Ethene (c2h4) can be halogenated by this reaction: c2h4( g) + x2( g) ∆ c2h4x2( g) where x2 can be cl2 (green), br2 (brown), or i
2 (purple). examine the three figures representing equilibrium concentrations in this reaction at the same temperature for the three different halogens. rank the equilibrium constants for the three reactions from largest to smallest.
1 answer:
Answer :The correct order for equilibrium constant = K₁ > K₂ >K₃
Ethene is an alkene and X2 is elemental halogen . The reaction between alkene and halogen is known as Halogenation reaction . This reaction involves electrophilic addition mechanism . In this reaction elemental halogen (X₂ ) produces a elctrophile , X⁺ . This electrophile is attracted by the double bond in alkene and attached to it .
Due to addition of electrophile , it produces carbocation intermediate .On this positive carbon , nucleophile attacks , hence , a resultant 1,2-di halide alkane is produced .( in image )
The reaction between ethene ( C₂H₄) and halogen ( X₂) produces 1,2 - dihalide ethane (C₂H₄X₂) . The image shows reaction between chlorine ,Cl₂ , a halogen (X₂) .
The expression for equilibrium constant of reaction between ethene and halogen is given as:
Product = C₂H₄X₂ Reactants = C₂H₄ and X₂
On applying this concept in figures :
Given : Green = Cl₂ Brown = Br₂ Purple = I₂ Grey = Ethene
Grey + Green = 1,2- dihalide ethane
a) Number of Cl₂(green ) = 2
number of ethane ( grey ) = 2
Number of 1,2-dichloride alkane( grey + green ) = 8
Plugging these number in equilibrium constant expression :
Hence K₁ for ethene + chlorine = 2
b) Number of Br₂(brown ) = 4
number of ethane ( grey ) = 4
Number of 1,2-dibromide alkane( grey + brown ) = 6
On plugging these values in K expression as :
Hence K₂ for ethen + Bromine = 0.375
c) Number of I₂(purple ) = 7
number of ethane ( grey ) = 7
Number of 1,2-diiodide alkane( grey + purple ) = 3
On plugging values in K expression as:
Hence K₃ for ethene + Iodine = 0.0612
So , the order of equilibrium constant (K) from largest to smallest :
K₃ = 0.0612 K₂ = 0.375 K₁= 2
K₁ > K₂ >K₃
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You don't need precisely 100 mL solution.
If the beaker is graduated, you can easily measure 100 mL of the stock solution.
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Answer : The correct answer is 1) AlCl₃ - CH₃Cl 2) HNO₃ -H₂SO₄ at room temperature 3) Fuming HNO₃ -H₂SO₄ at 90-100 ⁰ C heat .
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Step 1: Conversion of Benzene to Toluene .
Benzene can be converted to toluene by Friedel Craft Alkylation of benzene . In this reaction reagent AlCl₃ and Ch3Cl is used . Electrophile CH³⁺ is produced which attached on carbon of benzene and formation of Toluene and HCl occur.
Step 2 : Conversion of Toluene to dinitrotoluene.
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Toluene
Step 3) Conversion of Dinitro toluene to trinitrotoluene.
This reaction is extended nitration of toluene . Further nitration is done in extreme condition . The temperature of reaction is increased to 90- 100 ⁰ C . Due to which there is more production of NO²⁺ ion occurs from HNO₃ -H₂SO₄ and they attack on C6 carbon atom of dinitrotoluene which forms 2,4,6- trinitrotoluene.
Dinitrotoluene
So over all reaction uses three reagents in order :
