Question 39
Ethene is formed when ethanol at 443k is heated with an excess of concentrated sulphuric acid. What is the role of sulphuric acid in this reaction? Write the balanced chemical equation for this reaction.

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Solution

Sulphuric acid acts as a dehydrating agent. The reaction involved is:

${CH}_{3} {CH}_{2} OH {Hot Conc. H}_{2} {SO}_{4} ⟶ {CH}_{2} = {CH}_{2} + H_{2} O$

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Q. Question 39
Ethene is formed when ethanol at 443k is heated with an excess of concentrated sulphuric acid. What is the role of sulphuric acid in this reaction? Write the balanced chemical equation for this reaction.

Q. Ethene is formed when ethanol at 443 K is heated with an excess of concentrated sulphuric acid. The role of sulphuric acid in this reaction is :

Q. When ethanol is heated with excess of concentrated sulphuric acid at

170^{o} C

, it gets converted to ethene. In this reaction, concentrated sulphuric acid acts as:

When ethanol is heated with concentrated sulphuric acid at $170^{\circ} C,$ it gets converted into ethene. In this reaction, concentrated sulphuric acid acts as :
(a) oxidising agent (b) catalyst (c) dehydrating agent (d) reducing agent

(a) What happens when ethanol reacts with ethanoic acid in the presence of a little of concentrated sulphuric acid? Write an equation of the reaction involved.
(b) What happens when ethanol is heated with concentrated sulphuric acid at $170^{\circ} C$ ? Write the equation of the reaction which takes place.

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Question 39 Ethene is formed when ethanol at 443k is heated with an excess of concentrated sulphuric acid. What is the role of sulphuric acid in this reaction? Write the balanced chemical equation for this reaction.

Sulphuric acid acts as a dehydrating agent. The reaction involved is: CH3CH2OHHot Conc. H2SO4⟶CH2=CH2+H2O

Question 39
Ethene is formed when ethanol at 443k is heated with an excess of concentrated sulphuric acid. What is the role of sulphuric acid in this reaction? Write the balanced chemical equation for this reaction.

Sulphuric acid acts as a dehydrating agent. The reaction involved is:

${CH}_{3} {CH}_{2} OH {Hot Conc. H}_{2} {SO}_{4} ⟶ {CH}_{2} = {CH}_{2} + H_{2} O$