Step 1: Identify the compound that forms ethene
The compound that forms ethene as its major product when heated at 443 K in excess concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) is ethanol (\( C_2H_5OH \)).
This reaction is an example of a dehydration reaction, where ethanol undergoes dehydration to form ethene.
Step 2: Role of concentrated sulfuric acid
Concentrated sulfuric acid acts as both a catalyst and a dehydrating agent. It removes water from the ethanol molecule, causing it to break down into ethene and water.
Step 3: Chemical equation for the reaction
The chemical equation for the dehydration of ethanol to form ethene is as follows:
\[ C_2H_5OH \xrightarrow{\text{conc.} \, \text{H}_2\text{SO}_4, 443 \, \text{K}} C_2H_4 + \text{H}_2\text{O} \]
Step 4: Explanation of the reaction
- When ethanol is heated with concentrated sulfuric acid at 443 K, it undergoes elimination of water (\( \text{H}_2\text{O} \)) to form ethene (\( C_2H_4 \)).
- The concentrated sulfuric acid helps in removing water and facilitates the formation of the double bond in ethene.
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
A hydrocarbon which does not belong to the same homologous series of carbon compounds is
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