1100 atm

These conditions are characteristic for the partial oxidation of methane to methanol. According to NCERT, methane is oxidized to methanol in the presence of copper at 573 K and 100 atm. The higher pressure (1100 atm) mentioned in the question is also a very high pressure, which favors the formation of methanol.

The balanced chemical equation for the formation of methanol from methane is:

$2CH_4(g) + O_2(g) \xrightarrow{Cu \atop 573K/100atm} 2CH_3OH(l)$

Stoichiometry Consideration:

The question provides the stoichiometry as $4CH_4 + O_2$. From the balanced equation, 1 mole of $O_2$ reacts with 2 moles of $CH_4$. In the given reaction, we have 1 mole of $O_2$ and 4 moles of $CH_4$. This means $O_2$ is the limiting reactant. 1 mole of $O_2$ will react with 2 moles of $CH_4$ to produce 2 moles of $CH_3OH$. The remaining 2 moles of $CH_4$ will be unreacted. When asked for "the product" in such a reaction, it refers to the primary chemical species formed.

Product:

The primary product of this controlled oxidation of methane under these conditions is methanol.

Chemical Structure of Methanol:

Methanol (methyl alcohol) has the chemical formula $CH_3OH$. SMILES: CO

Explanation:

Methane undergoes controlled oxidation in the presence of a copper catalyst at 573 K and high pressure (100-1100 atm) to yield methanol. The given stoichiometry indicates that oxygen is the limiting reactant, leading to the formation of methanol, with some unreacted methane remaining.